Question

The length of the shadows of a vertical pole of height $h$, thrown by the sun's rays at three different moments are $h$, $2h$ and $3h$. The sum of the angles of elevation of the rays at these three moments is equal to

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is A $\frac{\pi }{2}$

Let $OA$ be the vertical pole of height $h$ and $O{P}_1,O{P}_2$ and $O{P}_3$ be the lengths of shadow.
In $\Delta$ $AO{P}_1$, we have
$\tan {\theta }_1=\frac{OA}{O{P}_1}=\frac{h}{h}=1\Rightarrow {\theta }_1=\frac{\pi }{4}$
In $\Delta AO{P}_2$, we have
$\tan {\theta }_2=\frac{OA}{O{P}_2}=\frac{h}{2h}=\frac{1}{2}$
${\theta }_2={\tan }^{-1}1/2$
Similarly, in $\Delta AO{P}_3$, we have
$\tan {\theta }_3=1/3$
${\theta }_3={\tan }^{-1}\left(1/3\right)$
Therefore, sum of the angles of elevation of the eyes
$={\theta }_1+{\theta }_2+{\theta }_3$
$=\frac{\pi }{4}+{\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}\left(\frac{1}{3}\right)$
$=\frac{\pi }{4}+{\tan }^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{3}\times \frac{1}{2}}\right)$ $[\because {\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)]$
$=\frac{\pi }{4}+{\tan }^{-1}\left(\frac{5/6}{5/6}\right)$
$=\frac{\pi }{4}+{\tan }^{-1}\left(1\right)=\frac{\pi }{4}+\frac{\pi }{4}=\frac{\pi }{2}$