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| 1. |
In any ΔABC, prove thatasin(B - C) + bsin(C – A) + csin(A - B) = 0 |
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Answer» LHS = ksinAsin(B - C) + ksinBsin(C – A) + ksinCsin(A – B) = k[sin(B + C)sin(B – C)+sin(C + A)sin(C – A) + sin(A + B) sin(A – B)] = K[sin2B – sin2C + sin2 C -sin2A + sin2A – sin2B] = k x 0 =0 = RHS. |
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