Q13 of 87 Page 10

In any triangle ABC, prove the following:

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get




Here we will consider LHS, so we get



Multiply and divide by , we get






Substituting corresponding values from sine rule in the above equation, we get





Hence proved


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b sin B – c sin C = a sin (B – C)

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a2 sin (B – C) = (b2 – c2) sin A

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

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