Question

If A and B are positive acute angles satisfying 3cos²A+2cos²B=4 and (3sin A)/(sin B)=(2cos B)/(cos A), then the value of A+2B is equal to:

• A. \(\dfrac{\pi}{6}\)
• B. \(\dfrac{\pi}{2}\)
• C. \(\dfrac{\pi}{3}\)
• D. (π)/(4)

Hint:

Convert products of sine and cosine into double-angle form.

Answer 1

The Correct Option is C

From the second equation: 3sin A cos A = 2sin B cos B ⟹ sin 2A = sin 2B Thus, A=B. Substituting in first equation: 3cos²A+2cos²A=5cos²A=4 ⟹ cos²A=(4)/(5) Hence, A=30^∘, A+2B=90^∘=(π)/(3)