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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
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Verified Answer
The correct answer is:
$2 \alpha$
$2 \alpha$
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,
$$
\begin{aligned}
& & \tan \theta & =\frac{h}{h / \sqrt{3}} \Rightarrow \tan \theta=\sqrt{3} \\
\Rightarrow \quad & & \theta & =\pi / 3 \Rightarrow \theta=2 \alpha
\end{aligned}
$$
$16^{\sin ^2 \theta}+16^{\cos ^2 \theta}=10$
$\Rightarrow \quad 16^{\sin ^2 \theta}+16^{1-\sin ^2 \theta}=10$
$\Rightarrow x+\frac{16}{x}=10$, 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,
$$
\begin{aligned}
& & \tan \theta & =\frac{h}{h / \sqrt{3}} \Rightarrow \tan \theta=\sqrt{3} \\
\Rightarrow \quad & & \theta & =\pi / 3 \Rightarrow \theta=2 \alpha
\end{aligned}
$$
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