Let $\alpha$ be the solution of $16^{\sin ^2 \theta}+16^{\cos ^2 \theta}=10$ in $\left(0, \frac{\pi}{4}\right)$. If the shadow of a vertical pole is $\frac{1}{\sqrt{3}}$ of its height, then the altitude of the sun is

• A

  $\alpha$
• B

  $\frac{\alpha}{2}$
• Correct answer
  C

  $2 \alpha$
• D

  $\frac{\alpha}{3}$

Solution

We have, $16^{\sin ^2 \theta}+16^{\cos ^2 \theta}=10$ $\Rightarrow \quad 16^{\sin ^2 \theta}+16^{1-\sin ^2 \theta}=10$ $\Rightarrow x+\frac{16}{x}=10$, where $x=16^{\sin ^2 \theta}$ $\Rightarrow \quad x^2-10 x+16=0 \Rightarrow x=2,8$ $\therefore \quad 16^{\sin ^2 \theta}=2,8 \Rightarrow 2^{4 \sin ^2 \theta}=2,2^3$ $\Rightarrow \quad 4 \sin ^2 \theta=1,3 \Rightarrow \sin ^2 \theta=\frac{1}{4},\left(\frac{\sqrt{3}}{2}\right)^2$ $\Rightarrow \quad \sin \theta=\frac{1}{2}, \frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{6}, \frac{\pi}{3}$
Let the altitude of the sun be $\theta$. Then, $$ \begin{aligned} & & \tan \theta & =\frac{h}{h / \sqrt{3}} \Rightarrow \tan \theta=\sqrt{3} \\ \Rightarrow \quad & & \theta & =\pi / 3 \Rightarrow \theta=2 \alpha \end{aligned} $$

Hence, the correct answer is option: 'C'.