triangles

please prove the following

a3 sin(B-C) + b3sin (C-A) + c3 sin (A-B) = 0

3 Answers

1
shahrukh ·

pls help

106
Asish Mahapatra ·

LHS = a22Rsin(B+C)sin(B-C) + b22Rsin(C+A)sin(C-A) + c22Rsin(A+B)sin(A-B)

can you continue now?

21
Shubhodip ·

\Sigma a^3 sin(B-C) = 8R^3\Sigma sin^2A sin(B+C)sin (B-C)=\Sigma [sin^2A sin^2B - sin^2A sin^2C]= 0

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