Question

If $A$ and $B$ are acute positive 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. $0$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{3}$

Answer 1

Answer: (B)

$\frac{\pi }{2}$

$3{\sin }^{2}A=\cos 2B$ -(i)
$\frac{3}{2}\sin 2A=\sin 2B$ -(ii)
Dividing the two equations we get,
$\tan A=\cot (\frac{\pi }{2}-2B)$
Therefore $A+2B=\frac{\pi }{2}$
Hence option B is the correct option