Help please

Asked by  | 27th Jul, 2009, 06:38: PM

Expert Answer:

If A, B ,and C are interior angles of a triangle ABC then A+B+C=180o

B+C=180o-A

(B+C)/2  =90o-A/2

Applying sine on both sides we get
sin (B+C)/2 =sin (90o-A/2)  = cos A/2  (since sin (90o -x)  = cos x )

Answered by  | 28th Jul, 2009, 10:19: AM

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