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Question

b sin Bc sin C=a sin (BC)

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Solution

sin Aa=sin Bb=sin Cc=kRHS,a sin (BC)= a sin B.cos C a sin C.cos B= a(bk).(a2+b2c22ab)a(ck).(a2+c2b22ac)=k.(a2+b2c2)2k(a2+c2b2)2=2k.(b2c2)2=b.(kb)c(kc)=b (sin B)c (sin C)LHS.


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