In any triangle ABC, prove the following:
a(sin B – sin C) + b(sin C – sin A) + c(sin A – sin B) = 0
Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get
⇒ a = k sin A, b = k sin B, c = k sin C
Here we will consider LHS, so we get
LHS = a(sin B – sin C) + b(sin C – sin A) + c(sin A – sin B)
Substituting corresponding values from sine rule in above equation, we get
⇒ = 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
Cancelling the like terms, we get
LHS = 0 = RHS
Hence proved