Question

If in a triangle ABC sines of angles A and B satisfy the

equation 4 x ² – 2 $\sqrt { 6 }$ x + 1 = 0, then cos (A – B) is equal to –

• A. 0
• B. 1/2
• C. 1/ $\sqrt { 2 }$
• D. $\sqrt { 3 }$/2

Answer 1

Answer: (B)

1/2

Answer 2

sin A + sin B = $\frac { \sqrt { 6 } } { 2 }$

sin A . sin B = ¼ , Let A > B [Q A ¹ B]

̃ sin² A + sin² B + 2 × ¼ = 6/4

̃ sin² A + sin² B = 1 ̃ sin² A = cos² B,

̃ cos B = sin A ̃ B = 90⁰ – A

̃ A + B = C = 90⁰

Also sin A sin B = cos B sin B = ¼ ,

̃ sin2 B = ½ ̃ 2B = 30⁰ or 150⁰

̃ B = 15⁰ or 75⁰

̃ B = 15⁰, A = 75⁰, ̃ A – B = 60⁰

̃ cos (A – B) = ½ .