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. $None\, of \, these$

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)