Exam

BITSAT

Year

2022

Subject

Mathematics

Chapter

Heights and Distances

BITSAT 2022 · Mathematics · Heights and Distances

Let α be the solution of 16^ ^2 θ +16^ ^2 θ =10 in (0, (π)/(4)) ....

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

Step-by-step 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, $$\begin{aligned}& & \tan \theta& =\frac{h}{h / \sqrt{3}} \Rightarrow \tan \theta=\sqrt{3}\\ \Rightarrow \quad & \theta& =\pi/ 3 \Rightarrow \theta=2 \alpha \end{aligned}$$