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PLANE AND SPHERICAL
T R I G O N O M E T R Y
G. A. WENTWORTH,A .M.
AUTHOR OF A SERIES OF TEXT-BOOKS IN MATHEMATICS
REVISED EDITION
BOSTON,
AND LONDON
G INN COMPANY, PUBL ISHERS1897
Entered , according to Act of Congress , in the year 1882, by
in the M oe of the Librarian of Congress , atWashington.
Copyright , 1895 , by G . A . WENTWORTH .
I
A m 2”(a
HARVARDCOLLEGELIBRARY
Gm OF
mssmm L. wsurwoam
8 1939
PRE F A C E .
N preparing this w ork the aim has been to fu rnish ju st so m u ch
of Trigonom etry as is actu ally tau ght in ou r best schools and
colleges. Consequ ently, all investigations that are important onlyfor the special stu dent have been om itted, except the developm ent offu nctions in series. The principles have been u nfolded w ith theu tmost brevity consistent w ith simplicity and clearness, and interesting problem s have been selected w ith a view to aw aken a reallove for the stu dy. Mu ch tim e and labor have been spent in devisingthe simplest proofs for the propositions, and in exhibiting the bestmethods of arranging the logari thm ic w ork .
The au thor is u nder particu lar obligation for assistance to G. A .
Hill, A .M., of Cambri dge , Mass , to Prof. Jam es L. Patterson, of
Schenectady, N.Y., to Dr. F . N . Cole, of Ann Arbor, Mich., and to
Prof. S. F . Norris, of Baltimore , Md .
G. A . WENTWORTH.
EXE '
I‘
ER , NJL, Ju ly, 1895 .
C ONT ENT S.
PLANE TRIGONOMETRY .
CHAPTER I. F UNCTIONS O F A CUTE ANGLESAngular measu re, page 1 ; trigonom etric fu nctions, 3 representation
of fu nctions by lines, 7 changes in the fu nctions as the angle changes,10 ; fu nctions of complementary angles, 11 relations of the fu nctionsof an angle, 12 ; formu las for finding all the other fu nctions of an
angle, w hen one fum tion of the angle is given, 15 ; fu nctions of17.
CHAPTER II. THE RIGHT TRIANGLE :
Given parts of a triangle, 19 . Solu tions w ithou t logarithms, 19 ;Case I.
,w hen an acu te angle and the hypotenu se are given, 19 ;
Case II . , w hen an acu te angle and the opposite leg are given,20 ;
Case III .,w hen an acu te angle and an adjacent leg are given, 20 ;
Case IV. , w hen the hypotenu se and a leg are given, 21 ;
Case V., w hen the tw o legs are given,
21. General m ethod of
solving a right triangle, 22 solu tions by logarithms , 24 area of the
right triangle , 26 ; the isosceles triangle, 31 the regu lar polygon, 33.
CHAPTER III. GONIOMETRY
Definition Of goniom etry , 36 angles of any magnitu de , 36 ; generaldefinitions of the fu nctions of angles, 37 algebraic signs of the fu nctions, 39 ; fu nctions of a variable angle, 40 fu nctions Of angles greaterthan 42 form u las for acu te angles extended to all angles, 43 ;redu ction of the fu nctions of all angles to the fu nctions of angles in thefirst qu adrant, 46 fu nctions of angles that differ by 48 ; fu nctionsof a negative angle , 49 ; fu nctions of the su m of tw o angles, 51 fu nctions of the difference of tw o angles, 53 fu nctions of tw ice an angle , 55 ;fu nctions of half an angle , 55 su m s and differences of fu nctions, 56 .
CHAPTER IV. THE OBLIQ UE TRIANGLELaw of Sines, 60 ; law Of cosines, 62 law of tangents, 62 . Solu
tions : Case I. , w hen one side and tw o angles are given, 64 Case
vi TRIGONOMETRY.
w hen tw o sides and the angle opposite to one of them are given, 66 ;
Case III., w hen tw o sides and the inclu ded angle are given, 71 Case
IV., w hen the three sides are given, 74 ; area of a triangle , 78 .
CHAPTER V. MISCELLANEOUS EXAMPLESPlane Trigonometry , 82 ; Goniometry , 99 .
EXAMINATION PAPERS, 106.
CHAPTER VI. CONSTRUCTION O F TABLES
Loga rithms, 117 ; exponential and logarithm ic series, 120 ; trigonometric fu nctions of small angles, 125 ; Simpson’
s m ethod of con
stru cting a trigonom etric table , 127 ; De Moivre ’s theorem ,
128 ;
expansion of sin 33, cos at , and tan in infinite series, 132 .
SPHERICAL TRIGONOMETRY .
CHAPTER VII. THE RIGIIT SPHERICAL TRIANGLE :
Introdu ction, 135 form u las relating to right spherical triangles,137 ; Napier’s ru les , 141. Solu tions : Case I. , w hen the tw o legs aregiven, 142 ; Case II. , w hen the hypotenu se and a leg are given, 142Case III . , w hen a leg and the opposite angle are given, 143 Case IV. ,
w hen a leg and an adjacent angle are given, 143 Case V. , w hen thehypotenu se and an Obliqu e angle are given, 144 ; Case VI. , w hen thetw o obliqu e angles are given, 144 . The isosceles spherical triangle, 149 .
CHAPTER VIII. THE OBLIQ UE SPHERICAL TRIANGLEF u ndam ental formu las, 150 ; formu las for half angles and sides,
152 Gau ss’s equ ations and Napier ’s analogies, 154 . Solu tions : CaseI.
,w hen tw o sides and the inclu ded angle are given, 156 ; Case II. ,
w hen tw o angles and the inclu ded side are given, 158 ; Case III. , w hentw o sides and an angle opposite to one of them are given ,
160 ; CaseIV. ,
w hen tw o angles and a side opposite to one of them are given,162 ; Case V. , w hen the three sides are given, 163 ; Case VI. , w henthe three angles are given, 164 . Area of a spherical triangle, 166.
CHAPTER IX. APPLICATIONS O F SPHERICAL TRIGONOMETRI' :To r edu ce an angle measu red in Space to the horizon, 170 ; to find
the distance betw een tw o places on the earth ’s su rface, w hen the
latitu des of the places and the difference in their longitu des are know n,
171 the celestial sphere , 171 spherical co-ordinates, 174 ; the astronom ical tr
"tronom ical problems, 177.
PLANE TRIGONOMETRY .
CHAPTER I .
TRIGONOMETRIC F UNCTIONS O F ACUTE
ANGLES .
1 . ANGULAR MEASURE.
As lengths are m easu red in term s of variou s conventionalu nits, as the foot, m eter, etc.
,so different u nits for m easu ring
angles are em ployed, or have been proposed .
In the comm on or sexagesima l system the circu m ference ofa circle is divided into 360 equ al parts. The angle at thecentre su btended by each of these parts is taken as the u nitangle and is called a degree. The degree is su bdivided into60 m inu tes
, and the m inu te into 60 second s. A right angle isequ al to 90 degrees .
NO '
l‘E . The sexagesimal system w as invented by the early Babylonianastronom ers in conform ity w ith their year of 360 days.
In the circu la r system an arc of a circle is laid off equ al inlength to the radiu s . The angle at the centre su btended bythis are is taken as the u nit angle and is called a radian .
The nu mber of radians in 360° is equ al to the nu mber
of times the length of the radiu s is contained in the circu m ference . It is proved in Geom etry that this nu mber is
for all circles ; therefore the radian is thesam e angle in all circles .
2 TRIGONOMETRY.
Since the circu m ference of a circle is 2 1r tim es the radi u s,
2 1r radians and 1r radians 180°
therefore, 1 radian 57°17 45"
1 degree radian 0 017453 radian.
By the last tw o equ ations the m easu re of an angle can bechanged from radians to degrees or from degrees to radians .
Thu s, 2 radians= 2 X180
2 X (57°17' 114
°
35 '301?
NOTE . The circu lar system cam e into u se early in the last centu ry .
It is fou nd more convenient in the higher mathem atics, w here the radiansare expressed simply as nu mbers. Thu s the angle 1r means 1r radians ,
and the angle 3 m eans 3 radians.
On the introdu ction of the m etric system of w eights and m easu res atthe close of the last centu ry, it w as proposed to divide the right angle into100 equ al parts called grades, w hich w e re to be taken as u nits. The
grade w as su bdivided into 100 m inu tes and the m inu te into 100 seconds .
This F rench or centesima l system , how ever, never cam e into actu al u se .
EXERCISE I .
[Assu m e 7x
1 . Redu ce the follow ing angles to circu lar m easu re, expressing the resu lts as fractions of 1r. 11
°
123°
37°
2 . How many degrees are there in21r radians‘7 1: radians
15 77: radians ‘P 1r radians
151r radians
3. What decim al part of a radian is 1° 1 '
4 . How m any seconds in a radian
TRIGONOMETRIC FUNCTIONS . 3
5. Express in radians one of the interior angles of a regu laroctagon dodecagon .
6. On a circle of 50 ft. radiu s an arc of 10 ft . is laid off ;
how m any degrees does the arc su btend at the centre7. The earth’s equ atorial radiu s is approximately 3963
m iles . If tw o points on the equ ator are 1000 m iles apart,w hat is their difference in longitu de8 . If the difference in longitu de of tw o points on the equ atoris w hat is the distance betw een them in m iles9. What is the radiu s of a circle, if an arc of 1 foot su btends an angle of 1° at the centre10. In how m any hou rs is a point on the equ ator carried
by the earth’s rotation on its axis throu gh a distance equ alto the earth’s radiu s11 The m inu te hand of a clock is 35' ft. long ; how far
does its extrem ity move in 25 m inu tes [Take 1r
12. A w heel m akes 15 revolu tions a second ; how longdoes it take to tu rn throu gh 4 radians [Take 1r
§ 2 . THE TRIGONOMETRIC FUNCTIONS .
The sides and angles of a plane triangle are so related thatany three given parts, provided at least one of them is a side,determ ine the shape and the size of the triangle .
Geom etry show s how , from three su ch parts, to constru ct
the triangle and find the valu es of the u nknow n parts.Trigonometry show s how to comp u te the u nknow n parts of
a triangle from the nu merical valu es of the given parts .Geom etry show s in a general w ay that the sides and
angles of a triangle are mu tu ally dependent. Trigonom etrybegins by show ing the exact natu re of this dependence inthe right triangle, and for this pu rpose employs the ra tios
4 TRIGONOMETRY.
MAN (Fig.
0 E GF IG . 1.
pair, are equ al . That is,
AC AE AG Ac
AB AD AF’B C DE F G
’
These ra tios, therefore, rema in u nchanged so long as the angle
A rema ins u nchanged.
’
Hence, for every valu e of an acu te angle A there are certainnu mbers that express the valu es of the ratios of the sides ina ll right triangles tha t have this acu te angle A .
There are all together six different ratios :I. The ratio of the Opposite leg to the hypotenu se is called
the Sine of A, and is w ritten sin A .
II . The ratio of the adjacent leg to the hypotenu se is calledthe Cosine of A, and w ritten cos A.
III. The ratio of the opposite leg to the adjacent leg iscalled the Tangent of A, and w ritten tan A.
IV. The ratio of the adjacent leg to the opposite leg iscalled the Cotangent of A, and w ritten cot A.
V. The ratio of the hypotenu se to the adjacent leg is calledthe Secant of A, and w ritten sec A .
VI . The ratio of the hypotenu se to the opposite leg is calledthe Cosecant of A , and w ritten 080 A.
These six ratios are called the Trigonometric F unctions of‘ ngle A.
acu te angle . If from any pointsB, D,
F, in one of its sides
perpendicu lars B C', DE, F G,
are let fall to the other side, thenthe right triangles ABC, ADE ,
AF G,thu s formed have the
angle A comm on, and are therefore m u tu ally equ iangu lar and
sim ilar. Hence, the ratios ofNtheir corresponding Sides, pair by
TRIGONOMETRIC FUNCTIONS . 5
To these six ratios are Often added the tw o follow ing fu nctions
,w hich also depend only on the angle A
VII. The versed sine Of A is 1 cos A and is w ritten u ers A .
VIII. The coversed sine of A 18 1—sin A and is w rittencovers A .
In the right triangle ABC (Fig.
2) let a,b, c denote the lengths of
the sides opposite to the acu te ah
gles A,B
,and the right angle 0,
respectively, these lengths be ing allexpressed in term s of a commonu nit. Then
,
a opposite legSin A
c hypotenu secos A
A
A 5—31
.22t A
vers A 1 covers A= 10
EXERCISE II .
1 . What are the fu nctions of the other acu te angle Bthe triangle ABC (F ig.
2 . If A B prove
sin A cos B, secA csc B,
cos A sin B, csc A secB,
tan A cot B,
vers A covers B,
cot A tanB, covers A vers B.
6 adjacent legc hypotenu se
b adjacent lega opposite leg
'
0 hypotenu sea
_Opposite leg
'
6 TRIGONOMETRY.
3. Find the valu es of the fu nctions of A,if a , b, c respec
tively have the foll ow ing valu es :
(i.) 3, 4, 5 . (iii.) s, 15, 17 . (v .) s,
(ii .) 5, 12, 13. (iv.) 9, 40, 41 . (vi.)
4 . What condition m u st be fu lfilled by the lengths of thethree lines a , b, c (F ig. 2) in order to m ake them the Sides ofa right triangle IS this condition fu lfilled in Examme 35. Find the valu es of the fu nctions of A, if a , b, c respec
tively have the following valu es :
(i.) Zmn, m3—n
’, m
’+ (iii .) pqr, qrs, rsp .
2x31 x9+ y
” mn m u nr11. cc+ 1 Iv.)
x—y’ J’
46—9)m
’sq
’
p s
6. Prove that the valu es of a,b,c,in (i.) and (IL) , Example
5, satisfy the condition necessary to m ake them the sides ofa right triangle .
7. What equ ations of condition mu st be satisfied by thevalu es of a , b, c, in (iii .) and Example 5, in order thatthe valu es may represent the sides of a right triangle
Compu te the fu nctions of A and B w hen,
8 .
9 . a = 0.264, c= 0.265. 12 . a vp=+pq, c=p + q.
10. b= 9.5, c= 19 .3. 13. c=p + q.
Compu te the fu nctions of A when,
14 . a= 2h. 16. a + b=ge.
15 . a go. 17. a—b Z.18 . Find a if sin A= § and19 . F ind 6 if cos A=O.44 and c= 3.5.
20. F ind a if tan A= 131 and b=2f5p
TRIGONOME'
I‘
RIC FUNCTIONS . 7
21 . Find 6 if cot A= 4 and a = 17.
22 . Find c if sec A= 2 and
23. Find c if csc A and a=
Constru ct a right triangle given,24 . c= 6, tan A= g. 26. sin A= 0.6.
25. a = 3.5, cos A=5. 27. csc A= 4 .
28 . In a right triangle, c= 2 .5 m iles, sin A=O.6, cos A
compu te the legs.29. Constru ct (w ith a protractor) the 4 and
determ ine the ir fu nctions by m easu ring the necessary lines,and compare the valu es Obtained in this w ay w ith the m orenearly correct valu es given in the follow ing table :
30. Find, by means of the above table, the legs of a. righttriangle if A c
'
= 1 also if A= c=4 .
31 . In a right triangle, given a = 3 and c= 5 ; find the
hypotenu se of a sim ilar triangle in w hich a = m iles .32 . By dividing the length of a vertical rod by the lengthof its horizontal shadow , the tangent of the angle of elevationof the su n at the tim e of observation w as fou nd to beHow high is a tow er, if the length of its horizontal Shadow atthe sam e tim e is yards
§ 3. REPRESENTATION OF THE FUNCTIONS BY LINES .
The fu nctions of an angle, being ratios, are numbers ; bu tw e m ay represent them by lines if w e first choose a u nit oflength, and then constru ct right triangles, su ch that the
8 TRIGONOMETRY.
denom inators of the ratios shall be equ al to this un it. The
m ost convenient w ay to do this is as follow sAbou t a point 0 (Fig. 3) as
a centre, w ith a radiu s equ a l toone u nit of length, describe a
circle and draw tw o diametersAA'
and BB' perpendicu lar to
each other.The circle w ith radiu s equ alto 1 is called a u nit circle, AA '
the horizonta l, and BB ' the
vertica l diameter.m ”3°
Let AOP be an acu te angle,and let its valu e (in degrees, etc.) be denoted by x. We may
regard the A a: as generated by a radiu s OP that revolvesabou t 0 from the position 0A to the position show n in thefigu re ; view ed in this w ay, OP is called the moving radiu s .Draw PM _1_to OA,
PN _Lto OB . In the rt. A GPM the
hypotenu se OP 1 ; therefore, sin a: PM ; 00 8 9: OM.
Since PM is equ al to ON , and ON is the projection of OPon BB ’
, and since OM is the projection of OP on AA therefore
,in a u nit circle,
sin a projection of m oving radiu s on vertical diameter ;cos a: projection of m oving radi u s on horizontal diam eter .Throu gh A and B draw tangents to the circle m eeting OP ,
produ ced in T and S, respectively ; then, in the rt . A OAT,
the leg OA 1 , and in the rt. A OBS, the leg OB 1 ; w hilethe 4 OSE 4 cc. Therefore
,
tan x=AT 5 00 t x= BS ; vers x=AM ;
sec a: GT ; 080 11? OS ; covemx=BN.
These eight line valu es (as they may be term ed) of thefu nctions are all expressed in term s of the radiu s of the circleas a. u nit : and It 18 clear that as the angle varies in valu e the
TRIGONOMETRIC FUNCTION S . 9
line valu es of the fu nctions w ill alw ays rem ain equ al nu m er
ica lly to the ratio valu es . Hence,in stu dying the changes in
the fu nctions as the angle is su pposed to vary, w e m ay employthe simpler line valu es instead of the ratio valu es .
EXERCISE III .
1 . Represent by lines the fu nctions of a larger angle thanthat show n in Fig. 3.
If a: is an acu te angle, Show that2 . sin s: is less than tan a .
3. see n: is greater than tan x .
4 . ese ce is greater than cot r .
Constru ct the angle a: if
5. tan x= 3. 7 . COS T 4. 9 . sinx = 2 cos w .
6. csc a: 2 . 8 . sin s; cosm. 10. 4 sin a: tan x .
11 . Show that the sine of an angle is equ al to one-half thechord of tw ice the angle .
12. Find x if sin ce is equ al to one-half the side of a regu larinscribed decagon .
13. Given a and y, 33+ g be ing less than constru ctthe valu e of sin (a: g sinx.
14 . Given a: and g, 55 + g being less than constru ctthe valu e Of tan (x g) sin (93 g) tanx sin ac.
Given an angle as ; constru ct an angle 3; su ch that15. sing 2 sina . 17 . tan g 3 tanx.
16. cosy 500 8 a . 18 . sec g cscx.
19 . Show by constru ction that 2 sinA sin 2A .
20. Given tw o angles A and B,A+B being less than 90
°
Show that sin (A B) sin A sinB .
21 . Given sin ce in a u nit circle ; find the length of a linecorresponding in position to sin a: in a circle w hose radiu s is r .
22 . In a right triangle , given the hypotenu se c,and also
sinA m,cosA n ; find the legs .
10 TRIGONOMETRY .
5 4 . CHANGES IN THE FUNCTIONS AS THE ANGLE CHANGES.
If w e su ppose the A AOP ,or a: (Fig. 4) to increase gradu
ally by the revolu tion of the m oving radiu s OP abou t 0,the point P w ill m ove along the arcAB tow ards B ,
T w ill m ove alongthe tangent AT aw ay from A
, S willmove along the tangent BS tow ardsB, andM w ill m ove along the radiu sOA towards 0 .
Hence, the lines PM AT, OT will
gradu ally increase in length, and thelines 0M BS, OS w ill gradu allydecrease . That is,As an a cu te angle increa ses, its
sine,tangent, and sécant a lso increase,
F IG , 4 ,w hile its cosine, cotangent, and case
cant decrease.
On the other hand, if w e su ppose a: to decrease gradu ally ,
the reverse changes in its fu nctions w ill occu r .
If w e su ppose a: to decrease to OP w ill coincide w ith GAand be parallel to BS. Therefore, PM and AT w ill vanish ,OM will become equ al to 0A,
w hile BS and OS w ill each beinfinitely long, and be represented in valu e by the symbol so.
And if w e su ppose a: to increase to OP w ill coincidew ith OB and be parallel to AT. Therefore, PM and OS w illeach be equ al to OB , OM and BS w ill vanish, w hile AT and
OT w ill each be infinite in length.
Hence, as the angle ac increases from 0° tosin e: increases from 0 to 1 ,cos a: decreases from 1 to 0,tan a: increases from 0 to co,cot e: decreases from so to 0,sec s: increases from 1 to co,cscx decreases from 00 to 1 .
12 TRIGONOMETRY.
EXERCISE IV.
1 . Express the following fu nctions as fu nctions of thecomplem entary angle :
Sin tan csc 18° 10 cot 82°
cos cot cos 37° csc 54°
2 . Express the follow ing fu nctions as fu nctions of an
angle less than
sin tan csc 69° 2 '
cos,cot cos 85°
Given tan 30° i V3 ; find cot 60°
Given tan A cot A ; find A.
Given cos A sin 2 A ; find A .
Given sin A cos 2 A ; find A .
Given cos A sin (45°
1} A) find A .
Given cot 1} A tan A ; find A.
Given tan (45°
A) cot A ; find A .
10. F ind A if sin A cos 4 A.
11 . Find A if cot A tan 8 A.
12. Find A if cot A tan nA .
p
a
n
a
w
a
w
§ 6. RELATIONS OF THE F UNCTIONS OF AN ANGLE .
F ormu la Since (F ig. 5) a2 b”: 0
2, therefore,
02+ 1 or
Therefore (sin A)2
(cos A)2 1
or,as u su ally w ritten for convenience,
sin’A cos’A 1 .
That is The su m of the squ ares of the sine
is equ a l to u nity .
TRIGONOMETRIC FUNCTIONS. 13
F orm u la [1] enables u s to find the cosine of an angle w henthe sine is know n, and vice versa . The valu es of sin A and
of cos A dedu ced from [1 ] are
sin A 1 coszA
,cos A 1 sinzA.
F ormu la Since
a b a c d
c'
c s c b b’
therefore tanA xi [2]
That is The tangent of an angle is equ a l to the sine divided
by the cosine.
F orm u la [2] enables u s to find the tangent of an anglew hen the sine and the cosine are know n.
F ormu la Since
1,
and
therefore sinA X 080 A 1
cosA X seeA 1
tanA X cotA 1
That is : The sine and the cosecant of an angle, the cosine
and the secant, and the tangent and the cotangent, p a ir by
p a ir , a re recip roca ls.
The equ ations in [3] enable u s to find an u nknow n fu nction contained in any pair of these reciprocals w hen the otherfu nction in this pair is know n.
14 TRIGONOMETRY.
EXERCISE V.
1 . Prove Form u las [1] u sing for the fu nctionsline valu es in the u nit circle given in 5 3.
Prove that2 . 1 tan°A sec
’A.
3. 1 cot°A csc’A.
NOTE . Equ ations 2 and 3 shou ld be remembered.
cos Acm; A
Sin A
5. sinA secA tanA.
6. sinA cotA cosA .
7. cosA cscA cotA.
8 . tanA cosA sinA .
9 . sinA secA cotA 1 .
10. 0OSA csc A tanA 1 .
11 . (1 sin’A) tanzA sin’A.
12 . m cotA cosA.
13. (1 tanzA) sin
zA tanzA.
14 . csczA (1 sin’A) cot’A.
15. tan’A cos’A cos
°A 1 .
16. (sin’A cos
’A)° 1 4 Sin°A cosf
'A.
17. (1 tan°A)°
sec‘A 4 tan’A .
18.
sm A COSAsecA cscA.
00 8A SinA
19 . sin‘A cos‘A sinzA cos
zA.
20. secA cosA SinA tanA.
21 . cscA sinA cosA cotA.
cosA 1+ sinA22 °
1 sinA
TRIGONOMETRIC FUNCTIONS . 15
7. APPLICATION OF FORMULAS [1]l
Form u las and [3] enable u s, w hen any one fu nc
tion of an angle is given, to find all the others . A given valu eof any one fu nction, therefore , determ ines all the others .
EXAM PLE 1. Given sinA g ; find the other fu nctions .
By cos A= x/1
2 1 3By
By cot A see A g«5, cscA
EXAMPLE 2. Given tan A 3 find the other fu nctions .
sin ABy cos AAnd by sin°A cos
’A 1 .
If w e solve these equ ations (regarding sin A and cos Atw o u nknow n qu antities) , w e find that,
Then by cot A= 1, oscA= § «fl—0.
EXAMPLE 3. Given see A m ; find the other fu nctions .
1
711.cos A
16 TRIGONOME'
I‘RY .
EXERCISE VI .
F ind the valu es of the other fu nctions, w hen
1 . sin A = ; g. 5 . tan A 9 .
2 . sin A = 0 .8 . 6 . cot A = 1 . 10. sin A = m .
2m3, cos A z g-O
r. 7. cot A = 0 .5 . 11 . 8 111 A
1 + m °
COS A= 0 .28 . 8 . sec A = 2 . 1° COS A 2m n
m + n
13. Given tan 45° 1 ; find the other fu nctions Of 45°
14 . Given sin 30°
5; find the other fu nctions Of 30°
15. Given csc 60° find the other fun ctions Of
16 . Given tan 15° 2 V3 ; find the other fu nctions of
17 Given cot 22° V2+ 1 ; find the other fu nctions
of 22°
18 . Given sin 0°
0 ; find the other fu nctions of19 . Given sin 90
° = 1 ; find the other fu nctions Of20. Given tan 90 co find the other fu nctions of
Express the valu es of all the other fu nctions in term s
Of Sin A.
22 . Express the valu es Of all the other fu nctions in term sof cos A.
23. Express the valu es of all the other fu nctions in term sof tan A.
24 . Express the valu es Of all the other fu nctions in term sOf cot A .
25. Given 2 sin A cos A find sin A and cos A .
26. Given 4 sin A tan A ; find sin A and tan A .
27 . If sin A : cos A find sin A and cos A .
28 . Transform the qu antity tan°A cot2A sinzA cos
’A
into a form containing only cos A .
29 . Prove that sin A cos A (1 tan A) cos A .
P rove that tan A cot A see A X csc A .
TRIGONOMETRIC FUNCTIONS . 17
AB C (F ig.
F IG . 6.
5 9 . F UNCTIONS OF 30° AND
Let ABC (F ig. 7) be an equ ilateral triangle, in w hich thelength of each side is equ al to 1 ; and let CD bisect the angleC. Then CD is perpendicu lar to AB and bisects AB ; hence,AD= A}, and on : —
4= di e ; «3.
In the right triangle ADC, the angle ACD and the
angle CAD Whence
sin 30° cos 60° 4.
cos 30° sin 60°
A} x/3.
1tan 30
° cot 60°V5?
cot 30° tan 60° V3.
2sec 30
° csc 60°x/s
csc 30° sec 60°
2 .
F IG . 7.
The resu lts for sine and cosine of and 60°m ay be
easily rem embered by arranging them in the follow ing form
FUNCTIONS OF
an isosceles right triangle,in w hich
the length of the hypotenu se AB isequ al to 1 ; then AC is equ al to B C,
and the angle A is equ al toSince 117 7
2
B_
C2
1 , therefore2 A75
2 = 1 , and AC : VA} : sx/‘
i .
Thereforesin 45
° cos 45° Q 2 .
tan 45° cot 45° 1 .
sec 45° csc 45° x/z.
18 TRIGONOMETRY.
l vs
5vs
EXERCISE VII .
Solve the follow ing equ ations
1 . 2 cos w see a . 7 . 3 tau ’x
2 . 4 sin x csc x . 8 . tan sc+ cot zc= 2.
3. tan a: 2 sin x.
'
9 . sin”a: cos a: i .
4 . see a: tan as. 10. tan“a: see a: 1 .
5 . sin“a: 3 cos’ x. 11 . Sin a: cos a: 2 .
6. 2 sin“a: cos
°zc 3 12 . tan“
a: csc“ a: 3.
13. 2 cos x+ sec x=
14 . cos”a: 81112 a: sin r .
sin x+ cot x= 1 + 2 cos x.
16. Sin2 a: tan2a: 3 cos’ a .
17 . tan x+ 2 cot a: g csc x.
NOTE . Wentw orth Hill’s F ive-place trigonom etric and logarithm ictables have fu ll explanations, and directions for u sing them . Before proceeding to Chapter II. the stu dent shou ld learn how to u se these tables.
Table VI. is to be u sed in solu tions w ithou t logarithms. This fou rplace table contains the natu ral fu nctions of angles at intervals of
The decimal point m u st be inserted before each valu e given,except
w here it appears in the valu es of the table .
20 TRIGONOMETRY.
26
41136
13712
a
CASE II.
Given A 13°
a find B ,o,c.
1 . B 90°
A 76°
b= a cot A .
0.
F IG . 9.
CASE III .
Given A= 27° b= 31 find B, a , c.
cosA ;
b‘
l e . 100
sin A
14 484
7160
4828
2332
B= 90°— A= 62°
a = btan A.
04
9
O
I
Q‘
THE RIGIIT TRIANGLE .
CASE IV.
Given a = 47,
find A,B, b.
B = 90°—A .
—a2
A
a 47, c 63 .
63)44 1
2 90
2 52
sinA 380
A 48°
15' 378
B 41°45' 2
CASE V.
Given a 121 , find A,B,c.
a
I. tanA —Z
F IG . 12.
21
31 , cos A
26 682
4 3180
3 5576
7604
22 TEIGONOMETRY.
a 121 , b 37 .
37) 121111
100
74
A 260
A 73°
259
B 17°
1
12. GENERAL METHOD OF SOLVING THE RIGHT TRIANGLE.
From these five cases it appears that the general m ethod Offinding an u nknow n part in a right triangle is as follow sChoose from the equ a tion A+B and the equ a tions tha t
define the fu nctions of the angles, an equ a tion in w hich the
requ ired p a rt only is u nknow n solve this equ a tion, if noces
sa ry, to find the va lu e of the u nknow n p a rt ; then comp u te the
va lu e.
NOTE . In Case IV.,if the given sides (here a and c) are nearly alike in
valu e , then A is near and its valu e cannot be accu rately fou nd fromthe tables, becau se the Sines of large angles differ little in valu e (as isevident from F ig. In this case it is better to find B first, by means ofa formu la proved later . See form u la 530 ; viz. ,
tan } B
EXAMPLE . Given a
c
C a
c a
tan i B
5° 44'
B 11° 28’
A 78° 32’
TH E RIGHT TRIANGLE.
EXERCISE VIII.
1 . In Case II . give another w ay of finding c, after b
been fou nd.
2 . In Case III . give another w ay of finding c, after a
been fou nd.
3. In Case IV. give another w ay Of finding 6, afterangles have been fou nd.
4 . In Case V. give another w ay of finding 6, after
angles have been fou nd .
5 . Given B and 0 ; find A,a, b.
6 . Given B and b ; find A,a,c. .
7 . Given B and a ; find A,b, c.
8 . Given 6 and c ; findA, B, a .
Solve the follow ing right triangles
GIVEN .
23
24 TRIGONOME '
I‘
RY.
§ 13. SOLUTION BY LOGARITHMS .
CASE I .
Given A 34°
c find B, a , b.
1 . B : 90°
A 55°
a
SIII A ;c
b z o cos A .
F IG . 13.
log 0 log sinA
10
CASE II .
Given A 62°
a 8 ; find B,b, 0 .
sinA .
a = c sinA, and c
F IG . 14.
log a log cotA
10
log 0 log cosA
300
10
log 6 - log a colog SinAlog 0.
colog sinA:
log 6
6
THE RIGHT TRIANGLE.
CASE III .
G iven A= 50° b 88 ; find B, a , c.
1 . B 90°
A 39°
2 . g tan A ;
-bcos A .
c
b = c cosA, and c
logb log tanA
0 10
CASE IV.
Given 0 a find A,B,b.
1 . sinA
2 . B = 90° — A.
3.
9cotA ; b= a cotA .
Cb
log sinA log a colog clog acolog c 10
log sinA 10
A 54°
31 '
B 35°
29’
F IG . 15 .
F IG . 16.
— logb+cologcosA
log a log cotA
10
26 TRIGONOMETRY.
CA SE V.
Given a 40, b 27 ; find A,B, c.
1 . tan A
2. B 90°
A.
3.
2 sinA .
6
a. 0 sinAF IG . 17.
log tanA log a colog6log 0.colog 6 10
log tanA 10
A 55°
59'
B 34°
1'
NOTE . In Cases IV. and V . the u nknow n side m ay also be fou nd fromthe equ ations
(for Case IV. ) 0 m d )(c a) ;
(for Case V. ) 6 \/a2
These equ ations express the valu es of b and 6 directly in terms of the
tw o given sides ; and if the valu es of the sides are simple nu mbers (e.g . 5,
12 , it is often easier to find 0 or c in this w ay. Bu t this valu e of c isnot adapted to logarithms, and this valu e of b is not so readily w orked ou t
by logarithms as the valu e of 0 given u nder Case IV. See also 512 , Note .
§ 14 . AREA OF THE RIGHT TRIANGLE.
It is show n in Geom etry that the area Of a triangle is equ alto one-half the produ ct Of the base by the altitu de .
Therefore , if a and b denote the legs Of a right triangle,and F the area, F 2 1}ab.
By m eans of this formu la the area m ay alw ays be foundw hen a and b are given or have been compu ted.
a
sin A
logc— loga+cologsinA
log acologsinA
log 0
c
28 TRIGONOMETRY.
REQUIRED.
a= 992, B 76°
a 73, B 68°
a 2 .189, B 45°
b 4, A 37°
c 8590, a 4476.
c= 86.53, a = 71.78.
C= 9 .35, a=
c= 2194, 0=
0 ba = 38.313, ba b
a= 0
a 0
6 a=
Compu te the u nknown parts
31 . a = 5,32 . a c 70.
33.
34 . c = 7,35. b 12, A : 29
°
Find the valu e of F in term s of c and A.
Find the valu e of F in term s of a and A.
Find the valu e of F in term s of b and A.
Find the valu e of F in term s Of a and 0 .
Given F =58, a 10 ; solve the triangle .
Given F = 18, b 5 ; solve the triangle .
Given F = 12, A solve the triangle .
Given F = 100, c 22 ; solve the triangle .
F ind the angles of a r ight triangle if the hypotenu seto three times one of the legs.
and also the area, having given
36. c 68, A 69°
37. c 27, B 44°4 '
38. a : 47, B 48°
39. b 9, B 34°
40. c B 86°
THE RIGHT TRIANGLE. 29
50. Find the legs of a right triangle if the hypotenu se 6,
and one angle is twice the other .
51 . In a right triangle given 0, and A nB ; find a and b.
52 . In a right triangle the difierence betw een the hypote
nu se and the greater leg is equ al to the difierence betweenthe tw o legs ; find the angles .
The angle of elevation of an object (or angle of depression,if the object is below the level of the observer) is the anglewhich a line from the eye to the object makes w ith a horizontal l ine in the sam e vertical plane .
53. At a horizontal distance of 120 feet from the foot of asteeple , the angle of elevation of the top w as fou nd to be60
° find the height of the steeple .
54 . From the top of a rock that rises vertically 326 feet ou tof the water, the angle of depression of a boat w as fou nd to be
find the distance of the boat from the foot of the rock.
55 . How far is a m onu ment, in a level plain, from the eye,
if the height of the monum ent is 200 feet and the angle ofelevation of the top 3° 30'
56. In order to find the breadth of a river a distance ABis m easu red along the bank , the point A being directly op
posite a tree 0 on the other side . The angle ABC is alsom easu red. If AB is 96 feet, and ABC is 21
°14
' find the
breadth of the river.If AB C w ere w hat w ou ld be the breadth of the river57 . Find the angle . of elevation of the su n when a tow er
a feet high casts a horizontal shadow 6 feet long. Find theangle w hen a 120, b 70.
58 . How high is a tree that casts a horizontal shadow 6 feetin length w hen the angle of elevation of the su n is A° Findthe height of the tree when 6 80, A
30 Tnu mxomnrnv .
59 . What is the angle of elevation of an inclined plane if itrises 1 foot in a horizontal distance of 40 feet
60. A ship is sailing du e north-east w ith a velocity of 10m iles an hou r. Find the rate at w hich she is m oving du enorth, and also du e east.
61. In front of a w indow 20 feet high is a flow er-bed 6 feetw ide . How long m u st a ladder be to reach from the edge of
the bed to the window
62 . A ladder 40 feet long m ay be so placed that it w ill reacha w indow 33 feet high on one side of the street, and by tu rning it over withou t m oving its foot it w ill reach a w indow 21
feet high on the other side . Find the breadth of the street.
63. From the top of a hill the angles of depression of tw o
su ccessive m ilestones , on a straight level road leading to the
hill, are observed to be 5°and Find the height of the hill .
64 . A fort stands on a horizontal plain. The angle of
elevation at a certain point on the plain is and at a point100 feet nearer the fort it is How high is the fort
65. From a certain point on the grou nd the angles of elevation of the belfry of a chu rch and of the top of the steeplewere fou nd to be 40° and 51° respectively. From a point 300feet farther ofi , on a horizontal line , the angle of elevation of
the top of the steeple is fou nd to be 33° Find the
distance from the belfry to the top of the steeple .
66 . The angle of elevation of the top 0 of an inaccessiblefort observed from a point A,
is At a point B, 219 feet
from A and on a l ine AB perpendicu lar to AC, the angle ABCis 61
°F ind the he ight of the fort .
THE IsoscELEs TRIANGLE . 31
§ 15 . THE IsoscELEs TRIANGLE .
An isosceles triangle is divided by the perpendicu lar fromthe vertex to the base into tw o equ al right triangles .Therefore, an isosceles triangle is determ ined by any tw o
parts that determ ine one of these right triangles .Let the parts of an isosceles triangle AB C (F ig. among
w hich the altitu de CD is to be in
c lu ded, be denoted as follow s
a one of the equ al sides,0 the base,h the altitu de,A one of the equ al angles,C the angle at the vertex.
F or example : Given a and c ;
qu ired A, 0, h.
m 18“
c
a
2 . 2A= 2 (90°— A) .
1 . cos A
3. It may be fou nd by any one of the equ ations
_a2 .
w hence h= (cx—565724 a sin A ; 50 tan A.
The area F of the triangle m ay be fou nd, w hen 0 and k are
given or have been compu ted, by m eans of the form u la
F = «Qoh.
32 TRIGONOMETRY.
EXERCISE X.
Solve the following isosceles tr iangles, the anglesto the nearest second
Given a and A ; find C, c, h.
Given a and C ; find A, c,
Given 0 and A ; find C, a , h.
Given a and C ; find A,a , h.
Given h and A ; find C, a , 0 .
Given h and C ; find A,a,0 .
Given a and h ; find A, C, 0 .
Given 0 and h ; findA, C, a .
Given a c= 11 find A, C, h.
10. Given a=0.295, A= 68 find 0, h, F .
11 . Given 0 C=69° find a , h, F .
12. Given h= 7.4847,A= 76°14' find a
,c, F .
13. Given a = 6.71, h=6.60 ; findA, C, c.
14 . Given c= 9 , findA, C, a .
15. Given o=z i 47, F = findA, C, a, h.
16. Given F = 43.68 ; find A, C, a , c.
17 Find the valu e of F in term s of a and c.
18 . Find the valu e of F in term s of a and C.
19. F ind the valu e of F in terms of a and A.
20. Find the valu e of F in term s of h and C.
21 . A barn is 40 X 80 feet, the pitch of the roof isfind the length of the rafters and the area of both sides of theroof.22 . In a u nit circle what is the length of the chord corre
Sponding to the angle 45° at the centre
23. If the radiu s of a circle is 30, and the length of a chordis 44, find the angle at the centre .
24 . Find the radiu s of a circle if a chord w hose length is5 su btends at the centre an angle of
THE REGULAR POLYGON . 33
25 . What is the angle at the centre of a circle if the corresponding chord is equ al to g of the radiu s26 . Find the area of a circu l ar sector if the radiu s of thecircle= 12, and the angle of the sector
16. THE REGULAR POLYGON .
L ines draw n from the centre of a regu lar polygon (Fig. 19)to the vertices are radii of the circu m scribed circle ; and linesdra w n from the centre to the m iddle points of the sides areradii of the inscribed circle . These lines divide the polygoninto equ al right triangles . Therefore
,a reg ular polygon is
dete rm ined by a right triangle w hose sides are the radiu s ofthe c ircum scribed circle, the radiu s of the inscribed circle, andhalf of one side of the polygon.
If the polygon has n sides, the angle of this right triangleat the centre is equ al to
1 360°
If, also, a side of the polygon, or one of the above-mentioned radii
,is given, this triangle m ay be solved, and the
solu tion gives the u nknow n parts of the polygon .
Let,
91. nu mber of sides,
0 length of one side,r radiu s of circu m scribed circle,11. radiu s of inscribed circle,17 the perim eter
,
F : the area .
Then, by Geom etry,F 5» hp . F IG . 19.
34 TRIGONou ETRv.
EXERCISE XI .
Given n= 10, c= l ; find r, h,
Given u = i 2, p = 70 ; find r, h, F .
Given n= 18, r = 1 ; find h, F .
Given n= 20, r = 20 ; find h, 0,
Given a = 8, h= 1 ; find r , c, F .
Given n= 11 , F = 20 ; find r,h, c.
7 . Given n= 7, F = 7 ; find r, h, p .
8 . Find the side of a regu lar decagon inscribed in a u nitcircle .
Q
P'
P
Q
N
!‘
9 . Find the side of a regu lar decagon circu mscribed abou ta u nit circle .
10. If the side of an inscribed regu lar hexagon is equ al to1, find the side of an inscribed regu lar dodecagon.
11 . Given n and c, and let 6 denote the side of the inscribedregu lar polygon having 2 12 sides ; find 6 in term s of n and c.
12 . Com pu te the difference betw een the areas of a regu l aroctagon and a regu l ar nonagon if the perim eter of each is 16.
13. Compu te the difference betw een the perim eters of aregu lar pentagon and a regu lar hexagon if the area of each is12 .
14 . From a squ are w hose side is equ al to 1 the corners arecu t aw ay so that a regu lar octagon is left . Find the area ofthis octagon.
15 . Find the area of a regu lar pentagon if its diagonal s areeach equ al to 12 .
16. The area of an inscribed regu lar pentagon isfind the area of a regu lar polygon of 11 sides inscribed inthe sam e circle .
CHAPTER III.
G O N IOM E TRY .
5 17. DEEIN rrION or GONIonETRv.
IN order to prepare the w ay for the solu tion of an obliqu etriangle , w e now proceed to extend the definitions of the
trigonom etric fu nctions to angles of all magnitu des, and to
dedu ce certain u sefu l relations of the functions of differentangles .That branch of Trigonom etry w hich treats of trigono
m etric fu nctions in general, and of the ir relations, is cal ledGoniometry.
5 18 . ANGLES on ANY MAGN ITUDE .
Let the radiu s OP of a circle (F ig. 20) generate an angle bytu rning abou t the centre 0 . This angle w ill be m easu red by
the are described by the point P ;and it m ay have any m agnitu de ,becau se the are described by P
m ay have any magnitu de .
Let the horizontal line 0A b ethe initial position of OP , and
let OP revolve in the directionshow n by the arrow ,
or oppositeto the w ay clock hands revolve.
Let, also, the fou r qu adrants intow hich the circle is divided by thehor izontal and vertical diam eters
AA '
,BB '
,be nu mbered I.
, II . , III . , IV.,in the direction of the
m otion.
F IG . 20.
GON IoM RTRv . 37
Du ring one revolu tion OP will form with OA all angles from0
° to Any particu lar angle is said to be an angle of the
q u adrant in w hich OP lies ; so that,Angles betw een 0
°
and 90°
are angles of Q u adrant I .
Angles betw een 90°and 180
°
are angles of Q u adrant II .An gles betw een 180° and 270° are angles of Qu adrant III .Angles betw een 270° and 360° are angles of Qu adrant IV.
If OP m ake another revolu tion, it w ill describe all anglesfrom 360
° to and so on.
If OP ,instead of m ak ing another revolu tion in the direc
tion of the arrow , be su pposed to revolve backw a rds‘
abou t 0 ,
th is backw ard motion tends to u ndo, or cancel, the originalforw ard m otion. Hence, the angle thu s generated m u st beregarded as a nega tive angle ; and this negative angle m ay,
obviou sly, have any magnitu de . Thu s w e arrive at the con
ception of an angle of any m agnitu de , positive or negative .
19 . GENERAL DEFIN ITIONS on THE F UNCTIoNs.
The definitions of the trigonometric fu nctions m ay be
extended to al l angles, by m ak ing the fu nctions of any angleequ al to the line valu es in a u nit circle draw n for the anglein qu estion, as explained in 4 . Bu t the lines tha t rep resent
the sine, cosine, tangent, and cotangent m u st be rega rded as
nega tive, if they a re opp osite in direction to the lines tha t rep ro
sent the corresp onding fu nctions of an angle in thefi rst qu ad
rant; and the lines tha t rep resent the secant and cosecant m u st
be rega rded as nega tive, if they a re opp osite in direction to the
m oving radiu s.
Figs. 21—24 show the fu nctions draw n“
for an angle AOP ineach qu adrant, taken in order. In constru cting them , it m u stbe rem embered that the tangents to the circle are a lw ays
draw n throu gh A and B,never throu gh A
'or
Let the angle AOP be denoted by x ; then, in each figu re ,
38 TRIGONOMETRY.
the absolu te valu es of the fu nctions (that is, their valu eswithou t regard to the signs and are as follow s‘
:
sinx=MP , tanz=AT, secz = OT,
cosa Oil/I, cote: BS, cscx OS.
F IG . 21. F IG . 22.
F IG . 23. F IG . 24.
Keeping in m ind the position of the points A and B , w e m ay
define in w ords the first fou r fu nctions of the angle 2: thu s :sina: the vertical projection of the m oving radiu s ;cosx the horizontal projection of the m oving radiu s ;
the distance m easu red along a tangent to the circletanx from the beginning of the first qu adrant to the
m oving radiu s produ ced ;
GON IOMETRY . 39
the distance measu red along a tangent to the circlecota: from the end of the first qu adrant to the moving
radiu s produ ced.
Seca: and cscx are the distances from the centre of thecircle m easu red along the m oving radiu s produ ced to thetangent and cotangent, respectively.
20. ALGEBRAIG SIGNS on THE FUNCTIONS .
The lengths of the lines, defined above as the fu nctions ofany angle, are expressed nu m erically in term s of the radiu so f the circle as the u nit. Bu t, before these lengths can bet reated as algebraic qu antities, they m u st have the Sign orprefixed, according to the condition stated in 19 .
The reason for this condition lies in that fu ndamentalrelation betw een algebraic and geom etric magnitu des
,in virtu e
of w hich contra ry signs in Algebra corresp ond to opp osite
d irections in Geom etry.
The sine MP and the tangent AT alw ays extend from the
horizontal diam eter, bu t sometimes up w a rds and som etim esdow nw a rds ; the cosine OM and the cotangent BS alw aysextend from the vertical diameter, bu t sometim es tow a rds the
r ight and som etim es tow a rds the left. The fu nctions of anangle in the first qu adrant are assu m ed to be positive . Therefore,1 . Sines and tangents extending from the horizontal
diam eter up w a rds, are positive ; dow nw a rds, negative .
2 . Cosines and cotangents extending from the verticaldiameter tow a rds the r ight, are positive ; tow a rds the lef t, arenegative .
The signs of the secant and cosecant are alw ays m ade toagree w ith those of the cosine and sine, respectively . Thisagreem ent is secu red if secants and cosecants extending fromthe centre , in the direction of the m oving radiu s, are consideredpositive ; in the opp osite direction, negative .
40 TRIGONOMETRv .
Hence, the signs of the fu nctions for each qu adrant are
In Q u adrant I . a ll the fu nctions are positive .
In Q u adrant II. the sine and cosecant only are positive .
In Q u adrant III . the tangent and cotangent only are positive .
In Qu adrant IV. the cosine and secant only are positive .
21 . F UNCTIONS or A VARIABLE ANGLE .
Let the angle a: increase continu ou sly from 0° to
w hat changes w ill the valu es Of its fu nctions u ndergo ?It is easy, by reference to Fig. 25, to trace these changesthrou ghou t all the qu adrants.
F IG . 25.
1 . The In the first qu adrant, the sine MP increasesfrom O to 1 ; in the second it remains positive, and decreasesfrom 1 to 0 ; in the third it is negative, and increases inabsolu te valu e from 0 to 1 ; in the fou rth it is negative , anddecreases in absolu te valu e from 1 to 0.
GON IOMETRY. 41
2 . The Cosine. In the first qu adrant, the cosine 0211 decreases from 1 to 0 ; in the second it becom es negative, andincreases in absolu te '
valu e from 0 to 1 ; in the third it isnegative, and decreases in absolu te valu e from 1 to 0 ; in thefou rth it is positive, and increases from 0 to 1 .
3. The Tangent. In the first qu adrant, the tangent ATincreases from 0 to cc ; in the second qu adrant, as soon as the
angle exceeds 90° by the smallest conceivable am ou nt, them oving radiu s OP '
,prolonged in the direction opposite to that
of OP '
,will cu t AT at a point T ’ situ ated very far below A ;
hence, the tangents of angles near 90°in the second q u adrant
have very large nega tive valu es . AS the angle increases, the
tangent AT ' continu es negative, bu t dim inishes in absolu tevalu e . When a: then T ' coincides w ith A, and tan 180
°
0 . In the third qu adrant, the tangent is positive, and
increases from 0 to oo ; in the fou rth it is negative, and
decreases in absolu te valu e from 00 to 0.
4 . The Cotangent. In the first qu adrant, the cotangent BSdecreases from co to O; in the second qu adrant it is negative,and increases in absolu te val u e from 0 to cc ; in the third andfou rth qu adrants it has the sam e Sign, and u ndergoes the sam e
changes as in the first and second qu adrants, respectively .
5 . The Secant. In the first qu adrant, the secant OT in
creases from 1 to cc ; in the second it is negative (be ingm easu red in the direction Opposite to that of and
decreases in absolu te valu e from 00 to 1 ; in the third itis negative, and increases in absolu te valu e from 1 to oo ; in
the fou rth it is positive, and decreases from 00 to 1 .
6. The Cosecant. In the first qu adrant, the cosecant OSdecreases from no to 1 ; in the second it is positive, and
increases from 1 to oo ; in the third it is negative, and
decreases in absolu te valu e from 00 to 1 in the fou rth itis negative, and increases in absolu te valu e from 1 to co.
42 TRIGONOMETRY.
The lim iting valu es Of the fu nctions are as follow s
180°
Sines and cosines extend from + 1 to — 1 ; tangents and
cotangents from on to cc ; secants and cosecants from+ 00 to + 1, and from —1 to 00 .
In the table given above the dou ble Sign ziz is placed before 0 and Go.
F rom the preceding investigation it appears that the fu nctions a lw ays
change sign in passing through 0 and co ; and the Sign or prefixedto 0 or co Simply show s the direction from w hich the valu e is reached .
Take, for example , tan The nearer an acu te angle is to the
greater the positive valu e of its tangent ; and the nearer an obtu se angleis to the greate r the negative valu e of its tangent. Whenthe angleis OP (F ig. 25) is parallel to A T, and cannot meet it. Bu t tan 90°
m ay be regarded as extending either in the positive or in the negativedirection ; and according to the view taken, it w ill be on or oo.
§ 22 . FUNCTIONS OF ANGLES LARGER THAN
The fu nctions Of 360°
+ w are the sam e in Sign and in
absolu te val u e as those of x ; for the m oving radiu s has thesam e position in both cases . If n is a positive integer
,
The fu nctions of (n X 360°
x) a re the sam e as those of a .
F or example : The fu nctions Of 2200°(6 X 360°
are
equ a l to the fu nctions of
44 TRIGONOMETRY.
EXAMPLE. Given sin a: 3, and tan a: negative ; find thevalu es of the other fu nctions .Since sin a: is positive, a: m u st be an angle in Qu adrant I .
‘orII. bu t, Since tan a: is negative, Q u adrant I . is inadmissible .
By
Since the angle is in Qu adrant II. the m inu s Sign m u st betaken, and w e have
cos a:By [2] and
tan s: cot s: sec s: —g,
EXERCISE XII.
1 . Constru ct the fu nctions of an angle inQ u adrant II .What are their signs2 . Constr uct the fu nctions of an angle in Qu adrant III .What are their Signs3. Constru ct the fu nctions of an angle in Q u adrant IV .
What are their signs4 . What are the Signs of the fu nctions of the follow ing
angles : 3800°
5. How many angles less than 360° have the valu e of th e
sine equ al to 9, and in w hat qu adrants do they lie6. How m any valu es less than 720
°can the angle a: have
if cos and in w hat qu adrants do they lie7 . If w e take into accou nt only angles less than how
m any valu es can a: have if Sin x= 9 if cos x=g if cos x=if tan x if cot a —7
8 . Within w hat lim its mu st the angle a: lie if cos a:if eot zs=4 if sec x= 80 if cso sz —3 (if a:9 . In w hat qu adrant does an angle lie if Sine and cosine
are both negative if cosine and tangent are both negativethe cotangent is positive and the sine negative
GON IOMETRY. 45
10 . Betw een 0° and 3600° how many angles are there w hoseSine s have the absolu te valu e g Of these Sines how m any
are positive and how many negative11 . In finding cos a: by m eans Of the equ ation cosa:
i 1— sin2x,w hen m u st w e choose the positive Sign and
w hen the negative Sign12 . Given cos a: i ; find the other fu nctions w hen
a: is an angle in Qu adrant II .13. Given tan x= 3 ; find the other fu nctions w hen cc is
an angle in Q u adrant III .14 . Given sec x=+ 7, and tan a negative ; find the otherfu nctions of x.
15 . Given cot z —3 ; find all the possible valu es of theother fu nctions .16 . What fu nctions Of an angle of a triangle m ay be nega
tive In w hat case are they negative17 What fu nctions of an angle of a triangle determ ine the
angle , and w hat fu nctions fail to do so18 . Why m ay cot 360
°
be considered equ al either to+ 00
or to co
19 . Obtain by m eans of F orm u las [1] the other fu nctions Of the angles given
(i.) tan 90°
00 . (iii cot 270°= 0 .
(ii.) cos 180°
1 . (iv. ) csc 360°
oo.
20. F ind the valu es of sin tan cos cot 7sin csc21 . For w hat angle in each qu adrant are the absolu te valu es
of the sine and cosine equ al‘7
Compu te the valu es of the follow ing expressions22 . a sin 0
°6 cos 90° c tan
23. a cos 90°
b tan 180°
c cot
a sin 90°
b cos 360°
(a b) cos
25 . (a2 b“) cos 360
°
4 ab sin 27
46 TRIGONOMETRY.
24 . REDUCTION OF FUNCTIONS To THE FIRST QUADRAN T .
In a u nit circle (F ig. 26) draw tw o diameters PR and OS
equ ally inclined to the hor izontal diameter AA'
,or so that the
angles AOP ,A 'OQ , A
'OR , and
AOS shall be equ al. F rom the
points P , Q , R, S let fall pe rpendicu lars to AA'
; the fou r r ighttriangles thu s form ed, with a
comm on vertex at O, are equ a l ;becau se they have equ al hypotenu ses (radii Of the circle) and
E1"2°equ al acu te angles at 0 . There
fore, the perpendicu lars PM, ON, RN, SM, are equ al . Now
these fou r lines are the Sines of the angles AOP ,AOQ , AOR ,
and AOS, respectively . Therefore, in absolu te va lu e,
sinA0P = SinAOQ sin AOR= sinAOS.
And from 23 it follow s that in absolu te va lu e the cosinesof these angles are also equ al ; and likew ise the tangents, thecotangents
,the secants, and the
Hence, for every a cu te angle (AOP ) there is an angle in each
of the higher qu adrants w hose fu nctions, in absolu te va lu e,are
equ a l to those of this a cu te angle.
Let A AOP = a~
, A P OB = y ; then a: 1 and the
fu nctions of a: are equ al to the co-nam ed fu nctions Of yand A AOQ (in Q u adrant II .) 180
° —:c 90
°1
A AOR (in Qu adrant III .) 180°
a: 270°
y,
A AOS (in Qu adrant IV.) 360°
a: 270°
y.
Hence, prefixing the proper Sign w e have
In fu tu re, secants, cosecants, versed Sines, and coversed Sines w ill bed isregarded . Secants and cosecants m ay be fou nd by versed since
and coversed Sines by VII. and VIII . , page 5, if w anted , bu t they are
M ‘d om u sed in compu tations.
GON IOMETRY. 47
Angle in Q u adrant III
sin (180°
x) sin at. sin (90°
y)cos (180
°x) cos 23. cos (90
°
y)tan (180
°x) tan as. tan (90
°
y)cot (180
°x) cot a . cot (90
°
y)
Angle in Qu adrant III.
sin sin x. sin (270°—
y)cos (180
°93) cos x. cos (270
°
y)tan (180
°
x) tan x. tan (270°
y)cot (180
°ac) cot a . cot (270
°
y)
Angle in Qu adrant IV.
sin (360°
as) sin 23. sin (270°
y) cos y.
cos (360°
it ) cos it . cos (270°
y) sin y.
tan (360°
a ) tan a . tan (270°
y) cot y.
cot (360°
x) cot a . cot (270°
y) tan y.
REMARK. The tangents and cotangents m ay be fou nd directly fromthe figu re , or by form u la
It is evident from these form u las,1 . The f u nctions of a ll angles can be redu ced to thefu nctions
of angles not grea ter than
2 . If an a cu te angle be added to or su btracted from 180°or
the fu nctions of the resu lting angle a re equ a l in absolu te
va lu e to the l ike-namedf u nctions of the a cu te angle bu t if an
a cu te angle be added to or su btra cted f rom 90°
or the fu nc
tions of the resu lting angle a re equ a l in absolu te va lu e to the
co-named fu nctions of the a cu te angle.
3. A given va lu e of a sine or cosecant determ ines tw o supp le
menta ry angles, one a cu te, the other obtu se ; a given va lu e of any
other fu nction determ ines only one angle : a cu te if the va lu e is
p ositive, obtu se if the va lu e is nega tive. [See fu nctions
(180°
48 TRIGONOM E'I‘RY.
5 25. ANGLES WHOSE DIFFERENCE ’ IS
The general form Of tw o su ch angles is a: and 90°+ z,and
they m u st lie in adjoining qu adrants. The relations betw een
the ir fu nctions were fou nd in 524 ,bu t only for the case w hen a: isacu te . These relations
,how eve r ,
may be shown to hold tru e for a llvalu es Of x.
In a u nit circle (F ig. 27) drawtw o diam eters PR and QS per
pendicu lar to each other, and letfall to AA' the perpendicu larsP ill , QH , EX, and SN. The
right triangles OMP , OHQ , OKR,
and ONS are equ al, becau se they have equ al hypotenu sesand equ al acu te angles P OLI, OQH , ROK, and OSN .
Therefore,and
Hence,taking into accou nt the algebraic Sign
,
sin AOQ cos AOP ; sin AOS cos AOR ;cos AOQ z — sm AOP ; cos AOS —sin AOR ;
sin AOR cos AOQ ; sin (360°
AOP) cos AOScos AOR sin AOQ ; cos (360
°
AOP) sin AOS.
In all these equ ations, if cc denote the angle on the righthand side , the angle on the left-hand side w ill be 9o°+ r .
Therefore, if a: be an angle in any one of the fou r qu adrants,sin (90
°cc) cos a , tan(90
°x) cot z ,
cos (90°
at) sin a , cot(90°
x) tan a .
In like m anner, it can be show n that all the formu las of
5 24 hold tru e, w hatever be the valu es of the angles a: and y.
Hence, in every case the a lgebra ic sign of the fu nction of theesu lting angle w ill be the sam e as w hen a: and y a re both acu te.
GON IOMETRY . 49
5 26. F UNCTIONS OF A NEGATIVE ANGLE .
If the angle AOP (F ig. 26) is denoted by a , the equ al angleAOS, generated by a backw a rd rotation of the m oving radiu sfrom the initial position OA,
w ill be denoted by —x. It isObv iou s that the position OS Of the m oving radiu s for thisangle is identical w ith its position for the angle 360
°it .
Therefore, the fu nctions of the angle a: are the same asthose Of the angle 360°— x ; or (5
sin x) sin a , tan x)cos a ) cos a
,cot x)
ExERCISE XIII.
1 . Express sin 250° in term s of the fu nctions of an acu teangle less than
Ans . sin 250°
Sin (270° cos
Express the follow ing fu nctions in term s of the fu nctionsof angles less than 45°
2 . sin 8 . Sin 14 . sin 163°
3 . cos 9. cos 15 . cos 195°
4 .
'
tan 10. tan 16 . tan269°
5 . cot 11. cot 17 cot 139°
6 . sec 12 . sec 244°
18 . sec 299°
7 . csc 13. csc 271° 19 . csc 92°
Express all the fu nctions of the follow ing negative anglesin term s of those of positive angles less than 45°
20. 75°
22. 24 .
21 . 23. 25 .
26 . F ind the fu nctions Of
Hm . 180° or , 120°
then apply 5 24 .
50 TRIGONOMETRY.
Find the fu nctions of the follow ing angles
27. 135° 29. 31 . 33.
—30°
28 . 30. 32 . 34 .
35. Given sin a: and cos a: negative ; find the otherfu nctions of as, and the valu e of x.
36. Given cot s: and cc in Qu adrant II. ; find theother fu nctions of a , and the valu e of x.
37 Find the fu nctions ofWhat angles less than 360° have a sine equ al to —5?
a tangent equ al to39 . Which of the angles m entioned in Examples 27—34have a cosine equ al to a cotangent equ al to40. What valu es Of as betw een 0° and 720° w ill satisfy the
equ ation sin a: n}41 . Find the other angle betw een 0° and 360° for w hich thecorresponding fu nction (Sign inclu ded) has the sam e valu e assin cos tan cot sin cos tan244
°
,cot357°
42 . Given tan find sin122°
43. Given cos 333° find tan 117°
Simplify the follow ing expressions :
44 . a cos (90°
cc) 6 cos (90°
cs) .
45. m cos (90°
to) sin (90°
x) .
46. (a b) tan(90°
a) (a b) cot (90°
a ) .
47. as b’ 2 ab cos (180
°x) .
48 . sin (90°
a ) sin(180°
x) cos (90°
23) cos (180°
49. cos(180°-He) cos (270
°
y) sin(180°—He) sin(270
°
y) .
50. tan a: tan y) tan (180°
y) .
51 . F or w hat valu es of a: is the expression sin a: cos a:positive , and for w hat valu es negative Represent the resu ltby shading the sectors corresponding to the negative valu es.52 . Answ er the qu estion of last exam ple for sin a: cos a .
53. F ind the fu nctions of (a: in fu nctions of a .
54 . F ind the fu nctions of (a in fu nctions of x .
52 TRIGONOMETRY.
If these form u las hold tru e for any tw o acu te angle s a: andy, they hold tru e w hen one of the angles is increased by 90
°
Thu s, if for a: w e w rite a ' then, by 5 25,sin (x
'
y) sin (90°
a: y) cos (a: y) ,cos (x
'
y) cos (90°
x y) sin (a: y) .
Hence , by sin (a'
y) cos 2: cos y sin a: sin y,
by cos (x'
y) sin a: cos y cos a: sin y.
Now ,by 5 25, cos as Sin (90
°
a )sin a: cos(90
°x)
Su bstitu te these valu es of cos a: and sin a,then
sin (x’
y) sin x' cos y cos x' sin y,cos (x
'
y) cos w'cos y sin x' sin y.
It follow s that F orm u las [4] and [5] hold tru e if / either
angle is repeatedly increased by 90° therefore they apply to
all angles w hatever .
By § 23.
cos cos 9: cos y sin a: Sin y
If w e divide each term of the nu m erator and denom inatorof the last fraction by cos 2: cos y, and again apply 523, w e
Obtain
l — tanx tany[6]
In like m anner, by dividing each term Of the nu m eratorand denom inator of the valu e of cot by sin s: sin y, w e
Obtain
coty+ cot z
GON IOMETRY. 53
5 28 . F UNCTIONS OF THE DIFFERENCE OF Tw o ANGLES .
In a u nit circle (F ig. 30) let the angle AOB x, COB y ;
then the angle AOC as y.
In order to express Sin (a: y)and cos (a: y) in term s
i
of the
Sines and cosines Of a: and y, drawOF _L OA, CD _L OB ,
DE _L OA,
DG _1_F C prolonged ; then CDsin y , OD= cos y, and the angleDC'G= the angle EDC= T . And,
sin (ax—y) : CF =DE CG.
F IG . 30.
sin as ; hence , DE sin a: X OD= sin 2: cos y.
CG
CD— cos x ; hence, CG = cos x >< CD= cos x sm y .
Therefore, sin(x y) sinx cosy cosx siny.
Again, cos(a: y) OF OE DC .
cos cc ; hence , OE cos a: X OD cos x cos y .
DG
CDSin cc ; hence , DG $111 a: X CD= 8 111 x Sln y .
Therefore, cos(x y) cosxcosy sinx siny . [9]
In this proof, both ac and y are assu m ed to be acu te angles ;bu t, w hatever be the valu es Of a' and y, the sam e m ethod of
proof w ill alw ays lead to F orm u las [8] and w hen du e
regard is paid to the algebraic signs.
The general application Of these form u las m ay be at onceshow n by dedu cing them from the general form u las established in 5 27 as follow s :It is obviou s that (x— y)+ y x. If w e apply F orm u las
[4] and [5] to (x— y)+ y, then
54 TRIGONOMETRY .
sin g(a or sin e: sin(x—y) cosy+ cos(.v
—y) sin y,
cos §(zc or cos a cos(.v—y) cos y— sin (a
—y) s in y.
Mu ltiply the first equ ation by cos y, the second by sin y,
Sin 2: cos y sin(a: y) cos’
y cos (x y) sin y cos y,cos a: sin y sin(a: y) Sin
°
y cos (a: y) sin y cos y ;
w hence,by su btraction,
sin a: cos y cos a: sin y sin (a: y) (sin’y coszy) .
sin°g coszy 1 therefore, by transposing,sin(a: y) sin a: cos y cos a: sin y.
Again, if w e m u ltiply the first equ ation by sin y, the secondequ ation by cos y, and add the resu lts, w e obtain, by redu cing,
cos (a: y)= cos a cos y+ sin a: sin y.
Therefore, Form u las [8] and like [4] and fromw hich they have been derived, are u niversally tru e .
From [8] and by proceeding as in 5 27, w e obtain
cot (x y)coty cotx
F orm u las [4] m ay be combined as follow s
sin (a: i y) sin 2: cos y t cos a: sin y,cos (a: j : y) cos 2: cos y 4: sin a: sin y,
tan x i tan ytan (x i y)
cot (w r y) :cot y zt cot x
GON IOMETRY. 55
5 29 . F UNCTIONS OF TWICE AN ANGLE .
If y= x,F orm u las [4] becom e
sin 2x 2 sinx cosx. [12] cos’x sin’
x. [13]2 tanx cotzx 1
tan 2 xl tan
’x
[14]2 cotx [15]
By these form u las the fu nctions of tw ice an angle are‘
fou nd w hen the fu nctions of the angle are given.
5 30. F UNCTIONS OF HAL F AN ANGLE .
Take the form u lascos
’x sin’
a: 1
cos’a sinza: cos 2 a2 3in’ = 1 — cos Z a
2 cos’x — 1+ cOS Z :c
1—cos
200 8 2:
These valu es, if z is pu t for 2 a , and hence
sin -
l —°g[16]
Hence, by division (5
l —cosz. /1+ cosz
tan i z z zt cot l z i \1— cosz[19]
By these form u las the fu nctions of half an angle m ay be
com pu ted w hen the cosine Of the entire angle is given.
The proper Sign to be placed before the root in each casedepends on the qu adrant in w hich the angle 5z lies . (5Let the stu dent Show from F orm u la [18] that
tan QB Z:
(See page 22, Note .)
for x, becom e
1 cosz.
2
56 TRIGONOMETRY.
5 31 . SUMS AND DIFFERENCES OF FUNCTIONS .
From [5] and by addition and su btraction
sin (ex_ y) 2 sin a cos y,sin (x+ y)
- sin (a y) 2 cos x Sin y,
cos (x-l—y) + cos (a — y)_ 2 cosx cos y,cos — cos (as—y) 2 Sin a Sin y ;
or, by mak ing m+ y= A, and m—y
=B,
and therefore, a: i (A B) , and y= é‘ (A B) ,
sinA+ sinB 2 sin 1} (A+B) cosflA B) [20]sinA—sinB —B) . [21]cosA+ cosB 2 (A+B) cos} (A B) . [22]cosA cosB 2 sin&(A+B) sin i}(A B) . [23]
F rom [20] and by division, w e ObtainsinA sinBsinA— sin B
—B) ,
or, since cotQ(A—B ) tan 1} (I4 B)
sinA+ sinB tan lsinA— sinB tanHA—B)
EXERC ISE XIV;
1 . F ind the valu e of sin (a y) and cos (a: y), w hen sin a
g, cos a: g, sin y T5,” cos y if :
2 . F ind sin(90°
y) and cos (90°
y) by mak ing ac= 90°
in F orm u las [8] andF ind, by F orm u las [4] the first fou rfu nctions of3. 90
°
y. 8 . 360°
y. 13. y.
4 . 180°
y. 9 . 360°
y. 14 . 45°
y.
5 . 180°
y. 10. a: 15. 45°
y.
6 . 270°
y. 11 . a: 16. 30°
y.
7 . 270°
y. 12 . a: 17. 60°
y.
GON IOMETRY. 57
18 . F ind sin 32: in term s of sin to.
19 . Find cos 3a: in term s Of cos x.
20. Given tan i s: 1 find cos cc.
~21 . Given cot i s : x/
‘
i ; find sin e .
22 . Given sin zc= 0.2 find sin fix and cos 5m.
23. Given cos 2: find cos 2x and tan 2x.
24 . Given tan 45° 1 ; find the fu nctions of 22°
25. Given Sin 30°
find the fu nctions of 15°
0sin 33
°Sin3
°
26 . Prove that tan 18cos33
° cos 3,
Prove the follow ing form u las :2 tan re
27 . Sln 2x1 tan’
a
29 . tan 4x
1 tan’a:
28 . cos 2 a:1 tan’
a:30. cot 5x
31 . sin i n j : cos 5x a: Sin a .
cot z zt cot y—
fi: tan x tan y.
o1 tan a
.33. tan (45 x) 1 tan a:
If A,B, C are the angles of a triangle, prove that
34 . sinA + SinB + sin C= 4 COS 1}A cos 5B cosQ C.
35. cosA+ cosB + cos C= 1 + 4 Sin § A sin QB sin Q C.
36. tanA+ tanB+ tan C==tanA tanB tan C.
37. C= cot § A cot } B X cot gC.
Change to form s more convenient for logarithm ic com pu tation :
38 . cot a: tan a . 43. 1 tan a: tan y.
39 . cot a: tan as. 44 . 1 tan a: tan y .
40. cot x+ tan y. 45 . cot x cot y+ 1 .
41 . cot a—tan y. 46. cot a cot y— 1 .
42 .
1 cos 2 a:1 + cos 2x cot a -l—cot y
58 TRIGONOMETRY.
5 32 . ANTI-TRIGONOMETRIC F UNCTIONS .
If y is any trigonoinetric fu nction of an angle a, then x is
said to be the corresponding anti-trigonom etric fu nction Of-y.
Thu s, if y= Sin a,a: is the anti-sine of y, or inverse sine
Of y. The anti-trigonom etric fu nctions of y are w rittensin
“ 1y, tan
“ 1y, see" l
y, vers“ 1
y,
cos—1y, cot‘ l
y, csc" ly, coveTS
" l
y.
These are read, the angle w hose Sine is y, etc.
F or exam ple , sin hence 30°= sin Sim ilarly1
cos‘ 10 Sin
“ 11 and 45°
tan“ 1 1 Sin
—l —z
The symbol —1 mu st not be confu sed w ith the exponent 1 . Thu s
sin—1a: is a very different expression from Sid— r,
’ w hich w ou ld be w ritten
(sin25)- 1
. On the Continent of Eu rope mathem atical w riters em ploy thenotation arcsin, arccos, etc. ,
for sin—1
, cos—1
, etc. Bu t the latter symbolsare most common in England and America.
There is an important difference betw een the trigonometricand the anti-trigonom etr ic fu nctions . When an angle is given,its fu nctions are all completely determ ined ; bu t w hen one
of the fu nctions is given the angle m ay have any one of anindefinite nu mber of valu es . Thu s, if sin y= y may be
or or either of these increased or dim inished by any
integral m u ltiple of 360° or 2 1r, bu t cannot take any othervalu es . Accordingly or 150
°1 2n1r, w here
n is any positive integer . Sim ilarly, tan‘ 1 1= 45° i 2ns or
225°
2 n1r ; i.e . , tan—11 45
°i u rr.
Since one of the angles w hose sine is a: and one of the anglesw hose cosine is 2: together make and since sim ilar relations hold for the tangent and cotangent, for the secant andcosecant, and for the versed sine and coversed Sine
,w e have
0 WS in“ lx COS
" la
23 see
" 1a: csc—l a
CHAPTER IV .
THE OBLIQUE TRIANGLE .
5 33. LAW OF SINES .
LET A,B, C denote the angles of a triangle AB C
and and a,b,c,respectively, the lengths Of the
Sides .Draw CD _LAB , and meeting AB (Fig. 31) or AB p rO
du ced (Fig. 32) at D. Let CD= h.
F IG . 31. F IG . 32 .
In both figu res, SinA.
In F ig. 31 , sinB.
In F ig. 32 , sin (180° B) sinB.
Therefore, w hether h lies w ithin or Withou t the triangle ,w e Obtain, by division
THE OBLIQUE TRIANGLE . 61
By draw ing perpendicu lars from the vertices A and B tothe opposite sides w e may obtain, in the same w ay,
b sinB sinA
0 sin C,
sin C
Hence the Law of Sines, Which m ay be thu s statedThe sides of a tria ngle a re p rop ortiona l to the sines of the
0
If w e regard these three equ ations as proportions, and takethem by alternation, it Will be ev ident that they may be
w ritten in the symmetrical form ,
a b c
sin A SIn B SIn C
Each of these equ al ratios has a Simple geom etrical meaningw hich w ill appear if the Law of Sines is proved as follows
Circu m scribe a circle abou t the triangle AB C (Fig.
and draw the radii OA, OB , 0 0 ;
these radii divide the triangle intothree isosceles triangles . Let R
denote the radiu s . Draw OM
_LB C. By Geom etry , the angleBOC = 2 A ; hence, the angleBOM=A, then BM=R sinBOM
R sinA.
BC or a = 2R sinA.
In like manner, b 2 R sin B,and c 2R sin C. Whence w e
F IG ‘ 33 ‘
obtaina b c
sinA sinB sin C
That is : The ratio of any side of a triangle to the sine of the
opposite angle is nu m erica lly equ a l to the diam eter of the cir
cu mscribed circle .
62 TRIGONOMETRY.
5 34 . LAW OF COSINES .
Thislaw gives the valu e of one side Of a triangle in term sof the other tw o Sides and the angle inclu ded betw een them .
In F igs . 31 and 32,In F ig. 31, BD 0 —AD ;
in F ig. 32, BD =AD c ;
in both cases, BI)2AD’
2 c X AD c“.
Therefore, in all cases, a” m AD”
02 2 c xAD .
Now ,h’+A
-_
D2
b’,
and AD 6 cosA.
Therefore, a9= b°+ 0
° 2 bc cosA.
In like manner, it m ay be proved thatb2= a
2+ c
°— 2 ao cos B ,
— 2 ab cos C.
The three form u las have precisely the sam e form , and the
law may be stated as follow s :The squ are of any side of a triangle is equ a l to the s um of
the squ a res of the other tw o sides,dim inished by tw ice their
p rodu ct into the cosine of the inclu ded angle.
5 35 . LAW OF TANGENTS .
By 5 33,
w hence , by the Theory Of Proportion,
a —b SinA— sinB
a + b sinA -i- SinB
Bu t by page 56,
sinA— sinB —B ).SinA+ sin B
Therefore,
a+ b
THE OBLIQUE TRIANGLE . 63
By m erely changing the letters,
a + o b+ c
Hence the Law Of TangentsThe dif erence of tw o sides of a triangle is to their su m as the
ta ngent of ha lf the difi’erence of the opp osite angles is to the
ta ngent of ha lf their su m .
NOTE . If in [27] b) a , then B > A . The form u la is still tru e , bu t toavoid negative nu mbers, the formu la in this case shou ld be w ritten
b+ a
EXERC ISE XVI .
1 . What do the formu las of 5 33 become w hen one of theangles is a right angle2 . Prove by m eans Of the Law .Of Sines that the bisectorof an angle of a triangle divides the opposite side into partsproportional to the adjacent sides .3 . What does F orm u la [26] becom e w hen A= 9O° w hen
A= 0° w hen A 180° What does the triangle becom e in
each Of these casesNO TE . The case w hen A 90° explains w hy the theorem of 534 is
som etim es term ed the Generalized Theorem of Pythagoras.
4 . Prove (F igs . 31 and 32) that w hether the angle B isacu te or obtu se, c a cos B + b cosA . What are the tw o symmetrical form u las Obtained by changing the letters Whatdoes the form u la becom e w hen B = 90°
5 . From the three follow ing equ ations (fou nd in the lastexample) prove the theorem of 534
c= a cosB + bcosA
b= a cos C+ c cosA,
a= b cos C + c cosB .
HINT . Mu ltiply the first equ ation by c, the second by b, the thirdby a ; then from the first su btract the sum Of the second and third .
64 TRIGONOMETRY .
6 . In F orm u la [27] w hat is the m axim u m valu e of (A—B )7 . F ind the form to w hich F orm u la [27] redu ces, and
describe the natu re of the triangle, w hen
(i.) C 90°
(ii .) A B and B C.
5 36. THE SOLUTION OF AN OBLIQUE TRIANGLE .
The form u las established in 55 33—35, together w ith theequ ation A+B + C are su fficient for solving everycase of an obliqu e triangle . The three parts that determ inean Obliqu e triangle m ay be
1 . One side and tw o angles ;II . Tw o sides and the angle Opposite to one of these sides ;III . Tw o sides and the inclu ded angle ;IV. The three sides .Let A, B, C denote the angles, a , b, c the sides respectively.
5 37. CASE I .
Given one side a , and tw o angles A and B ; find the remain
ing p a rts C, b, and c.
1 . C= 180°
6 sinB a sinB a2 °
a sinA’
bsinA sinA
x sm B '
c sin C a sin C a
a sinA’
sinA sinAX 8m 0 °
EXAMPLE . a A= 45° B 22°
The w ork m ay be arranged as follow sa log a 8
A 45°18 ' colog SinA
B 22°
11 ' log sinB log sin CA B 67
°29' log b log c
C 112°31
' b c
NOTE . When 10 is om itted after a logarithm or a cologarithm,it
be rem embered that the log or the colog is 10 too large .
THE OBLIQUE TRIANGLE . 65
EXERCISE XVII .
1 . Given a 500 A 10° B 46
°
36'
find C 123 b c
2 . Given a 795, A 79° B 44
°41 '
find C 55°
b 0
3. Given a 804 , A 99° B 45
°
find C 35°
b c
4 . Given a 820, A 12° B 141
°
find C 25°
b 0
5 . Given c 1005,
A 78° B 54
°
find C 47°
a b
6 . Given b B= 13° C 57°13'
find A 108°
a 0
7 . Given a 6412 , A 70°
C= 52°
find B 56°
b 0
8 . Given b 999, A 37°
C 65°
find B a 0
9. In order to determ ine the distance of a hostile fort Afrom a place B, a line B C and the angles AB C and B CA
w ere m easu red, and fou nd to be yards, 60°
and
56°
respectively. Find the distance AB .
10. In m ak ing a su rvey by triangu lation,the angles B and
C of a triangle AB C w ere fou nd to be 50° 30' and 122°
re spectively, and the length B C is know n to be 9 m iles .F ind AB and AC.
11 . Tw o Observers 5 m iles apart on a plain,and facing
each other, find that the angles of elevation of a balloon inthe sam e vertical plane w ith them selves are 55
°
and
respectively . Find the distance from the balloon to eachObserver
,and also the height Of the balloon above the plain.
12 . In a parallelogram given a diagonal d and the anglesa: and y w hich this diagonal m akes w ith the sides . F ind thesides . F ind the Sides if d a : 19
°
and y= 42
°
66 TRIGONOMETRY .
13. A lighthou se w as Observed from a ship to bearN. 34°E. ;
after the ship sailed du e sou th 3 m iles, it bore N . 23°E. Find
the distance from the lighthou se to the ship in both positions .
NOTE . The phrase to bear N . 34° E . m eans that the line Of Sight tothe lighthou se is in the north-east qu arter of the horizon, and makes,w ith a line du e north, an angle of
14 . In a trapezoid given the parallel Sides a and b, and the
angles a: and y at the ends of one of the parallel Sides . Findthe non-parallel sides . Compu te the resu lts w hen a 15,
b= 7, a: y= 40°.
Solve the follow ing examples w ithou t u sing logarithm s
15. Given b= A C find a and c.
16. Given 0 A B find a and b.
17 The base of a triangle is 600 feet, and the angles at thebase are 30
°
and Find the other sides and the altitu de .
18 . Tw o angles of a triangle are, the one the otherF ind the ratio of the opposite sides .19 . The angles of a triangle are as 5 10 21 , and the Sideopposite the smallest angle is 3. Find the other sides .20. Given one side of a triangle equ al to 27 the adjacentangles equ al each to F ind the radiu s of the circumscribed circle . (See 5 33, Note .)
5 38. CA SE II .
Given tw o sides a and b,and the angle A opp osite to the
side a ; find the rem a ining p a rts B, C, c.
This case , like the preceding case, is solved by m eans ofthe Law of Sines .
O B 0
SinceSin 2
, therefore 8 111Bsm A a
C= ISO°
68 TRIGONOMETRY.
These resu lts, for convenience, m ay be thu s stated
Tw o solu tions ; if A is acu te and the valu e of a lies betw een6 and 6 sinA.
No solu tion ; if A is acu te and a < bsinA ;
if A is obtu se and a < 6.
One solu tion ; in all other cases .The nu mber of solu tions can Often be determ ined by insPection. In case of dou bt, find the valu e of bsinA.
Or w e m ay proceed to com pu te log sinB . If log SinB= O,the triangle requ ired is a right ,
triangle. If log sinB > O, thetriangle is impossible . If log sinB 0, there is one solu tionw hen a b ; there are tw o solu tions w hen a b.
When there are two solu tions, let B
’, C c
', denote the
u nknow n parts of the second triangle ; then,
B ’= 180°—B, C '= 180° (A+B
'
)=B—A
,
a sin C '
sinA
EXAMPLES .
1 . Given a 16, b 20, A find the remainingparts .
In this case a < b, and A therefore the triangle is impossible.
2 . Given a = 36, b= 80, A= find the remainingparts.
Here w e have bainA 80 x 40 ; so that a < bsinA , and the
triangle is impossible .
3. Given a 72630, b 117480,A 80
°find B
, c, c.
a 72630 colog a Here logsinB > 0.
b 117480 logb no solu tion.
A 50 log sinA
log sin B
THE OBLIQUE TRIANGLE . 69
4 . Given a b A 57°13
'15" find B, c, c.
a colog a c bcosA
b logb log b
A log sinA log cosA
Here logsinB 0, log sin B log 6
a right triangle. B 90° c
C
5 . Given a 767, b 242,A 36
°
53' find B, c, c.
a 767 colog a log a
5 242 log 0 log sin C
A 36° 53’
2 log sinA colog sinAHere a > b, log sinB log c
and log sinB < 0. B 10° 54’58 c
one solu tion. C 132° 12’0”
6 . Given a b= 216.45, A= 35°36' find the
other parts.a colog a log a6 log b cologsinAA 35° 36
’20
”log sin A log sin C
Here a b, log sin B log c
and log sinB 0. B 45° 23’28 c or
tw o solu tions. or 134° 36’32
C 99° 0’12
”
or 9° 47’
8”
EXERCISE XVIII.
1 . Determ ine the number of solu tions in eachfollowing cases :
A
A
A
A 77°
A 60°
B 52°
9 ’ 11
A 30°
70 TRIGONOMETRY.
2 . Given a 840, b 485 A 21°
find B 12°13' c : 146 15' c
3. Given a b A 120°
find B 57°23' c 2
°1 ' 20 c
4 . Given a b= A 51°9'6”
find B 41°
C== 87°
37'54 c
5. Given b= 66.66, B = 77°
find A 54°31 ' c 47
°44 ' c
6. Given a = 309, b= 360 A= 2 1°
14 '
find B 24°
57 ' C 133 c=
B ’= 155°2 ' C
'= 3° 43'
7. Given a b A 38°14 '
find B 44°1 ' c 97
°44 ' c
B '= C’= 5° 47' c'=
8. Given a b A 57°
find B C : 32°22 ' c= 2 .79 .
9 . Given a 34 b 22 B 30°
find A 51 18 ’ C : 98 21 ' c
A ' 128°41 ' C’= 20
°58 ' c
'
10. Given 6 19,
c 18, C : 15°
find B = 16°43
'13 A 147
°27 ' a
B ' 163°
A ’ 0°
54 ’ 13 a'
11 . Given a = 75,b= 2 9 , B = 16
°15' find the differ
ence betw een the areas of the tw o corresponding trianglesw ithou t finding their areas separately .
12 . Given in a parallelogram the side a, a diagonal d,
and the angle A m ade by the tw o diagonals ; find the otherdiagonal . Special case : a 35 , (l= 63, A 21
°
36 ' 30"
THE OBLIQUE TRIANGLE . 71
5 39 . CASE III .
Given tw o sides a and b and the inclu ded u ngle C ; find the
rem a ining p arts, A,B,and c.
SOLUTION I . The angles A and“
B may both be found bymeans of Form u la 5 35, w hich may be w ritten
u— b
tan § (A—B)=
Since 4(A B) 4(180°
C) , the valu e of Q(A B) isknown ; so that this equ ation enables u s to find the valu e of4(A
—B) . We then have
and t (A+ B)
After A and B are know n, the Side 6 may be fou nd by theLaw of Sines, w hich gives its valu e in tw o w ays, as follow s :
a sin Cor c
6sin C
8 111A sm B
SOLUTION II . The third side c m ay be fou nd directly fromthe equ ation (5 34)
C
and then, by the Law of Sines, the follow ing equ ations forcompu ting the valu es of the angles A and B are Obta ined
sinA= a X8 1
20
; sinB= bX
SOLUTION III . If, in the triangle ABC(F ig. BD draw nperpendicu lar to the side AC, then
tan ABD BD
AD AC—DCBD a sin C
DC= a cos C.
a sin Cb a 00 8 0 F IG . 35.
TRIGONOMETRY.
By m erely changing the letters,6 Sin C
tau Ba —beos C
It is not necessary, how ever, to u se both formu las . Whenone angle
,as A, has been fou nd, the other, B, m ay be found
from the relation A B C
When the angles are know n, the third side is fou nd by theLaw Of Sines, as in Solu tion I .
NOTE . When all three u nknow n parts are requ ired , Solu tion 1. is the
most convenient in practice . When only the third Side 0 is desired, Solu .
tion II. may be u sed to advantage , provided the valu es of a2 and b2 can
be readily obtained w ithou t the aid of logarithms. Bu t Solu tions 11. and
III. are not adapted to logarithm ic w ork.
EXAMPLES .
1 . Given a = 748, b= 375, C= 63°35
' find A, B,
and c.
a b 1123
a b 373 log (a b)=2 .57171 logb
(A+B) 116° 24’30
”colog(a 94962 log sin C
58° 12’15
”log tan i}(A+B)=O.2O7G6 colog sinB
1}(A—B) 28° 10’54
”log tan §(A log 6
A 86° 23’
9"
a} (A B) 28° 10’ 54
”c
B 30°
NOTE . In the above Example w e use the angle B in finding the side6 , rather than the angle A , because A is near and therefore the u seof its sine Shou ld be avoided.
2 . Given a = 4,c= 6, B find the third side b.
Here Solu tion II. may be u sed to advantage . We havec2—2 ac cosB = 36— 24 V273,
log 28 logV223 Viz—s
that is, b
THE OBLIQUE TRIANGLE . 73
EXERCISE XIX.
1 . Given a b C 72°
find A 51° B 56
°c
2 . Given 6 c A
find B 60°
C 39°
a
3. Given a = 17 b= 12, C : 59
°
find A 77 12' B 43°30' c
4 . Given b «5, c v3, A 35°
w efind B 93
°
28 ' C 50°
38 ' a
5 . Given a b C 33°
7'
find A 132°18 ’ B 14
°
34 ’c
6 . Given a c B= 15° 22 ' 36find A 118
°55’ C : 45
°41 ’ b
7 . Given 6 c A 86°4 '
find B 65°13' C 28
°42 ’ a
8 . Given a 4527, b 3465, C 66°
6' 27
find A 68°29
' B 45°24' c 4449.
9 . Given a b C 30°
findA 117°24 ' B 32
°11
'c
10 . Given a b C 175°19 ' 10
find A= 2° 46' B : 1°54 ’
c= 81 .066 .
11 . If tw o Sides of a triangle are each equ al to 6, and the
inclu ded ang le 18 find the third side .
12 . If tw o sides of a triangle are each equ al to 6, and the
inclu ded angle is find the third side .
13. Apply Solu tion I . to the case in w hich a is equ al to b ;that is, the case in w hich the triangle is isosceles .14 . If tw o sides of a triangle are 10 and 11 , and the inclu dedangle is find the third side .
15. If tw o sides of a triangle are and 25, and the
inclu ded angle is find the third side .
16. In order to find the distance betw een tw o objects Aand B separated by a sw am p
,a station C w as chosen, and the
74 TRIGONOME '
I‘
RY .
distances CA= 3825 yards, CB :yards, together w ith
the angle AOB = 62° w ere m easu red. F ind the distancefrom A to B.
17 Tw o inaccessible objects A and B are each view ed fromtw o stations C and D on the same side of AB and 562 yardsapart . The angle ACE is 62° BCD 41
°ADB 60
°
and ADC 34° requ ired the distance AB .
18 . Tw o trains start at the sam e tim e from the sam e station ,
and m ove along straight tracks that form an angle of one
train at the rate Of 30 m iles an hou r, the other at the rate of
40 m iles an hou r. How far apart are the trains at the end Of
half an hou r ?19 . In a parallelogram given the tw o diagonals 5 and 6 ,
and the angle that they form 49°
F ind the sides:20. In a triangle one angle 139
°and the sides form ing
the angle have the ratio 5 9 . F ind the other two angles .
5 40. CASE IV.
Given the three sides a,b, c; find the angles A, B, C.
The angles m ay be fou nd directly from the form u las established in 5 34 . Thu s, from the form u la
— 26c cosA
2 2
w e have6 + 0 a
From this equ ation form u las adapted to logarithm ic w orkare dedu ced as follow sF or the sake Of brevity, let a b c= 2 3 ; then 6 c
— 2 (s—h
a ) , a—b+ c= 2 (s
—b) , and a + b— c= 2 (s—
c) .Then the valu e of 1 cosA is
b°+ c"—
a2 2bc— b2— c
°+ a
2a2
(b— c)
’
2bc 2 bc 2bc
76 TRIGONOMETRY .
It is not necessary to compu te by the form u las m ore thantw o angles ; for the third m ay then be fou nd from the equ ation
There is this advantage,how ever, in compu ting all three
angles by the formu las, that w e m ay then u se the su m of theangles as a test of the accu racy of the resu lts.In case it is desired to compu te all the angles, the form u las
for the tangent may be pu t in a more convenient form .
The valu e of taa m ay be w ritten
s (s—a )
2
Hence, if w e pu t
w e have tan5A8
Likew ise, tan5B 1 tan 4 C
EXAMPLES .
1 . Given d = 3.4 l,b= 2 .60, c= 1 .58 ; find the angles.
Using F orm u la and the corresponding form u la for tan 5B,w e
may arrange the w ork as follow sa colog s colog s 9 .42079 10
b colog (s a) log(3 a )= 10
c log (3 b) colog(s b)= 10
2 3 log (3 c) log (3 c)=
3 2) 19 27425 20
s—a = log tan i}A log tan } B—10
s—b= } A { B
3 c A 106° 46'
40 B 46°
A B : and
THE OBLIQUE TRIANGLE . 77
2 . Solve Example 1 by finding all three angles by the u se
of F orm u las [31] and
Here the w ork m ay be compactly arranged as follow s, if w e find
log tan i A , etc. , by su btracting log (s a) , etc. , from log r instead of
adding the cologarithm :
a log (3 a ) log tan 1»A 10. 12903
b log (8 b) 737 log tan5» B
c log (3 c) log tan 5 C
2 8 colog s 5A 53° 23’
20
s log r2 9 42899 i B 23° 26
’
37
s a log r 5C 13° 10’3
s b= A : 106° 46’
40
s c B 46° 53’14
2 s (proof) . C 26° 20’6
Proof, A B + C = 180° 0’0
NOTE . Even if no m istakes are m ade in the w ork, the su m of the
three angles fou nd as above m ay differ very Slightly from 180° in cou sequ ence of the fact that logarithm ic compu tation is at best only a m ethodOf close approximation . When a difference of this kind exists it shou ldbe divided among the angles according to the probable amount Of errorfor each angle .
EXERCISE XX.
Solve the follow ing triangles, tak ing the three sides as thegiven parts
78 TRIGONOMETRY .
11 . Given a = 6, b= 8, c= 10 find the angles .12. Given a = 6 b= 6, c= 10 ; find the angles.13. Given a=6, b= 6, c= 6 ; find the angles .14 . Given a = 6, b 5, c= i 2 ; find the angles.15. Given a = 2
, b= 46, 6 : V3 1 ; find the angles .16 . Given a = 2, b= V6, 0= «3 1 ; find the angles .17. The distances betw een three cities A,
B,and C are as
follow s : AB 165 m iles, AO= 72 m iles, and B C= 185 m iles.
B is du e east from A . In w hat direction is C from A Whattw o answ ers are adm issible18 . Under w hat visu al angle is an Object 7 feet long seen
by an observer w hose eye is 5 feet from one end of the Objectand 8 feet from the other end ?
19 . When F orm u la [28] is u sed for finding the valu e of anangle
,w hy does the ambigu ity that occu rs in Case II . not exist ?
20. If the sides of a triangle are 3, 4 , and 6, find the sins
of the largest angle .
21 . Of three tow ns A, B , and C, A is 200 m iles from B
and 184 m iles from C, B is 150 m iles du e north from C ; how
far is A north of C ?
v
5 41 . AREA OF A TRIANGLE .
CASE I . When tw o sides and the inclu ded angle a re given
In the triangle AB C (F ig. 31 or the areaF v} c X CD
By 5 11, CD= a sin B.
Therefore, F a} ac sinB.
Also, ab Sin C and F = § bc sin A .
CASE II. When a side and the tw o adja cent angles a re given
By 5 33,
Therefore, ca sin 0
sin A
THE OBLIQUE TRIANGLE. 79
Pu tting this valu e of c in Form u la w e have
Fa2sinB sin0
2 sin(B 0)
CASE III . When the three sides of a tri angle a re given
By 5 29, sin B= 2 sin 5-B COSfi-B .
By su bstitu ting for Sin 1}B and cos 5B their valu es in term sOf the sides given in 5 40,
sinB —2
ac
By pu tting this valu e of sinB in w e have
— a) (s—b) (s
— c) . [35]
CASE IV. When the three sides and the radiu s of the circu m
scribed circle, ar the radiu s of the inscribed circle, a re given
If R denotes the radiu s of the circum scribed circle, w e have,from 5 33,
bSIn B
273°
By pu tting thi s valu e Of sinB in w e have
abc
4 B
If r denotes the radiu s of the inscribed circle,
divide the triangle into three triangles by lines from the
centre of this circle to the vertices then the altitu de ofeach Of the three triangles is equ al to r . Therefore
,
= rs. [37]By pu tting in this form u la the valu e of F given in
r :
8
w hence r,in [31] 5 40, is equ al to the radiu s of the inscribed
c ircle .
80 TRIGONOMETRY.
EXERCISE XXI.
F ind the area
1 . Given a b C 116°
30'
2 . Given b 6 A 66°4 ' 19"
3. Given a 510, c 173, B 162°30' 28"
4 . Given a 408, b 41,
c 401 .
5. Given a 40, b 13, c 37
6. Given a 624 , b 205, c 445.
7 . Given b 149, A 70°42 ’ B 39
°
18 ' 28"
8 . Given a c A 25°
9 '31”
9 . Given b 8, c 5, A
10. Given a 7 c 3 A
11 . Given a = 60, B 40°
35 ' area 12 find the
radiu s of the inscribed circle .
12 . Obtain a form u la for the area of a parallelogram in
term s of tw o adjacent sides and the inclu ded angle .
13. Obtain a form u la for the area of an isosceles trapezoidin term s of the tw o parallel sides and an acu te angle .
14 . Tw o sides and inclu ded angle of a triangle are 2416,
1712 , and and tw o sides and inclu ded angle Of anothertriangle are 1948, 2848, and find the su m Of their a reas.
15. The base of an isosceles triangle is 20, and its area is100 V3 find its angles .16. Show that the area of a qu adrilateral is equ al to one
half the produ ct of its diagonals into the sine of their inclu dedangle .
EXERCISE XXII .
1 . F rom a ship sailing dow n the English Channel the EddyStone w as Observed to bear N . 33
°45'W. and after the ship
had sailed 18 m iles S. 67 30'W. it bore N . 11
°15 ’E. Find
its distance from each position of the ship.
THE OBLIQUE TRIANGLE . 81
2 . Tw o objects, A and B,w ere Observed from a Ship to be
at the sam e instant in a line bearing N. 15°E. The ship then
sailed north-w est 5 m iles, w hen it w as fou nd that A bore du eeast and B bore north-east. F ind the distance from A to B.
3. A castle and a m onu m ent stand on the sam e horizontalplane. The angles of depression of the top and the bottom ofthe m onu m ent view ed from the top of the castle are 40
°
and
the height of the castle is 140 feet. Find the height ofthe m onu ment .4 . If the su n’s altitu de is w hat angle m u st a stick
make w ith the horizon in order that its shadow in a horizontalplane m ay be the longest possible ?5 . If the su n’s altitu de is find the length of the longestshadow cast on a horizontal plane by a stick 10 feet in length .
6 . In a circle w ith the radiu s 3 find the area of the partcompri sed betw een parallel chords w hose lengths are 4 and 5.
(TWO solu tions .)7 . A and B, tw o inaccessible objects in the sam e horizontalplane , are Observed from a balloon at C, and from a point Ddirectly u nder the balloon and in the sam e horizontal planew ith '
A and B. If CD= 2000 yards,4 B CD= 6° 7 ' A ADB= 49° 34 ’ find AB .
8 . A and B are tw o objects w hose distance, on accou nt ofinte rvening Obstacles, cannot be directly m easu red . At the
su m m it C of a hill,w hose height above the comm on horizontal
plan e of the Objects is know n to be yards, L ACB isfou nd to be 15° The angles of elevation of C view edfrom A and B are 21
°
9 ' 18"and 23° 15'34”respectively . F indthe d istance from A to B.
CHAPTER V.
M IS CEL LA NEO U S EXAMPLES .
PROBLEMS IN PLANE TRIGONOMETRY.
1 . THE angu lar distance of any object from a horizontalplane
,as Observed at any point of that plane, is the angle
w hich a line draw n from the Object to the point of observation makes w ith the plane . If the object observed is situ atedabove the horizontal plane (that is, if it is farther from the
earth’s centre than the plane is) , its angu lar distance fromthe plane is called its angle of elevation. If the Object isbelow the plane , its angu lar distance from the plane is calledits angle of dep ression. These angles are evidently verticalangles .If tw o objects are in the same horizontal plane w ith thepoint Of observation, the angu lar distance of one object fromthe other is called its bear ing from that object .If tw o objects are not in the sam e horizontal plane With
e ither each other or the point of Observation, w e m ay su pposevertical lines to be passed throu gh the tw o Objects, and to
m eet the horizontal plane of the point of observation in tw opoints . The angu lar distance of these tw o points is the
bearing Of either of the Objects from the other . It m ay
also be called the hori zonta l distance of one object from the
other .
NOTE . Problems in Plane Trigonometry are selected from those
pu blished by Mr. Charles W. Sever, Cambridge, Mass. The fu ll se t canbe Obtained from him in pamphlet form .
84 TRIGONOMETRY.
12. From a tow er 58 ft. high the angles of depression of
two Objects situ ated in the sam e horizontal line w ith the baseof the tower, and on the sam e side, are 30
°13' 18”and 45
°
46’ Find the distance betw een these tw o objects.
13. Standing directly in front of one corner of a fiat-roofedhou se, which is 150 ft. in length, I observe that the horizontalangle w hich the length su btends has for its cosine Vi, andthat the vertical angle su btended by its height has for its sine3
What is the height of the hou seV34
14 . A regu lar pyram id, w ith a squ are base, has a lateral edge150 ft. in length, and the length Of a side of its base is 200 ft.Find the inclination of the face Of the pyram id to the base.
15. From one edge of a ditch 36 ft. Wide , the angle of
elevation of a wall on the opposite edge is 62° 39 ' Findthe length of a ladder w hich will reach from the point ofobservation to the top of the w all.
16. The top of a flag-staff has been broken off, and tou chesthe grou nd at a distance of 15 ft. from the foot of the Staff.The length of the broken part being 39 ft.
,find the w hole
length of the staff.17 From a balloon
,w hich is directly above one town, is
observed the angle of depression of another tow n, 10°14 ’
The'
tow ns being 8 m iles apart, find the height of the balloon.
18. From the top of a m ou ntain 3 m iles high the angle ofdepression of the m ost distant object w hich is visible on the
earth’s su rface is fou nd to be 2° 13’ Find the diam eterof the earth.
19 . A ladder 40 ft. long reaches a window 33 ft. high, onone side of a street. Being tu rned over u pon its foot, itreaches another w indow 21 ft. high, on the Opposite side Of
the street. F ind the w idth of the street.
MISCELLANEOUS EXAMPLES . 85
20. The height of a hou se su btends a right angle at aw indow on the other side of the street ; and the elevation ofthe top of the hou se, from the same point, is The streetis 30 ft. w ide. How high is the hou se
21 . A lighthou se 54 ft . high is situ ated on a rock. The
elevation of the top of the lighthou se, as Observed from a ship,is 4° and the elevation of the top of the rock is 4°
Find the height of the rock, and its distance from the ship.
22. A m an in a balloon observes the angle of depression ofan Object on the grou nd, bearing sou th, to be 35° the
balloon drifts 21} m iles cast at the same height, when the angleof depression of the same object is 23
° Find the heightof the balloon.
23. A man standing sou th of a tow er, on the same horizon
tal plane, observes its elevation to be 54°16' he goes east
100 yds.,and then finds its elevation is 50° Find the
he ight of the tow er.24 . The elevation of a tow er at a place A sou th of it is
and at a place B, west of A, and at a distance of a fromit, the elevation is Show that the height of the tow er is
the tangent of 18° being5 1
25. A pole is fixed on the top of a m ou nd, and the anglesof elevation of the top and the bottom of the pole are 60
°and
30° respectively. Prove that the length of the pole is twice
the height of the mou nd.
26. At a distance (a ) from the foot of a tow er, the angleof elevation (A) of the top of the tow er is the complem ent ofthe angle of elevation of a flag-staff on top Of it. Show thatthe length of the staff is 2 a cot 2A .
27. A line of tru e level is a line every point of w hich isequ al ly distant from the centre of the earth. A line draw n
86 TRIGONOMETRY.
tangent to a line of tru e level at any point is a line ofapp a rent level. If at any point both these lines are draw n,and extended one m ile, find the distance they are then apart .
28 . In Problem 2, determ ine the effect u pon the com pu tedheight of the tow er, of an error in either the angle of elevationor the m easu red distance .
OBLIQUE TRIANGLES .
29 . TO determ ine the height of an inaccessible objectsitu ated on a horizontal plane, by Observing its angles ofelevation at tw o points in the same line w ith its base , andm easu ring the distance betw een these tw o points .30. The angle Of elevation Of an inaccessible tow er, situ ated
on a horizontal plane, is 63°
at a point 500 ft. fartherfrom the base of the tow er the elevation of its top is 32°
Find the height Of the tow er.31 . A tow er is situ ated on the bank of a river. From the
Opposite bank the angle of elevation of the tow er is 60°
and from a point 40 ft. m ore distant the elevation is 50°
F ind the breadth of the river.
32 . A ship sailing north sees tw o lighthou ses 8 m iles apart,in a line du e w est ; after an hou r’s sailing, one lighthou sebears S.W., and the other S.S.W. Find the ship’s rate .
33. To determ ine the height of an accessible object situ atedon an inclined plane.
34 . At a distance of 40 ft. from the foot of a tow er on an
inclined plane, the tow er su btends an angle of 41° at a
point 60 ft. farther aw ay, the angle su btended by the tow eris 23° F ind the height of the tow er .
35. A tow er m akes an angle of 113° 12 ' w ith the inclinedplane on w hich it stands ; and at a distance of 89 ft. from its
base, m easu red dow n the plane , the angle su btended by thetow er is 23° 27 F ind the height Of the tow er.
M ISCELLANEOUS EXAMPLES . 87
36. From the top of a hou se 42 ft. high,the angle Of
e levation Of the top of a pole is 14° at the bottom of the
hou se it is 23° Find the height of the pole .
37 The sides of a triangle are 17, 21, 28 ; prove that thelength of a line bisecting the greatest side and draw n fromthe opposite angle is 13.
38 . A privateer, 10 m iles SW. of a harbor, sees a ship sailfrom it in a direction S. 80
°E .
, at a rate of 9 m iles an hou r.In What direction, and at what rate, m u st the privateer sailin order to com e u p w ith the ship in 1 1} hou rs ?39. A person goes 70 yds. u p a slope of 1 in 34} from the
edge of a river, and observes the angle Of depression of anObject on the Opposite bank to be Find the breadth Ofthe river.40. The length of a lake su btends, at a certain point, anangle of 46° and the distances from this point to the twoextrem ities of the lake are 346 and 290 ft. Find the lengthOf the lake .
41 . Tw o ships are a m ile apart. The angu lar distance ofthe first ship from a fort on shore, as observed from the secondship, is 35
°the angu lar distance Of the second ship
from the fort, observed from the first ship,is 42° 11 ' 53"
F ind the distance in feet from each Ship to the fort.42. Al ong the bank Of a river is draw n a base line of 500feet. The angu lar distance Of one end of this line from an
Object on the Opposite side of the r iver, as Observed from the
other end of the line, is that of the second extrem ityfrom the sam e object, observed at the first, is 79
° Findthe perpendicu lar breadth of the river .
43. A vertical tower stands on a declivity inclined 15° to
the horizon. A m an ascends the'
declivity 80 ft. from the baseof the tow er, and finds the angle then su btended by the tow erto be Find the height of the tow er.
88 TRIGONOMETRY.
44 . The angle su btended by a tow er on an inclined plane is,at a certain point, 42
°325 ft. farther dow n, it is 21° 47
The inclination of the plane is 8° Find the he ight of thetow er.
45. A cape bears north by east, as seen from a Ship. The
ship sails northw est 30 m iles, and then the cape bears east.HOW far is it from the second point of observation
46. Tw o observers, stationed on opp osite sides of a clou d,observe its angles of elevation to be 44° 56' and 36° Theirdistance from each other is 700 ft. What is the linear heightof the clou d
47 From a point B at the foot of a m ou ntain, the elevationOf the top A is After ascending the m ou ntain one m ile ,at an inclination of 30° to the horizon, and reaching a point C,the angle AOB is fou nd to be Find the height Of them ou ntain in feet.
48 . From a ship tw o rocks are seen in the sam e right linew ith the ship, bearing N . 15
°
E. After the ship has sailednorthw est 5 m iles , the first rock bears east, and the secondnortheast. Find the distance betw een the rocks .
49 . F rom a w indow on a level w ith the bottom of a steeplethe elevation Of the steeple is and from a second Window18 ft. higher the elevation is 37° Find the height Of thesteeple .
50. To determ ine the distance betw een tw o inaccessibleobjects by Observing angles at the extrem ities of a line ofknow n length .
51 . Wishing to determ ine the distance betw een a chu rch Aand a tow er B,
on the opposite side of a river,I m easu re a
line CD along the river (C being nearly opposite A) , and
Observe the angles ACB ,58
°ACD, 95
°
ADB , 53°
BDC, 98°
CD is 600 ft. What is the distance requ ired ?
MISCELLANEOUS EXAMPLES . 89
52 . Wishing to find the height of a su mm it A,I m easu re a
horizontal base line CD,440 yds. At C, the elevation of A is
37 and the horizontal angle betw een D and the su mm it is76
° at D,the horizontal angle between C and the su mm it
is 67° Find the height.
53. A balloon is observed from tw o stations 3000 ft. apart .
At the first station the horizontal angle of the balloon and the
other station is 75° and the elevation of the balloon isThe horizontal angle of the fir st station and the balloon
,
measu red at the second station, is 64° Find the height
of the bal loon.
54 . Tw o forces, one of 410 pou nds, and the other of 320pou nds
,m ake an angle of 51° Find the intensity and the
direction of their resu ltant .
55. An u nknow n force, combined w ith one of 128 pou nds,produ ces a resu ltant of 200 pou nds, and this resu ltant m akesan angle of 18° 24 ’
w ith the know n force. Find the intensityand direction of the u nknow n force .
56. At tw o stations, the height of a kite su btends the sameangle A. The angle w hich the line joining one station and
the kite su btends at the other station is B ; and the distancebetw een the tw o stations is a . Show that the height of thekite is 5 a sinA see B.
57 Tw o tow ers on a horizontal plane are 120 ft . apart. A
person standing su ccessively at their bases observes that theangu lar elevation of one is dou ble that of the other bu t, w henhe is hal f-w aybetw een them , the elevations are complem entary .
Prove that the heights of the tow ers are 90 and 40 ft .58 . To find the distance of an inaccessible point C from
either of tw o points A and B, having no instru m ents tomeasu re angles . Prolong CA to a
, and CB to b, and join
AB,Ab, and Ba . Measu re AB, 500 ; aA
,100 ; dB
, 560 ;
bB, 100 ; and Ab, 550.
TRIGONOMETRY.
59 . Tw o inaccessible points A and B, are visible from D,
bu t no other point can be fou nd w hence both are visible.
Take some point C, whence A and D can be seen, and m easu re CD,
200 ft. ; ADC, ACD, 50°
Then take somepoint E, whence D and B are visible, and m easu re DE ,
200 ;
BDE, 54°
BED, 88°
At D measu re ADB , 72°
Compu te the distance AB .
60. To compu te the horizontal distance between tw o inaccessible points A and B, w hen no point can be fou nd whenceboth can be seen. Take tw o points C and D, distant 200 yds .
,
so that A can be seen from C, and B from D. From C m easu re CF ,
200 yds. to F ,whence A can be seen ; and from D
m easu re DE, 200 yds. to E ,w hence B can be seen. Measu re
AF C, A CD, 53°
A CF , 54°
BDE, 54°
BDC, 156° DEB, 88
°
61 . A colu mn in the north temperate zone is east-sou theastof an Observer, and at noon the extrem ity of its Shadow isnortheast of him . The shadow is 80 ft . in length, and the
elevation of the colu mn, at the observer’s station, isFind the height of the colu mn.
62. F rom the top of a hill the angles of depression of twoobjects situ ated in the horizontal plane of the base of the billare 45
°and and the horizontal angle betw een the tw o
objects is Show that the height of the hill is equ al tothe distance betw een the objects.
63. Wishing to know the breadth of a river m A to B,I take AC, 100 yds. in the prolongation of BA, and then takeCD, 200 yds. at right angles to AC. The angle BDA is 3718 ' 30 Find AB.
64 . The su m of the sides of a triangle is 100. The angleat A is dou ble that of B , and the angle at B is dou ble that atC. Determ ine the Sides .
92 TRIGONOMETRY .
80. TWO sides Of a triangle are ch. and eh.,and
the inclu ded angle is 46° F ind the area .
81 . TWO sides Of a triangle are ch. and ch.
,
and they form a right angle . F ind the area .
82 . TWO angles of a triangle are 76°
5 1’ and 57°33’ 12
and the inclu ded Side is 9 ch. F ind the area.
83. TWO Sides of a tr iangle are ch. and ch.
The first bears N . 82°
30'W . ; the second S. 24
°15 ' E. Find
the area.
84 . The three Sides of a tr iangle are 49 eh., ch.
, and
ch. F ind the area .
85. The three sides Of a triangle are ch.
, eh.
,
and 9 ch. F ind the area .
86. The sides Of a triangu lar field, Of w hich the area is14 acres, are in the ratio Of 3, 5, 7 . Find the sides .
87. In the qu adrilateral ABCD w e have AB , ch. ;
AD, eh. ; CD, eh. ; B C, eh. ; and the diagonalAC, ch. Find the area .
88 . The diagonals of a qu adrilateral are a and b, and theyinte rsect at an angle D. Show that the area of the qu ad rilateral is } ab sinD.
89 . The diagonals of a qu adrilateral are 34 and 56, intersecting at an a ngle of lh
‘
nd the area .
90. The diagonals of a qu adrilateral are 75 and 49, intersecting at an a ngle at F ind the area .
91 . Show that the aw e of a regu lar polygon Of n sides, of0m” 18 0
°
w hi ch one Is a , is4
ou t
92 . One side of a regu lar pentagon is 25 . F ind the area .
93. One side of a regu lar hexagsm is 32
. F ind the area.
MISCELLANEOUS EXAMPLES . 93
94. One side of a regu lar decagon is 46.
-Find the area.
95. Find the area Of a circle w hose circu m ference is 74 ft.
96. Find the area of a circle w hose radiu s is 125 ft.
97. In a circle w ith a diam eter of 125 ft. find the area of asector with an arc of
98. In a circle w ith a radiu s of 44 ft. find the area of asector with an arc of
99 . In a circle w ith a diam eter of 50 ft. find the area of asegment w ith an arc of
100. Find the area of a segm ent (less than a sem icircle) , ofw hich the chord is 20, and the distance of the chord from the
middl e point of the sm aller are is 2 .
101 . If r is the radiu s of a circle, the area of a re
circum scribed polygon of n sides is nr’ tan18
7?n
0r2Sln
H
The area of a regu lar inscribed polygon is
102 . If a is a side of a regu lar polygon of n sides, the area2 o
of the inscribed circle isf
f cot“12
10
The area of the circu m scribed circle is If103. The area Of a regu lar polygon inscribed in a circle isto that Of the circu m scribed polygon of the sam e nu mber ofSides as 3 to 4 . Find the nu mber of Sides .
104 . The area of a regu lar polygon inscribed in a circle isa geom etric m ean betw een the areas of an inscribed and acircu m scribed regu lar polygon of half the nu mber of sides .
105 . The area of a circu m scribed regu lar polygon is an
harm onic m ean betw een the areas Of an inscribed regnl
94 TRIGONOMETRY.
polygon of the same nu mber of sides, and of a circu m scribedregu lar polygon of half that number.
106. The perimeter of a circum scribed regu lar tr iangle isdou ble that of the inscribed regu lar triangle .
107. The squ are described abou t a circle is fou r-thirds theinscribed dodecagon.
108. Tw o sides of a triangle are 3 and 12, and the inclu dedangle is Find the hypotenu se Of an isosceles right triangle of equ al area.
PLANE SAILING.
109. Plans Bailing is that branch of Navigation in w hichthe su rface of the earth is considered a plane . The problem swhich arise are therefore solved by the methods of PlaneTrigonom etry.
The follow ing definitions will explain the technical tefm s
which are employed
The difi’erence of la titu de of tw o places is the arc of a
m eridian comprehended betw een the parallels of latitu depassing throu gh those places .
The dep artu re betw een tw o m eridians is the arc of aparallel of latitu de comprehended betw een those m er idians .It evidently dim inishes as the distance from the equ ator at
w hich it is m easu red increases.
When a Ship sails in su ch a m anner as to cross su ccessivem er idians at the sam e angle, it 18 said to sail on a rhu mb-line.
The constant angle w hich this line makes w ith the m eridiansis called the cou rse, and the distance betw een tw o places ism easu red on a rhu mb-line .
If w e neglect the cu rvatu re of the earth, and consider the
r tu re, and difference of latitu de of tw o places to
96 TRIGONOMETRY.
point left to the point reached? (Take earth’s radiu s,statu te m iles .)
PARALLEL AND MIDDLE LATITUDE SAILING .
118. The difference of longitu de of tw o places is the angleat the pole m ade by the m eridians of these tw o places ; or, itis the arc of the equ ator com prehended betw een these tw o
m eridians .119. In Parallel Bailing, a vessel is su pposed to sail on aparallel of latitu de ; that is, either du e east or du e w est. The
distance sailed is,in this case, evidently the departu re m ade ;
and the difference of longitu de m ade depends on the solu tionof the follow ing problem120. Given the departu re betw een any tw o m eridians at
any latitu de, find the angle w hich those m eridians make, orthe difference of longitu de of any point on one m eridian fromany point on the other. (The earth is considered to be a.
perfect Sphere, and the solu tion depends on simple geom etricand trigonom etric principles . Cf . Problem The solu tiongives the follow ing form u la :
Difi . long. depart. x sec. lat.
121 . A ship in latitu de N .,longitu de W. ,
sails du e east a distance of 149 m iles . What is the positionof the point reached ?122 . A Ship in latitu de 44
°49
'S.
, longitu de 119° 42 ' E. ,
sails du e w est u ntil it reaches longitu de 117° 16 ' E. Find thedistance m ade .
123. In Middle Latitu de Sailing, the departu re betw een twoplaces, not on the sam e parallel of latitu de
,is considered to
be, approxim ately, the departu re betw een the m eridians ofthose places, m easu red on that parallel of latitu de w hich lies
b etw een the parallels of the tw o places . Except In
M ISCELLANEOUS EXAMPLES . 97
very high latitu des or excessive ru ns, su ch an assu mptionprodu ces no great error. By the form u la of Example 120,then, w e shall have
Difi . long. depart. x sec. m id . lat.
124 . A ship leaves latitu de 31° 14 ' N.,longitu de 42° 19'W. ,
and sails E.N.E. 325 m iles . F ind the position reached.
125. Find the bearing and distance of Cape God fromHavana . (Cape Cod, 42
°2 ' N 70
°
3'W. Havana, 23° 9 'N .,
82°22 'W .)
126 . Leaving latitu de 49° 57 N . longitu de 15° 16'W. , aship sails betw een S. and W. till the departu re is 194 m iles,and thelatitu de is 47 18
'N . Find the cou rse, di stance, andlongitu de reached.
127 . Leaving latitu de 42° 30' N ., longitu de 58° 51 'W., a
ship sails S.E. by S. 300 m iles . F ind the position reached.
128 . Leaving latitu de 49° 57 N longitu de 30° W. , a shipsails S. 39
° W. , and reaches latitu de 47 44 'N . F ind thedistance, and longitu de
.
reached.
129. Leaving latitu de 37 N.,longitu de 32° 16'W.
, a shipsails betw eenN . andW. 300 m iles, and reaches latitu de 41
°N.
F ind the cou rse, and longitu de reached.
130. Leaving latitu de 50° 10' S.
, longitu de 30°E ,a ship
sails making 160 m iles’ departu re. Find the distance,and position reached.
131 . Leaving latitu de 49° 30'N. , longitu de 25° W . , a ship
sails between S. and E. 215 m iles, m ak ing a departu re of 167miles. Find the cou rse, and position reached.
132 . Leaving latitu de 43° S. , longitu de 21° ship sails
273 m iles, and reaches latitu de 40° 17 S. What are the tw ocou rse s and longitu des, either one of
'
w hich w ill satisfy thedata ?
98 TRIGONOMETRY.
133. Leaving latitu de 17 N longitu de 119° E. , a ship sails219 m iles, m aking a departu re of 162 m iles. What fou r setsof answ ers do w e get
134 . A Ship in latitu de 30° sails du e east 360 statu te m iles.What is the shortest distance from the point left to the pointreachedSolve the same problem for latitu de etc.
TRAVERSE SAILING.
135. Traverse Bailing is the application of the principlesof Plane and Middle Latitu de Sailing to cases when the shipsails from one point to another on tw o or m ore differentcou rses . Each cou rse is w orked by itself, and these independent resu lts are combined, as may be seen in the solu tionof the follow ing example
136 . Leaving latitu de 37 16' S. , longitu de 18°42 'W. , a
ship sails N E. 104 m iles, then N.N .W. 60 m iles, then W. by
S. 216 m iles . Find the position reached, and its bearing anddistance from the point left.
We have, for the first cou rse, difference of latitu de N .,
departu re 7 E.
We have , for the second cou rse, difference of latitu de,N . , departu re 23 W.
We have, for the third cou rse, difference of latitu de S.
departu re W.
On the w hole, then, the ship has m ade m iles of northlatitu de, and m iles of sou th latitu de. The place reachedis therefore on a parallel of latitu de m iles to the north ofthe parallel left ; that is, in latitu de 35
°
S.
The departu re is, in the sam e w ay, found to be m ilesW. ; and the m iddle latitu de is 36° With these data,
100 TRICONOME'rRx.
Prove that
14 . 2 sec 2a: see(a: see(a:cos a: sin x
15 . tan 2 z + sec 2x cos a:—SIn a:
2 tan x
sin2 2x— sin’z
sin x
_ 3 tan x—tan’
z tan 2x+ tan a=19. tan 3z
1—3 tan’x20.
tan 2as—tanx21 . sin (a: y) cos (a: y) 2 sin(a: i n) Sin (y+ i n) .
22 . sin(x+ y)— cos(z
—y)=
—2 Sin(z—1 1r) .
8 111
23. tan a: tan 3; cos 3: cos y
cos Zm+ cos2y
sin x+ cosy
Sim s— cos y -y)
Sin 2 z + sin 4x= 2 sin 3x cosx.
sin 4x= 4 sinx cos x— Ssin’x cos a:8 cossx sin a: 4 cos a: sin x.
cos 4m= 1 = 1—8 Sin’x -l-Ssin‘x.
cos 2x cos 4x : 2 cos 3x cos x.
sin sin az 2 00 8 sin cesinsa: Sin 3m cosh:a cos 3xcos
‘x sin’
a: cos 2x.
cos‘x sin‘
a: 1 1} sin’ 2x.
cossx sin6x cos 2 a: (1 sin’w cos’x) .
cos°z sin‘b: 1 3 sin’
x cos’x.
sin 3x sin 5x= cot x.
cos3x cos 5 a:
M ISCELLANEOUS EXAMPLES . 101
Prove that
sin 3x sin 5 2:37 '
sin a: sin 3az2 cos 2m.
38 . csem—2 cot 2 1: cos x= 2 sin cc.
39. (sin 2x sin 2y) tan (a: y) 2 (sin’x sin’
y) .
csc a:40. (1 cot x tanx) (sin a: cos x)
sec’z
41 . Sin a: sin33: sin 5a:
3 cos zc+ cos 3x
3 Sin x— sin 3a:
43. (60° —ai) .
sin 4x= 2 sinx 008 3m+ sin 2 az
45. Sin x+ sin(x
46. cos assin(y
47. cos(at-Hy) Sin g—cos(x+ z) sin z
Sin(a + y) cos y— sin (x -l-z) cos z .
48.
—y+ z)
+ cos (y+ z cosy cos z.
49. sin cos (x—y)+ sin (y+ z) cos(y—z)+ sin (z+x) cos(z sin 2 z.
sin 75°
sin 15°
sin 75°
sin 15°
51 . cos 20° cos 100° cos 140° 0.
52. cos 36° sin 36° «5 cos
53. tan 11°15' 2 tan 22
°30' 4 tan 45
° cot 11°
50. tan
If A,B, C
’are the angles of a plane triangle, prove that
54 .
55. cos 2A+ cos 2B+ cos 2 C 1 — 4 cosA cosB cos 0 .
102 TRIGONOMETRY.
If A, B, C’ are the angles of a plane triangle, prove that
56. sin 3A+ Sin 3B + sin 3 C —4 cos§
2£ $
5
15 3
2
0
57 cos’A cos
’B cos“0
’1 2 cosA cosB cos
If A B C' prove that
58 . tanA tau B + tanB tan C+ tan 0 tanA= 1 .
59. sin’A sin’B sin’ 0’1 2 sinA sinB sin
'
C’.
60. sin 2A + sin 2B sin 2 0 =4 cosA cosB cos 0.
Prove that
61 . sin (sin“ 1a: sin
- l
g) a: ya
ym .
w y,62 . tan (tan az+ tan y) 1—xy
2x
64 . 2 sin—ls
65. 2 cos—1x Cos" l
(2 a“
—l —l66. 3 tan x—tan1 —3a:"
67. Sin- 1
68 . sin“ 1
1 1—I69 ' tan1 1+ 2 x+ 4 z
’
70 sin—1a:
Vl —x’
— 1
104 TRIGONOMETRY.
Solve the following equ ations
104 .
105.
106.
108.
109 .
110.
111 .
tana: tan 3xsin (45
°x) cos (45
°x) 1 .
tanx secx a . 107. cos 2a: a (1 cosa).
cos 2 x (1 tanz ) a (1 tanaz) .sin6 a: cos
°x 1
7, sin
2 2 23.
cos 3a: 8 00 833: = 0.
see (a: see (a: 2 cos cc.
112 . cscx cota +V3. 114 . cos a: cos 2 a:= 1 .
113.
116.
117.
118.
119 .
120.
121 .
123.
124 .
125 .
126.
127.
128 .
129 .
130.
133.
134 .
135.
136.
137
138 .
4 cos 2x+ 6 8inx= 5. 115. sin 4x—sin 2 z2 sin’x sin’ 2a: 2 .
sec x— cot x= csc x— tanx.
tan2x+ cot2x 1
50
.
sin 4m—cos3x=sin 2 z .
sinx+ cosz = secx. 122 . 2 cosaccos3z + 1 0.
cosSx
tan 2 a: tana: 1 .
sin (a: sin (a: sin 20°
tan (60°
x) tan (60°
x) 2.
sin (a: sin (a: a.
sin (a: sin (a:
sin‘x cos‘a g. 131. tan (a: 2 COSx.
sin‘z cos
‘x 132 . sec a:=2 tanx i .
sin (a: y) cos a , cos (a: y) sin a .
tan x+ tan y= a,cot x+ cot y=b.
Sin (a: cos (a: cos 33° sin 57sin
- 1a: sin
“ 1i n:
tan—Iw tan
" 1 2 a: tan“ 1 3V3.
sin a: 2 cos- 1 a: gs .
MISCELLANEOUS EXAMPLES . 105
Solve the following equ ations
139. sin“ 1a: 3 Cos
" la:
140. tan—1x 2 cot—1m
141 . tan—1(a: 1) tan
" 1
(a: 1) tan“ 1 2as.
142 . tanx+ 1
+ tan
143. tan" 1
2x
Find the valu e of
144 . a see a: bcsca , when tana:
145. sin 3x, when sin 2 a: 1 m ’.
esc’as sec’x
146.
csc’x sec
°:c
’ when tan ac V5.
147. sin 33, when tan’
z 3 cot’ 4 .
148. cos a , when 5 tana: secx 5.
a
V2 a + 1.
Simplify the following expressions2 i
150.
(cosx cosy ) (Sinai: Siny )00 8
”i (w y)
151 .
8111 (2:+ 231) 2 8111 (a: y) Sln as.
cos (x+ 2 y) -2 cos
152sin (a: z ) 2 sina: sin (a: z )sin (y z) + 2 sin y+ sin (31+ z)cos 6m— cos 4xSin 6m+ sin 4 x
154 . tan" l(2x 1) tan“ l
(2 a:
1551 1 1 1
1'
+ Sin’x 1+ cos’a: 1 + sec
’x 1 + csc
’x
156. 2 sec’x sec‘a: 2 csc’ a: csc‘x.
149. secx, w hen tan zc=
153.
ENTRANCE EXAMINATION PA PERS.
PLANE TRIGONOMETRY AND LOGARITHMS.
(Cornell, Ju ne,
(One qu estion may be om itted . )
1 . Prove thatcos co-0 sin 0
sec (if-n 0) csc 0 ;tan 0)
— tan 0 ;
csc (1: 0) csc 0.
2 . Draw the cu rve of tangents, and Show the changes inthe valu e of this fu nction as the arc increases from 0
° to
3. In term s of fu nctions of positive angles less thanexpress the valu es of sin cscH r , tan
— 1301 113 Al so
find all the valu es of 0 in term s of a w hen cos 0 : VSin’a .
4 . (a ) Given cos a: find cos 2 a: and tan(b) Prove that vers (180
°
A) vers (360°
A) 2 ,
5. Prove the check form u laea + b zo cos § (Aa —b zc Sin } (A
Norm. In these papers, as in m any text-books, the Greek lettersa (alpha ) , 5 (bayta) , 7 (gamma) , 6 (delta) , 0 (thayta) , and 4»(phee) , areoccasionally u sed to denote angles.
108 TRIGONOMETRY .
plane of the Objects is known to be 517.3 yds., angle ACE is
fou nd to be 15° 13' The angles of elevation of C viewedfrom A and B are 21
°9'18"and 23° 15'34"respectively. Find
the distance from A to B.
(Cornell, September,
1 . Trace the valu e of tan 0 and that of csc 0, as 0 increasesfrom 0
° to
2 . (a ) F ind the remaining fu nctions Of 0w hen cos 0 V3.
(6) Determ ine all the valu es of 0 that w ill satisfy therelation cot 0 2 cos 0.
3. Prove the identity2 2
tanA—cotAsm
.
A cos A — 2 cot 2 A.
8111 A cos A
4 . Derive an expression for the sine of half an angle in a
triangle in term s of the sides of the triangle .
5. Constru ct a figu re and explain fu lly (giving form u lae)how you w ou ld find the height above its base, and the distancefrom the Observer, of an inaccessible vertical object that isvisible from tw o points w hose distance apart is known, andw hich can be seen from one another.
6 . Given tw o sides Of a plane triangle equ al respectively toand and the inclu ded angle 47 to find
the rem aining parts of the triangle .
7 . In a right triangle, if the difference Of the base and the
perpendicu lar is 12 yds. , and the angle at the base is 38° 1 '
w hat is the length of the hypotenu se
ENTRANCE EXAM INATION PAPERS. 109
(Cornell, June,
1 . By means of an equ ilateral triangle, one of whose anglesis bisected, find the nu merical valu es of the fu nctions of 30°
and
2. If 0 be any angle, prove thatsin 0= tan 0 VI tan2 0, cos 0 : Vosc’ o 1 csc 0.
sin 0 sin 0‘
3. Prove thatcos 0— cos 0’
cot 4} (0 w here 0 and 0'
are any angles .
4. Find Sin 2 0, cos 2 0, and tan 2 0, in term s of functions of 0.
5. Assu m ing the law of Sines for a plane triangle, prove that
(a + b) c=cos } (A—B) sin é C
'
,
(a—b) c sin } (A
—B) cos } 0 .
6. At 120 feet distance, and on a level w ith the foot.
of asteeple , the angle of elevation of the top is 62
°27 find the
height.
7 . Solve the plane triangle given the three sides,a=48 .76,
V.
(Harvard, June,1. In how many years w ill a su m of money dou ble itself at
4 per cent. , interest being compou nded sem i-annu allyl -i-
‘
x/l —m 2
2. Given sink:
3. Find all valu es of as,u nder w hich satisfy the
equ ationV8 c0 8 2x= 1—2 Sinx.
110 TRIGONOMETRY.
4 . What is alw ays the valu e of
2 Sin’x sin’
g 2 00 822: cos’y cos 2 a: cos2y
5. Find the area of a parallelogram , if its diagonals are2 and 3, and intersect each other at an angle of
6. Find the bearing and distance from Cape Horn (55°55'S. ,
67 40'W.) to Falkland Island (51
°
40' S. , 59°
(Harvard, June,
1 . In a certain system of logarithms is the logarithmof 15. What is the base
Be carefu l to remember w hat means.
2 . Find the tangent of 3a: in term s of the tangent of a .
3. One angle of a triangle is and one Of the sidesinclu ding this angle is 24 . What are the smallest
'
valu es the
other Sides can have
4 . Find all valu es of a,u nder w hich satisfy the
equ ationtan 2x (tan
za: 2 sec’x 6.
5 . Tw o ships leave Cape Cod (42°N .
, 70°
one sailingE. , the other sailing N .E. How m any m iles m u st each sail toreach longitu de 65° W.
6 . If A B C’ find the valu e of
tanA+ tan B + tan C— tanA tan B tan C.
TRIGONOMETRY .
4 . One angle of an Obliqu e-angled triangle is and an
adjacent side is V2 . What is the smallest valu e w hich theOpposite Side can have Solve the triangle w hen the Oppositeside is f .
5. A ship leaves Cape Cod (42°2'N 70
°4 'W.) and sails
200 knots on a cou rse S. 40°E. Find the latitu de and longi
tu de reached.
*6. If 2 tan 2 a tan 2b sin 2 6, find the relation betweenthe tangents of a and 6.
(Harvard, June,
(Take the problems in any order. One of the starred problems may beom itted . )
1 . What is the base of the system of logarithm s, when
log3
*2 . Solve the right-angled triangle in w hich one angle isand the difference of the legs is 4 .
*3. F ind it , given see a: 2 tan a: 2 .
*4 . One angle of a triangle is dou ble another angle . The
side opposite the first angle is three-halves of the Side oppositethe second angle. F ind the angles .
5 . Find, by Middle Latitu de sailing, the cou rse and
distance from F u nchal (32°38 ' N .
, 16°54 'W.) to Gibraltar
(36°
7'N ., 5
°21 '
*6. Redu ce to its simplest form cos 2 a: tan (45°
x) sin 2 x.
ENTRANCE EXAMINATION PAPERS. 113
X.
(Harvard, September,
(One of the starred problems may be om itted. )
1 . If the base of ou r system of logarithm s w ere 20 insteadof 10, what w ou ld be the logarithm of one tenth?
‘2 . The area of a right triangle is 6, and the su m of thethree sides is 12. Solve the triangle.
*3. Redu ce to its simplest formcos2B SinaBcos 2 A sin’A cos 2 B.
*4 . Tw o angles Of a triangle are 40°14 ' and 60
°The
sum of the tw o opposite sides is 10. Find these sides .
5. A ship leaves Cape of Good Hope (34°22'S. , 18
°
30'
and sails N. 40°W. to Latitu de 30° S. Find
,by Middle Lati
tu de Sailing, the Longitu de reached and the distance sailed.
*6. The base angles of a triangle are 22°30' and 112°
F ind the ratio betw een the base and the height of the triangle .
XI .
(Harvard, Ju ne,
(Arrange you r w ork neatly. )
1 . What is m eant by the logarithm of a nu mber n in thesystem w hose base is 8 What w ill be the logarithm of 4 inthis system
2 . Establish the formu la
sinas i (1+ 2 cos x)
Which Sign Shou ld be u sed w hen a: lies in the first qu adrantWhen a: lies in the second qu adrant?
114 TRIGONOMETRY.
3. In a triangle tw o angles are equ al to 32°47 ' and 49
°28’
respectively and the length of the inclu ded side isSolve the triangle .
4 . A circu lar tent 30 feet in diam eter su btends at a certainpoint an angle of Find the distance of this point fromthe centre of the tent.5. A Ship leaves Latitu de 42° 2 'N .
,Longitu de 70° 3'W. ,
and sails N. 40°E. a distance of 420 m iles . Find by Middle
Latitu de Sailing the position reached.
XII .
(Sheflield Scientific School, September,
1 . Express an angle of 60° in radians .2 . Represent geom etrically the different trigonometricfu nctions of an angle. State the Signs of each fu nction foreach qu adrant.3.
‘
Express tan fl’ and see d; in term s of sin <b
4 . Derive the form u lasin a + sinfl= 2 sin 501+ B) 008 5(a —fi) .
5. Show that, if a , b and c are the sides of a triangle and
A is the angle opposite the side a,then a
“: 62+ 0
2 260 cos A .
6. Given cos 2 x= 2 sin a , to find the valu e Of sinx'
.
7. Given tw o sides of a triangle a = 450.2, b= 425.4
, and
the inclu ded angle 0 62°
find the rem aining parts.
XIII.
(Shefiield Scientific School, Ju ne,1 . Express an angle of 15° in radians .2 . Write the simplest equ ivalents for Sin tan(21’cos (31r sec (11
'
40
116 TRIGONOMETRY.
7. The sum of two sides, a and b, of a triangle is ft.,
the sum of the Opposite angles, a and B, is and sin a : sin Bfind the angles and sides Of the triangle .
8. Given0 a and log cot a to determ ine a .
XV.
(Shefl’ield Scientific School, Ju ne,
1 . Express (a) an angle Of 2 radians in degrees ;
(6) an angle of 30° in rad ians.
2 . Give simple equ ivalents for the follow ing fu nctionstan x) , cosec x), sin(a: i t ) , sin(a: i n) , tan(i tsin (2 1r x) .
a
3. Given tan a:
b, to express Sin x, cos a , cot as, see a , and
cosec a: in term s of a and 6.
a: b4 . Show that tan a t tan b
em a
cos a cos b
5. Derive the formu la
cos ila
, sin—g§ 2 °
6. Given 180° 27 and log cot find
7. The sides Of a triangle are a = ft., 6 ft. .
0 ft. : calcu late the area of the triangle and the angle 0Opposite the side 0
,u sing the follow ing form u lae :
in w hich S denotes the area of the triangle , and
CHAPTER VI.
CONSTRUCTION O F TABLES .
42 . LOGARITHMS .
Properties of Logarithms. Any positive nu mber beingselected as a base, the logarithm of any other positive numberis the exponent of the pow er to which the base m u st be raisedto produ ce the given nu mber.
Thu s, if a"=N , then n logaN .
This is read, n is equ al to logN to the base a .
Let a be the base, M and N any positive nu mbers, m and n
their logarithm s to the base a ; so that
=M, a” : M
m loga lll , n= logaNZ
Then, in any System of logarithm s
1 . The logarithm of 1 is o.
For,
-a°= 0= log
61 .
2. The logarithm of the base itself is 1 .
For,
a1
a . 1= loga a .
3. The loga rithm of the recip roca l of a p ositive nu mber is the
negative of the logarithm of the number .
1 1For
,If a
” N; thenIV (17‘
n logaN.
118 TRIGONOMETRY.
4 . The loga rithm of the p rodu ct of tw o or m ore positive
numbers is fou nd by adding together the loga rithms of the
severa l factors.
For,
am “
Sim ilarly for the produ ct of three or m ore factors.
5 . The logarithm of the qu otient of tw o positive nu mbers
is fou nd by subtracting the loga rithm of the divisor from the
logarithm of the dividend.
For,M a
_N a
“
m n= logaM logaN.
6. The logarithm of a p ow er of a p ositive nu mber is fou nd bym u ltip lying the logarithm of the number by the exp onent ofthe p ow er .
For,
log, (NP) np =p loga lV.
7 . The loga rithm of the rea l p ositive va lu e of a root of a
positive nu mber is fou nd by dividing the logarithm of the
nu mber by the index of the root.
Change of System . Logarithm s to any base a m ay be
converted into logarithm s to any other base b as follow sLet N be any nu mber, and let
n= loga N and
N= a"and
a“
120 TRIGONOMETRY.
3. Given logw e=0.43429 find
loge2, log, 3, logc5, log, 7, log, 8,108439: logo?) loge?) logeH:
4 . Find a: from the equ ations
§ 43. EXPONENTIAL AND LOGARITRMIC SERIES.
Exponential Series. By the binom ial theorem
1 1 - 1 11+ nx x
a+nfi%— 2X
?
This equ ation is tru e for all real valu es of 173, since thebinom ial theorem may readily be extended to the case Of
incommensu rable exponents (College Algebra , it is,how ever, tru e only for valu es of n numerically greater than
1 , since 1 mu st be nu m erically less than 1 (College Algebra ,1b
s
As (1) is tru e for all valu es Of x, it is tru e w hen x= 1 .
1
L?
CONSTRUCTION or TABLES . 121
Hence, from (1) and
1
1+ s + —
|g
This last equ ation is tru e for al l valu es of n nu mericallygreater than 1 . Taking the lim its Of the tw o members as nincreases w ithou t lim it w e Obta in
1 a: £81
L-
Z lg(3)
and this is tru e for all valu es of a . It is easily seen that bothseries are convergent for all valu es Of x.
The su m of the infinite series in parenthesis is the natu ralbase e.
Hence by (3)
To calcu late the valu e of e w e proceed as follow s
e
To ten places, e
122 TRIGONOMETRY.
Limit of l By the binom ial theorem ,
3
“ M EA L— 2“
x
1
1+ x+
This equ ation is tru e for all val u es of n greater than a:
(College Algebra , Take the lim it as n increases w ithou t lim it, x remaining finite ; then
limit 33
n infinite1+ x+
Logarithmic Series.
Let y log, (1 x)
1 a: e'limit
1
If n is merely a large nu mber, bu t not infinite,
s +
w here c is a variable nu mber w hich approacheswhen n increases withou t limit. Hence
“ 54
‘
: e
y= nV1 + x+ e
124 TRIGONOMETRY.
Su btracting (2) from since
log.(1 31) v) log.
w e find loge
1+ y2
1 + y z+ 1
l —y z
log, (z 1) log
ez
1 1
2 z+ 1 5(2 z+ 1)3
This series is convergent for all positive valu es of z .
Logarithm s to any base a can be cal cu lated by the serieslog
“ (z+ 1) loga z
2 1 1 1
log,a 1
Calcu late log,2 to five places of decimals .
Let z=1 ; then z + l =2 , 2 z + l =3,2 2 2 2
and
The w ork may be arranged as follow s :
1
3
5
7
9
-E11
log!2 0.693147
NOTE . In calcu lating logarithms the accu racy Of the w ork m ay be
tested every tim e w e come to a composite nu mber by adding together thelogarithms of the several factors. In fact, the logarithms of composite
nu mbers are best fou nd in this w ay , and only the logarithms of prim e
numbers need be compu ted by the series.
CONSTRUCTION OF TABLES. 125
EXERCISE XXIV.
1 . Calcu late to five places of decimals loge3, loge5, loge7.
2 . Cal cu late to ten places of decimals log,10.
3. Calcu late to five places of decimals log,o2, log“,e, log,o11 .
§ 44 . TRIGONOMETRIC FUNCTIONS OF SMALL ANGLES .
Let A OP be any angle less than 90°
and 2: its circu larmeasu r e . Describe a circleof u nit radiu s abou t 0 asa centre and take 4 AOP '
A A OP . Draw the
tangents to the circle at Pand P '
, meeting 0A in T.
Then from Geometrychord PP '
arcPP '
P T P 'T,
or, dividing by 2 F IG . as.
MP < arcAP < PT,
sin a: < se < tan a .
Hence, dividing by sin a:
sec 23,
cos re. (1)
Then lies betw een cos a: and 1 . If now the angle a: is
constantly dim inished, cos a: approaches the valu e 1 .
sin cc
2:
as 2: approaches 0, is 1 ;Accordingly, the lim it of
8111 93or
,In other w ords, If a: 18 a very small angle
a:
differs from
1 by a small valu e e, w hich approaches 0 as 9: approaches 0.
TRIGONOMETRY.
To find the sine and cosine of
If a: is the circu lar m easu re of2 sr
360 X 60 10800
the next figu re in a: being either 7 or 8 .
Now sin 2: 0 bu t a: hence sin 1’ lies betw een 0 andAga in
a:
Hence cos 1’
Bu t, from sin a: 2: cos a:sin 1 ' X
(1
Hence sin 1' lies betw een and
that is, to eight places Of decimalssin 1
'
the next figu re being 6, 7, or 8 .
EXERC ISE XXV.
'
Given 1r
1 . Com pu te sin cos and tan 1 ' to eleven places ofdecimals .2 . Com pu te Sin 2 ' by the sam e m ethod
, and also by the
form u la sin 2 a: 2 sin 3: cos x. Carry the operations to nine
places of decimals . Do the tw o resu lts agree3. Compu te Sin 1
° to fou r places of decimals .
4 . F rom the form u la cos a: 1 2 sin”822Show that
cos zc > 1
128 TRIGONOMETRY.
This process need be carried only as far as Forsin (30
°x) sin (30
°
x) 2 sin 30° cos 2: cos a:
cos (30°
x) cos (30°
x) 2 Sin 30°
sin a:
sin (30°
it ) cos a: Sin (30°
x) ,cos (30
°
at) sin a: cos (30°
Moreover the sines and cosines need be calcu lated only toSince
Sin (45°
x) cos (45°
cc) ,cos (45
°
x) sin (45°
cc) .
In u sing this method the m u ltiplication by COS w hichoccu rs at each step, can be simplified by noting that
cos 1 ' 1
Simpson’s m ethod is su perseded in actu al practice by m u ch m ore rapid
and convenient processes in w hich w e employ the expansions of the
trigonometric functions in infinite series.
EXERCISE XXVI.
1 . Compu te the Sine and cosine of 6' to seven decimal places .2 . In the form u la (1) let y=1
°. Assu m ing sin 454
cos compu te the sines and cosines from degreeto degree as far as
§ 46. DE MOIVRE’S THEOREM .
Expressions Of the formcos a: i sin is
,
w hen i VT 1 , play an im portant part in m odern analysis .
Given tw o su ch expressionscos x+ i sin a
'
, cos y+ i sin y,their produ ct is
(cos a: i sin a ) (cos y i sin y)cos x cos y— Sin w sin y+ i (cos zvsin y+ sin :vcos y)cos
(fr —L i sin (x+ y) .
CONSTRUCTION OF TABLES . 129
Hence , the produ ct of two expressions Of the form cos x
i sin x, cos y+ i sin y is an expression Of the same form in
w hich x or y is replaced by x y. In other w ords, the anglew hich enters into su ch a produ ct is the su m of the angles Ofthe factors .If x and y are equ al, w e have at once from the preceding
(cos x+ i sin 2 x i sin 2x ;
and again
(cosx i Sinx)8
(cos x i sin x)“
(cos x i Sin x)(cos 2x i Sin 2x) (cos x i sin x)cos 3x i sin 3x.
Sim ilarly (cos x i sin x)‘ cos 4x i sin 4x,
and in general if n is a positive integer
(cosx+ i sin nx. (1)HenceTo ra ise the exp ression cos x i sin x to the nth p ow er w hen
n is a p ositive integer, w e have only to m u ltip ly the angle x by n.
Again, if n is a positive integer as before,
cos -72+
i sm cos x+ zsm x
x
(cosx i sm x)"= cosa+
Since, how ever, x m ay be increased by any integral m u ltipleof 2 sr w ithou t changing cos x i Sin x, it follow s that all then expressions
x xcos t Sln i sm
n+
n,
x 4 x 4+9,
+7,
130 TRIGONOMETRY.
are nth roots of cos x i sin x. There are no other roots, since
x x x xcos —+ 2 1r + i sin —+ 2 1r 00 85+
t Sln -7
n
and cos — L+ 2 7r+ i
'
Sin— J — l—x+ 2”
+ i sin”
22"
n n
and so on.
Hence, if n is a positive integer,1
(cosx+ i sin x);
— cosx+ 2 kr
+ i sin n—1)
From (1) and (2) it follow s at once that if m and n
positive integers
(cos 2 "1" (cos x i sinx);
Finally, if 18 a negative fraction,
(cos x i Sin x) n
1
cosx+ i sin x
1
(cos x i sin x)»
cosx—i sin x
(cos x+ i Sin x) (cos xcos x— i sin x
coszx+ i sin
zx
’
cos x— i sin x ,
cos x) i Sin x) .
32 TRIGONOMETRY.
Equ ating now the real parts and the imaginary partsseparately, w e Obtain
cos n 0 cos"015 9
21 12cos“"0sin’ 0
n(u— l )(n
—2 )(n—3) H 4
Ifcos 08m 0
sinn0=n cos" 1 0sin 0—
n—Qz—t -1BLQ—i—t gl cosfl 0sin’ 0
n(n—l )(n
-2 )(n— 3)(n
—4 ) H 6
iécos 0SIn 0
EXERCISE XXVII .
1 . Find the six 6th roots Of —1 of + 1 .
2 . Find the three cu be roots of i.
3. F ind the fou r 4th roots Of i.
4 . Express sin 4 0 and 003 4 0 in terms Of sin0 and cos0.
§ 47. EXPANSION or SIN x, COS x, AND Tu x IN
INFIN ITE SERIES .
Let one radian be denoted simply by 1, and letcos l + i sin 1= k .
Then cos x i sin x (cos 1 i sin k
and pu tting —x for xcos —x)+ i sin x) cos x
That iscos x+ i sin x= k:
and 00 8 x i sin x It"
By taking the su m and difference of these two equ ations,and dividing the sum by 2 and the difference by 2 i, w e have
009 x t(kt
Sin x5
1
; (1c:t k
—z»
k" e—m i
,
CONSTRUCTION or TABLES . 133
IE
9
It only rema ins to find the valu e of k, and thisobtained by dividing the last equ ation throu gh byletting x approach 0 indefinitely, w hen w e have
limit sin xx 5 0 x
log h= i, k= e".
Therefore w e have4 6
cosx — x
Ex31 x
“x7
a II
F rom the last two series w e obtain by division
x3 2x5
tan x
By the aid of these series the trigonometric fu nctions Ofany angle are readily calcu lated. In the compu tation itmu st be rem embered that x is the circu la r m easu re of thegiven angle .
134 TRIGONOMETRY.
EXERCISE XXVIII.
Verify by the series ju st Obtained that
1 . Siu ’x cos
’x 1 .
2 . Sin x) sin x, COS x) cosx.
3. Sin 2x= 2 3in x cos x. 4 . cos 2x= 1 — 2 sin’x.
5. F ind the series for see x as far as the term containingthe 6th pow er of x.
x6. Find the ser ies for x cot x, noting that x cotx SIB;
cosx.
7. Calcu late sin 10°
and cos 10° to 6 places of decimals.
8 . Calcu late tan 15° to 5 places of decim als .
From the exponential valu es of sinx and cosx Show that
9 . cos 3x 4 cos’x 3 cos x.
10. Sin 3x= 3 siu x
136 SPHERICAL TRIGONOMETRY .
Spherical triangles are said to be isosceles, equ ilateral,equ iangu lar, right, and obliqu e, u nder the same conditions asplane triangles . A right Spherical triangle, how ever, m ay
have one, tw o, or three right angles.When a spherical triangle has one or m ore Of its Sides equ al
to a qu adrant, it is called a qu adrants] triangle.
It is show n in Solid Geometry, that in every spher icaltriangleI . If tw o sides of a sp her ica l tr iangle a re u nequ a l, the
angles opp osite them are u nequ a l, and the grea ter angle is
opposite the grea ter side ; and conversely.
II . The su m of the sides is less than
III . The su m of the angles is grea ter than 180°and less
than
IV. If, from the vertices as p oles, a rcs of grea t circles a re
described, another sp heri ca l triangle is formed so rela ted to the
fi rst triangle tha t the sides of each tria ngle a re supp lem ents ofthe angles opp osite them in the other triangle.
Tw o su ch triangles are calledp olar triangles, or supp lem enta ltriangles.Let A, B, C (Fig. 37) denote the angles of one triangle ;
a , b, c the sides Opposite theseangles respectively ; and let A '
,
B '
, C’and a
'
,b', c
' denote thecorresponding angles and sidesof the polar triangle. Then theabove theorem gives the six
follow ing equ ationsA a
'
B b'
C c’
A'
a
F IG . 87. B! l)
0 + c = 180°
THE RIGIIT SPHER ICAL TRIANGLE . 137
EXERCISE XXIX.
1 . The angles of a triangle are 7 and find the
sides of the polar triangle .
2. The sides of a triangle are and find the
angles of the polar triangle .
3. Prove that the polar Of a qu adrantal triangle is a righttriangle.
4 . Prove that, if a triangle has three right angles, the sidesof the triangle are qu adrants .5. Prove that, if a triangle has tw o right angles, the sides
opposite these angles are qu adrants, and the third angle ismeasu red by the number of degrees in the opposite side .
6. How can the sides of a spherical triangle, given in
degrees, be fou nd in u nits of length, w hen the length of theradiu s Of the Sphere is know n?7. Find the lengths of the sides of the triangle in Example 2 ,if the radiu s of the Sphere is 4 feet.
49. FORMULAS RELATING TO RIGHT SPHERICAL TRIANGLES .
As is evident from 48, Examples 4 and 5, the only kindof right spherical triangle requ iring fu rther investigation isthat w hich contains only one right angle .
Let ABC (F ig. 38) be a right spherical triangle havingonly one right angle ; and let A,
B , C
denote the angles of the triangle ; a , b, 0,respectively, the opposite sides .Let C be the right angle ; and for thepresent su ppose that each of the otherparts is less than and that the radiu sof the sphere Is 1 .
Let planes be passed throu gh the sides, F IG 38
intersecting In the radii 0A, 0B , and CC.
138 SPHERICAL TRIGONOMETRY .
Also, let a plane perpendicu lar to OA be passed throu gh B,
cutting OA at E and 0 0 at D. Draw BE ,
BD,and DE .
the plane AO (Geom .hence BD,
which is the intersection of the planes
BDE and BOC, is _l_to the plane AOG
(Geom .therefore _1_to OC and
DE .
Now cos c OE OD x cos b,
and OD cos a .
cos c cos 9. cosb .
sin a = BD= BE x sinA,
and BE sin 0 .
Therefore, sin a sin 0 sinA
changing letters, sinb sin 0 sinB
cos ADE OE tan b
BE OE tan c
Hence, cosA tanb cot
changing letters, cosB tan a cot
Again,DE OD sin 6
BE sin o
By su bstitu ting forSin 6 its valu e from w e obtain8 111 c
cosA cos a sinB
changing letters, cosB cosb sinA[41]
Also, sin b g—g EPOn
t—é tan a cotA.
Hence , sinb tana cotA
Changing letters, sina tanb cotB[42]
If in [38] w e su bstitu te for cos a and cos b their valu es from
w e Obtaincosc cotA cotB. [43]
“
econd form u las in [39] geometriIssed throu gh A _L to OB .
140 SPHERICAL TRIGONOMETRY.
~
EXERCISE XXX .
1 . Prove, by aid of Form u la that the hypotenu se ofa right Spherical triangle is less tha n or grea ter than
according as the tw o legs are a like or u nl ike in kind.
2 . Prove, by aid of Form u la that in a right Sphericaltriangle each leg and the Opposite angle are alw ays alike ink ind.
3. What inferences may be drawn from Form u las [38]respecting the valu es Of the other parts : (i.) if c= 90°
(ii .) if a = 90°
; (iii .) if c= 90°and a = 90° ; (iv.) if a = 90
and b 90°
Dedu ce from [38] and [18] the follow m g
form u las :
4 . tanfi b= tan i}(c a ) tan 1} (c a ) .
HINT. Use F orm u la [18] and su bstitu te in it the val u e of cosb in
tan"(45° = tan } (c a ) cot .) (c+ a ) .
6 . isni.) B sin(c a ) csc (c a ) .
tan2+c cos (A+B) see (A—B) .
tan2 § a = tan [2(A B ) tan [2(A B)
inn2
(45°
i s)= tan i (A a ) cot l (A a ) .
tan“
(45°
i t) Sin (A a ) csc (A a ) .
tan’
(45°
+B) tan 2(A a ) tan 1) (A a ) .
THE RIGHT SPHERICAL TRIANGLE . 141
§ 50. NAPIER’S RULES .
The ten form u las dedu ced in § 49 express the relationsbetw een five parts of a right triangle, the three sides andthe tw o obliqu e angles . All these relations m ay be show n tofollow from tw o very u sefu l Ru les, devised by Baron Napier,the inventor of logarithm s .F or this pu rpose the right angle (not entering the form u las)
is left ou t Of accou nt, and instead of the hypotenu se and the
tw o Obliqu e angles, their respective comp lements are employed ;so that the five parts considered by the Ru les are : a
, b, co. A,
co. o, co. B. Any one Of these parts m ay be called a m iddle
part ; and then the tw o parts imm ediately adjacent are calledadjacent parts, and the other tw o are called opposite parts .Ru le I . The sine of the m iddle p a rt is equ a l to the p rodu ct
of the tangents of the adjacent p a rts.
Ru le II . The sine of the m iddle p art is equ a l to the p rodu ct
of the cosines of the opp osite p a rts.
These Ru les are easily rem embered by the Sxpressions,tan. ad. and cos. op.
The correctness of these Ru les may be show n by tak ingeach of the five parts as m iddle part, and com
paring the resu lting equ ations w ith the equ ations contained in Form u las [38]F or example, let co. 0 be taken as m iddle
part, then co. A and co. B are the adjacent parts,and a and b the opposite parts, as is very plainlyseen in F ig. 42 . Then, by Napier
’s Ru les
Sin(co. c) tan (co. A) tan(co. B) ,COS c cotA cotB ;
sin(co. c) cos a cos b,cos c cos a cos b ; n o . 42.
resu lts w hich agree w ith F orm u las [43] and [38] respectively .
142 SPHERICAL TRIGONOMETRY .
EXERCISE XXXI.
1 . Show that Napier’s Ru les lead to the equ ations con
tained in Formu las and
2 . What w ill Napier ’s Ru les become, if w e take as the five
parts of the triangle,the hypotenu se, the two obliqu e angles,
and the comp lements of the tw o legs
§ 51 . SOLUTION or RIGHT SPHERICAL TRIANGLES .
By means of Formu las [38] w e can solve a right triangle in all possible cases . In every case two parts besidesthe right angle m u st be given.
CASE I . Given the tw o legs a and b.
The solu tion is contained in F orm u las [38] and viz
cos c cos a cos 6,tanA tan a csc 6,tan B tan b csc a .
F or example, let a 27°
28 ' b 51°12 ' then the
solu tion by logarithm s is as follow s
log cos alog cos blog cos c
c 56°
13'41"
log tan a log tan blog csc b log csc alog tanA log tan B
A 33°42' 51" B 69
°
38 '54"
CASE II . Given the hypotenu se c and the leg a .
F rom F orm u las and [40] w e obtaincos b cos c see a
,
sinA sin a csc 0,
cos B tan a cot c.
144 SPHERICAL TRIGONOMETRY.
CASE V. Given the hyp otenu se c and the angle A.
From F ormu las and [43] it follows that
sin a sin 0 sin A,
tan b tan c cos A,
cotB cos c tan A.
Here a is determ ined by sin a , since a and A mu st bein kind (see Exercise XXX .
,Example
CASE VI . Given the tw o angles A and B.
By m eans of Form u las [43] and [41] w e obtain
cos c cot A cot B,cos a cos A csc B,cos b cos B csc A.
NOTE 1 . In Case I . (a and b given) the form u la for compu ting c failsto give accu rate resu lts w hen c is very near 0° or in this case it m aybe fou nd w ith greater accu racy by first compu tingB, and then compu ting c,as in Case IV.
NOTE 2 . In Case II. (c and a given) , if b is very near 0° or it
m ay be com pu ted m ore accu rately by m eans Of the derived formu latan i (c
—a) . (EX.
And if A is so near 90° that it cannot be fou nd accu rately in the Tables ,it may be compu ted from the derived form u la
tan2 (45°i A ) tan 1} (c a) cot 1} (c a) . (Ex. 5, 3
In like manner, w hen B cannot be accu rately fou nd from its cosine w em ay m ake u se of the formu la
tan2 1} B sin (c a) csc (c (EX. 6, 5
NOTE 3. In Case III. (a and A given) , w hen the form u las for the
requ ired parts do not give accu rate resu lts , w e m ay employ the derivedform u las
tan2 (45°
i c) tan 1} (A—a ) cot } (A a) , (Ex. 9 , 5
tan2 (45°
i b) sin (A a) csc (A a) , (EX. 10, 5
tan2 (45°
i B) tan 1} (A a) tan 1} (A a ) . (Ex. 11, 5
THE RIGHT SPHERICAL TRIANGLE . 145
NOTE 4 . In Case IV. (a and B given) , if A is near 0° or it may
be more accu rately fou nd by first compu ting b and then finding A .
NOTE 5. In Case V. (c and A given) , if a is near it m ay be fou ndby first compu ting b, and then compu ting a by m eans of F orm u la
NOTE 6. In Case VI. (A and B given) , for unfavorable valu es Of thesides greater accu racy m ay be obta ined by m eans of the derived form u las
tan2 1} c cos (A B) see (A B) , (EX. 7,
tan2+a tan (A B) tan [45°
i (A (EX. 8,
tan2 5b tan [5(A B) tan [45°
NOTE 7. In Cases I. , IV. , and V. , the solu tion is alw ays possible.In the other Cases, in order that the solu tion may be possible , it isnecessary and su fiicient that in Case II. sin a sin c ; in Case III. , thata and A be alike in kind , and sinA > Sin a ; in Case VI. , that A + B + Cbe > and the difference of A and B be
NOTE 8 . It is easy to trace analogies betw een the form u las for solvingright spherical triangles and those for solving right plane triangles. The
former, in fact, become identical w ith the latter if w e su ppose the radiu sof the Sphere to be infinite in length ; in w hich case the cosines of thesides becom e each equ al to l , and the ratios of the sines of the Sides and
of the tangents of the sides m u st be taken as equ al to the ratios of the
sides themselves.
NOTE 9 . In solving Spherical triangles, the algebraic Sign of the
functions m u st receive carefu l attention. If the Sign of each fu nction isw ritten ju st above it, the Sign Of the fu nction in the first m ember w ill be
or according to the ru le that like Signs give and u nlike signsgive
If the fu nction is a cos, tan, or cot, the Sign show s that the angle isless than the Sign Show s that the angle is greater than and
the supp lement of the angle Obtained from the table m u st be taken.
If the fu nctipn is a sine , Since the Sine of an angle and its su pplem entare the same , the acu te angle obta ined from the table and its su pplem entmu st be considered as solu tions, u nless there are other conditions thatremove the ambigu ity . F or the conditions that remove the ambigu ity ,
in case of right spherical triangles see examples 1 and 2 in Exercise XXX. ,
and in case of Obliqu e Spherical triangles see I. of 5 48 .
146 SPHERICAL TRIGONOMETRY .
NOTE 10. The solu tions of a spherical triangle may conveniently betested by su bstitu ting them in the form u la containing the three requ iredparts.
If the formu la requ ired for any case is not rem embered, itis alw ays easy to find it by m eans Of Napier’s Ru les . In
a pplying these Ru les w e m u st choose for the m iddle part thatone of the three parts considered the tw o given and the one
requ ired— which w ill m ake the other tw o either adjacentparts or opposite parts .
F or example given a and B ; solve the triangle.
First, represent the parts as in Fig. 42, and to preventm istakes m ark each Of the given parts w ith across . To find b, take a as the m iddle part ;then b and co. B are adjacent parts ; and by
Ru le I .sin a = tan bcot B ;
whence, tan b sin a tan B .
To find 0,take co. B as m iddle part ; then a
and co. c are adjacent parts ; and by Ru le I .
cos B : tan a cot c ;w hence, tan 6 tan a sec B .
To find A,take co. A as m iddle part ; then a
the opposite parts ; and by Ru le
cos A cos a sin B.
In like manner, every case of a right Spherical triangle m aybe solved.
EXERCISE XXXII .
Solve the follow ing right triangles, tak ing for the givenparts in each case those printed in columns I . and
148 SPHERICAL TRIGONOMETRY .
23. Define a qu adrantal triangle, and Show how its solu tionmay be redu ced to that of the right triangle .
24 . Solve the qu adrantal triangle w hose sides area 174
°
12 ' b 94°
8'
c 90°
25. Solve the qu adrantal triangle in w hichc A 110
°
47 B 135°35 '
26. Given in a spherical triangle A, C, and 0 each equ al tosolve the triangle .
27 Given A C and c solve the triangle .
28 . Given in a right Spherical triangle,A 42
°24 '
B 9°4 ' solve the triangle .
29 . In a right spherical triangle, given a 119°
B 126° solve the triangle .
30. In a right Spherical triangle, given c=50°, b=44°
l 8 '39
solve the triangle.
31 . In a right spherical triangle,given A= I56
°
a= 65° 15' solve the triangle .
32 . If the legs a and b of a right spherical triangle are
equ al , prove that cos a cot A Vcos c.
33. III a right Spherical triangle prove thatCOS’A singe= sin(c
—a) sin (c+ a ) .34 . In a right spherical triangle prove that
tan a cos c sin b cot B.
35. In a right spherical triangle prove thatsin2A cos”B sin2 a sin2B.
36. In a right spherical triangle prove thatsin (b c) 2 cos“} A cos 6 sin c.
37 . In a right Spherical triangle prove thatsin (c b) 2 sin2 1} A cos 6 sin c.
38 . If,in a right spherical triangle, p denotes the arc of the
great circle passing throu gh the vertex of the right angle and
perpendicu lar to the hypotenu se, m and n the segments of thehypotenu se m ade by this are adjacent to the legs a and b
,
prove that (i .) tan2a tan c tanm
, (ii . ) sin2p tanm tan n.
THE RIGHT SPHERICAL TRIANGLE . 149
§ 52 . SOLUTION on THE ISOSCELES SPHERICAL TRIANGLE .
If an arc of a great circle is passed throu gh the vertex of anisosceles spherical triangle and the m iddle point of its base, thetriangle w ill be divided into tw o symmetrical right sphericaltriangles . In this w ay the solu tion of an isosceles sphericaltriangle may be redu ced to that of a right spherical triangle .
In a sim ilar manner the solu tion of a regu lar sphericalpolygon“ may be redu ced to that of a right spherical triangle .
Arcs of great circles, passed throu gh the centre of the polygonand its vertices, divide it into a series of equ al isosceles triangles ; and each one of these m ay be divided into tw o equ ivalent r ight triangles .
EXERCISE XXXIII .
1 . In an isosceles spherical triangle , given the base b andthe side a ; find A the angle at the base, B the angle at the
vertex, and h the altitu de.
2 . In an equ ilateral spherical triangle, given the side a ;
find the angle A .
3. Given the side a Of a regu lar spherical polygon of nsides ; find the angle A of the polygon, the distance .R fromthe centre of the polygon to one of its vertices, and the distance r from the centre to the m iddle point of one of itsSides .4 . Compu te the dihedral angles m ade by the faces of the
five regu lar polyhedrons.5. A spherical squ are is a regu lar Spherical qu adrilateral .
F ind the angle A of the squ are , having given the side a .
A regu lar spherical polygon is the polygon formed by the intersections of the Spherical su rface by the faces of a regu lar pyram id w hosevertex is at the centre of the sphere.
CHAPTER VIII .
THE OBLIQUE SPHERICAL TRIANGLE .
5 53. FUNDAMENTAL F ORMULA S .
LET AB C (Fig. 45) be an Obliqu e spherical triangle , a , b, c
its three sides, A,B, C the anglesopposite to them , respectively .
Throu gh C draw an arc CD
of a great circle , perpendicu larto the side AB , m eeting AB atD. F or brevity let CD p ,
AD= m , BD= n, A ACD= x,
A BCD y.
1 . By § 49 in the righttriangles BDC and ADC,
F IG . 45 . Sinp sin a sin B,
and Sinp sin b Sin A .
Therefore, sin a sinB sinb sinA
sim ilarly, sin a sin 0 sin 0 sinA
and sinb sin C= sin c sinB
These equ ations m ay also be w ritten in the form Of
proportionssin a : sin b : Sin e sin A sin B : Sin C.
That is, the sines Of the sides of a spherical triangle are
proportional to the Sines of the Opposite angles .In F ig. 45 the arc of the great circle CD cu ts the side AB
w ithin the triangle . In case it cu ts AB produ ced w ithou t thetriangle, sin (180
°—A) , sin(180°—B) , or sin(180
°
C ) , w ou ld
152 SPHERICAL TRIGONOMETRY.
Formu las [45] and [46] are also u niversally tru e ; for thesam e equ ations are Obtained w hen the arc CD cu ts the sideAB w ithou t the triangle .
EXERCISE XXXIV.
1 . What do F orm u las [44] becom e if A= 90° if Bif C = 90° if a = 90° if if2. What does the first of [45] becom e if A=O° ? if A= 9O°?if A 180
°
3. F rom Form u las [45] dedu ce Form u las by m eans ofthe relations betw een polar triangles
§ 54 . FORMULAS F OR THE HALF ANGLES AND SIDES .
From the first equ ation of
cos a cos b cos 0cos A
SIn b SIn c
w hence,sin b sin c cos a
1 cos A
sin b sin 0
Sin b sin c cos 6 cos c cos a1 cos A
8 111 b 8 111 c
Sin 6 sin c
Hence, by 30 [16] and and 31
sin°i A= sin i (a + b
sin } (b+ c—a ) cscb csc c.
Now let
w hence, i (b c a )101
— M C)i <a + b
— c>
THE OBLIQUE SPHERICAL TRIANGLE . 153
Then, by su bstitu tion and extraction of the squ are root,
tani A csc s csc(s a) sin(s—b) sin(8 c)
In like manner, it may be show n that
sini B= sin(s—a) sin (s
— c) csca cscc
tuni c Vcsc scsc(s—c) sin(s
—a) sin(s—b)
Again,from the first equ ation of
cosB cos C + cosA.cos a
SInB 8 111 C
w hence ,sinB sin C — cosB cos C — cosA
1 cos aSInB Sin C
SinB sin 0 cos B cos C cosA1 cos a
sinB Sin C
If w e place § (A+ B + O)= S, and proceed in the samemanner as before , w e Obtain the follow ing resu lts
154 SPHERICAL TRIGONOMETRY.
And, in like manner,
cosi b= Vcos(S—A) cos(S
—C) cscAcscC
tani b=V cosScos(8—B) sec(8
—A) sec(8—O)
sin+c=V—cosScos (8—0) cscA cscB
003 151 3: Vcos(S—A) cos(S
—B) cscA cscB
tuni c : —cos8 cos(8—0) sec(S
—A) soc(8 —B)
s55. GAUSS’S EQUATIONS AND NAPIER’S ANALOGIES .
By i 27
cos .) (A B)= cos i A cos i B—sin i A sin QB ;
or, by su bstitu ting for COSQA, cos i B , sin i A, sin QB ,their
valu es given in 54, and redu cing,
sin s sin(s—
a )sin 6 sin 0 sin a Sin 0
Sin 6 sin 0 sin a sin c
sin a sin b
This valu e, by applying 29 31 and Observingthat the expression u nder the radical is equ al to sin} C, becom es
_2 sinfic cos(s00 8 1HA +B)_ 2 sin § c cos § csm i C ,
and this, by cancelling common factors, clearing Of fractions ,and Observing that s—1}c= i (a + redu ces to the form
COS 1}(A+B) cos to cos sin i C.
By proceeding in like manner with the valu es ofsing, (A B) , cos“ A B) , and sin (A B) ,
three analogou s eq u ations are obtained.
156 SPHERICAL TRIGONOMETRY.
56. CASE I .
Given tw o sides, a and b, and the inclu ded angle C.
The angles A and B m ay be fou nd by the first tw o ofNapier’s Analogies. ; viz.
b)c°t f 0 °
After A and B have been fou nd, the side c m ay be fou ndby [44] or by bu t it is better to u se for this_pu rpose
Gau ss’s Equ ations, becau se they involve fu nctions Of the
sam e angles that occu r in w ork ing Napier’s Analogies . Any
one of the equ ations m ay be u sed ; for exam ple, from the
first w e havecos i} (a b)
cos i c 8111 1} C.
EXAMPLE .a 73° 58
’
54 therefore , 4} (a b) 17°36’
57
b=38°45’ 0C 46° «i C 23° 16
'
log cos 1} (a b) log sin 5(a b)log see i (a b) log csc g.(a b)log cot s} C
log tan } (A B)log sec 1} (A B) “5(A B) 75°
log cos i (a b) 5,(A B) 40° 11’
A 116° 9’
log cos i c B 35° 46’15. l
i c c 51° 2’
If the side 0 only is desired, it m ay be fou nd fromwithou t previou sly compu ting A and B. Bu t the Form u las
[45] are not adapted to logarithm ic w ork. Instead of changingthem to form s su itable for logar ithm s, w e m ay u se the follow ingm ethod, w hich leads to the sam e resu lts, and has the advantagethat, in applying it, nothing has to be rem embered exceptNapier’s Ru les
THE OBLIQUE SPHERICAL TRIANGLE . 157
Make the triangle (F ig. as in 53, equ al to the sum
(or the difference) of tw o righttriangles . F or this pu rpose,throu gh B (or A,
bu t not C)draw an arc of a great circleperpendicu lar to AC, cu ttingAC at D. Let BD=p , CD=-1m ,
AD= n ; and mark w ith crossesthe given parts .By Ru le I .
cos C tanm cot a ,w hence tanm tan 0. cos C.
‘
Pio . 46.
By Ru lecos a cos m cosp , w hence cosp cos a secm .
cos c cos n cosp , w hence cosp cos csec n.
Therefore, cos c sec n cos a secm ;
or, Since b m
,cos c cos a secm cos (b m ) .
It is evident that c m ay be compu ted, w ith the aid oflogarithm s, from the tw o equ ations
tanm tan a cos C,cos c cos a sec m cos (b m ) .
EXAMPLE . Given a 97 30' b 55°
C 39°
find 0 .
log tan a (n) log cos a (n)log cos C log cos (b m )log tanm = (n)
m 99°
45 ' 14" log cos cb—m —44° 33' 4" c 56
°40
'20
EXERCISE XXXV.
1 . Write form u las for finding, by Napier
’s Ru les, the Sidea, w hen 6, c, and A are given, and for finding the side b w hen
a, c, and B are given.
158 SPHERICAL TRIGONOMETRY.
2. Given a 88°
12' b 124°7' C 50
°
find A 63°15' B 132
°17'59"c 59
°4 ' 18
3. Given a 120°55' b= 88 12 '
find A 129°
58' B 63°15' c 55
°
52 '
4 . Given 6 63°15' c 47
°A 59
°4 '
find B 33°12 ' C 55
°
52 ' a 50°1 '
5. Given 6 69°
25' c 109°46' A 54
°
54 ' 42
find B 56°11 ' C 123
°a 67
°
§ 57. CASE II .
Given the side 0 and the tw o adjacent angles A and B.
The sides a and b m ay be fou nd by the third and fou rth ofNapier’s Analogies,
AW tan t c,
sin 1} (A Btan t} (a b) tal i i c,
and then the angle C m ay be fou nd by by Napier’ssecond Analogy, or by one of Gau ss
’s equ ations, as, for instance,the second, w hich gives
cos C1} cos 5. (a b)
EXAMPLE .A 107° 47
’
7
B 38° 58’27
c 51° 41’14
log cos t(A B)log see ;, (A B)
c
logtan i (a b)logsin i (A B) i (a b) 54° 24
’
log sec t(a b) 5(a b) 15°
log costc a 70°
log 00 8 5C 9 95272 b 380 27159
to 26° 14'
C 3 2° 29'
45
if (A B) 34°
f, (A B) 73°
to 25° 50’
37
log sin 4(A B)log csc (A B)log tan c
160 SPHERICAL TRIGONOMETRY.
4 . Given B 153°17 ' C’= 78
°
43'36 a = 86° 15' 15
find 6 152°
43'
c 88°12 ' A 78 15' 4s"
5 . Given A 125°41 ' C 32
°47
'z, 52
°
37'
find a 128°41 ' c 33
'B 55
°47'40
58 . CASE III.
Given tw o sides a and b, and the angle A opp osite to a .
The angle B is fou nd from w hence w e havesin B= sinA sin bcsc a .
When B has been fou nd, C and c may be fou nd from the
fou rth and the second of Napier’s Analogies, from w hich w e
ObtainA
tan i c —b) ,
cot i C — B) .
The third and first of Napier’s Analogies m ay also be u sed.
NOTE 1 . Since B is determ ined from its sine, the problem in generalhas tw o solu tions ; and, moreover, in case sin 3 ) 1, the problem is
impossible . By geom etric constru ction it may be Show n, as in the
corresponding case in Plane Trigonom etry , u nder w hat conditions theproblem really has tw o solu tions, one solu tion, and no solu tion. Bu t in
practical applications a general know ledge of the shape of the triangle isknow n beforehand ; so that it is easy to see , w ithou t special investigation,
w hich solu tion (if any) corresponds to the circu mstances of the qu estion .
It can be show n that there are tw o solu tions, w hen A and a are alikein kind and sinb> sin a > sinA sin b no solu tion w hen A and a are
u nlike in kind (inclu ding the case in w hich either A or a is and
Sin b is greater than or equ al to sin a , or w hen sin a sinA Sin b ; and
one solu tion in every other case .
NO TE 2 . The side c or the angle C may be compu ted , w ithou t firstfinding B, by m eans of the form u las
tan m cosA tan b, and cos (c m ) cos a see bCOSm ,
cotx tanA cos b, and cos (0 x) cot a tan bcosx.
THE OBLIQUE SPHERICAL TRIANGLE . 161
These form u las m ay be obtained by resolu tion of the triangle into righttriangles, and applying Napier’s Ru les ; m is equ al to that part of theside c inclu ded betw een the vertex A and the foot of the perpendicu larfrom C , and x is equ al to the corresponding portion of the angle C.
NOTE 3. After the tw o valu es of B have been obtained , the number
of solu tions may read ily be determ ined by 48 I. If log sin B is positive
,there w ill be no solu tion.
EXAMPLE . Given a 57° b 31° A 104°
A > log sinA
and a b log sin b
therefore, A B
hence , B log sin B
and only one solu tion . B 36°
a b 88° 50’
a b 26° 26’
A B 140° 51’
53”
A B 67° 59’7”
logsin 5(A B)log csc i (A B)
log tan i c
5c 21° 35’
38
c 43° 11’
16
EXERCISE XXXVII .
1 . Given a 73°49' b 120
°53' A 88
°
52 '42
find B 116°42 ' c 120
°57 ' C 116
°
47 '
2 . Given a 150°57 b 134
°15' A 144
°
22 '42
find B1 120°47 c, 55
°42
'C, 97 42 '
B, 59°
c2 23°
57 17 C2 29°
8'39
3. Given a 79°
0' b 82°17' A 82
°
9'
find B c 45°
C 45°
4 . Given a 30°
52 ' b 31°
A 87°34 ' 12
Show that the triangle is im possible .
“ a b)
i (a b) 13° 13’
“A B) 70°
g(A B) 33° 59’
5”
log sin ; (a b)log csc } (a b)
log cot C
i C 25° 51’15
”
C
162 SPHERICAL TRIGONOMETRY .
§ 59. CASE IV.
Given tw o angles A.
and B ,and the side a opposite to one of
them .
The side 6 is fou nd from w hence
sin b sin a sinB cscA.
The valu es of c and C may then be fou nd by m eans ofNspier’s Analogies, the fou rth and second of w hich give
tan § c= :—i:—§E—jig§—b) ,
coti C=$ tan i (A—B) .
NOTE 1. In this case the conditions for one , tw o, or no solu tions canbe dedu ced directly by the theory of polar triangles from the correspond
ing conditions of Case III. There are tw o solu tions, w hen A and a are
alike in kind and Sin B > SinA sin a sin B ; no solu tion w hen A and a
are u nlike in kind (inclu ding the case in w hich either A or a is and
sin B is greater than or equ al to sinA , or w hen sinA < sin a sinB and
one solu tion in every other case.
NOTE 2 . By proceeding as indicated in Case III. , Note 2 , form u lasfor compu ting c or C , independent of the side b, may be fou nd ; viz. :
tan m tan a cosB, and sin (c m ) cotA tanB sinm ,
cot x cos a tan B, and sin (C x) COSA see B sinx.
In these form u las m BD, x L BCD, D being the foot of the per
pendicu lar from the vertex C.
NOTE 3. As in Case III. , only those valu es of b can be retained w hichare greater or less than a , according as B is greater or less than A . Iflog sin b is positive, the triangle is impossible .
EXERCISE XXXVIII .
1 . Given A 110°
B 133°
a 147°
5' 32 find
6 155°
0 33°1 ' C 70
°20
'
164 SPHERICAL TRIGONOMETRY.
EXAMPLE 2 . a 124° 12’
31 s a 13° 39’5”
5 54° 18’
16 s b 83° 33'
20”
c 97° 12’
25 s c 40° 39’11
”
2 s 275° 43’
12 log tan 5A3 137° 51
'
36 log tan 5R
log sin (3 a)log Sin (s b)log sin (3 c)
log tanzr A 127° 22
'
7
log tan r B 51° 18'
11
C 72° 26’
40
EXERCISE XXXIX .
1 . Given a = 120° 55 ' b=59° 4 ' c= 106° 10'22
find A 116°44 ' B 63
°C 91
°7 ' 22"
2 . Given a = 50° 12 ' b 116°44 ' c= 129° 11 ' 42
find A 59°4 ' B 94
°23' C 120
°4 '
3. Given a = 131° 35' b= 108° c= 84° 46'
find A 132°14 ' B 110
°10' C : 99
°42 '
4 . Given a = 20° 16' 38" 6 : 56°19
'c= 66° 20' 44
find A 20°9 ' B 55 52 ' C= 114°
61 . CASE VI .
Given the three angles, A, B,and C.
The sides are compu ted by m eans Of Form u las The
form u las for the tangents are in general to be preferred.
If w e m u ltiply the equ ation
tanya V cos S 00 8 (S—A) sec (S B) sec(S C )
by the equ ation 7 and pu tsec(S A)
V cos S sec (S A) sec (S B) sec(S C )= tanR,
and also make analogou s changes in the equ ations for tan i band tan i s, w e obtain
THE OBLIQUE SPHERICAL TRIANGLE . 165
tan i d tanR cos (S A) ,tan a} b= tanR cos (S B) ,tan .) c= tanR cos (S C) ,
w hich are the m ost convenient form u las to u se in case allthree sides have to be compu ted.In Example 1, after w e find the valu es of S, S—A, S B,
S C, w e w rite the form u la for tan 4a w ith the algebraicsign w ritten above each fu nction as follow s :
tan ska V—‘
cos S cbs (S—A) see (S—B) sec (S C ) .
EXAMPLE 1 . A 220° log cos 3 (n)B 130° log cos (S A )C 150° log sec (S B) (n)2 S 500° log sec (S C) (n)S 250°
S A 30° log tan a
S B 120° {y a 61° 34’6
S — C = 100° a =123°
NOTE . Here the efiect, as regards algebraic Sign, of three negativefactors, is cancelled by the negative Sign belonging to the w hole produ ct.
In Example 2, after w e find the valu es Of S, S —A, S
—B,
S C, w e w rite the form u la for tanR w ith the algebraic signw ritten above each fu nction as follow s
tanB ‘l—COS S sbc(S—A) sdc (S—B) 830 (S C) .
EXAMPLE 2 . A 20° 9'
56 s 95° 11'
21”
B 55° 52’32 S A 75° 1
’
25”
C 114° 20’
14 S B 39° 18’
49
2 S = 190° 22’
42 S— C — 19°
log cos S (71) log tan a
log sec (S A ) log tana, b 9 . 72867
log sec (S B) log tan 21c
log sec (S C) i a 10° 8’
log tan‘l h
’b 28° 9
’
log tan R 21c 33° 10
’
a 20° 16’
38"
b )6° 19’
41”
c 66° 20’
43
166 SPHERICAL TRIGONOMETRY.
EXERCISE XL .
1 . Given A B C=
find a 139°21 ' b= 126° 57 ' c= 56° 51 ' 48
2 . Given A 59°55 ' B 85
°36
'C 59
°55' 10
find a 51°17
' b 64°2 ' c 51
°17'31
3. Given A 102°
14'
B 54°
32 ' C 89°
5'46
find a 104°25' 5 53
°49 ' c 97
°44
'19
4 . Given A 4°
23'35" B 8°28
'C'= 172
°17'56
find a 31°
b 84 18'
c 115°
§ 62 . AREA OF A SPHERICAL TRIANGLE .
I . When the three angles, A, B, C, a re given.
Let It radiu s of sphere,E the spherical excess A B C
F : area of triangle .
Three planes passed throu gh the centre of a sphere, eachperpendicu lar to the other tw o planes, divide the su rface ofthe sphere into eight tri-rectangu lar triangles .It is convenient to divide each Of these eight triangles into
90 equ al parts, and to call these parts spherical degrees . The
su rface Of every sphere, therefore, contains 720 sphericaldegrees .Since in spherical degrees, the A AB C E
, and the entiresu rface of the sphere is equ al to 720 spherical degrees,
A ABC : su rface of the Sphere E : 720 ;
or, since the su rface of a sphere 4 5 132,
A AB C : 4 1rR°= E z 720 ;
whence
168 SPHERICAL TRIGONOMETRY.
If w e su bstitu te , in equ ation for A and B, the valu es
5(a b) and t o, and al so su bstitu te s for 5(a + b c) and
s c for 5(a b c), equ ation (b) w ill becom e— cos .} c
tan gs tan § (s c) .
Comparing (a) , (c) , and (d) , w e Obtaincot i (2 C
— c) . (e)
By beginning w ith the second equ ation of and treatingit in the Sam e w ay, w e Obtain as the resu l t,
tan i (2 C— b) . (f)
By taking the produ ct Of (e) and (f) , w e Obtain the elegantform u la,
stunt(s—a) tana,(s
—b) tan5(s—c) , [52]
w hich is know n as l ’Hu ilier’s F orm u la .
By m eans of it E may be com pu ted from the three Sides,and then the area of the triangle may be fou nd by
III . In a ll other cases, the area m ay be fou nd by firstsolving the triangle so far as to Obtain the angles or the sides,w hichever may be m ore convenient
, and then applying [51]or
EXAMPLE 1 . A 102° 14'
12
B 54° 32’
24
C 89° 5'
46
245° 52’
22”
65° 52’
22”
237142
180° 2 648000
If,therefore , w e know the radiu s of the sphere, w e can
express the area Of a spherical triangle in the ordinary u nitsOf area.
See Wentw orth Hill ’s Tables, page 20.
logE
64800010
log F logR2
F R2
THE OBLIQUE SPHERICAL TRIANGLE . 169
EXAMPLE 2 . a 133° 26’
19"
b 64° 50'
53"
c 144°
2s 342° 30’
57”
s a 37° 49’
s b 106° 24’
s c 27°
EXERCISE XLI.
1 . Given A 84° B 27
°22' C 75
°33'
find E F R”
2 . Given a 69°
15'
b 120°
42 ' c 159°18'33
find E=216° 40'28
3. Given a 33°1 ' b 155
°5' C 110
°10' 5
find E = 133°48
'53
4 . Find . the area of a triangle on the earth’s su rface
(regarded as spherical) , if each side of the triangle is equ alto (Rad iu s of earth 3958 m iles .)
i s 85°37'
44"
i (s a) 18° 54'
35”
“ s b) 53° 12'18
”
log tan i slog tan i (s a)log tan (s b)log tan 3 c
log tan24E
log tan i i:
4B 50° 11’43
E 200° 46’
52”
CHAPTER IX .
APPLICATIONS O F SPHERICAL TRIGONOMETRY .
63. PROBLEM .
To redu ce an angle m easu red in sp ace to the horizon.
Let 0 (F ig. 48) be the position of the Observer on the grou nd,AOB = h, the angle m easu red inspace, (for example, the anglebetw een the tops of tw o chu rchspires) , OA
'and OB
'the projec
tions of the sides of the angleu pon the horizontal plane HR,
A OA ' m and B OB 'n, the
angles of inclination Of 0A and
OB respectively to the horizon.
Requ ired the angle A'OB 'x
made by the projections on the
horizon.
The planes of the angles of inclination AOA'and BOB '
produ ced intersect in the line 0 0 , w hich is perpendicu lar tothe horizontal plane (Geom .
F rom 0 as a centre describe a Sphere, and let its su rface cu t
the edges of the trihedral angle O—AB C in the points JV, N ,
and P . In the Spherical triangle MNP the three sidesMN h
,MP 90
°m
,NP 90
°n, are know n, and the
Spherical angle P is equ al to the requ ired angle x.
F rom 48 w e Obtaincos l x=Vcos s cos (s h) secm sec n,
w here A = s.
F IG . 48.
172 SPHERICAL TRIGONOMETRY .
The Celestial Meridian of an Observer is the great circle inw hich the plane of his terrestrial m eridian produ ced m eetsthe su rfaee of the celestial sphere .
Hou r Circles, or Circles of Declination, are great circlespassing throu gh the poles, and perpendicu lar to the equ inoctial .
The Horizon of an Observer is the great circle in w hich aplane tangent to the earth’s su rface, at the place w here he is,m eets the su rface of the celestial Sphere .
The Zenith of an Observer is that pole of his horizon w hichis exactly above his head.
Vertical Circles are great circles passing throu gh the zenithof an observer, and perpendicu lar to his horizon.
The vertical circle passing throu gh the east and w est points ofthe horizon is called the PrimeVertical ; that passing throu ghthe north and sou th points coincides w ith the celestial m er idian.
The Ecliptic is a great circle of the celestial sphere ,apparently traversed by the su n in one year from w est to east
,
in consequ ence of the m otion of the earth arou nd the su n .
The Equ inoxes are the points w here the ecliptic cu ts theequ inoctial . They are distingu ished as the Verna l equ inoxand the Au tu mna l equ inox ; the su n in his annu al jou rneypasses throu gh the form er on March 21
, and throu gh thelatter on September 21 .
Circles of Latitu de are great circles passing throu gh the
poles of the ecliptic, and perpendicu lar to the plane of theecliptic .The angle w hich the ecliptic m akes w ith the equ inoctial iscalled the obliqu ity of the ecliptic ; it is equ al to 23
°
nearly, and is Often denoted by the letter e.
These definitions are illu strated in F igs . 50 and 51 . In
F ig. 50,AVB U is the equ inoctial, P and P ' its poles
,NP Z S
the celestial m eridian of an observer, NESW his horizon, Zhis zenith, M a star, PMP '
the hou r circle passing throu ghthe star, ZMDZ ’ the vertical throu gh the star.
APPLICATIONS . 173
In Fig. 51, AVB U represents the equ inoctial, E VEU the
ecliptic, P and Q their respective poles, Vthe vernal equ inox,U the au tu mnal equ inox, M a star , PMR the hou r circlethrou gh the star, QMT the a rch of latitu de throu gh the star,and A TVR = e.
F IG . 50. F IG . 51.
The earth’s diu rnal motion cau ses all the heavenly bodiesto appear to rotate from east to west at the u niform rate Of
15°
peThou r. If in Fig. 50 w e conceive the observer placed atthe centre 0 , and his zenith, horizon, and celestial m eridianfixed in position, and all the heavenly bodies rotating arou ndPP '
as an axis from east to w est at the rate of 15° per hou r,w e form a correct idea of the apparent diu rnal m otions ofthese bodies. When the su n or
'
a star in its diu rnal m otioncrosses the m eridian, it is said to m ake a transit across themeridian ; when it passes across the part NWS of the horizon,it is said to set and w hen it passes across the part NES, it issaid to rise (the effect of refraction being here neglected) .Each star, as M , describes daily a small circle of the Sphereparal lel to the equ inoctial, and called the Diu rnal Circle ofthe star. The nearer the star is to the pole the sm aller is thediu rnal circle ; and if there w ere stars at the poles P and P '
,
they w ou ld have no diu rnal motion. To an Observer north of
174 SPHERICAL TRIGONOMETRY .
the equ ator, the north pole P is eleva ted above the horizon
(as Shown in Fig. to an Observer sou th of the equ ator,the sou th pole P ' is the elevated pole .
§ 66 . SPHERICAL CO-ORDINATES .
Several system s of fixing the position of a star on the su r
face Of the celestial sphere at any instant are in u se . In eachsystem a great circle and its pole are taken as standards ofreference, and the position of the star is determ ined by m eansof tw o qu antities called its spherica l coo rdina tes.
I . If the horizon and the zenith are chosen, the co-ordinatesof the star are called its altitu de and its azimu th .
The Altitu de of a star is its angu lar distance, m easu red ona vertical circle, above the horizon. The complem ent of thealtitu de is called the Zenith Distance.
The Azimu th of a star is the angle at the zenith form ed bythe m eridian of the observer and the vertical Circle pas singthrou gh the star, and is measu red therefore by an arc of thehorizon. It is u su ally reckoned from the north point of thehorizon ip north latitu des
,and from the sou th point in sou th
latitu des ; and east or west according as the star is east orw est of the meridian.
II. If the equ inoctia l and its p ole are chosen, then the
position of the star m ay be fixed by m eans of its declinationand its hou r angle .
The Declination of a star is its angu lar distance from the
equ inoctial, m easu red on an hou r circle . The angu lar distanceOf the star, measu red on the hou r circle, from the eleva ted p ole,
is called its Polar Distance .
The declination of a star, like the latitu de of a place on theearth’s su rface, m ay be e ither north or sou th ; bu t, in practicalproblem s, w hile latitu de is alw ays to be considered positive ,declination, if of a different nam e from the latitu de
,m u st be
regarded as nega tive.
176 SPHERICAL TRIGONOMETRY .
In many problem s, a Simple w ay of representing the m ag
nitu des involved, is to project the sphere on the plane of thehorizon, as show n in F ig. 52 .
NE SW is the horizon,Z
the zenith, N Z S the m eridian,WZE the prime vertical, WAEthe equ inoctial projected on theplane of the horizon, P the
elevated pole, M a star, DMits altitu de, ZM its zenith distance, A PZM its azimu th
,MR
its declination, PM its polarF m . 52,
distance,4 ZPM its hou r angle .
§ 67. THE ASTRONOMICAL TRIANGLE .
The triangle ZPM (F igs . 50 and 52) is often called theastronom ica l triangle, on accou nt Of its im portance in problem sin Nau tical Astronomy.
The side P Z is equ al to the complem ent of the latitu de ofthe observer. F or (F ig. 50) the angle Z OB betw een the zenithof the Observer and the celestial equ ator is obviou sly equ al tohis latitu de
,and the angle P OZ is the complem ent of Z OB .
The are NP being the complem ent of PZ , it follow s that thea ltitu de of the eleva ted p ole is equ a l to the la titu de of the p la ce
of observa tion.
The triangle ZPM then (how ever m u ch it m ay vary in
shape for different positions of the star M ) alw ays containsthe follow ing five m agnitu des
PZ co-latitu de of Observer 90°
l,
ZM = zenith distance of starP ZM : azim u th Of starPM polar distance of starZPM : hou r angle of star
APPLICATIONS . 177
A very Simple relation exists betw een the hou r angle Of thesu n and the local (apparent) tim e Of day . Since the hou rlyrate at w hich the su n appears to move : from east to w est is
and it is apparent noon w hen the su n is on the m eridianof a place, it is evident that if hou r angle etc. ,
tim e of day is noon, 1 O’clock P .M .
,11 O’clock A .M .
, etc.
In general, if t denotes the absolu te valu e of the hou r angle ,
tim e of day -
tP .M . , or 12
12according as the su n is w est or cast of the m eridian.
5 68 . PROBLEM.
Given the la titu de of the observer and the a ltitu de and the
azim u th of a sta r, tofind its declina tion and its hou r angle.
In the triangle ZPM (F ig.
given PZ 90°
l co-latitu de,ZM 90
°h co-altitu de
,
A PZM a azim u th,to find PM 90
°d polar distance,
hou r angle .
Draw MO_J_to NS, and pu t Z Q m,
then, if a P Q 90°
(l m ) ,and if a > 90
°
,PO= 9O
° —(l—m ) ;
and,by Napier
’s Ru les,cos a zt tan m tan h,
sin d cos P Q cos MO,sin h= cos m cosMO ;
w hence, tanm= zt: cot h cos a,
sin d sin h sin(l i m ) sec 777,in t Oh the Sign is to be u sed if a > The hou r anglem ay then be fou nd by m eans of w hence w e have
sin t sin a cos h sec d .
SPHERICAL TRIGONOMETRY.
69 . PROBLEM.
angle of a heavenly body w hen its declina
tion, its a ltitu de, and the la ti
tude of the p la ce a re know n .
In the triangle ZPM (F ig .
given PZ 90°
l,
PM = 90° d=p ,
ZM= 90° h ;
requ iredA ZPM t.
If,in the first form u la
F m. 53.
Vsiu (s—b) Sin(s— c) cse b csc c,
w e pu t
a = 9o°—h,
c= 90°— l,
w e have
s—b= 90° c= t(l+ p— h) ,
and the form u la becomes
sin i t : ziz [cos (l+p + h) sin4-(l+ p h) sec l CSCp ] i ,
in w hich the sign is to be taken w hen the body is east o fthe m eridian .
If the body is the su n, how can the local tim e be fou ndw hen the hou r angle has been compu ted (See
180 SPHERICAL TRIGONOMETRY .
Then, by Napier’s Ru les, cos t = tan m tan d ,
sin h cos n cosMO,
sin d= cosm cosMO;
w hence,
cot d cos t,cos m sin h csc d,
and it is evident from the figu rethat
l= 90°
in w hich the Sign or the Signis to be taken according as
the body and the elevated pole are on the sam e side of theprim e vertical or on opp osite sides.
In fact, both valu es of l m ay be possible for the sam e altitu de and hou r angle ; bu t, u nless n is very sm all, the tw o
valu es w ill differ largely from each other, so that the Observerhas no difficu lty in deciding w hich of them shou ld be taken .
F IG . 54.
72. PROBLEM .
Given the declina tion, the right ascension of a star, and the
obliqu ity of the eclip tic, to find the
la titu de and the longitu de of the
star .
Let M (F ig. 55) be the star, P bethe pole of the equ inoctial, and Qthe pole of the ecliptic.Then
,in the triangle PMQ ,
given P Q e= 23°
PM 90°
d,
AMP Q= 90° r (see Fig.
requ ired QM 90°
u and A P QM 90°
v (see Fig.
F IG . 55 .
APPLICATIONS . 181
In this case, also, tw o sides and the inclu ded angle are
given. Draw ME _J_to PO, and m eeting it produ ced at H,
and let PH n.
By Napier’s Ru l es,Sin r = tan n tan d,sin u cos (c n) cos ME ,
sin d cos n cos MH,
sin (e n) tan 0 tan MH,
sin n tan r tan ME ;
whence, tan n cot d sin r,
sin u sin d cos (c n) sec n,tan v tan r sin (e n) csc n.
To avoid obtaining u from its sine w e may proceed as
follow s
From the last two equ ations w e have, by division,
sin n tan v cot (e n) sin d cot r tan n.
By taking ME as m iddle part, su ccessively, in the trianglesMOH and MPH , w e obtain
cos u cos v cos d cos r ;
w hence, cos u see v cos d cos r .
From these valu es of sin u and cos n w e Obtain, by division,
tan u sin v cot (e n) tan d csc r tan n.
F rom the relation
sin r tan n tan d,
it follow s that tan d csc r tan n 1 .
Therefore tan u sin v cot (e n) ,
a formu la by w hich u can be easily fou nd after v has beencompu ted .
182 SPHERICAL TRIGONOMETRY.
EXERCISE XLII .
1 . Find the dihedral angle m ade by adjacent lateral facesof a regu lar ten-sided pyram id ; given the angle V= m adeat the vertex by tw o adjacent lateral edges .
2 . Throu gh the foot of a rod w hich m akes the angle A w itha plane , a straight line is draw n in the plane . This line m akesthe angle B w ith the projection of the rod u pon the plane .
What angle does this line m ake w ith the rod?
3. Find the volu m e V of an Obliqu e parallelopipedon ;given the three u nequ al edges a
,b, c, and the three angles
l, m ,
n,w hich the edges make w ith one another .
4 . The continent of Asia has nearly the shape of an
equ ilateral triangle, the vertices being the East Cape , CapeRomania, and the Prom ontory of Babs . Assu m ing each sideof this trIangle to be 4800 geographical m iles, and the earth’sradiu s to be 3440 geographical m iles, find the area of thetriangle : (i.) regarded as a plane triangle ; (ii .) regarded as aspherical triangle .
5 . A ship sails from a harbor in latitu de l, and keeps onthe arc of a great circle . Her cou rse (or angle betw een the
direction in w hich She sails and the m eridian) at starting is a .
F ind w here she w ill cross the equ ator, her cou rse at theequ ator, and the distance she has sailed .
6 . Tw o places have the sam e latitu de l, and their distance
apart, m easu red on an arc of a great circle, is d. How m u chgreater is the arc of the parallel of latitu de betw een the placesthan the arc of the great circle Compu te the resu lts for7 : d = 90
°
7 . The shortest distance d betw een tw o places and the irlatitu des l and l’ are know n. F ind the difference betw eentheir longitu des .
184 SPHERICAL TRIGONOMETRY .
tim e of the year is it a maxim u m at a given p la ce? (Givensin h sin l sin d.)
15. .Given the latitu de of a place north of the equ ator , andthe declination of the su n ; find the tim e of day when the su n
bears du e east and du e west. Compu te the resu l ts for thelongest day at St. Petersbu rg (l 59
°
16. Apply the general resu lt in Example 15 (cos t cot l
tan d) to the case when the days and nights are equ al inlength (that is, when Why can the su n in su mm ernever be du e east before 6 A .H . , or du e west after 6 R M 9
How does the tim e of bearing du e east and du e west changew ith the declination of the su n? Apply the general resu ltto the cases w here l < d and l= d . What does it becom e atthe north pole
17. Given the su n’s declination and his altitu de when he
bears du e east ; find the latitu de of the observer .
18. At a point 0 in a horizontal plane MN a staff 0A isfixed, so that its angle of inclination AOB w ith the plane isequ al to the latitu de of the place, 51
°30'N . and the direction
0B is du e north. What angle w ill 0B make w ith the shadowof on the plane, at 1 P .M .
,w hen the su n is on the equ i
noctial ?
19 ..What is the direction of a w all in latitu de 52° 30'N .
w hich casts no shadow at 6 A .H . on the longest day of the year
20. At a certain place the su n is observed to rise exactlyin the north-east point on the longest day of the year ; findthe latitu de of the place.
21 . Find the latitu de of the place at w hich the su n sets at10 o’clock on the longest day .
22 . To w hat does the general form u la for the hou r angle ,in 69, redu ce w hen (i.) 7b (ii .) l 0
°
and at : (iii .)l or d 90
°
APPLICATIONS . 185
23. What does the general form u la for the azim u th of acelestial body, in 70, become w hen t= 90
°
6 hou rs
24 . Show that the form u las of 71 , if t lead to theequ ation sin l= sin h csc d ; and that if d=O
°
,they lead to
the equ ation cos I: sin it see t.
25. Given latitu de of place 52° declination of starits hou r angle 28° find its altitu de .
26 . Given latitu de of place 51° polar distance ofstar 67 59' its hou r angle 15° find its altitu de and
its azim u th .
27 Given the declination of a star 7° its altitu de22
° its azimu th 129° find its hou r angle and
the latitu de of the observer .
28 . Given the longitu de v of the su n, and the obliqu ity of
the ecliptic e 23°27 find the declination d, and the right
ascension r .
29 . Given the obliqu ity of the ecliptic e= 23°
the
latitu de of a star its longitu de find its declinationand its right ascension.
30. Given the latitu de of place 44° 50' the azim u th of astar 138° 58 ' and its hou r angle find its declination.
31 . Given latitu de of place 51° 31 ' altitu de of su n w estof the m eridian 35° 14 ' its declination 21
°
find the
local apparent tim e .
32 . Given the latitu de of a place l, the polar distance p of
a star,and its altitu de h ; find its azim u th a .
A P P ENDIX .
F ORMULA S .
PLANE TRIGONOMETRY.
1 . sin°A cos’A 1 .
sin A2 . tanA
cos A
sinA cscA= 1 .
cosA secA= 1 .
tanA X cotA= 1 .
sin (a: y) sin a: cos y+ cos a: sin y.
cos (a: y) cos a: cos y Sin a: sin y.
tan x+ tan y6.
cot avcot y“ 1
7.
cot y+ 00 t w
sin (a: y) sin it: cos y cos a: sin y.
cos (a: y) cos a: cos y sin a: Sin y.
tan a: tan yta'n(w
cot x cot y+ 1cot (x—y) cot y— cot z
sin 2 x= 2 sin x cos a:
cos 2 a: cos’a:
188 FORMULAS.
29 . cos 15A
s(s—
a )
32. tan”8
33. F = § a e sinB.
a’sin B sin 0
34 ' FZ Sin(B+ 0 )
_ i‘flc .
4B
37 .
SPHERICAL TRIGONOMETRY
38 cos 0 cos a cos 6.
sm a sin 0 sinA .
39sin b sin 0 sinB.
cosA tan b cot 0 .
cos B tan a cot 0 .
cos A cos a SinB.
cosB cos 6 sin A .
sin b tan a cotA.sin a tan b cot B.
cos c cotA cotB.
sin a sinB sin 6 sinA .
sin a sin 0 sin 0 sinA .
sin 6 sin 0 sin 0 sinB.
51 .
52 .
FORMULAS .
cos 0. cos 6 cos c sin 6 sin 0 cosA .
cos b cos a cos c Sin a sin 0 cos B.
cos 0 cos a cos b sin a sin 6 cos 0 .
— cosB cos C + Sin B sin C’cos a .
cos A cos 0 sinA Sin C cos 6.
— cos A cosB sinA sinB cos c.
cos l c sin l C.
Sin § (A + B) cos l o cos § (a b) Cos i’ C.
cos§ (A B) sinare sin QC.
sin%(A—B) sin l o sin%(a —b) cosi C,
tan § (A—B )
”H im—b)
w R’EF _
180
tanfi E tan l s tan§(s—a )tan —b)tan
—c) .
189
190 FORMULAS .
PROF. BLAKSLEE’S constru ction by w hich the direction ratios
for plane right triangles give directly from a figu re the analogies for a right trihedral or for a right spherical triangle
The constru ction consists of tw o parts .
(a ) Lay off from the vertex Va u nit’s distance on each edge .
(6) Pass throu gh the three extrem ities of these distance sthree planes perpendicu lar to one of the edges, as VA . Now
these three parallel planes w ill cu t ou t three sim ilar righttriangles . The first being constru cted in either of the tw o
u su al w ays, the constru ction of the others is evident.Since the plane angles A1 , A2 , As all equ al the dihedral A ,
and the nine right triangles in the three faces give the valu e sin the figu re, w e have
(1) sin A Sin a sin h ; Sim ilarly, sin B sin b sin h .
(2) cos A tan b tan h ; sim ilarly, cos B tan a tan h .
(3) tanA tan a. sin b ; sim ilarly, tanB tan b Sin a .
(4) cos it cos a cos b ; (by 3) cot A cot B.
(5) sinA cosB : cos 6 ; sin B cos A cos a .
NOTE . If a Sphere of u nit radiu s be described abou t V as a centre ,the three faces w ill cu t ou t a right spherical triangle , having the Sides a ,
b, and h, and angles A ,B, and H . The above form u las are thu s seen to
be the analogies of
A NSWERS.
PLANE TRIGONOMETRY .
EXERCISE I .
6 7:1. 60° 45°
69
11 1: 6 7x
16 24
3 7:o
15 7:o
40
112 30 ,16
168 45 ,
1° radian. radian.
1 radian 7. 14° 27’28 10. 3 hr. 49 m in. 11 sec .
5
6
”8 . m iles. 11 . 9 s . 2 m.
11° 27’33
”9 . 57 ft. in. 12 . Th sec.
EXERCISE II .
sin B . cos B tan B ga cot B o secB go csc B
(i. ) Sin 3, cos g, Sin etc. (v. ) Sin
tan : 3, cot i , (iii. ) sin etc. (vi. ) Sin
sec Q, csc 3, (iv. ) Sin fir, etc.
The requ ired condition is that a2 b2 c“. It is.
2 mn6. (i. ) Sin "12 + ”2
, etc. (iii. ) Sin etc.
M ysm etc. (Iv. ) em etc.
In (iii. ) zfiq2
qfs2 p
23“ ; in (iv. ) mg'nfis2 m2p
2v2
c 145 ; w hence , Sin A 12,
cos B ; cosA {fltanA : cot A = secA H
TRIGONOMETRY .
b=o.o23 ; w hence, tanA=cot B=W ; cotA =tanB=f£n etc.
a= w hence, sinA m cos B, etc.
« W e
c=p + q ; w hence, sm A cos B ; etc.
w hence, tan Aq
a=17—q ; w hence, sinA fi—I—g cosB ; etc .
etc. 15. sin A =g; etc.
SinA H5 V?) etc.
008 4 : 1) sinA 1) etc.
a= 20. a =9. 22 .
t =i .54. 21. b=68. 23. c=229.62 .
Constru ct a rt. A w ith legs equ al to 3 and 2 respectively ; thenconstru ct a Similar A w ith hypotenu se equ al to 6.
In like manner, 26, 26, 27 may be solved .
a m iles ; b 2 m iles. 31. m iles.
a b a b 32 . yards.
EXERCISE III.
Throu gh A (F ig. 3) draw a tangent, and take A T : 3 ; the angleA OT
.is the requ ired angle .
F rom 0 F ig. 3) as a centre, w ith a radiu s 2 , describe an are cu t
ting at T the tangent draw n throu gh B ; the angle A OT is the
requ ired angle.
In F ig. 3, take OM t}, and erect MP _L 0A and intersecting the
circumference at P ; the angle POM is the requ ired angle .
Since Sin a: cos 45, OM PM (F ig. and a: hence , con
stru ct a:
Constru ct a rt. A w ith one leg tw ice the other ; the angle Oppositethe longer leg is the requ ired angle .
Divide 0A (F ig. 3) into fou r equ al parts ; at the first point of division from O erect a perpendicu lar to m eet the circu m ference atsom e point P . Join OP the angle A OP is the requ ired angle .
r Sin 22 . Leg adjacent to A no, leg opposite to A me.
sec A
23 sin A :
tanA1_
cot A’
cosA _cotA
srn A z g. 5,
s1n A =%V1—5,
1 z =45° 5. a:
2 m=3o°. 6. a;
3 7. a:
4 33‘
8 . a:
cosA ; c
sm A ; c
cosA
cos A ;-. c
5. A =90°—B ,
a c cos B ,
b=osin B .
TRIGONOMETRY.
tanA
tanA
cscA
cosA1
csc AtanA
9. a:
10. a:
11 . a: 30°
12 . a:
EXERCISE VIII .
a OSO A
9 OOt A
13. a: or 60°
14. z
15. a: or 45°
16. z
17. a: 60°
27. sin A491 7
28.
1 3 cos2A 3cos4A
COSQA —COB‘ A
6 A =90°—B,
a =bcot B,
b
sin B
7 A =90° — B,
b=a tan B,
ca
cos B
8 COS A 2:B =90° —A
ANSWERS . 5
EXERCISE IX .
31. c A 39° B 50° F 15.
32 . b A 30'
B 89° 29'48 F
33. a A 43° B 46° F
34. 5 c B 71° F
35. a c B 80° F
36 . a b B 20° F
37. a b A 45° F
38 . b c A 41° F
39. a c A 55° F
40. a b A 3° F
41. F 1} (c? SinA cosA ) . 43. F : 1} (b2 tan A) .
42 . F i (a2 cotA ) .
45. b c A 40° 45'48 B =49° 14' 12
48 . a c B 34° 46’40
”A 55° 13
'20
47. a b c B
48. a b A 27° B 62°
49. 19° 28 17”and 70° 31
’43
”
57. 590 441
50° 3 °° I9°L58. feet.
51 . a c 00811
9-
(
rl 7 59. 1° 25’
60. m iles in each direction.
“ mn 1 81 . feet.
36° 52’12
”and 53° 62 . feet.
feet. 63. yards.
feet. 64. feet.3270 feet. 65. 140 feet.feet, 96 feet. 66 . feet.
EXERC ISE X.
1 . (90°—A ) , c=2 a cosA , h=a sinA .
2 . c=2 a cos A , h= a sinA .
3. (90°—A ) , a c + 2 cosA ,
h=a sinA .
TRIGONOMETRY.
- C) , a c -E-2 cosA ,
C’=2 (90°—A ) , a=h-Z- SinA ,
a =h -2—SinA ,
sinA =h—Z- a , —A ) ,—A ),
A 67° C 45° 14’
c 11
a h
a c
A 79° 36’
C 20°
A 77° 19 C 25° 21'
38”
A 47 C'=129° 4' 28
A 81° 12'
C 17° 35'
F =
F 012 SinA cosA .
F h2 tan 50 .
feet, sq. feet.
EXERCISE XI.
h a SinA .
c 2 a cosA .
c 2 a cosA .
c 2 a cosA .
a h sinA .
h
F
F
F
c
a
a
a 17,
22 .
23. 94°
24.
25. 38° 56’
33"
26.
71 F :
h F =
p F :
c F : 1238 .
c F =
17 0
11 p
12.
13.
14.
15.
16 .
TRIGONOMETRY.
Sin 163° 49’
sin 16° 17. cot 139° 17’
cos 195° 33’
cos 15° 18 . sec 299° 45'
tan 269° 15’
cot 0° 19. csc 92° 25’
Sin Sin 75° cos l 5°,
cos( cos 75° sin etc.
sin sin 127° cos
cos( cos l 27° sin etc .
sin Sin 160° Sin
cos cos200° cos20°, etc .
Sin sin 345° sin
cos cos345° cos etc.
Sin 52° Sin 52° 37’
cos 37°
cos( 52° cos 52° 37’
sin 37° etc.
sin 196° sin 196° 54'
sin 16°
cos( 196° cos 196° 54'
cos 16° etc.
sin 120° n/S, cos 120° 1, etc.
sin §V2, cos 135° {V2} etc.
sin 150° 4, cos p /fi, etc.
Sin 210° cos 210° iV3, etc.
sin 225° §V2 , cos 225° §V2, etc.
Sin 240° 1}V3, cos 240° etc.
sin 300° cos 300° 4, etc.
sin —4, cos i V3, etc.
sin 4V2 , cos 75 , etc.
cos a: {V2 or V3, etc. , at:
tan s:—V§ , 008 2:
—}V3, 2 :
Sin 3540° Sin 300° Sin 60° 1}V3, cos 3540° 1, etc.
210° and 120° and
and 150° and
and
sin cos tan cot
Sin cos tan cot 177°
(Hint : tan 238° tan sin 122° Sin
45. m Sin 2: cos x.
(a— b) sin z . 46. (a
— b) cot z
ANSWERS . 9
a2 + bz+ 2 ab cos sa 49. cos z sin y—
sin z cos y.
0. 50. tan 3 .
Positive betw een a: 0° and a: and betw een a: 315° and
z negative betw een a; 135° and a:
Positive betw een a; 45° and a; negative betw een £0 0° and
a: and betw een a: 225° and a:
Sin (a: cos 73, 008 (27 sin 23, etc.
sin (a: sin 23, cos (a: cos 73, etc.
EXERCISE XIV.
sin (2: y) gg, cos (a: y) 44. 2 . cos y, Siny .
Sin 90° y) cosy , cos (90°
y) sin y , etc.
Sin (180°
y) Sin 11, cos (180°
y) cos etc.
Sin (180°
y) sin 11, cos (180°
y) cos etc.
sin (270°
y) cos11, cos (270°
y) sin y , etc .
sin (270°
y) cosy , cos(270° y) sin 11, etc.
sin (360°
y) Sin 11, cos (360°
y) cos 11, etc.
Sin (360°
y) sing , cos (360°
y) cosy , etc.
sin (a: cos 23, cos (a: sin 23, etc.
sin (a: sin 23, cos (a: cos 73, etc.
sin (a: cos it , cos (a: sin 23, etc .
sin y) Sin 11, cos y) cos 11, etc.
Sin(45°—y) l,V2(cosy—siny) , cos(45
°—y) QV2(cosy+siny) , etc,
V2(cosy+siny) , V2(cosy—siny) , etc.
§(cosy+V3Siny) , cos(30°+ 11) cos y
—Sin etc.
d raw—y)=ite m —ems , m ew
—y)=4<cosy+V3sin etc.
3 Sin a: 4 Sin3z . 19 . 4 00831: 3 cosx. 20. 0. 21. V3.
1+0.4V8cos z
2
23. 808 2 7: —4, tan 2 z
—V3.
10 TRIGONOMETRY.
I
24. sin V2 V2 cos V2 V2
tan22§° V2 1 cot 225
° V2 1
25. sin 15° V2 V3 cos 15° V2 V3
tan 15° 2 V3 cot 15° 2 V3
27—33. The tru th of these equ ations is to be established by expressing
the given fu nctions in terms of the same function of the same angle .
Thu s, in Example 27,sin 2 2:=2 Sin z cos z ,
sin a: 1a 1 + tan2z =seczz
cos a: cos a:
By making these su bstitu tions in the given equ ation its tru th w illbe evident.
34.
B) cos } (A
cos lA cos éB, (88 8 55 29 and 31).
Bu t
Therefore,35. Proof Similar to that for 34 .
36. tan A + tanB + tan C
and 2 tan z =2
Sin C+
8111 0 Sin C cos C + cosA cosB Sin C
cosA cosB cos C' cosA cos B cos C
| cosA cosB
cosA cosB cos C cdsA cosB cos C'
Sin A Sin B Sin 0
cosA 008 B cos C
Proof Sim ilar to that for 36.
tan A tan B tan C'.
2
38.
2 42 . tan 3 .
mn 2 z
39 . 2 cot 2 2 . cos it cosy
40°
sinxcos y cosxcos y
Sin z cosy Sinx sin y
12
11.
TRIGONOMETRY.
EXERCISE XVIII .
420. 12 .
EXERCISE XIX.
15. 25. 18. m iles.
18 . 3800 yards . 19. and
17. yards. 20. 26° 0’10
”and 14° 5
’
50
EXERC ISE XX.
A 36° 52'
12 B 53° 7'
48 C' 18 .
A =B 33° 33’
0 112° 53'
17. 4° 23'W . ofN. , orW. of s.
18.
Impofl ble . 20.
15° 21. miles.
EXERCISE
EXERCISE
miles ; m iles.
m iles.
feet.
ab Sin A .
11, b“) tanA .
2421000.
120°
XXII.
5. 20 feet.6. or
7. yards.
8 . yards.
ANSWERS . 13
MISCELLANEOUS EXAMPLES.
2 . feet 21 . feet ; 46. feet.feet. feet. 47. feet.
3. feet. 22 . m iles. 48. m iles.
37° 34’
23. yards. 49. feet.m iles. 27. 8 inches. 51. feet.m iles. 30. feet. 52. yards.
m iles. 31 . feet. 53. feet.32 . m iles. 54 . pou nds ;
34. feet. 22° 23’47
”w ith first
35. feet. force .feet. 36 . feet. 55. pou nds ;
feet. 38 . N. 76° 56’
E. ; 45° 37’
16”w ith
45 feet. m iles an hr . know n force.26° 39. yards. 58.
feet. 40. feet. 59. feet.
75 feet. 41 . feet 60. yards.
m iles. feet. 61 . feet.m iles . 42 . feet. 63. yards.
feet . 43. feet. 64.
feet. 44. feet. 65. zls i 135°
45. miles.
66 . cos A
4 2 9
73. Abe sin A .
74. 402 Sin A Sin B csc (A
75.
14 TRIGONOMETRY .
199 A . 3 R. 8. P. 94 . 114 . S. 56° 7’30
”E . ;
210 A . 3 R. 26 P. 95. sq . ft. m iles.
12 A . 3 R. 36 P. 96. sq. ft. 115. N. 17° 25’ W.
3 A. 0 R. 6 P. 97. sq. ft. 37° 46’
N.
12 A . 1 R. 15 P. 98 . sq . ft. 116. S.
4 A. 2 R. 26 P. 99. sq . ft. 117. m iles.
14 A . 2 R. 9 P. 100. 121 . Long. 68° 55’ W.
61 A . 2 R. 103. 6. 122 . m iles.
4 A . 2 R. 26 P. 108 . 6. 124 . 33° 18
’N. ;
feet. 36° 24'
w .
ch . 111. 5° 25’
S. ; 125. N. 28° 47’E . ;
9 A . 0 R . 1 P. miles. 1293 m iles.
112 . miles ; 126 . S. 50° 40’ W . ;
m iles. 20° 9’ W.
113. 229 m iles ; 127. 38° 21’
N. ;
lat. 11° 39
’
S. 55° 12’ W.
171 m iles ; 32° 44’ W. 129. N. 36° 52’ W. 36° 8
’ W.
173 m iles ; 51° 16
’
s. 34° 13'
E.
s. 50° 58’
E . 47° 15'
N. 20° 49'w .
N. 53° 20'
E. , 18° 7
' W. or N. 53° 20' W . ,
25° 53' W.
N. 47° E., 19
° 27'N. 121° 51
'E . or N. 47° W . ,
19° 27'N.
118° 9'E. or S. 47° E . , 14
° 33'
N. ,121° 48
’
E . or S. 47°
W. , 14° 33
'
N. ,116° 12
'E .
Lat. m iles ; lat. m iles ; lat. m iles .
N. 72° 33’
E . 45 m iles ; 42° 15’
N. , 69° 5
’ W.
N. 72° 4’ W. , 287 m iles ; 32° 54’
S. , 13° 2
’
E.
16 TRIGONOMETRY.
z
2: tan“ 1V2
a: :1:
a; :l: :1:
a: :1: :1: zt
i
2: :l: sin—1 4.
1a: 30° 150° “ 1
1 008
(1 2
a: tan- 1 tan
“ 1 44.
y cc indeterm inatezt V; j; 1}V3
2: 94
0
a: y3 ’
a; 71
3 2 y 2
150. 4 .
151. tan (a:
1Sin 2:
y tansin y
z “an 3
3; 4
1
1, :t V5. tan
—1
-
7§ivg.
2 .
a: l‘fii . cotfi a: tan2 a .
EXAMINATION PAPERS.
7. feet.
ANSWERS . 17
III.
1
2 . tan 0==p V8,
3—34 080 0 i 2 .
(b) 150°
6. 33° 34'
5 99° 4'
43 7. yds.
IV.
6 . feet. 7. A 37° 24’
58 B 51° 37’
52 C -90° 57’
10
l . 17} years. 4. 1 .
2 . 8111 2 23 : i m ’ tan 2 z =zt 5.
6. N. 50° 18'
E. , 399 m iles.
3. a: 44° 25'
135° 34'30
1 , 16,4 . 116° 33
'
54
3 tan a: tan8 3:296° 33
'
1 3 tan”.5. F irst ship , 223 m iles ; second
3 . Third side, any valu e ; opposite ship , 306 m iles.
side,
1 . 25. 4 . :l:
2 . 2 . 5 S. 83° 41'E . 1907 m iles .
3 .fis‘l, 43
° 43’
10 106° 16’50
VIII.
27. 2 . a V2 F tan A , b=x/2 F cotA .
a i : zt b a:
Smallest valu e of Opposite side , 1 63° 7’48 81° 52
’12 or
126° 62’
12 8° 7’
5 . 39° 29’N. , 67
° 14’ W. 6. tan a tan2 b or cot2 b.
18 TRIGONOMETRY.
1.
2 . B =60°
3. 155° 42'
114° 17'
4. 41° 24'
82° 49 55° 40'15"
5. N. 59° 55'E . 609 m iles.
1. 4 .
2 . a 4 , b 3, c 5, A 53° 7’
5. 14° 10’ E . 342 miles .
B 36° 52’
12 6. 2.
3. cos2 A .
1 . log, 4 4. feet.3. 97° 5. 47° 24
’
N 63° 43’ W.
7. 61° 37’
56° 14’
30”
XIII.
7. 8 . 44° 35'
40
XIV.
1. 200° 32'7 5 . 1.
7. a b c
u = 62° 9’ 42 B=61° 50
’
8. 42° 49'
1 . (a) 114° 35
’
(b) g 8 . 205° 24'47
7. 590
20 SPHERICAL TRIGONOMETRY.
V3 i V3 i
2 2
3. cos 674° i sin cos 1574
° i Sin cos2474° i sin
cos i sin
4 . Sin 40 : 4 0083 0 Gino—4 008 0 81118 0.
cos 4 0=cos4 0—6 0082 0 8in9 0+ sin4 0.
EXERC ISE XXVIII.
2 o
5.
4 720
t—1“ M ’
3 45 945
Sin 10° cos 10° 0. 984808 .
tan 15°
SPHERICAL TRIGONOMETRY.
EXERCISE XXIX .
1 . 80° 2 . 7.
EXERCISE XXX.
3. (i. ) Either a. or 5 m u st be equ al to A B b.
EXERCISE XXXI .
2 . I. The cosine of the middle part the produ ct of the cotangents of
the adjacent parts.
11.
‘
The cosine of the m iddle part the produ ct of the sines of Oppo
site parts.
ANSWERS . 21
EXERCISE XXXII .
A : 175° 57’10 B = C = 135° 34
’
7
C’ 104° 41
’39 a. 104° 53
’ b 133° 39’
48
a b and B are indeterm inate.
b= B=90 .
The triangle is impossible .b 130° 41
’
c 71° 27’43 A 112° 57
’2
a =26° 3’
51 A B=65°
Impossible .
EXERCISE XXXIII .
1. cosA =cot a tan 4b, Sin § B =csc a sin -; b, cos h z cos a sec i b.
2. sin § A =4sec 4 a .
3. sin 4A =sec § a cos 8111 R =81n%0. 080
sin r =tan .1. a m t-1&n
4 . Tetrahedron, 70° 31
’
46 octahedron, 109° 28’14 icosahedron,
138° 11’
cu be , dodecahedron, 116° 33’
44
5. cot 4A a .
EXERC ISE XXXV.
l . (i. ) tanm tanbcosA , (ii. ) tan m tan c cos B ,
cos a cos b secm cos (c m) ; cosb cos 0 sec m cos (a m) .
EXERCISE XXXVI .
1. (i. ) cot a: tan B csc (1 , (ii. ) cot a: tanC’ cscb,
cosA cosB csc ccsin (C cosB=cos 0 csc z Sin (A x) .
EXERCISE XLI.
4 . squ are m iles.
22 SPHERICAL TRIGONOMETRY.
EXERCISE XLII .
If 2: denotes the angle requ ired , sin 42: cos 18° sec 2: 148°
cos a: cos A cos B.
Let w the inclination of the edge c to the plane of a and b. Thenit is easily show n that V=abc sin Isin 10 . Now , conceive a sphereconstru cted having for centre the vertex of the trihedral anglew hose edges are a , b, c. The spherical triangle, w hose vertices
are the points w here a , b, c meet the su rface of this sphere, hasfor its sides I, m ,
n ; and w is equ al to the perpendicu lar are fromthe side 1 to the Opposite vertex. Let L , M, N denote the anglesof this triangle . Then, by means of [39] and w e find that
sin w sin m sin N =2 sin m sin c os 4N
w here
4 . (i. ) squ are m iles ; squ are m iles.
5. Let m longitu de of point w here the ship crosses the equ ator, Bher cou rse at the equ ator, d distance sailed . Then
tau m sin l tan a ,
6. Let k arc of the parallel betw een the places, a: difierence requ iredthen sin 4 h=sin 4d sec z. c : 90°(V2
7. tan 1} (m— d) sin (8
— l) sin(8— l w here 2 3
l l’
d, and m and m’
are the longitu des of the places.
9 . 44 m in. past 12 O’clock. 10.
11. cos t —tan d tan l ; tim e of su nrise 12i
t—so’clock A .H . tim e
of su nset1
t—
5o
’clock p . r1 . cos a sin d sec 1. F or longest day
at Boston : tim e of su nrise , 4 hrs. 26 m in. 50 sec. A .M . tim e ofsu nset, 7 hrs. 33m in. 10 sec. p . r1 . Azim u th of sun at these tim es ,57° 25
’15
” length of day , 15hrs. 6m in . 20 sec. for shortest day ,times of su nrise and su nset are 7 hrs. 33 m in. 10 sec. A .M . and
4 hrs. 26 m in. 50 sec. P . M . azim u th of su n, 122° 34
’ lengthof day , 8 hrs. 53m in . 40 sec.
12. The problem is impossible w hen cot d < tan l that is , for places in
the frigid zone .
