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Prove that in a triangle ABC, summation of sinAsin(B-C)=0

Prove that in a triangle ABC, summation of sinAsin(B-C)=0

Grade:11

1 Answers

Arun
25750 Points
5 years ago
SinA . Sin(B - C) + SinB . Sin(C - A) + SinC . Sin(A - B)=0
=>SinA . (SinBCosC - CosBSinC) + SinB . (SinCCosA - CosCSinA) + SinC . (SinACosB - CosASinB)
=>SinASinBCosC -SinACosBSinC + SinBSinCCosA - SinBCosCSinA + SinCSinACosB - SinCCosASinB
All get cancelled
=>0
Hence Proved
 
 

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