Question

In a triangle ABC, if A, B, C are in A.P. and b : c = $\sqrt{3}$: $\sqrt{2}$

, then –

• A. sin (2C – A) = sin (B/4)
• B. sin (A – C) = cos (B/2)
• C. sin (A + C) = cos 2B
• D. cos (A – C) = sin (B/2)

Answer 1

Answer: (A)

sin (2C – A) = sin (B/4)

Answer 2

A, B, C are in A. P. ⇒ B = 60⁰.

⇒ $\frac{\sin B}{b}$= $\frac{\sin C}{c}$

⇒ sin C = $\frac{c}{b}$ sin B = $\frac{\sqrt{2}}{\sqrt{3}} \times \frac{\sqrt{3}}{2}$=$\frac{1}{\sqrt{2}}$

⇒ C = 45⁰ and A = 75⁰

Hence sin (2C – A) = sin 15⁰ = sin B/4

and other relations are not true.