Question

A vertical pole subtends an angle ${\tan }^{-1}\frac{1}{2}$ at point $P$ on the ground. The angle subtended by the upper half of the pole at the point $P$ is__

• A. ${\tan }^{-1}\frac{1}{4}$
• B. ${\tan }^{-1}\frac{2}{9}$
• C. ${\tan }^{-1}\frac{1}{8}$
• D. ${\tan }^{-1}\frac{2}{3}$

Answer 1

The correct option is A ${\tan }^{-1}\frac{1}{4}$

$\angle ABP={90}^{\circ }$ and $\angle APB=ta{n}^{-1}\left(\frac{1}{2}\right)$ (given)

Let height of the pole be $h$

In $\Delta ABP$

$\frac{AB}{BP}=\tan \left(\angle APB\right)$

$\Rightarrow \frac{h}{BP}=\tan \left({\tan }^{-1}\frac{1}{2}\right)$

$\Rightarrow \frac{h}{BP}=\frac{1}{2}$

$\Rightarrow BP=2h$ ---(i)

in $\Delta DBP$

$\frac{DB}{BP}=\tan \left(\angle DPB\right)$

$\Rightarrow \frac{\frac{h}{2}}{2h}=\tan \left(\angle DPB\right)$ (from (i))

$\Rightarrow \frac{h}{4h}=\tan \left(\angle DPB\right)$

$\Rightarrow \tan \left(\angle DPB\right)=\frac{1}{4}$

$\Rightarrow \angle DPB={\tan }^{-1}\left(\frac{1}{4}\right)$

Angle subtended by upper half of the pole at point $P$ $=\angle APD=\angle APB-\angle DPB={\tan }^{-1}\left(\frac{1}{2}\right)-{\tan }^{-1}\left(\frac{1}{4}\right)$

$={\tan }^{-1}\left(\frac{\frac{1}{2}-\frac{1}{4}}{1+\frac{1}{2}\times \frac{1}{4}}\right)$ $[\because {\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\left(\frac{x-y}{1+xy}\right)]$

$={\tan }^{-1}\left(\frac{2}{9}\right)$