Question

guyzz help me out plz​

Answer 1

In a triangle,

A+B+C=180

B+C=180-A

(B+C)/2=90-(A/2)

sin(B+C/2)=sin(90-A/2)

sin(B+C/2)=cos(A/2)

Answer 2

$\huge \bold{hello \: mate}$

If A, B and C are the interior angles of a triangle, then

A+B+C=180° ---> Angle sum property of a triangle

B+C=180°-A

Dividing by 2 from both sides,

=>B+C/2=180°-A/2

=>B+C/2=90°-A/2

Now, taking sin from both the sides,

sin(B+C/2)=sin(90°-A/2)

sin(B+C/2)= cos(A/2)-----------{sin(90-theta)=cos theta)

Hence, proved.