Question

If $A$ and $B$ are acute angles such that $\sin A={\sin }^{2}B,2{\cos }^{2}A=3{\cos }^{2}B$; then

• A. $A=\frac{\pi }{6}$
• B. $A=\frac{\pi }{2}$
• C. $B=\frac{\pi }{4}$
• D. $B=\frac{\pi }{3}$

Answer 1

Answer: (A), (C)

The correct options are
A $A=\frac{\pi }{6}$
C $B=\frac{\pi }{4}$

From the given conditions

$2(1-{\sin }^{2}A)=3(1-{\sin }^{2}B)=3(1-\sin A)$

$\Rightarrow 2{\sin }^{2}A-3\sin A+1=0$

$\Rightarrow (2\sin A-1)(\sin A-1)=0$

$\Rightarrow \sin A=1$ or $\sin A=\frac{1}{2}$

$\Rightarrow A=\frac{\pi }{2}$ or $\frac{\pi }{6}$

But, since $A$ is acute, we have $A=\frac{\pi }{6}$.

$\Rightarrow {\sin }^{2}B=\sin A=\sin \left(\frac{\pi }{6}\right)=\frac{1}{2}$

$\Rightarrow \sin B=\frac{1}{\sqrt{2}}\Rightarrow B=\frac{\pi }{4}$