Question

of triangle ABC then show that sin b + c by 2 is equal to Cos A by 2​

Answer 1

Step-by-step explanation:

In a triangle:

‹A + ‹B + ‹C = 180

‹A= 180-‹B-‹C

dividing both sides by 2

‹A /2= (180-‹B-‹C)/2

‹A/2= 90- (‹b+‹c)/2

cos a/2=cos[90-(‹b+‹c)/2]

cos(90-x) = sin x

cos a/2=sin (‹b+‹c)/2

hence proved