if A,B,c are interior angles of triangle ABC then show that sin (B+C/2) = cos A/2 Share with your friends Share 32 Varun.Rawat answered this -5 View Full Answer 123456789 answered this A + B + C = 180^{o}Therefore, B + C = 180 - A(B + C) / 2 = (180 - A) / 2(B + C) / 2 = 90 - ( A / 2 )Hence R.H.S :- Cos ( A / 2 ) = Sin [ 90 - ( A / 2 ) ] = Sin [ (B + C) / 2 ] .................... As 90 - ( A / 2 ) = (B + C) / 2Therefore,Cos ( A / 2 ) = Sin [ (B + C) / 2 ]Hence proved. 54 Ayush Maskara answered this A+B+C=180So, b+c=180-Ab+c/2=90-a/2 ( Dividing the L.H.S and R.H.S by 2)sin(b+c/2)= sin ( 90 -a/2) ( Substituting b+c/2 value)= cos a/2Hence Proved Thumbs up plz..................... 97 Ujwal Uttarwar answered this since A B and C are angles of a triangletherefore, A+B+C=180and B+C=180-Anow substitue the value of B+C as 180-A=sin(180-A)/2=sin(180/2 - A/2)=sin(90-A/2)since we konw that sin(90-A)=cosAtherefore sin(90-A/2) = cosA/2 = R.H.S 37