In any triangle ABC, prove the: a (sin B – sin C) + b (sin C – si

1.	In any triangle ABC, prove the: a (sin B – sin C) + b (sin C – sin A) + c (sin A – sin B) = 0
Answer» On using the sine rule we know, a/sin A = b/sin B = c/sin C = k a = k sin A, b = k sin B, c = k sin C Let us consider the LHS a (sin B – sin C) + b (sin C – sin A) + c (sin A – sin B) On substituting the values of a, b, c from sine rule in above equation, we get a (sin B – sin C) + b (sin C – sin A) + c (sin A – sin B) = k sin A (sin B – sin C) + k sin B (sin C – sin A) + k sin C (sin A – sin B) = k sin A sin B – k sin A sin C + k sin B sin C – k sin B sin A + k sin C sin A – k sin C sin B Then, upon simplification, we get = 0 = RHS Thus proved.

In any triangle ABC, prove the: a (sin B – sin C) + b (sin C – sin A) + c (sin A – sin B) = 0

Answer»

On using the sine rule we know,

a/sin A = b/sin B = c/sin C = k

a = k sin A, b = k sin B, c = k sin C

Let us consider the LHS
a (sin B – sin C) + b (sin C – sin A) + c (sin A – sin B)

On substituting the values of a, b, c from sine rule in above equation, we get

a (sin B – sin C) + b (sin C – sin A) + c (sin A – sin B) = k sin A (sin B – sin C) + k sin B (sin C – sin A) + k sin C (sin A – sin B)

= k sin A sin B – k sin A sin C + k sin B sin C – k sin B sin A + k sin C sin A – k sin C sin B

Then, upon simplification, we get

= 0

= RHS

Thus proved.