Question

If A and B are positive acute angles satisfying the equations $3{\sin }^{2}A+2{\sin }^{2}B=1$ and $3\sin 2A-2\sin 2B=0$, then A + 2B =

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. 0

Answer 1

Answer: (C)

Given $3{\sin }^{2}A+2{\sin }^{2}B=1$
$\Rightarrow 3{\sin }^{2}A=1-2{\sin }^{2}B$
$\Rightarrow 3{\sin }^{2}A=\cos 2B$
$\therefore \cos \left(A+2B\right)=\cos A\cos 2B-\sin A\sin 2B$
$=\left(\cos A\right)\left(3{\sin }^{2}A\right)-\sin A\left(\frac{3}{2}\sin 2A\right)$
$\left(\because 3\sin 2A-2\sin 2B=0\therefore \sin 2B=\frac{2}{3}\sin 2A\right)$
$=3{\sin }^{2}A\cos A-\frac{3}{2}\left(\sin A\right)\left(2\sin A\cos A\right)=0$
$\Rightarrow A+2B=\frac{\pi }{2}.$