Question

In any triangle ABC, prove that a sin (B – C) + b sin (C – A) + c sin (A – B) = 0.​

Answer 1

Answer:

i am class five so i can't answer

Answer 2

Given :-

A triangle ABC

Required To Prove :-

a sin (B-C) + b sin (C- A) + c sin (A-B) = 0

Proof :-

Given that

ABC is a triangle.

We know that

in a triangle sin A /a = sin B /b = sin C /c

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

sin A/ a = k => sin A = ak

sin B / b = k => sin B = bk

sin C/ c = k => sin C = ck

Now,

sin(B-C) = sin B cos C - cos B sin C

=> sin (B-C) = bk . cos C - cos B .ck-----(1)

sin (C- A) = sin C cos A - cos C sin A

=> sin (C -A) = ck . cos A - cos C . ak ---(2)

sin (A-B) = sin A cos B - cos A sin B

=> sin (A-B) = ak. cos B - cos A . bk-----(3)

On adding (1),(2)&(3)

a sin (B-C)+ b sin (C- A) + c sin (A-B)

= a(bk . cos C - cos B .ck)+b(ck . cos A - cos C . ak)+ c( ak. cos B - cos A . bk)

= abk. cos C - ack .cos B + bck. cos A - abk. cos C + ack. cos B - bck. cos A

= (abk.cos C -abk.cos C)+(ack.cos B - ack.cos B)+( bck. cos A - bck. cos A)

= 0+0+0

= 0

Therefore,

a sin(B-C) +b sin (C- A)+c sin (A-B) = 0

Hence, Proved.