Question

If A, B and C are interior angles of a triangles ABC, then show

that Sin B + C/2 = cos (A/2) ​

Answer 1

$\Huge{\color{pink}{\fbox{\textsf{\textbf{Solution:-}}}}}$

We will be using the trigonometric ratios of complementary angles 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

Answer 2

Answer:

We will be using the trigonometric ratios of complementary angles 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