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in a triangle ABC if B=3C prove that in(A/2) =(b-c)/2c
B = 3C , sin B = sin3C => sinB = sinC(4cos2C – 1) Using sin law for sinB and sinC => b = c(4cos2 C– 1) => cos2C = (b+c)/4aA+B+C=180 => A = 180 – 4C =>sin(A/2) = cos2C = 2cos2C – 1 => sin(A/2) = (b – c)/2c
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