Question

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

Answer 1

Use sin formula

a

sinA

=

b

sinB

=

c

sinC

=k

sinA=ak,sinB=bk,sinc=ck

Also, sin(A−B)=sinA..cosB−cosA−sinB

=akcosB−cosA.bk

k(acosB−bcosA)

Similarity, sin(B−C)=k(bcosC−ccosB)

sin(C−A)=k(ccosA−acosc)

LHS=asin(B−C)+bsin(C−A)+csin(A−B)

=ak(bcosc−cosB)+bk(ccosA−acosC)+ck(acosB−bcosA)

=k(bccosA−bccosA)+k(accosB−accosB)+(abcos−abcosc)

=0+0+0

=RHS

Answer 2

Question :

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

Answer :

$\sf \small \bullet \small \frac{sin A}{a}=\frac{sin B}{b}=\frac{sin C}{c}$

• sinA=ak,sinB=bk,sinc=ck

• Also, sin(A−B)=sinA..cosB−cosA−sinB

• =akcosB−cosA.bk

• k(acosB−bcosA)

• Similarity, sin(B−C)=k(bcosC−ccosB)

• sin(C−A)=k(ccosA−acosc)

• LHS=asin(B−C)+bsin(C−A)+csin(A−B)

• =ak(bcosc−cosB)+bk(ccosA−acosC)+ck(acosB−bcosA)

• =k(bccosA−bccosA)+k(accosB−accosB)+(abcos−abcosc)

• =0+0+0

• =RHS