# If A, B and C are interior angles of a triangle ABC, then show that:

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

**Solution:**

We use the standard trigonometric identities to solve the given question.

sin (90° - θ) = cosθ

We know that for ΔABC,

∠A + ∠B + ∠C = 180° (Angle sum property of triangle)

∠B + ∠C = 180° - ∠A

On dividing both sides by 2, we get

(∠B + ∠C)/2 = (180° - ∠A)/2

(∠B + ∠C)/2 = 90° - ∠A/2

Applying sine angles on both the sides:

sin {(∠B + ∠C)/2} = sin (90° - ∠A/2)

Since, sin (90° - θ) = cos θ, we get

sin (∠B + ∠C)/2 = cos A/2

**Video Solution:**

## If A, B and C are interior angles of a triangle ABC, then show that: sin (B + C)/2 = cos A/2

### Maths NCERT Solutions Class 10 - Chapter 8 Exercise 8.3 Question 6:

If A, B and C are interior angles of a triangle ABC, then show that: sin (B + C)/2 = cos A/2

If A, B, C are interior angles of a triangle ABC, then sin [(B+C)/2] = cos (A/2)