Question

in a triangle ABC, cos (B+C÷2) in terms of angle A

Answer 1

Sol: In a ΔABC, ∠A + ∠B + ∠C = 180° [Angle sum property] ⇒ ∠A + ∠B = 180° - ∠C ⇒ (∠A + ∠B) / 2 = (180° - ∠C) / 2 ⇒ (∠A + ∠B) / 2 = (180° - ∠C) / 2 ⇒ (∠A + ∠B) / 2 = (90° - ∠C/ 2) ⇒ sin (∠A + ∠B) / 2 = sin (90° - ∠C/ 2) ⇒ sin (∠A + ∠B) / 2 = cos ∠C/ 2. [sin (90 - θ) = cos θ)]Therefore, sin (∠A + ∠B) / 2 in terms of angle c is cos ∠C/ 2.

Answer 2

Sol: In a ΔABC, ∠A + ∠B + ∠C = 180° [Angle sum property] ⇒ ∠A + ∠B = 180° - ∠C ⇒ (∠A + ∠B) / 2 = (180° - ∠C) / 2 ⇒ (∠A + ∠B) / 2 = (180° - ∠C) / 2 ⇒ (∠A + ∠B) / 2 = (90° - ∠C/ 2) ⇒ sin (∠A + ∠B) / 2 = sin (90° - ∠C/ 2) ⇒ sin (∠A + ∠B) / 2 = cos ∠C/ 2. [sin (90 - θ) = cos θ)]Therefore, sin (∠A + ∠B) / 2 in terms of angle c is cos ∠C/ 2.

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