Question

if abc are the interior angels of a triangle ABC, show that :sin C+B/2=cosA/2​

Answer 1

Answer:

Step-by-step explanation:

Given ,

A,B,C are angles of triangle ABC

So ,

A + B + C = 180° —(1) ( angle sum property)

Divide equation (1) by 2

A/2 + B/2 + C/2 = 90°

=> A/2 + B+C/2 = 90°

=> B+C/2 = 90°- A/2 —(2)

Apply sin ratio on both sides of the equation (2)

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

Sin (B+C/2) = Cos A/2 [sin(90-A) = cos A]

Hence proved