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2015-07-21T10:51:50+05:30

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 Sin[(B+C)/2]
Since A+B+C=180 for interior angles of triangle ABC.
then B+C=180-A.

NOW   Sin [(180-A)/2]
              =Sin[90-(A/2)]                      since Sin(90-A)=CosA
               =Cos(A/2)



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2015-07-21T10:58:35+05:30

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