if abc are interior angles of
a triangle then show that sin(b+c)/2 =cosA/2
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A+B+c=180
B+C=180-A
B+C/2 =90-A/2
therefore sin(B+C)/2=sin (90-A)/2
sin(B+C)/2=cos -A/2
B+C=180-A
B+C/2 =90-A/2
therefore sin(B+C)/2=sin (90-A)/2
sin(B+C)/2=cos -A/2
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