In any ∆ A B C , prove that c - b cos A b - c cos A = cos B cos C

Need to prove: c - b cos A b - c cos A = cos B cos C L.H.S = c - b cos A b - c cos A = c - b b 2 + c 2 - a 2 2 b c b - c b 2 + c 2 - a 2 2 b c = 2 c 2 - b 2 - c 2 + a 2 2 c 2 b 2 - b 2 - c 2 + a 2 2 b = c 2 + a 2 - b 2 2 c b 2 + a 2 - c 2 2 b = c 2 + a 2 - b 2 2 a c b 2 + a 2 - c 2 2 a b [Multiplying the numerator and denominator by 1 a ] = cos B cos C = R.H.S Hence, proved.

In any ∆ A B C , prove that 2 ( b c cos A + c a cos B + a b cos C ) = a 2 + b 2 + c 2

Need to prove: 2 ( b c cos A + c a cos B + a b cos C ) = a 2 + b 2 + c 2 L.H.S = 2 ( b c cos A + c a cos B + a b cos C ) = 2 b c b 2 + c 2 - a 2 2 b c + c a c 2 + a 2 - b 2 2 c a + a b a 2 + b 2 - c 2 2 a b = b 2 + c 2 - a 2 + c 2 + a 2 - b 2 + a 2 + b 2 - c 2 = a 2 + b 2 + c 2 = R.H.S Hence, proved.

In any ∆ A B C , prove that 4 b c cos 2 A 2 + c a cos 2 B 2 + a b cos 2 C 2 = ( a + b + c ) 2

Given ∆ A B C with the length of sides opposite to ∠ A , ∠ B and ∠ C as a , b and c respectively, we have to prove that 4 b c cos 2 ⁡ A 2 + c a cos 2 ⁡ B 2 + a b cos 2 ⁡ C 2 = ( a + b + c ) 2 . L.H.S = 4 b c cos 2 ⁡ A 2 + c a cos 2 ⁡ B 2 + a b cos 2 ⁡ C 2 = 4 b c cos 2 ⁡ A 2 + c a cos 2 ⁡ B 2 + a b cos 2 ⁡ C 2 = 4 b c cos 2 ⁡ A 2 + 4 c a cos 2 ⁡ B 2 + 4 a b cos 2 ⁡ C 2 = 2 b c ( cos A + 1 ) + 2 c a ( cos B + 1 ) + 2 a b ( cos C + 1 ) .... ( cos 2 x = 2 cos 2 x - 1 ) = 2 b c ( b 2 + c 2 - a 2 2 b c + 1 ) + 2 c a ( a 2 + c 2 - b 2 2 a c + 1 ) + 2 a b ( a 2 + b 2 - c 2 2 a b + 1 ) (by cosine rule) = 2 b c ( b 2 + c 2 - a 2 + 2 b c 2 b c ) + 2 c a ( a 2 + c 2 - b 2 + 2 a c 2 a c ) + 2 a b ( a 2 + b 2 - c 2 + 2 a b 2 a b ) = b 2 + c 2 - a 2 + 2 b c + a 2 + c 2 - b 2 + 2 a c + a 2 + b 2 - c 2 + 2 a b = a 2 + b 2 + c 2 + 2 a b + 2 b c + 2 a c = ( a + b + c ) 2 = R.H.S Hence, proved.

In any ∆ A B C , prove that sin ( A - B ) sin ( A + B ) = a 2 - b 2 c 2

Given ∆ A B C with the length of sides opposite to ∠ A , ∠ B and ∠ C as a , b and c respectively, we have to prove that sin ⁡ ( A − B ) sin ⁡ ( A + B ) = a 2 − b 2 c 2 By sine rule, we have a sin A = b sin B = c sin C = 2 R where R is circumradius. L.H.S = sin ⁡ ( A − B ) sin ⁡ ( A + B ) = sin ⁡ ( A − B ) sin ⁡ ( A + B ) sin 2 ( A + B ) = sin 2 A − sin 2 B sin 2 ( π - C ) [ ∵ sin 2 x - sin 2 y = sin ( x - y ) sin ( x + y ) ] = sin 2 A − sin 2 B sin 2 C = a 2 4 R 2 − b 2 4 R 2 c 2 4 R 2 = a 2 − b 2 c 2 = R.H.S Hence, proved.

In any ∆ A B C , prove that ( b - c ) a cos A 2 = sin ( B - C ) 2

Need to prove: ( b - c ) a cos A 2 = sin ( B - C ) 2 We know that, a sin A = b sin B = c sin C = 2 R where R is the circumradius. L.H.S = 2 R sin B - 2 R sin C 2 R sin A cos A 2 = 2 cos B + C 2 sin B - C 2 sin A cos A 2 = 2 sin B - C 2 cos π 2 - A 2 sin A cos A 2 ∵ A + B + C = π = 2 sin A 2 cos A 2 sin B - C 2 sin A = sin A sin B - C 2 sin A = sin B - C 2 = R.H.S Hence, proved.

Solve the triangle in which a = 2 cm , b = 1 cm and c = 3 cm

Given ∆ A B C with the length of the sides opposite to angles A , B and C as a = 2 cm , b = 1 and c = 3 cm . All the angles of a triangle are less than 180 ° cos A = b 2 + c 2 - a 2 2 b c = 1 2 + 3 2 - 2 2 2 × 1 × 3 = 0 2 3 = 0 ⇒ ∠ A = 90 ° So, the remaining two angles of the triangle are acute angles. By Sine rule, we have a sin A = b sin B = c sin C ⇒ 2 sin 90 ° = 1 sin B = 3 sin C ⇒ 2 1 = 1 sin B = 3 sin C ⇒ sin B = 1 2 , sin C = 3 2 ⇒ ∠ B = 30 ° , ∠ C = 60 ° Thus, the angles of a triangle opposite to the sides, a = 2 cm , b = 1 cm and c = 3 cm are 90 ° , 30 ° and 60 ° .

E M B I B E

Senior Secondary School Mathematics FOR CLASS 11>Trigonometry>Chapter 18 - Solution of Triangles>EXERCISE

R S Aggarwal Solutions for Exercise 1: EXERCISE

Author:R S Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Exercise 1: EXERCISE

Attempt the practice questions from Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. Senior Secondary School Mathematics FOR CLASS 11 solutions are prepared by Experienced Embibe Experts.

Questions from R S Aggarwal Solutions for Exercise 1: EXERCISE with Hints & Solutions

EASY

11th CBSE

IMPORTANT

Senior Secondary School Mathematics FOR CLASS 11>Chapter 18 - Solution of Triangles>EXERCISE>Q 4

In any $∆ A B C$ , prove that $\frac{c - b \cos A}{b - c \cos A} = \frac{\cos B}{\cos C}$

MEDIUM

11th CBSE

IMPORTANT

Senior Secondary School Mathematics FOR CLASS 11>Chapter 18 - Solution of Triangles>EXERCISE>Q 5

In any $= \frac{\frac{c^{2} + a^{2} - b^{2}}{2 a c}}{\frac{b^{2} + a^{2} - c^{2}}{2 a b}}$ , prove that $2 (b c \cos A + c a \cos B + a b \cos C) = (a^{2} + b^{2} + c^{2})$