Question

Let $A$ and $B$ be acute angles such that $\sin A={\sin }^{2}B$ and $2{\cos }^{2}A=3{\cos }^{2}B.$ Then, $A$ equals to

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

Answer 1

Answer: (B)

Suppose, $A=\frac{\pi }{6}$ and $B=\frac{\pi }{4}$
Now, $\sin A={\sin }^{2}B$
$\Rightarrow$ $sin\frac{\pi }{6}={\sin }^2\left(\frac{\pi }{4}\right)$
$\Rightarrow$ $\frac{1}{2}={\left(\frac{1}{\sqrt{2}}\right)}^2$
$\Rightarrow$ $\frac{1}{2}=\frac{1}{2}$ (true)
Now, $2{\cos }^{2}A=3{\cos }^{2}B$
$\therefore$ $2{\cos }^2\frac{\pi }{6}=3{\cos }^2\left(\frac{\pi }{4}\right)$
$\Rightarrow$ $2\times {\left(\frac{\sqrt{3}}{2}\right)}^2=3{\left(\frac{1}{\sqrt{2}}\right)}^2$
$\Rightarrow$ $2\times \frac{3}{4}=3\times \frac{1}{2}$
$\Rightarrow$ $\frac{3}{2}=\frac{3}{2}$ (true)