Question

if A,B and C are interior angles of triangle ABC then show that;
sin B±C/2= COS A/2

Answer 1

Answer:

Step-by-step explanation:

sin(B+C)/2=cosA/2

L.H.S

=sin(B+C)/2

We know that in triangle ABC , A+B+C=180°

Therefore B+C= 180° -A.

=sin(180°-A)/2

=sin(90°-A/2)

=cosA/2 , proved.

hope it helps you

Answer 2

Answer:

Step-by-step explanation

A+B+C=180° Angle sum

A+B=180-C

A+B/2=90-C/2

SINA+B/2=SIN90-C/2

SIN A+B/2= COSC/2