Wikimama Class 11 Ch 17 Solution of Triangles

8/4/2019 Wikimama Class 11 Ch 17 Solution of Triangles

1/4

1

ST

In a triangle ABC, the angles are denoted bycapital letters A, B, and C and the lengths of thesides opposite these angles are denoted by a, b,

c respectively. Semi-perimeter of the triangle is

written asa b c

s ,2

and its area by

sinB sinC

b cS or .

A

C

b

aB

c

Let R be the radius of the circumcircle of triangle ABC.

Basic Formulae

(i) Sine rule:

sinA sinB sinC 1

a b c 2R .

(ii) Cosine rule:

2 2 2 2 2 2 2b c a c b b ccosA , , .

2bc

2 2a a

cosB= cosC=2ac 2ab

(iii) Projection rule:a = b cosC + c cosB,b = c cosA + a cosC,c = a cosB + b cosA.

(iv) Napier's analogy:B C b c A c a B a b C

tan cot , cot , .cot .2 b c 2 c a 2 a b 2

C A A Btan tan

2 2

(v) Trigonometric ratios of half - angles:

s b s c s c s a s a s bA B Csin , sin , sin ,

2 bc 2 ca 2 ab

s s a s s b s s c, , ,

bc ca ab

A B Ccos cos cos

2 2 2

s b s c s c s a s a s bAtan , , .2 s s a s s b s s c

B Ctan tan

2 2

Since, every angle of a triangle is less than 1800, half of each angle will be less

than 900, and thus, all the trigonometric ratios of half the angles are positive.

(vi) Area of a triangle:

1 abcbc c a s b s c

2 4R

1 1sin A = a sinB = b sin C = s s-a

2 2


2/4

2

Solution of Triangles:

The three sides a, b, c and the three angles A, B, C are called the elements of thetriangle ABC. When any three of these six elements (except all the three angles) of atriangle are given, the triangle is known completely; that is the other three elements canbe expressed in terms of the given elements and can be evaluated. This process is

called the solution of triangles.(i) If the three sides a, b, c are given, angle A is obtained from

s b s cAtan

2 s s aor

2 2 2b c acos A .

2bcB and C can be obtained in the

similar way.(ii) If two sides b and c and the included angle A are given, then

B C b c Atan cot

2 b c 2gives

B C.

2Also

B C A90 ,

2 2so that B and C can

be evaluated. The third side is given by a =b sinA

sinB

or a2

= b2

+ c2 2bc cosA.

(iii) If two sides b and c and the angle B (opposite to side b) are given, thenc

sinC sinBb

, A = 180 (B + C) andbsinA

bsinB

give the remaining elements.

If b < c sinB, there is no triangle possible (fig1). If b = c sinB and B is an acuteangle, then only one triangle is possible (fig 2). If c sinB < b < c and B is anacute angle, then there are two value of angle C (fig 3). If c < b and B is an acuteangle, then there is only one triangle (fig 4).

A

B

c

D

bc sinB

fig. 1

A

B

c

D

b c sinB

fig. 2

A

B

c

D

bb

C2 C1

c sinB

fig. 3

A

B

c

C1

b

C2

b c sinB

Fig. 4

This is, sometimes, called an ambiguous case.

If one side a and angles B and C are given, then A = 180 (B + C), andasinB asinC

b , csinA sinA

.

If the three angles A, B, C are given, we can only find the ratios of the sides

a, b, c by using the sine rule (since there are infinite number of similar triangles).


3/4

3

Circles Connected with Triangle(i) Circum-circle:

The circle passing through the

vertices of the triangle ABC is calledthe circum-circle. Its radius R is calledthe circum-radius. In the triangleABC,

a b c abcR

2sinA 2sinB 2sinC 4.

A

B C

O

R

(ii) In-Circle:

The circle touching the three sidesof the triangle internally is called theinscribed or the in-circle of thetriangle. Its radius r is called the in-radius of the circle. In the triangleABC,

A

B C

Or

A B C A B Cr s a tan s b tan s c tan 4Rsin sin sin .

s 2 2 2 2 2 2

Remark:

From r = 4R sinA

2sin

B

2sin

C

2, we find that r 4R.

1

8

2r R. Here equality holds for the equilateral triangle.

(iii) Escribed circles:

The circle touching BC and the two sides

AB and AC produced of ABC externallyis called the escribed circle opposite A. Itsradius is denoted by r1. Similarly r2, and r3denote the radii of the escribed circlesopposite angles B and C respectively.

r1, r2, r3 are called the ex-radii of ABC.

Here

CA

B

O1r1

1A A B C

r s tan 4Rsin cos coss a 2 2 2 2

,

2B B C A

r s tan 4Rsin cos cos ,s b 2 2 2 2

3C C A B

r s tan 4Rsin cos coss c 2 2 2 2

,

r1 + r2 + r3 = 4R + r,


4/4

4

2 1 2 31 2 2 3 3 1

r r rr r r r r r s

r.

Documents

Wikimama Class 11 Ch 17 Solution of Triangles