Solutions Manual of Geometry and Discrete Mathematics

Solutions_Manual_of_Geometry_and_Discrete_Mathematics_/HM12_Geometry_and_Discrete_Mathematics_Solutions_Manual/sm_files/ch01.pdfSolutions for Selected Problems

Exercise 1.1

4. c. Clearly if n 41 the expression becomes 412 41 41 or 41 43, which is composite.Actually, if n 40, we have 402 40 41 40 (40 1) 41 412.The expression gives a prime number for n 1, 2,3, , 39, but not for 40.

6. The expression to be tested is n2 9, n even.For n 4, n2 9 7, which is not composite.The statement is not true.

7. The expression to be examined is n2 9, n odd.For n 5, n2 9 16, which is divisible by 8.For n 7, n2 9 40, which is divisible by 8.For n 9, n2 9 72.For n 13, n2 9 160.For n 15, n2 9 216.All are divisible by 8.

8. The expression to be tested is n2 3.For n 5, n2 3 22.For n 6, n2 3 33.For n 7, n2 3 46.For n 8, n2 3 61, which is prime.The expression is not a composite number for all values of n.

Exercise 1.2

2. Every odd integer n can be written as 2p 1, where p 1, 2, 3, . Now, p itself can be either even or odd. If p is even, set p 2k, k 0, 1, 2, 3, . If p is odd, set p 2k 1, k 0, 1, 2, 3, .Then every odd number can be written as n 4k 1or n 4k 3, k 0, 1, 2, 3, .If n 4k 1, then n 7 4k 8 4(k 2).If n 4k 3, then n 5 4k 8 4(k 2).Hence one of n 5, n 7 is always divisible by 4.

3. n3 4n n(n2 4) n(n 2)(n 2).

If n is even, then set n 2k, k 1, 2, 3, .Now n3 4n 2k(2k 2)(2k 2)

8k(k 1)(k 1).Since k 1, k, k 1 are three consecutive integersfor all values of k, one of them is divisible by 2 andone (possibly the same one) is divisible by 3.Then k(k 1)(k 1) is divisible by 6.Therefore n3 4n is divisible by 8 6 48.

4. Every odd integer n can be written as n 2p1,k 0, 1, 2, .Then n2 4p2 4p 1

4p (p 1) 1.Now p and p 1 are consecutive integers, so one ofthem is even. Then p(p 1) 2k. Therefore n2 8k 1, where k is an integer.

5. Since 71 ends in 7, 72 ends in 49, 73 ends in 43, 74

ends in 01, then 75 71 74 ends in 07, 76 72 74 ends in 49, 77 73 74 ends in 43, and 78 74 74 ends in 01.Then 79 ends in 07, 710 ends in 49, 711 ends in 43, and712 ends in 01.This cycle continues.Hence 74k+1 ends in 07, 74k+2 ends in 49, and so on.Since 201 4 50 1, 7201 ends in 07.

6. If x and y are integers, then 2x is even and 4y is even.Hence the left side of the equation is divisible by 2,which is impossible if the right side is 5. Then it isimpossible for x and y both to be integers.

7. n5 5n3 4n n(n4 5n2 4) n(n2 1)(n2 4) (n 2)(n 1)n(n 1)(n 2).

This is the product of 5 consecutive integers, so one ofthem is divisible by 5, and at least one is divisible byeach of 4, 3, and 2, provided that n 3.Then the expression is divisible by 5 4 3 2 120, for n 3 and an integer.

Chapter 1: Introduction to Proof 1

Chapter 1 Introduction to Proof

8. Since p and q are odd primes, then p q is even, so 2divides p q.Now, if q p, then

p 2

q lies between p and q. It is

larger than p and smaller than q.But every number between p and q is composite,

since p and q are consecutive odd primes, so p

2q

has at least two divisors.Then p q has at least three divisors.

9. Every integer can be written as 3k, 3k 1, or 3k 2.Hence there are two possibilities. The first is that thereare three of a

1, a

2, a

3, a

4, a

5, of the same form. That

is, three of them are of the form 3k1, 3k

2, 3k

3, or they

are 3k1

1, 3k2

1, 3k3

1, or they are 3k1

2,3k

2 2, 3k

3 2. In each case their sum is divisible

by 3.The second case is that there are not three of the sameform, so there must be at least one of each form, say3k

1, 3k

2 1, 3k

3 2. But now this sum is 3(k

1 k

2 k

3 1), which is divisible by 3.

Hence there is always at least one subset of threenumbers whose sum is divisible by 3.

10. Consider n2 4 (n 2)(n 2).If n 3, then each of n 2 and n 2 is greater than 1, and n2 4 is a composite number.

11. Every odd integer n can be written as n 2k 1,k 2, if we wish integers greater than 5.Now n2 25 (2k 1) 2 25

4k2 4k 24 4(k)(k 1) 24.

Now k and k 1 are consecutive integers, so one ofthem is even. Then 4k(k 1) is divisible by 8.Since 24 is also divisible by 8, then n2 25 is divisible by 8 for n 5.

Exercise 1.3

3. Let the angles in any quadrilateral have measures a,b, c, d, and let the exterior angles at oppositevertices have measures x and y, as shown.Then a b c d 360 (angles in aquadrilateral).Also b x 180 and d y 180 (straight angles).Then b x d y 360.Therefore a x c y 0 by subtraction and a c x y.The sum of the exterior angles at opposite vertices is equal to the sum of the interior angles at the othertwo vertices.

4. Let ABC be 2x and ACB be 2y.Then 2x 2y 90 180 (angles in a triangle).

Therefore x y 45.Now BDC x y 180 (angles in a triangle).

Therefore BDC 135.BDC 135.

5. In PQR, PQR PRQ (isosceles triangle).In QRS, QRS QSR (isosceles triangle).

Then SQR 2 QSR 180 (angles in a triangle).Also SQR PQR 180.

Then PQR 2 QSR.Then PQS 3 QSR.

P

Q R

S

B

A C

x x

Dy

y

y

a

d

c

b x

2 Chapter 1: Introduction to Proof

6. If the polygon has n sides and n angles each of size x,180(n 2) nx180n nx 360

n 18

3060

x .

Note that (180 x) must divide evenly into 360 if thepolygon is constructable.

7. Mark angles as shown, using the given information.Since AB AD, ADB b.Since AB BE, BEA a.Now 2x a 180

so x 90 12

a.

Also 2y b 180

so y 90 12

b.

In ABD, b b a 90 12

a 180

a 4 b 180.

In ABE, a a b 90 12

b 180

4a b 180.Adding 5a 5b 360

a b 72.In ABC, a b c 180

Then c 180 72 108.

BCA is 108.

8. Mark angles as shown.Since b 2x 180

x 90 b2

.

In ABD, a2

b 90 b2

t 180 (angle sum)

t 12

(a b) 90

2t (a b) 180

In ABC, c (a b) 180 (angle sum)Subtracting 2t c 0

t 12

C

or APB 12

ACB.

9. a. For the polygon ABCDEFG the sum of the exteriorangles is 360.But each exterior angle occurs twice.Hence P Q RST UV720 7 180.Then P Q R S T U V 540.

VA

P

B Q

C

R

D

SE

T

FU

G

B

A

C

P

a2

bxx

t

c

a2

A

a xx

cC

byy

B

E

D


b. In general, if there are n sides then there are nsmall triangles surrounding the inner polygon. Letthe sum of the angles at the tips of the star be S.Then S 720 n 180

S 180(n 4).

Exercise 1.4

6. Name the vertices as in the diagram, using propertiesof the rectangle.

Then AC a2 b2and BD a2 b2Therefore AC BD.Note that positioning the rectangle makes the proofsimple.

7. Name the vertices as in the diagram. Since we usemidpoints, we name vertices appropriately.Since W is the midpoint of A(0, 2b)and B(2a, 2b), its coordinates are (a, 2b).Similarly X has coordinates (2a, b), Y has coordinates (a, 0), and Z has coordinates (0, b).

Now WX (2a a)2 (b 2b)2 a2 b2XY (2a a)2 (b 0)2 a2 b2YZ a2 b2ZW (0 a)2 (b 2b)2 a2 b2.

These are all equal. Then WXYZ is a rhombus.

8. Let ABC be any triangle and let its coordinates be (2b, 2c), (0, 0), and (2a, 0), as shown.If D is the midpoint of AB, its coordinates are (b, c).If E is the midpoint of AC, its coordinates are (a b, c).The slope of BC is 0 and the slope of DE is 0, so theyare parallel.The length of BC is 2a and the length of DE is a,

so DE 12

BC.

The line segment joining the midpoints of two sides of a triangle is parallel to the third side and equal toone-half of it.

9. Let the parallelogram be positioned as shown, and letthe coordinates of A be (b, c), D be (0, 0), and C be(a, 0). Then since AB is equal and parallel to DC, thecoordinates of B are (a b, c).Now AB2 BC2 CD2 DA2 (b2 c2) [(a b b) 2 02] [(a b a)2 c2] a2

b2 c2 a2 b2 c2 a2

2 (a2 b2 c2).Also AC2 BD2 (a b)2 c2 (a b)2 c2

2(a2 b2 c2).The sum of the squares of the sides is equal to thesum of the squares of the diagonals.

A(b, c) B(a + b, c)

D(0, 0) C(a, 0)

y

x

y

A(2b, 2c)

D(b, c)

B(0, 0) C(2a, 0)x

E(a + b, c)

y

A(0, 2b) W(a, 2b) B(2a, 2b)

X(2a, b)

x

Z(0, b)

D(0, 0) Y(a, 0) C(2a, 0)

y

A(0, b) B(a, b)

D(0, 0) C(a, 0)x


10. Let the triangle be positioned as shown, and let thecoordinates of B be (0, 0), C be (2a, 0), and A be (b, c).Since D is the midpoint of BC its coordinates are (a, 0).

Then AB2 AC2 b2 c2 (2a b)2 c2

4a2 2b2 2c2 4abAlso 2BD2 2AD2 2(a2) 2 ((b a)2 c2)

4a2 2b2 2c2 4ab.The statement is true.

11. a. Let the rectangle be positioned as shown. Let P(x, y) be any point in the interior. Then 0 x aand 0 y b.Now PA2 PC2 x2 y2 (a x) 2 (b y2)

2x2 2y2 a2 b2 2ax 2by.

Also PB2 PD2 x2 (y b)2 (a x)2 y2

2x2 2y2 a2 b2 2ax 2by.

Then PA2 PC2 PB2 PD2.

b. Through P draw XY AD and BC, as shown.Then AY BX and YD XC.Now PA2 PY2 AY2

and PC2 PX2 XC2

so PA2 PC2 PY2 PX2 AY2 XC2.Also PB2 PD2 PX2 BX2 PY2 YD2

PX2 AY2 PY2 XC2.Then PA2 PC2 PB2 PD2.

12. Position the triangle as shown and let the coordinatesof the vertices be A (0, a), B (b, 0), and C (c, 0).Then the perpendicular from A meets BC at O (0, 0),and the equation of AO is x 0.

The slope of AB is ab

, so the slope of CE is ba

, and

the equation of CE is y 0 ba

(x c).

The intersection of AO and CE is the point P and forthe coordinates of P we set x 0 in the equation of CE.Then y

bac.

P is the point 0, bac.bac 0

Then the slope of the line BF is ac

.0 b

The slope of AC is ac

.

Now ac

ac 1.Then BF AC, and the altitudes are concurrent.

y

A(0, a)

F

xC(c, 0)OB(b, 0)

E

P

B X C

p

A Y D

B(0, b) C(a, b)

P(x, y)

A(0, 0) D(a, 0)

y A(b, c)

B(0, 0) D(a, 0) C(2a, 0)x


13. Position the triangle as shown and let the coordinatesbe A(2a, 2b), B(2c, 0) and C(2c, 0).Then the median from A meets BC at D (0, 0).The median from C meets AB at E (a c, b).

The equation of AD is y ba

x.

The equation of CE is y 0 a

b3c

x 2c.These lines intersect at P, and for intersection

ba

x a

b3c

x 2c(a 3c) x a (x 2c)

3cx 2ac

x 23a

Then y ba

x 23b.

P has coordinates 23a, 23b.The median from B meets AC at F (a c, b).

The equation of BF is y a

b3c

x 2c.We check that P lies on BF.

For y 23b, the right side is

a b

3c 23a 2c

a

b3c

2a 3 6c

23b.

Point P is on BF, and the medians are concurrent.

Chapter 1 Test

1. c(x, y) must lie on the right bisector of AB. Theequation of the right bisector is x 2.

2. In a convex hexagon there are six vertices at each ofwhich there is an interior and an exterior angle. Thesum of the interior and exterior angles at a vertex is180. Then the sum of the six interior and six exteriorangles is 6 180 1080.The sum of the six interior angles is 180(6 2) 720.Therefore the sum of the exterior angles is 1080 720 360.

3. Let the trapezium be positioned as shown and let thecoordinates of the vertices be A(2a, 2b), B(2c, 2b),C(2e, 0), and D(0, 0).Then the coordinates of P, Q, R, S, are P(a, b),Q(c e, b), R(c, b), S(a e, b).All of these points are b units above the x-axis, so allof them lie on the line whose equation is y b.

4. Let the equal parts of ABC be x and let the equalparts of ACD be y. Now let ACB be z.Because BCD 180, 2y z 180,

In ABC, 2x z 58 1802x z 122

Then 2x 2y 2z 302x y z 151.

In EBC, E x y z 180Therefore E 29.

58

xx

BZ

yy

CD

EA

y

A(2a, 2b) B(2c, 2b)

C(2e, 0)x

D(0, 0)

PR S

Q

E

y

A(2a, 2b)

F

C(2c, 0)DB(2c, 0)x


5. Solution 1.Let the three consecutive even numbers be 2k2, 2k,2k2.Then (2k 2)2 (2k)2 (2k 2)2

4k2 8k 4 4k2 4k2 8k 4 12k2 8 4(3k2 2) 22(3k2 2).The expression is always divisible by 2, 2, and at leastone other number.

6. Let D be the midpoint of AB. D has coordinates

0, a2.The right bisector of AB has equation y

2a

.Let E be the midpoint of BC.

E has coordinates b2, 2c.The slope of BC is

bc

.

Then the right bisector of BC has slope bc

.

The equation of the right bisector of BC is

y 2c

bc

x b2or y

bc

x 2bc

2

2c

bc

x b2

2c

c2.

These right bisectors intersect at P. For thecoordinates of P solve the equations:

y a2

and y bc

x b2

2c

c2

a2

bc

x b2

2c

c2

bc

x b2

2c

c2

a2

b2

2c2

c ac

x b2

2c2

b ac

P has coordinates b2 2c2

b ac,

a2

.We now show that if F b2, a 2 c is the midpoint ofAC, then PF is the right bisector of AC. This requiresonly that PF AC, or that the product of the slopesof PF and AC be 1.

The slope of AC is c

ba

.

The slope of DF is

2

c

-

b2 b2

b

c2 ac

c

ac

b

c2

a

b

c.

The product of these slopes is c

b

a

a

b

c 1.

Then PF is the right bisector of AC.Therefore the right bisectors are concurrent.

7. a. D 92 4(8)(10) 239

Since D 0 the equation has imaginary roots.

b. For the equation (n 1) x2 nx (n 1) 0,D n2 4(n 1)(n 1)

n2 4(n2 1)3n2 4.

Since the coefficients are positive integers, n 2.Then D 0 for all n.Therefore the equation has imaginary roots for all n.

y

A(0, a)F

C(b, c)

P

E

xB(0, 0)

D

a

2c

a2

b2

b2

2c2

b ac



Exercise 2.1

4. a. In ADC and CBAAD CB parallelogramDC BAAC is commontherefore ADC CBA (side-side-side).

b. In PST, PS PTtherefore S T (isosceles triangle)and PSQ PTR (supp. to S and T).Now in PSQ and PTRPS PT (given)PSQ PTR (proven)SQ TR (given)therefore PSQ PTR (side-angle-side).

c. In ABC and DCB,AB DC (given)ABC DCB (given)BC is commontherefore ABC DCB (side-angle-side).

d. Diameters of a circle bisect each other. HenceOA OB OC OD.In AOB and CODOA OC radiiOB ODAOB COD (vert. opp.)therefore AOB COD (side-angle-side).

6. In PQR, PQ PRtherefore PQR PRQ (isosceles)and SQR TRQ (supp. angles)Now PS PT (given)

PQ PR (given)therefore PS PQ PT PR and QS RT.In QRS and RQT : QS RTSQR TRQ, QR is common.Therefore QRS RQT (side-angle-side)and QRS RQT.

7. In ABC and ADCAB AD (given)BAC DAC (given)AC ACtherefore ABC ADC (side-angle-side)therefore BC DC and BCA DCAtherefore AC bisects BCD.

8. In ABC and DBEAB BE (given)ABC DBE (opp. angles)CAB DEB 90 (given)therefore ABC EBD (angle-side-angle)and AC ED.

9. In PQS and PRSPQ PRQPS RPSPS is common.Therefore PQS QRS (side-angle-side)and PQS PRS x.In PQR, PQ PRtherefore PQT PRThence PQT x PRT xand SQT SRT.In AEB and CEDAB CD (given)BAE DCE (proven)AEB CED (opp. angles)therefore AEB CEDand AE CE, BE DEand diagonally bisect each other.

10.

In PQR and SRQPQ SRPR SQ (given)QR is commontherefore PQR SRQ (side-side-side)and PQR SRQ.

Q

P

RS

8 Chapter 2: Plane Figures and Proof

Chapter 2 Plane Figures and Proof

11.

Given: Quadrilateral ABCD with AB DCand AD BC. Prove AC and DB bisect each other;i.e., AE EC and DE BE.Proof: In ABC and CDA

AB CDBC DA

and AC CAtherefore ABC CDA (side-side-side)and BAC DCA.

12.

In QTS and RWSQST RSW (opp. angles)QS RSQTS RWS 90therefore QST RSW (angle-side-angle) and QT RW.

13. Place vertex D of DEF on vertex A of ABC, and letDE coincide with AB. Since DE AB, E will fall onB. Since EDF BAC, DF will fall along AC andsince DF AC, F falls on C. Therefore DEFcoincides with ABC and the triangles are congruent.

14. a. Two angles and three corresponding sides.

b. Three angles and two corresponding sides.

c. No similarity for sides only.

15. Since A and B are the mid-points of equal sides SRand UV.SA AR UB BVin ASP and BVW

SA VBASP BVW 90

PS WVtherefore ASP BVW (side-angle-side)

and AP BW.In ASX and BUP

AS BUSX UP

ASX BUP 90therefore ASX BUP (side-angle-side)

therefore AX BP.

16. Since the faces are equilateral triangles,ABX ACY 60

and AB AC.Now in ABX and ACY

AB ACABX ACY 60

BX CYtherefore ABX ACY (side-angle-side)

and AX AY.

Exercise 2.2

5. Since A and B are the midpoints of PQ and SR,PA AQ SB BR a.Let the distance between PQ and SR be h.

Now gm PQRS 2ahgm ASBQ ah

therefore gm PQRS 2gm ASBQ

or gm ASBQ 12

gm PQRS.

x

o

y

o x

Q

T

RS

W

P

A B

D C

E

Chapter 2: Plane Figures and Proof 9

6.

PS is a median therefore QS SR a.Let height of PQR be h.Therefore RQS 1

2 ah

PSR 12

ah

and PQS PSR.Therefore a median bisects the area of the triangle.

7. Since AD is a median ADC 12

ABC.

Since BE is a median BEC 12

ABC.

Therefore ADC BEC.But quad DFEC is common to both triangles thereforeADC quad DFEC BEC quad DFECTherefore AEF BFD.

8. AD is a median of ABCtherefore ABD ADCsimilarly BED DECtherefore ABD BED ADC DEC and ABE ACE.

9. Since the diagonals of a parallelogram bisect eachother, PT TR.Therefore PTQ RTQ aand PST TSR b.Also ST TQ therefore STP TQPhence a b and PQT PTS TSR TRQ.

10.

Since PTS STR and both have the same heightthen the bases are equal. Then PT RT. SimilarlyPTS PTQ and both have the same height.Therefore ST QT.

Now in PTS and QTRTS TQPT TR

PTS QTRtherefore PTS RTQ (side-angle-side)therefore PS RQ.Also PST TQR therefore PSQR (all s).Since PS QR and PSQR.Therefore PSQR is a parallelogram.

11.

Construct a line through T that is parallel to PQ andSR meeting PS at X and RQ at Y.Now PXYQ is a parallelogram

and PTQ 12

gm PXYQ.

Similarly TSR 12

gm XSRY.

But gm PSRQ gm PXYQ gm XSRY.

Therefore

PTQ TSR 12

gm PXYQ 12

gm XSRY

12

gm PSRQ.

12.

Since ADC ABC and AC is common, thealtitudes of ADC and ABC are equal, i.e., DX BY.But DX and BY are both perpendicular to AC thereforeDXBY. Hence BYDX is a parallelogram and thediagonals of a parallelogram bisect each other.Therefore BT DT, i.e., AC bisects BD.

A

B C

D

Y

TX

P Q

X YT

S R

P

S

T

Q R

P

Q RS

h


13.

CF is a median of ABC therefore

CFB 12

ABC

gm BCE 12

ABC.

Therefore CFB BCE. But COB is commontherefore CFB COB BEC COBand EOC FOB a.OE is a median of AOCtherefore EOC OEA aOF is a median of AOBtherefore AOF OFB aand AOE AOF.

14.

AC bisects parallelogram ABCD.Therefore ABC ACD. The base AC is common.Therefore the height of each triangle is the same, say h.

Therefore ADX 12

Ax h

ABX 12

Ax h.

Therefore ADX ABX.

15.

AOD 12

gm ABCD

gm ABCD AOD 12

gm ABCD

gm ABCD AOD AOB DOC

therefore AOB DOC 12

gm ABCD.

16.

Given quad ABCD construct through A a line parallelto BD meeting CD extended at E. Join EB.Then EBC quad ABCD.Proof: BDC is common to both BEC and quad ABCD.BDE BDA (same base, same altitude).Therefore BDE ADC ABD ADC orBEC quad ABCD.

17.

Since AB AC x

ABC 12

CD x 12

BE x

Therefore CD BE.In DBC and BEC, BC is common, CD BEBDC 5BEC 90therefore BDC CEB (hyp. side)therefore BDC CEBnow BC is common; therefore altitudes from D and Eare equal. Therefore DEBC.

18.

Extend BC to F so that EF BF.Draw DG BC.

A

D

BG K

E

C F

A

D E

B C

A

E

D

B C

A D

B O C

A

B

D

C

X

C

E

A BF

O


In DGB and EFC,BD CD (given)

DBG ECF (ABC ACB ECF)DGB EFC (90)DGB EFC

then DG EF.Now in DGK, EFK,

DG EFDGK EFK (90)DKG EKF (vert. opp.)

then DGK EFK (right-angled triangles)therefore DK EK.

Exercise 2.3

6. Join S1

and S2. Construct the right bisector of S

1S

2meeting the circle at two points, A and B. Gates arelocated at A and B. Since they lie on the right bisectorof S

1S

2they will be equidistant from S

1and S

2.

7. Approximate the second road as a straight line andextend the roads to meet. The pumping station shouldlie on the river at the point determined by the bisectorof the angle formed by the two roads.

8. Bisect the angle formed by the intersecting lines. All points on this line are equidistant from the twointersecting lines. Construct the right bisector of AB.All points on this line are equidistant from A and B.Then the intersection of these bisectors gives therequired point.

9. Since a circle can always be drawn through three non-collinear points, the fourth point is restricted to be onthe circle if the four points are to be concyclic.

10. Construct the right bisector of YZ meeting the circle atX

1and X

2. X

1YZ and X

2YZ are isosceles triangles.

These two are always possible. If YZ is not a diameterthere are two others, X

3found by YZ YX

3and X

4by

ZY ZX4.

11.

Let the right bisectors of AB and AC intersect at O.Then OB OA and OC OA therefore OB OC.Since O is equidistant from B and C, it lies on the rightbisector of BC. Therefore the right bisectors of thesides of a triangle pass through a common point.

13.

Let the bisectors of B and C meet at T. Draw aline through T parallel to BC meeting AB and AC at Dand E respectively. Now DTBC. Therefore DTB TBC a, hence DT DB.Similarly ET CE.Therefore DT TE DB CE

DE DB CE.

14.

Let the midpoint of BC be X. The right bisector of AXmeets AB at D. D is the required point.

15. If the angles form an arithmetic sequence, then let theangles be a d, a, a d.Now a d a a d 180

3a 180a 60

therefore one of the angles is 60.If one of the angles is 60 then one must be smallerthan 60 and the other larger than 60 (or else all are 60). Let the angles be (60 x), 60, and (60 y).Now 60 x 60 60 y 180

x y 0x y

Therefore the angles are in arithmetic sequence.

A

D

B X C

A

D T E

B C

A

BO

C

YX


16. Let the roots be p1

p and q.Now (x p)(x p)(x q) 0

(x2 p2)(x q) 0x3 qx2 p2x p2q 0

x3 ax2 bx c 0c p2q, but p2 b

and q a.Therefore c ab.Now if c abx3 ax2 bx c 0becomes

x3 ax2 bx ab 0x2(x a) b(x a) 0

(x1 a)(x2 b) 0x a 0 or x2 b 0

x2 bx b

Therefore one of the roots is the negative of the other.

Exercise 2.4

1. Assume that DBE DEB.Now ACE DEB (DEAC)and AC AB (isosceles triangle)but AB > AC, therefore there is a contradiction, andhence DBE DEB.

3. Either the line intersects the curve or it doesnt.Assume that it does. Thenx4 3x2 2x 2x 1x4 3x2 1 0

Since the left side is always greater than 1, there areno values of x that satisfy the equation. We have acontradiction. Therefore there is no intersection.

4. Either a, b, c are consecutive terms of a geometricsequence or they are not. Assume they are. Then wecan represent a, b, and c as a, ar, and ar2. Theequation becomesax2 arx ar2 0

x2 rx r2 0, since a 0

then x r

2r2 4r2

r

2 3r2

.

This equation has no real roots. Then a, b, and ccannot form a geometric sequence.

5.

Either CD is parallel to EF or CD is not parallel toEF. Assume that CD is not parallel to EF and let PTintersect CD, EF, and AB as in the diagram at Q, R,and S. If CD is not parallel to EF then DQR FRS. Since CD parallels AB, then DQR BST.Since EF parallels AB, then FRS BST. But thenDQR FRS and it is a contradiction that they arenot equal. Then since DQR FRS, CDEF.

6.

Suppose that ABC does not have two equal angles.Then ABC BAC. Since CEAB, BAC ACE(alt. angles) and ABC ECD (corr. angles).Therefore ACE ECD.But this is a contradiction, because CE bisects ACD.Then ABC BAC and the triangle has two equalangles.

7.

In ADX and CDBXD DB (given)

ADX CDB (vert. opp.)AD DC (median)

then ADX CDB (side-angle-side)therefore DAX DCBtherefore AXBC (alt. angles).Similarly AY BC.Since AXBC and AYBC and the line AX and AY havethe common point A, then X, A, and Y are collinear.

AY X

E D

B C

A

BC

E

D

a

b ca

b

P

Q

R

S

F

B

T

C

E

A

D


8.

Since PXQYPXM QYN x (alt. angles).In PMX, PX PM (radii)therefore PMX PXM x (isosceles triangle).In QNY, QY QN (radii)therefore QNY QYN x (isosceles triangle)hence PMX YNQ x.Extend QN to T.TNM QNY x (vert. opp.)therefore TNM PMXand TQPM (corr. angles)i.e., PMQN.

9.

Let l1

be the right bisector of BC and l2

be the rightbisector of CA. Either l

1and l

2meet or they do not

meet. Assume that they do not meet. Then l1l

2.

Since l1BC and l

2CA if l

1l

2then BCCA.

But this is impossible since BC and CA are sides of atriangle. It is impossible for l

1to be parallel to l

2, and

the lines must meet.

10. Originally there are 32 white and 32 coloured squareson the checkerboard. Each domino tile will cover onewhite and one coloured square. By removing twowhite squares we are left with 30 white and 32 blacksquares. To cover the board we must have the samenumber of each colour, so it is not possible.

11. Either x < 1.1, or x 1.1.If x 1.1, x9 2.357 and 7x 7.7. Then x9 7x >10.057. This is a contradiction, since x9 7x < 10.Then x < 1.1.

Exercise 2.5

5.

In BFC, FD is a median.Therefore BFD DFC x, say.Similarly in ABD and ADC, BF and CF aremedians, hence ABF BDF x and AFC DFC x.

a. Therefore

A

A

B

B

C

F

4

x

x

1

4

b.

A

A

F

B

C

C

4

x

x

1

4

c. and ABF AFC x.

6. a. If a

b

d

c

then a

b 1

d

c 1

a

b

b

c

d

d.

b. If a

b

d

c

then m

nb

a

m

nd

c

and m

nb

a 1

m

nd

c 1

Therefore ma

nb

nb

mcnd

nd

multiplying by n gives ma

bnb

mc

dnd

.

7.

ADE and EDB have the same height ED and basesAD and DB.

Therefore

A

E

D

D

E

B

2

1.

A

E D

C B

2

1

3

3k

4k

2k

A

B CD

F

x x

l1

l2A

BC

Y

TN

x

xx

x

x

X

PM

Q


Since EDCB, EDAB, and ADE is isosceles,so ED 2. EDB and ECB have the same height DB and basesED 2 and CB 3.

Therefore

E

E

D

CB

B

2

3.

Let EDB 2k (to avoid fractions).Then ADE 4k and ECB 3k.Therefore trapezoid DECB:ADE 5:4.

8. In ABD, EGDB

therefore A

E

E

B

G

AG

D.

In ADC, GFDC

therefore G

AG

D

F

A

C

F.

Then in ABC, A

E

E

B

F

A

C

F and EFBC.

10.

Join RX meeting YQ at A.

In XZR, YAZR

therefore XYZ

Y

XA

AR.

In XPR, XPAQ

Therefore X

A

A

R

Q

PQ

R

hence XYZ

Y

PQ

QR

11.

Let BX and DY meet AC at M and N. X and Y are themidpoints of AD and BC henceAX XD BY CYalso XDBYtherefore XDYB is a parallelogram so BXYD.In AND, MXND, AX XDtherefore AM MN.In BMC, BMYN, BY YCtherefore MN NC.Therefore AM MN NC and AC is trisected by BX and DY.

12.

Join BR. Since A

P

P

B

3

4, then

A

B

P

P

R

R

3

4.

Let BPR 12kthen APR 9k.

Similarly

B

A

R

B

C

R

3

2, so ABR 21k and BRC

14k.

Join AZ.

C

AZ

Z

R

R

3

2.

Let CZR 2xthen AZR 3x.

Now

A

B

P

P

Z

Z

3

4

therefore 9

2

k

6k

3x

2x

3

4

36k 12x 78k 6x6x 42kx 7k

therefore RCZ 14k BCR.Since both triangles have the same height their baseswill be equal hence BC CZ and C is the midpoint of BZ.

A

P

B C Z

R 3x

2x

9k

12k

14k

A X D

M

N

B Y C

X

Y

Z

A

P

Q

R


13.

Extend DA and CB to meet at P.Let AX XD x

BY YC y.Let AP a and BP b.

In PDC, ABDC

therefore A

A

D

P

B

P

C

B

or 2

a

x

2

b

y from which

a

x

b

y.

In PXY, a

x

b

y or

P

A

A

X

B

P

Y

B

hence ABXYtherefore ABXY DC.

14.

Join AD.In ABC, KEAB

therefore C

K

K

B

C

EA

E

In BDC, KFBD

therefore C

K

K

B

F

C

D

F

now in CAD, C

EA

E

F

C

D

F

therefore EFAD.

Exercise 2.6

6. a. ADE ~ ABC (angles equal)

therefore

AA

DBC

E

AA

DB2

2

499

therefore ABC 81 499

ABC 441.

b. Quad DBCE 441 81 360.

8.

a. Given ABC and DEF with A D, B Eand C F.Therefore ABC ~ DEF.Now C and F are bisected.Therefore ACM DFN and AMC ~ DNF(angles equal)

therefore CF

MN

FC

DA.

9. Given ABC ~ DEF.

Therefore DAB

E

DAC

F

BE

CF k, say.

Now AB k DEAC k DF

and BC k EF.Adding: AB AC BC k (DE DF EF)

DAB

E

AD

CF

BECF

k

i.e., the perimeter of ABC (AB AC BC)to the perimeter of DEF (DE DF EF) is equalto k which in turn is equal to

DAB

C

DAC

F

BE

CF.

10. In PQO, ABPQ so OAP

A

OBQ

B.

In OQR, BCQR so OBQ

B

OC

CR and

in ORS, CDRS therefore OC

CR

OO

DS

therefore OAP

A

OO

DS hence in OPS, ADPS.

A

M

B C

xx

D

N

FEx

x

A

BK

C

F

E D

P

A B

X Y

D C


11. a. ABD ~ ACD ~ BCA.

b. (i) ABD ~ ADC

Therefore BA

DD

DAD

C and (AD)2 BD DC.

(ii) ABC ~ ADC

Therefore BA

CC

DAC

C and (AC)2 BC DC.

(iii) ABC ~ ABD

Therefore BA

DB

BA

CB and (AB)2 BC BD.

c. From parts (ii) and (iii) addingAC2 AB2 BC DC BC BD

BC [DC BD] BC BC

AC2 AB2 BC2.

12.

Join ED.Since E and D are midpoints of AB and AC, EDBC.Now AED ~ ABC

therefore AA

EB

EB

DC

12

.

Now EDF ~ BCD (angles equal)

therefore FE

CF

FB

DF

EB

DC

12

.

Therefore FC 2EF and BF 2FD.

13.

PAB ~ PQS EDF ~ BCD

Therefore PP

QA

QAB

S Therefore

PP

CR

BSR

C

and PAC ~ PQR therefore PP

QA

PP

CR

QAC

R

therefore QAB

S

BSR

C and

BA

CB

QSR

S.

14.

Since quad ABCD ~ quad EFGH,

EAD

H

AE

BF

FB

GC

GCD

H

and A E, B F, C G, D H.

In ABD, EFH, EAB

F

AE

DH and A E

therefore ABD ~ EFH.Similarly BCD ~ FGH.

Then

EA

FB

HC

AE

BF

2

2 k

and

FB

GCD

H

BE

CF2

2

k

ABC k EFH and BCD k FGHABC BCD k (EFH FGH)quad ABCD k quad EFGH

or qquuaadd

EA

FB

GCD

H k

GD

HC2

2.

A

B

D C

EF

GH

P

A CB

QS R

A

E D

B C

y x

F

x y

A

B DC

y x

x y


15. The proof is similar to question 14. Divide thepentagons into one triangle and a quadrilateral and usethe result from question 14.

16.

Trap DBCE 2245 ABC

therefore if ABC 25 then ADE 1.Now ADE ~ ABC (angles equal)

therefore

AA

DBC

E AA

DB

2

215.

Therefore AA

DB

15

and AD

DB

14

.

Review Exercise

7. a. BED 15

ABD

ABD 25

ABC

Therefore BED 15

25 ABC 225 ABC.

b. Using the fact that areas of triangles areproportional to bases with constant heights, weassign areas to various triangles as shown.

Therefore

AB

DED

C

7.x5x, x 0.

8. a. Since XD BY and XDBY, XDYB is aparallelogram therefore BXYD and BX YD.

b. Since AX XD and XHDM (from a.), therefore

XA

DX

HAH

M.

c. Since AX XD and XHDM, therefore AH HM.d. Exactly the same reasoning as in previous parts.e. Using c. and d., AH HM CM, BX and DY

trisect AC.

9.

28..28

13a.2 and

34

3b.6

Therefore 13.2 4a 4b 10.8a 3.3 b 2.7

thus a b 3.3 2.7 6.0therefore y 6.0.

10. a. 1:2

b. All areas are equal.

c. rec

tAADBECD

18

rec

tAABBFCD

16

12.

Since ABE ~ CDE (angles equal)

therefore CA

DB

CAE

E

DBE

E

12

thus, CE 2AE and DE 2BE.Therefore the diagonals of the trapezoid trisect eachother.

A B

D C

E

x

x

3.6

2.2

6.6

a b

13.2

k

3k

A

BD

E

C

4x7.5x

x

A

D E

B C


13. In ABD, AA

DP

BE

DP.

In ADC, AA

DP

DPF

C.

Therefore BE

DP

DPF

C

or (EP)(DC) (BD)(PF) but DC BD, therefore EP PF.

15. (i) If the three altitudes are equal then the triangle isequilateral. Let the three altitudes be h, and let thethree sides be a, b, and c as shown.

ABC 12

ha 12

hb 12

hc.

Thus ah hb hc or a b c (h 0).

(ii) If the triangle is equilateral then the three altitudesare equal.

Calculating the altitudes, if the sides are 2a inlength then the three altitudes must each be a3.

16.

Extend AD to meet the perpendicular from C at F.Draw the perpendicular from A to meet BC at E. AE isthe height of ABC. CF is the height of ACD. SinceADBC, AE CF h.If AD a and BC b,

then area trap ABCD 12

ah 12

bh

12

(a b)h.

17.

GBD GDC x (equal bases, same height)and GAF GBF y (equal bases, same height)and GAE GCE z (equal bases, same height).

Since ADB ADC and GBD GCDtherefore ADB GBD ADC GCD

or AGB AGCtherefore 2y 2z

y z.Similarly, x y and x y z. Each of the smaller triangles have equal areas.

18. Let AE k and ED 2ktherefore BC 3k.Since AEF ~ CBF

therefore

CAE

BFF

3kk2 19.

Let AEF x and CBF 9x.

Since EF

FB

13

, AFB 3x.

Since diagonal AC divides parallelogram into twotriangles of equal area, quad EFCD 11x.

Therefore qua

dAEBFFCD

131xx

131.

Thus ABF:quad EFCD 3:11.

20. Let the area of parallelogram AHIE be P, the area ofparallelogram EIFB be Q, the area of parallelogramHIDG be R, and the area of parallelogram IGCF be S.Let AI:IC x:y.Since HIDC, then AH:HD x:y and since IGAD,then DG:GC x:y. Now parallelogram AEGD andparallelogram EGCB have the same height so

P R xy

(Q S).

Similarly using parallelogram AHFB and

parallelogram HDCF, P Q xy

(R S).

Then R Q xy

(Q R)

or R(x y) Q(x y)

R Q (since (x y 0) or parallelogram HIDG parallelogram EIFB.

A

F

B CD

G

Ey

y

x x

z

z

A

BE C

D F

A

c b

CBa

h


21.

Since

CBD

DEE

34

and BDE 6, CDE 8.

Since

CC

AB

DD

12

and CBD 14, CAD 7.

The total area is 14 7 21.

22.

Using parallelogram properties, mark equal angles as shownADE ~ FBE (equal angles)

then AE

EF

AB

DF.

Also ABF ~ GDA (equal angles)

then AA

GF

AB

DF

therefore AE

EF

AA

GF.

Chapter 2 Test

1. a. Converse of Theorem If AD is a median thenADC ADB.

b. If AD is a median then ADC and ADB have thesame bases and equal heights. Thus, ADC ADB.

c. ABC ADB if and only if AD is a median.

2. a. A 12

220

120.

b. ABC 24ADB 12

Therefore DEB 6.

c. APB CPD 12

(10)(8 x) 12

10x

5 [(8 x) x] 40.

d. 120 d 8 212

d 12.

3. AM 12

AB therefore AM 5

AMY ~ ACB

AA

MC

AA

BY

MCB

Y

256

A10

Y

M24

Y.

Therefore 26AY 50 and 26MY 120.

AY 11123 MY 4

183

4.

AEC 36therefore CED 108

then ADC 144therefore ABC 288.

A

Ea

3a

BD C

A B

D C10

xp

(8x)

A

E

C D B

A B

D C

E F

G

z

yx

x y

C

AD

E

Bx 2x

3m

4m


5.

Draw PQ parallel to AB.Then ABPQ is a parallelogram and AP is a diagonal.Therefore APQ ABP.Also QPCD is a parallelogram with DP as a diagonal.Therefore QPD PCD.Then APQ QPD APD ABP PCD.

6.

Mark angles as shown.Then APD ~ CBA

therefore

PP

QBA

A

PC

QB2

2

P

3Q6

A

49

therefore PQA 16.

Also, AA

QB

23

(from above)

therefore

PP

QBA

A

23

therefore P

16BA

23

therefore PBA 24.Therefore trap BCQ has an area of

16 36 24 24 100.

7. Either the line from A to D is parallel to BC or it isnot. Assume that AD is not parallel to BC. There mustbe a point G on BD or BD extended so that AGBC.

If AGBC then ABC GBC (same base, equalheight).Since we are given ABC DBC, therefore GBC DBC. This is clearly not possible so theassumption that AD is not parallel to BC must befalse. Then ADBC.

A

B C

D

G

P Q

B C

A

y x

x y

A

Q

D

B

P

C



Exercise 3.1

5. Draw XY to cut AB at P. Since B is on the right bisector of XY (BX BY) and A is on the right bisector of XY, AB is the right bisector of XY and XPB 90. In right-angled triangle XAP let XP 4 and AP x. In XPB, XP 4 and PB y. Then x2 16 25, so x 3. Also y2 16 64, so y 43. Then AB 3 43.

6. Let the centre of the circle be O. Draw OA PQ and OB RS as in the diagram. Then PA 4 (perpendicular bisects chord), and similarly RB 6. Let OA be h units; then OB is 10 h. In OAP, OP is the radius r, and r2 42 h2

similarly, in ORB, r2 62 (10 h)2subtracting these equations, O 20 100 20h

h 6then r2 16 36 52

r 231.The radius is 231.

P QA

R SB

O

X

A BP

Y

5 8

22 Chapter 3: Properties of Circles

Chapter 3 Properties of Circles

8. Draw a line from O, the centre of the circle, to P. A P. Draw a line perpendicular to OP, meetingthe circle at A and B. Then OP bisects AB.Proof: Since OP AB, AP PB (perpendicular to chord). It is always possible to draw a line through P that isbisected at P.

9. Let O be the centre of the circle and AB be a chord.OC is perpendicular to AB and is extended to meet thecircumference at D. Since OC AB, AC CB (chord bisector). In OAC and OBC, OA OB (radii)

AC BC (proven) OCA OCB (right angles).

Then OAC OBC (right-angled triangles).Therefore AOC BOC.

Now arc AD 3A6O0

C (2r)

and arc DB 3B6O0

C (2r)

since AOC BOC, arc AD arc DB.

10. Let O be the common centre and let OA XY. Since PQ is a chord of the inner circle, PA AQ(perpendicular to chord). Similarly, XY is a chord of the outer circle,so XA AY. Then XA PA AY AQ

or XP QY.

O

A BC

D

OB

P

A

11. Draw CA and CB from the centre C to meet PQ at Aso that CA PQ and XY at B so that CB XY. Join OC. In OAC, OBC, AC BC (perpendiculars to equal chords)

OC OC (common side) OAC OBC (right angles)

then OAC OBC (hypotenuse and side of right-angled triangles)therefore OA OB. Now QA YB (half of equal chords)then OA QA OB YBor OQ OY.

12. Since XP XQ (radii)X is on the right bisector of PQ.Similarly YP YQ (radii)Y is on the right bisector of PQ. Then XY is the rightbisector of PQ.

13. From O draw OM AB and ON CD as in thediagram.Since AB CD; then OM ON (equal chords).In OMX, ONX,

OX OX (common)OM ON (proven)

OMX ONX (right angles)then OMX ONX (hypotenuse, side)Therefore OXC OXB.

14. As in question 12, PQ is the right bisector of AB. Let AT have length h and PT have length x. Then TQ has length 21 x.

in PAT, x2 h2 132in QAT, (21 x)2 h2 202

subtracting x2 (21 x)2 132 202

x 2 441 42x x2 169 40042x 210

x 5.Then h2 169 24

h 12444(h > 0).Therefore AB 24.

A

P QT

B

AD

NX

MC

O

B

P

X Y

Q

O

QA

P

YB

X

C

X

Y

O

AP

Q

Chapter 3: Properties of Circles 23


Exercise 3.2

4. a.

In OAP, OP OAtherefore OPA OAP x (isosceles triangle)

AOB 2x (exterior angle).

b.

In OAP, OP OA so OPA OAP x (isosceles triangle)

then AOC 2xin OPB, OP OB

so OPB OBP y (isosceles triangle) then COB 2y

therefore AOB 2x 2y 2(x y)

and APB x ytherefore AOB 2 APB.

5. Join OA and OB. Draw AM AB.

Then BM 2r

.

In OMB, OM2 OB2 BM2

r2 r4

2

34

r2

OM

23

r.

Then MOB 30 and OBM 60therefore AOB 60 and reflex AOB 300.

Then ACB 12

AOB

30

and ADB 12

reflex AOB

150therefore ADB 5 ACB.

6. In gm ABCD, A C (opp. angles in gm).Let A C a.Join OB and OD.Then on arc BAD, BOD 2 C 2a.Also, on arc BCD, BOD 2 A 2a.Then 4a 360

a 90.Then A 90 and ABCD is a rectangle.

7. Since CD < AB, arc CED < arc CAD.

Then CED 12

reflex COD.

But reflex COD > 180therefore CED > 90.

A OB

DC

E

A D

B C

a

aO

O

A Bm

r

C

D

r2

P

O

AB

x

x

y

yC

P

O

AB

x

x

8. Join BD. Since BOD 2 A,BOD 180.

Then BD is a diameter of the circle.Then BD2 AB2 AD2

8Now BED 90 (angle in semicircle)Therefore BE2 DE2 BD2

8By joining AC, in like fashion, AE2 CE2 8.Then AE2 BE2 CE2 DE2 16.

9. This is a generalization of question 8. The prooffollows the same steps.

10. Let BAP be x. Join PL.Since BAL 90, BL is a diameter.Then BPL 90 (angle in semicircle)

also BLP BAP x (subtended by chord BP)therefore PBL (90 x) (angles in BPL).Then QBK (90 x) (opposite PBL)therefore KAQ (90 x)since KAQ QBK (subtended by chord QK).But KAB 90then QAB 90 (90 x)

xtherefore QAB BAP and AB bisects QAP.

11. Join CD and CE. Let the intersection of DE and BCbe F, and let BAC be 2x, so DAC is x.Then DEC is x (angles subtended by chord DC).Now EFC is 90then EFC (90 x) (angles in EFC).Now BAD x (half of BAC)

so BCD x (angles subtended by chord BD)then ECD (90 x) x

90

Then DE is a diameter of the circle.

12. Let BOC be .Arc BC subtends BOC at the centre and BAC atthe circumference.

Then BAC 2 (angle at the circumference).

In ACX, XAC 2 and AXC 90.

Then ACD (90 2) (angles in a triangle).

Now AOD 2 ACD

(180 ) (subtended by arc AD).

Then AOD BOC (180 ) 180.

13. Let D be the midpoint of BC.Since AB is a diameter, ADB 90 (angle in asemicircle).Similarly ADC 90.In ADB, ADC, AD AD (common)

BD CD (given)ADB ADC (right angles)

then ADB ADC (two sides, contained angle).Therefore AB AC and the triangle is isosceles.

A

C

B

D

O

x

AE

B F

D

C

x xx

AK L

B

QP

AE

B

D C

O


Exercise 3.3

6. Extend BC to E. BAD BCD 180 (cyclic quad)DCE BCD 180 (straight angle).Therefore BAD DCE.An exterior angle of a cyclic quadrilateral is equal tothe interior opposite angle.

7. Using the diagram given,ADB ABD x (isosceles triangle).Then BAD (180 2x).In BDC, BCD DBC 2x (isosceles triangle).Then BAD BCD (180 2x) 2x

180Therefore ABCD is cyclic (opposite anglessupplementary).

8. Consider the trapezoid ABCD in a circle.Since AB CD, ABC BCD 180,so BCD 180 ABC.Since ABCD is cyclic, ABC ADC 180,so ADC 180 ABC.Therefore ADC BCD. The base angles are equal.Join AC and BD. In ADC, BCD, ADC BCD (proven)

DAC DBC (angles subtendedby chord DC)

Then ACD BDC (third angles intriangles)

Since DC is commonthen ADC BCD (two angles and one side).Therefore AC BD.The diagonals are equal.

9. Using the diagram given, RSQ is an exterior angle of QST. Then RSQ SQT STQ (exterior angle).Since RSQ 2 SQT, SQT STQ.Now since PR PT, STQ PRT (isoscelestriangle).Then PRS SQP, angles subtended by SP.Therefore PQRS is a cyclic quadrilateral.

10. Join SR and QT. In PQR, PQ PRThen PQR PRQ (isosceles triangle)

Let PQR . Then PRQ .Since ST QR, PST .Then TSQ (180 ).Therefore TSQ TRQ 180 and STRQ is cyclic.In cyclic quadrilateral STRQ, chord SQ subtends equalangles at T and R. Then QTS SRQ.

11. Join ST and QR. Since OT OS, PQ PR (equalchords).Then PQR PRQ (isosceles triangle).

Now PT 12

PQ and PS 12

PR, so ST RQ.

Then PST PRQ PQR. But PST TSR 180,so PQR TSR 180.

Therefore STQR is a cyclic quadrilateral (oppositeangles supplementary).

P

S T

Q R

A B

D C

A B

D C

E

A

B CD


12. Join OA and OB and let X1

be on arc AB.

Since the octagon is regular, AOB 18

(360)

45. Then reflex AOB 315.

Chord AB gives AX1B

where AX1B

12

reflex AOB

31

25.

But this is true for all given angles.

Hence AX1B BX

2C HX

8A 8312

5

1260.

13. Join OS and OT. Since ST is fixed and OS OT areradii, the triangle OST is fixed and SOT is constant.

Since M is the midpoint of ST, SOM 12

SOT,and SOM is constant as ST moves.Now SMO 90 (property of midpoint of chord),

and SPO 90.Then SPOM is cyclic (opposite angles supplementary).Therefore SPM SOM (subtended from chord SM).Since SOM is constant, SPM is constant.Note: This is a surprising result. Investigation withGeometers Sketchpad is advised.

Exercise 3.4

7. Since DF is tangent to the circle, DE and FE aretangents.Now DE DB (tangents from D)also FE FC (tangents from F)also AB AC (tangents from A).

Then AD DE AB and AF FE AC.

The perimeter of the triangle is AD DE FE AF AB AC 2AB 20.

8. Let the tangent contact points be W, X, Y, Z as in the diagram.Since tangents from an external point are equal,AW AZ a, say,BW BX b,CX CY c,DY DZ d.Then AB CD a b c d

and AD BC a b c dTherefore AB CD AD BC.

A

B

C

D

W

X

Y

Z

BD

A EO

F C

TM

S

A BP O

A

H

G

F

E

D

C

B

O

TP Q

SO

R


9. Since BD is a tangent and PB is a radius,DBP 90.Join PC. Since AD is a tangent and PC a radius,ACP 90.In ACP, AC2 AP2 PC2

36 4 32

AC 42Now AB 8, AP 6, PC 2.

In ACP, ABD, CAP BAD (same angle)ACP ABD (right angles)

Then ACP ~ ABD (equal angles).

Therefore AA

CB

BP

DC.

Substituting, 4

82

B2D

BD 4162

22.

10. Using the given diagram,MA MX (tangents from external point)MX MB (same)

then MA MBtherefore M is the midpoint of AB.

11. Join XO, PQ, POXP XQ (tangents from external point)OP OQ (radii).Then XO is the right bisector of PQ.Therefore XRP 90.In XRP, XPR 90 PXR 180.In XOP, XOP 90 PXO 180.Therefore XPQ XOP.

12. PR2 PQ2 QR2

162 302

1156PR 34.Let I be the centre of the incircle. From I drawperpendiculars to the three sides of the triangle, andjoin IP, IQ, IR.

Now PQR 12

(30)(16)

240.Also PQR IPQ IPR IQR

12

(16r) 12

(30r) 12

(34r)

40rthen 40r 240


Exercise 3.5

6. CBE BAC (tangent-chord property).Since AC is a diameter, ABC 90 (subtended by diameter)then ABD CBE 90.In ADB, ABD BAD 90 (ADB 90)then CBE BADtherefore BAC BAD and AB bisects CAD.

A

O

C

EBD

P

R30Q

16 rr

rI

P

XR

O

Q

D

BPO

A

EC


7. Join PQ, AQ, and BQ. Using the circle with AB as chord, CAP AQP(tangent-chord property).Similarly, CBP BQP.In CAB, CAB CBA ACB 180.Then AQB ACB AQP BQP ACB

CAB BAC ACB 180.

Therefore AQBC is a cyclic quadrilateral.

8. Join PQ. YXP XQP (tangent-chord property)XYP YQP (same)YXP XYP XPY 180 (angle sum)then XQP YQP XPY 180or XQY XPY 180.

9. a. Let AB and AC be secants intersecting the circle atD and E. Join CD and BE.In ABE, ACD, EAB CAD (same angle)

ABE ACD (subtended byarc DE)

Then ABE ~ ACD (equal angles)

Therefore AA

CB

AA

DE or

AB AD AC AE.

b. Let AB be a secant cutting the circle at E, and letAC be a tangent meeting the circle at D.We wish to prove that AD2 AB AE. Join DE.

In AED, ADB, EAD DAB (same angle)ADE ABD (tangent-

chord property).Then AED ~ ADB (equal angles).

Therefore AA

DE

AA

DB

or AD2 AB AE.

10. Join AT and BT. PT 2 PA PB (intersecting secantsproperty).Now PA PO OT (OA OT) and PB PO OT (OB OT).

Then PT2 (PO OT)(PO OT) PO2 OT2

or PO2 PT2 OT2.Therefore OTP 90 (Pythagorus).Then PT OT.

Review Exercise

11. Since AB CD, CAB ACD 180.Since ABDC is cyclic, CAB BDC 180.Then ACD BCD. Join AD and BC.In ACD, BDC, CD CD (common)

ACD BDC (proven)CAD DBC (subtended by

chord CD)then ACD BCD (two angles and side)therefore AC BD.

P A O B

T

B

E

AD

C

AD B

E

C

X

Y

P

Q

B

C

A

Q

Py

x

xy


12. Since ABC 60 and ACB 45,BAC 75 (angle sum)

AH AE (tangents from external point)then AHE AEH

therefore AHE 12

(180 75)

52 12

.

Similarly, BHD 12

(180 60)

60.

Then EHD 180 60 52 12

67 12

.

Similarly, CED 12

(180 45)

67 12

.

Then HED 180 67 12

52 12

60

and HDE 180 67 12

60

52 12

.

13. Let the isosceles triangle have AB as diameter of thecircle and let BC intersect the circle at D.We wish to prove that D is the midpoint of BC.Since AB is a diameter, ADB 90.Then ADC 90.In ABD, ACD, AB AC (given)

AD AD (common)ADB ADC (right angles).

Then ABD ACD (hypotenuse, side, right triangles)Therefore BD CD and D is the midpoint

of BC.

15. Assume that x is in degrees.Since TA is a tangent, OAT 90.Then AOT (90 x).Now OAB is isosceles since OA OB, and AOT isan exterior angle of the triangle.Then AOT OAB OBA

2OAB.

Therefore OAB 12

(90 x).

16. Using the given diagram, in ADP we have ADP 180 4x.In AQB, ABQ 180 5x.Since DBCA is a cyclic quadrilateral,

ADC CBA 180 or 180 4x 180 5x 180

x 20.

17. Since RPT is an angle in a semicircle,RPT TPX 90.

Similarly, TQX 90.We must now show that PTQ 90.Draw TQ to AB, meeting PQ at Y.Then TY is tangent to both circles.Therefore YP YT YQ (tangents from external point).Then YPT is isosceles, so YPT YTPSimilarly YQT is isosceles, so YQT YTQ.Then PTQ PTY QTY YPT YQT.But PTQ QPT PQT 180 in PQT.Therefore PTQ 90.Then the angles in PXQT are all 90 and PXQT is a rectangle.

A

BO

Tx

A

BD

C

A

HE

B DC

x xA B

C D


18. Draw PR OX and QS OX and let QS have length r.Since P is equidistant from the arms of the angle,PO bisects O and POR 30.Then POR has angles of 30, 60, and 90.Therefore PO 2PR 20.Since the circles are tangent, PQ 10 r and QPOis a straight line.Then in OQS, OQ 30 r.Now OPR ~ OQS (angles equal)

Then OPR

P

OQ

QS

or 2100

30r r


19. Let the tangent at A meet the second circle at C. Join OB.Since OA and OB are radii, OAB OBA.Since OA is a chord in the circle with centre Q and CAis tangent, CAO OBA (tangent-chord property).

Then CAO OAB.Therefore OA bisects CAB.

20. Join BR, QR, and RP.Since BAQR is cyclic, BRQ BAQ 180.Since QAD 180, BAD BAQ 180.Then BAD BRQ.Since RBCP is cyclic, BRP BCP 180.Since DCP 180, BCP BCD 180.Then BCD BRP.But since ABCD is cyclic, BAD BCD 180Therefore BRQ BRP 180.Therefore QRP is a straight line.

21. Bisecting A, B, and C gives BAX XAC 30, ABY YBC 25,and BCZ ZCA 35.Now, using arc BX, BAX BYX 30 and using arc BZ, BCZ BYZ 35.Then XYZ 65.Similarly YXZ 60 and XZY 55.

22. Using the given diagram, join AB. Then chord ABsubtends APB in one circle and AQB in the othercircle and the circles are identical. Therefore theangles are equal. Then BPQ is isosceles and BP BQ.

A

Y

C

X

B

Z

A

QR

PCD

B

A

C

O

B

Q

O

P

R

Q

S X

X

P

Y

Q

RA T B

S


23. Let AD and CB intersect at E. Since C and D arepoints on a semicircle, ACB ADB 90.Then FCB FDA 90.Since FCB FDA 180, FCED is a cyclicquadrilateral.Join CD. Then EFD ECD (subtended by chord ED).Now since chord BD subtends equal angles in thesemicircle, BAD BCD.Therefore BAD EFD.In BAD, FJB, BAD JFB (proven)

DBA FBJ (same angle)then BAD ~ FJBtherefore ADB FJBbut ADB 90 (angle in semicircle)therefore FJB 90then FJ AB.

Chapter 3 Test

1. a. x 50 (tangent-chord property).b. a2 4.12 (secant property)

a 43.c. 3b 2 8 (secant property)

b 136.

d. x 120 (exterior angle of cyclic quadrilateral).

y 100 180 (cyclic quadrilateral)y 80.

2. Using the given diagram, join AD.Then CAD 90 because it is an angle in a semicircle.Then BAD 140.Since BADC is cyclic,BCD BAD 180Therefore BCD 40.

3. Using the given diagram, M is the midpoint of ABbecause OM AB, so AM MB.In right-angled triangle OMB, OM 8 and OB 17.Then MB2 172 82

225MB 15.

In right-angled triangle OMC, OM 8 and OC 10.Then MC2 102 82

36MC 6.

Then AC MA MC 15 6 9.

4. Using the given diagram, and joining OD,ODC 90 (angle in a semicircle).Then OD CE.Therefore CD DE (perpendicular to chord)

5. RP is tangent and RB is a secant in one circle.Then RP2 RA. RB (tangent-chord property) and RQ2 RA. RB.

Then RP2 RQ2

or RP RQand BR bisects PQ.

6. a. Let AB and CD be any two of the equal chords in acircle with centre O. Let X and Y be the respectivemidpoints. Then AX CY. In AXO, CYO, AX CY (proved)

OA OC (radius)AXO CYO 90

(midpoint of chords).Then AXO CYO (side, hypotenuse in

right triangles).Therefore OX OY.

By a similar argument the lines joining everymidpoint of the equal chords to the centre areequal. Hence a circle with centre O and radius OXpasses through the midpoint of every one of theequal chords.

b. Since every one of these chords meets the circle ata point such that the line connecting this point isperpendicular to the chord, each chord is a tangent.

P

R

Q

A

B

F

C D

A BJ

E


7. Let AP extended meet tangent XTY at X. Join TP and TB.Then XTP PAT (tangent-chord property)

and YTB BAT (same)but PAT BAT (bisected)

therefore XTP YTB.Let YTB be x, so XTP x.In ATB, ATB 90 since it is subtended bydiameter AB.Then ABT (90 x) using the angle sum property.Now PTBA is a cyclic quadrilateral, and XPT is anexternal angle, so XPT ABT (90 x).In XPT, XPT XTP (90 x) 90.Therefore PXT 90, and XTY APX.

Cumulative Review Chapters 13

13. Let AC and BT intersect at X. Join AT and BC.In SBC, NTA, SC AN (given)

SB NT (given)BSC TNA (opposite angles

in parallelogram)Then SBC NTA (side-angle-side).Therefore BC AT and SCB NAT.Now, since MN SR, NAC SCA.

Then NAC NAT SCA SCBor TAX BCX.

In BXC, TAX, BC AT (proved)BCX TAX (proved)BXC TXA (opposite)

then BXC TAX (side-angle-angle)therefore BX XTtherefore AC bisects BT.

14. Extend BA and DC to meet at X.By the secant property, XB XA XD XCor (XA AB)XA (XC CD)XCor XA2 XA AB XC2 XC CDor XA2 XC2 AB(XAXC) (since AB CD) or (XA XC)(XA XC) AB(XA XC).Now either XA XC 0 or XA CX AB.It is not possible that XA XC AB, sinceXA XC > 0.Therefore XA XC 0, and XA XC.Then ABC BDC since XBC is isosceles.Since ABDC is a cyclic quadrilateral,ABD ACD 180.Since BDC ABD, BDC ACD 180.Then AC BD.

15. Let AB have length 2. Then BC 1.Let BF have length x. Then FC 1 x.Now ABC ~ EFC since C is in both triangles andEFC ABC 90.

Then FB

CC

EA

FB,

or 1

1x

2x

2 2x x

x 23

.

Then square DEFB : ABC x2 : 12

(2)(1)

49

: 1

4 : 9.

AD

B

E

C

F

AX

C

D

B

x

M A N

T

RCS

B

X

P

A B Y

T

X

A

B

O

C

YD


16. In a regular pentagon each angle is 3 18

50 108.

In ABC, B C A 180.

Also A 12

B and C B.

Then 52

B 180

B 72.Then XBC 36 and C 72,

so BXA 108 (exterior angle of triangle).Now we must prove that XA XB.Since B 72, A 36.Also XBA 36.Therefore XAB is isosceles and XA XB.Then XA and XB are consecutive sides of a regularpentagon.

17. Using the given diagram and the property of tangentsfrom an external point, AX AZ, BX BY, and CY CZ. Then AX BY CZ BX YC ZA.

18. Let the polygon have n sides. Then the sum of allangles is 180(n 2).There are n angles in the polygon, so the sum of the angles is 100 140 120(n 2) 120n.

Then 180(n 2) 120n60n 360

n 6.There are six sides to the polygon.

19. Using the given diagram, ABC ACB since thetriangle is isosceles.Also ABC BDE 180 since DE AC.Then ACB BDE 180, and DBCE is a cyclicquadrilateral.

20. Since CD is a diameter, CBA 90 30 120.Since BCDA is cyclic, CBA ADC 180.Therefore ADC 60.

21. Part 1. We prove that AE BF if AE BF. Since AE BF, FTE 90.Then FCE FTE 180 (FCE is an angle in a square).Then FCET is a cyclic quadrilateral.Therefore TFC TEC 180.

Now TEB TEC 180 (straight angle).Then TEB TFC.In ABE, BCF, AB BC (sides of square)

ABE BCF (right angles)AEB BFC (proved).

Then ABE BCF (side-angle-angle).Therefore AE BF.

Part 2. We prove that if AE BF then AE BFIn ABE, BCF, AB BC (sides of square)

AE BF (given)ABE BCF (right angles).

Therefore ABE BCF (side, hypotenuse in right-angled triangle).Then AEB BFC.Now AEB TEC 180 (straight angle).Therefore BFC TEC 180.Then FTEC is a cyclic quadrilateral.Therefore FCE 90 (angle in a square).Therefore FTE 90 and AE BF.

D F C

ET

A B

A

B C

X


22. Since M and N are midpoints of AB and AC, MN BC.Then BNC and BMC have common base BC andequal height.Therefore BNC BMC.Since BYC is common,BNC BYC BMC BYC

or BYM CYN.Since N is the midpoint of AC,ABN NBC.Then ABN BMY NBC CYN

Therefore quad AMYN BYC.

23. Join QT and RT. Since RP RS, RPS is isosceles.Then RPS RSP.Then PRQ 2 RPS (external angle).Since PQ PR, PQR is isosceles.Then PQR PRQ 2 RPS.Now chord RT subtends RPT RQT.

Therefore RQT 12

PQR, and QT bisects PQR.

24. Using the given diagram and recalling the CosineLaw, x2 22 32 2(2)(3) cos where is ABC.Since ABCD is cyclic,ADC 180 ABC 180 .Then x2 42 62 2(4)(6) cos (180 )

42 62 2(4)(6) cos .Then 13 12 cos 52 48 cos

60 cos 39

cos 1230.

Therefore x2 13 12 cos

13 359

20.8x 4.6.

25. Note that AHE is incorrectly labelled as a rightangle. In fact BHC is 90. Using tangent TAN andchord AC, NAC CBA.Since BHC BEC 90, a circle with BC asdiameter will pass through both H and E.Then HBCE is a cyclic quadrilateral, and HEA is an external angle.Therefore HEA HBC NAC.Since HEA NAC, then TAN HE.

T

A

N

HE

BC

P

T

Q R S

A

M N

B C

Y



Exercise 4.1

12. a.

The first leg of the trip is OA. 240 km/h for 114

hours gives OA 240 54

300 km.

The second leg of the trip is AB and AB 240 2 480 km.

In OAB, OAB 70 10 80.

From the cosine law OB2 3002

4802 2 300 480 cos 80 OB 519.99.Length of the third leg is 520 km.

Let ABO then from the sine law

s3in00

si

n

OB80

34.62.

The displacement vector for the third leg, BO,has a magnitude of 520 km at S 45 W.

b. The total distance the aircraft travelled is 300 480 520 1300 km. The time taken is

1234000

5.42 hours or 5 hours 25 minutes.

13. Since (k 2)v < 4vk 2v < 4v.

Since v 0, v 0 and k 2 < 44 < k 2 < 4

and 2 < k < 6.

14. If u kv, u and v have the same direction for k > 0

and opposite directions for k < 0. In either case u will

be parallel to v.

If u is parallel to v, u and v will have the same

direction or opposite directions hence u is some

multiple of v and u kv.

Exercise 4.2

7. a.

In the diagram u v AC and

u v BD.

AC and BD are the diagonals of gm ABCD.

Now if u v u v , the diagonals of the parallelogram are equal, hence the parallelogram is

a rectangle and u v.

b. If u v > u v, the angle between u and vwill be acute.

c. If u v < u v, the angle between u and v willbe obtuse.

9.

Let R 5a 2b. Since a 2 and b 3,5a 10 and 2b 6. From the cosine lawR2 102 62 2 10 6 cos 50R 7.672AOC then from the sine law

sin

6

sin

R5

0

, 36.80

5a 2b has a magnitude of 7.7 and makes an angleof 37 with 5a and 93 with 2b.

b

a2b

B C

R

A

b

505a

50 10O

B C

A D

v

u

B

A

O

10

20300 km

480 km

36 Chapter 4: Vectors

Chapter 4 Vectors

12. a 3x 2y

b 5x 4y Solving the two equations for x and y

2

11x 2a b

x 121a

111b.

Substitute into a 161a

131b 2y

11a 6a 3b 22y

y 252a

232b.

16. Since a and b form the sides of a parallelogram and

since a b, the parallelogram will be a rhombus.a b and a b will be the diagonals of this

rhombus and since the diagonals of a rhombus are

perpendicular to each other so will (a b) be

perpendicular to (a b).

17.

AB ED. Then ABDE is a parallelogram

and AB AE AD.

AF CD. Hence ACDE is a parallelogram

and AC AF AD.

Now AB AC AD AE AF

AB AE AC AF AD

AD AD AD

3AD.

But O is he midpoint of AD and ABO is equilateral

therefore AO 1 and AD 2.Now AB AC AD AE AF 3AD

6.

Since the sum is equal to 3AD, the direction of the

sum is along AD which makes an angle of 60 with

AFand 60 with AB.

18.

In ABC, ABC 180 and from the cosine law x y2 x2 y2 2xy cos (180 ). But cos (180 ) cos .

Hence x y2 x2 y2 2xy cos . From ABD, ABD and from the cosine law

x y2 x2 y2 2xy cos . Adding and

x y2 x y2 2x2 2y2. Now x 11, y 23, x y 30. Hence x y2 302 2 112 2 232 x y 20.

19. Vectors u v and u v are the diagonals of a

parallelogram where u and v are adjacent sides. Since

u v < u v the angle between u and v will be obtuse. Since the diagonals of a parallelogram

bisect each other, draw u v and u v bisecting

each other.

AC u v

DB u v

with AB u, AD v.

A

B

C

D

u

v

A

C

D

B

y

yx y

x

y

y

x + y

A F

E

DC

B O

Chapter 4: Vectors 37


20.

a. FG FH HG

AC DA

j k.

b. BH BG GH

j k

DH DE EH

j i

CH CG GH

i k

FE FH HE

j i

EG EH HG

i k.

c. BD BG GC CE

j i k

BE i j k.

d. AH i j k

CF i j k

GD i j k.

e. Face diagonal is FG.

FG 1 1 2.

Body diagonal is AH.

AH 1 1 1 3.

21.

In OAB, OAB 180 and cos (180 ) cos .

From the cosine law

u v2 u2 v2 2uv cos (180 )u v2 u2 v2 2uv cos . In OAC, AC v, AC v v, OAC .From the cosine law

u v2 u2 v2 2uv cos u v2 u2 v2 2uv cos . Adding and

u v2 u v2 2u2 2v2.

u v2 u v2 2(u2 v2).

Exercise 4.3

9.

Represent the tensions in the cords by T1

and T2

asshown in the diagram.

98 N

v

v u + v

u

u v

C

v

A

v

B

180

O

D E

FH

B

A

G

C

ij

k


From the force diagram and the sine law,

sin

T610

sin

T425

sin

9875

T1 98

sisnin75

60

T2 98

sisnin75

45

T1 87.9

T2 71.7.

The tension in the cord making an angle of 45 withthe ceiling is approximately 87.9 N and the tension inthe other cord is approximately 71.7 N.

10.

Represent the magnitude of the forces by a.From the cosine law we have

R2 a2 a2 2 a a cos 120

2a2 2a2 12

302

3a2 900a2 300a 103.

The magnitude of each force is 103 N.

11. An object will be in a state of equilibrium when theresultant of all the forces acting on it is zero. This meansthat the sum of any two magnitudes must be greater thanor equal to the magnitude of the third force.

a. Since 5 2 7 < 13, equilibrium cannot beachieved.

b. 7 N, 5 N, and 5 N can be arranged to produceequilibrium.

c. 13 14 17, hence equilibrium can be achieved.In this case the three forces would be collinear.

d. Since 12 13 24 < 26, equilibrium cannot be achieved.

12. a.

b. From the triangle of forces and the cosine law

82 52 72 2 5 7 cos

cos 52

2

75

2

782

82.The angle between the 5 N and 7 N forces will be 180 82 98.

13.

Represent the tensions in the cords by T1

and T2

as

shown in the diagrams. From the triangle of forces

and the sine law,

sin

T710

sin

T625

sin

68465

T1 68

s6insi

4n570

T1 911.6

T2 68

s6insi

4n565

T2 879.3.

70

686 N

6525

2025

20 25

70 kg

686 N

8

7

5

7

82 98

8

5

aR

60 120

a

The tension in the rope making a 25 angle with thehorizontal is approximately 911.6 N and in the otherrope is approximately 879.3 N.

14.

The 20-, 15-, and 25- metre lengths form a right-angled triangle as shown in the diagram. Since the 375 N force is collinear with the 15 m steel wire, itwill have a tension of 375 N and the tension along the25 m steel wire will be 0 N.

15.

Let T1

represent the tension in the wire and T2

thecompression in the steel brace as in the diagrams.

Now sin 65 8

T51

0

T1

sin85

605

T1 937.9

tan 65 8

T52

0

T2

ta8n5605

T2 396.4.

The tension in the wire is approximately 937.9 N andthe compression in the steel brace is approximately396.4 N.

16.

Let px and p

y represent the components of p along

the x-axis and y-axis respectively.

Now uy 5, u

x 0

vx 9 cos 40 6.9

vy 9 sin 40 5.8

wx 12 sin 65 10.9

wy 12 cos 65 5.1.

If p w v u then px 12 sin 65 9 cos 40

3.98

and py 9 sin 40 12 cos 65 5

10.86.

17.

The horizontal component moving the log is 1470 cos 15 N 1420 N.

18. a.

Let Fp and F

n represent the parallel and

perpendicular components respectively.

Fp 196 sin 28 F

n 196 cos 28

92 173

The component parallel to the plane is 92 N andperpendicular to the plan 173 N.

28

28

20 kg

196 N

151470 cos 15

1470 N

y

x

|u| = 5

|v| = 9|w| = 12

6540

65

850 N

65

25

850 N

25

375 N

15 m

20 m

25 m


b. The component normal to the ramp pushes downagainst the ramp and it in turn pushes back with anequal but opposite force. The component parallel to and down the ramp contributes to the luggage

sliding down the ramp. If Fp is greater than the

force of friction opposing Fp

, then the luggage willslide down the ramp.

19.

Let F represent the horizontal force and T the tensionin the rope. is the angle the rope makes with thehorizontal.

Now cos 15.5

72.54

sin 3

T4

3

T sin

37423.54

T 359.6

tan 3

F4

3

F tan

37423.54

F 107.9.A force of 107.9 N will hold the girl in this positionand the tension in the rope is 359.6 N.

20. a.

Let Fv represent the vertical component and F

H

be the horizontal component.

Now Fv 66 cos 8

54.5F

H 55 sin 8

7.7.

b. The vertical component is approximately 54.5 kNand is the component that gives the helicopter lift.The horizontal component is approximately 7.7 kNand is the component that moves the helicopter in ahorizontal plane.

21.

The component that is parallel to the ramp is 2450 sin 25 1035.4. The force of friction,to oppose this, must have a magnitude of at least1035.4 N.

22.

If Fx is the horizontal component then

Fx 320 cos 42 237.8.

The horizontal component causing the roller to moveis approximately 238 N.

320 N

42

50 kg

490 N

25

250 kg

2450 N

8

55 kN

5

1.5

T

F

35 kg

343 N

T

F

343 N


23.

Since the forces are perpendicular to each other,consider them acting along the edges of a rectangularsolid with dimension 15 by 10 by 6. Now the

magnitudes of the forces AD 6 N, AE 10 N,and AB 15 N. AG will be the sum of these forces

where AG2 62 102 152

AG 19.In AGB, GAB

and cos 1159

38.

In AEG, EAG

and cos 1109

58.The magnitude of the resultant is 19 N and it makesangles of approximately 58 and 38 with the 10 Nand 15 N forces respectively.

24.

Let R represent the vector along which the shipmoves. From the parallelogram and cosine law, we

have R2 F2 4F2 2 F 2F cos 150

5F2 23F2

R 5 23 F

From the sine law

s

i

F

n

sin

R15

0

sin

s

5

in

15

2

0

3

9.89.The ship will move approximately 20 9.89 10off the starboard bow.

25. a.

Let T1

and T2

be the tension in each length of

string. Since the mass is suspended from the

midpoint of the cord, T1 T

2. From diagram 1

the vertical components of T1

and T2

are T1 sin

and T2 sin . For equilibrium the sum of these

vertical components will be 400.

Therefore T1 sin T

2 sin 400.

But T1 T

2 therefore 2T

1 sin 400.

From the 100, 80, 60 triangle, sin 45

hence 85

T1 400

T1 250.

The tension in each length would be 250 N hencethe string will support the weight.

OR

From diagram 2 and the sine law

sin (9

T0

1 )

si4n020

.

But sin (90 ) cos , sin 2 2 sin cos

and sin 45

.

Hence T1

24s0in0

cocsos

s2in00

250.

Conclusion as above.

100 cm

400 NDiagram 1 Diagram 2

60 cm

80 cm

400 N

|F|

2 |F|

1020

R

150|F|

H

E

G

F

C

B

D

A

10

15

6


b.

Represent the tensions as T1

and T2

and the angles

in ABC and and as shown in the diagram.Since AC AB 120, ACB ABC .From the cosine law

802 1202 1202 2 120 120 cos

cos 22112200

2

128002

38.94.Also 2 180

therefore 70.53.

From the sine law

sin (9

T0

1 )

sin (9

T0

2 )

sin (

4

00 )

90 51.0690 19.47

109.47

T1 40

s0in

s1in09

5.14.706

330.0

T2 40

s0in

s1in09

1.94.747

141.4.

Since the tension T1 330 N > 300 N, the string

will not support the 400 N weight.

Exercise 4.4

2. a.

Let the angle to the bank be . The componentperpendicular to the bank will be 2 sin , the speedthat takes him across the river, and the componentparallel to the bank is 2 cos . For the man to swimdirectly across the river then 2 cos 1

cos 12

and 60.The man must swim at an angle of 60 to the bankif he is to reach a point directly across from hisstarting point.

b. If the speed of the current if 4 km/h, 2 cos 4,cos 2 which is not possible since cos 1.He will not be able to swim to a point directlyacross the river in this case. As long as the currentis less than 2 km/h, he will be able to swim to apoint directly across the river.

3.

Let vs, v

b, and v

trepresent the velocities of the

streetcar, bus, and taxi respectively and vs

35,v

b 42, v

t 50 where north is positive and

south is negative.

N

W E

S

v tv s

v b

2 cos 2 km/h

2 sin

1 km/h

400 N

90

90

400 N

120 cm

120 cm

A B

C

80 cm


a. The velocity of the streetcar relative to the taxi,v

s v

t 35 (50) 15, is 15 km/h south.

b. The velocity of the streetcar relative to the bus,v

s v

b 35 (42) 77, is 77 km/h north.

c. The velocity of the taxi relative to the bus,v

t v

b 50 (42) 92, is 92 km/h north.

d. The velocity of the bus relative to the streetcar,v

b v

s 42 (35) 77, is 77 km/h south.

4. a.

The distance downstream will be the distance

travelled in 6 min at 6 km/h, 110 6 0.6, 0.6 km.

He will touch the bank 0.6 km downstream fromthe marina and will be there in 6 minutes.

b. The boat will proceed across the river at a speed of20 km/h regardless of the speed of the current.Hence the time it takes to cross the river will be thetime it takes to travel 2 km at 20 km/h,

220

110, 6 minutes.

5.

Let the resultant velocity be v.

Now v2 4502 1002

v 460.9772

and tan 140500

12.53.

a. The plane will travel a distance of

3 v 1383 km in 3 hours.b. The direction of the plane is approximately N 13 E.

6.

Adding the vectors creates OAB where OA is thevelocity of the aircraft, OA 175, AB is the velocity of the wind, AB 40, and BAO 90 10 8

72.

OB v is the resultant velocity and BOA .From the cosine law

v2 402 1752 2 40 175 cos 72v 167.03.

From the sine law

si4n0

sin

v7

2

sin 40 s

i

vn

72

13.17.

The ground velocity is approximately 167 km/h in adirection N 5 W (13 8 5).

W E

S

8

B10

A

N8

O

N

W E

S

100

450 v

2 km

6 km/h

20 km/h


7.

Adding the vectors creates OAB where OA is theboats velocity, OA 3; AB is the currents velocity,AB 2, and BAO 55, BOA .From the cosine law

v2 32 22 2 3 2 cos 55v 2.473.

From the sine law

sin

2

sin

v5

5

sin 2 si

n

v55

41.48.The velocity is approximately 2.5 m/s in a direction ofN 56 W (41.48 15 56.48).

8.

Adding the vectors forms OAB where the planesteering east at 240 km/h is represented by OA,

the wind from the northwest is represented by AB,

AB 65, and the planes actual velocity is v whereAOB .From the cosine law

v2 2402 652 2 240 65 cos 135v 289.63.

A

45

Bv

O

A

B

4015

40

v

O


From the sine law

si6n5

sin

v135

sin 65 si

n

v135

9.1.The planes actual direction is approximately S 81 E.

9.

a. The horizontal component is 215 cos 18 204 km/h. The vertical component is 215 sin 18 66 km/h.

b. The horizontal component is the speed that the jetadvances. The vertical component is the speed atwhich the jet gains vertical altitude.

10.

OA represents the vector along which the plane steers.

OA 520 km/h, BOA hence the plane steersat S (20 ) E. AB represents the wind velocity,

AB 46 km/h and ABO 80 20 100.OB v represents the velocity with respect to the

ground.

From the sine law

si4n6

sin

521000

sin v

O

AB

sin 46 s

5in20

100

4.997.

N

W E

SB

A46

520

1080

20

v

O

18

215 km/h


OAB 180 100 75

v 520si

snin

100OAB

v 510.04.The pilot should steer in a direction S 25 E and theplanes ground speed will be approximately 510 km/h.

11.

The destroyer travels in a direction as in the diagramand will intercept the sub in t hours. Hence thedistance DB 30t nautical miles and SB 20tnautical miles. OSB 135 and from the sine law

s2in0t

sin3103t5

sin 20 si

3n0135

28, 0 90The destroyer should travel in a direction of N 62 Eto intercept the submarine.

12. a.

Represent the velocity of the aircraft as v and thewind velocity as w, v > w. Let the distance betweenToronto and Vancouver be x km. The speed ingoing from Vancouver to Toronto with the wind is(v w) km/h and from Toronto to Vancouver willbe (v w) km/h. The time to go from Vancouver

to Toronto will be v

xw

h and from Toronto to

Vancouver v

xw

h.

Total time is Ta

v

xw

v

xw

x v(v

ww

)(vv

ww)

v22

xvw2

.

b. When there is no wind, the time required to travel

from Vancouver to Toronto is xv

h and from

Toronto to Vancouver is xv

h.

Total time is Tb

2vx.

Now Ta

Tb

v2

2

xvw2

2vx

2xv2

v

(v2(v

2

w2)w2

v(v22x

w2

w2) > 0

Therefore Ta

Tb

> 0

Since Ta

Tb

> 0, Ta

> Tb, it takes longer to travel

from Vancouver to Toronto and back when there is

a wind.

13.

The speed relative to the ocean floor is represented byOA, a diagonal of a rectangular solid with sides oflength 0.5, 3, and 12, as shown in the diagram.

OA2 (0.5)2 122 32

OA 12.379The speed of the sailor relative to the ocean floor isapproximately 12.4 m/s.

14. Let vc

represent the velocity of the car and vtthe

velocity of the truck. Vector vR, the velocity of the

truck relative to the car, is such that vR

vt v

c.

80 km/h

50 km/h

vc

vtvt

vc vR

A

N

E

0.5 m/s

3 m/s

12 m/s

O

V T

w

D S

B

30t20t

458

vR2 802 502

vR 94.34

is the angle between vR

and vt.

tan 8500

58.The velocity of the truck relative to the car isapproximately 94.3 km/h in a direction N 32 E.

Review Exercise

7.

Represent the tension in the string by T and the anglethe string makes with the vertical by as shown in the diagrams. Since the system is in equilibrium the sum of the threeforces will be O as shown in the triangle diagram.

Now T2 29.42 122

T 31.75

tan 2192.4, 22.2.

The tension in the string has a magnitude of 32 N andthe string makes an angle of 22 with the vertical.

8. a.

Represent the resultant by R. From the cosine law

R2 F12 F

22 2F

1F

2 cos 125

542 342 2 54 34 cos 125

R 78.601.The magnitude of the resultant is approximately 79 N.

b. F1 21, F

2 45, 140

R2 F12 F

22 2F

1F

2 cos 40

212 452 2 21 45 cos 40

R 31.909The magnitude of the resultant is approximately 32 N.

9.

From the sine law

sin

R5

0

si

n

F

71

5

sin

F

52

5

R 480 N

F1 R

s

insi

5n075

F1 605.2

F2 R

s

insi

5n055

F2 513.3.

The magnitudes of the two forces are approximately 605 N and 513 N.

R

75

7555

F2

F1

180 130= 50

F 2

F 1

R

55180 55= 125

12 N

3 kg

29.4 N

29.4

12

TT


10.

Resolve the 2 N and 5 N forces into rectangularcomponents along and perpendicular to the 12 N force.

Let R be the resultant of F1

and F2

where

F1 12 2 cos 20 5 cos 40 17.7096

and F2 5 sin 40 2 sin 20 2.5299.

Now R2 F12 F

22

R 17.8894

tan

F

F

2

1

8.13.The resultant has a magnitude of approximately 17.9 N and makes an angle of 8 with the 12 N forceand 32 with the 5 N force.

11.

Let T1

and T2

represent the tensions in each string and

and be the angles that the strings make with the

ceiling as shown in the diagram. In OAB, OA 7,OB 5, AB 10. From the cosine law

72 102 52 2 5 10 cos

cos 102

2

55

2

1

072

cos 1295, 40.5.

Also 52 72 102 2 7 10 cos

cos 3315, 27.7.

From the sine law

sin (

98 )

sin (9

T0

1

)

sin (9

0

T

2

).

But sin (90 ) cos , sin (90 ) cos

and 68.2

T1 98

sicno6s82.72.7

93.5

T2 98

sicno6s84.02.5

80.3.

The tension in the 5 m string is 93.5 N and in the 7 mstring is 80.3 N.

90

90

98 N

T1

T2

A B10 m

T2

T1

10 kg

98 N

O

2 sin 20

5 sin 40

202 N

4012 N

5 N

F2

F1

R


12.

To fly due east let the bearing of the plane be northof east, and v represent the velocity due east. From thesine law

si8n0

sin

801035

si

n

v

sin sin

11035

4.1 180 135 4.1

40.9.

Now sin

4

v0

.9

sin

801035

v 800sin

sin13

450.9

v 740.8.a. The planes heading should be N 85.9 E.

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Solutions Manual of Geometry and Discrete Mathematics