Question

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

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

Answer 1

The correct option is D ${\tan }^{-1}\left(\frac{2}{9}\right)$

as shown in the figure

$tan\left(ta{n}^{-1}\right(\frac{1}{2}\left)\right)=\frac{1}{2}=\frac{2h}{b}$ .......................(1)

$tan\left(ta{n}^{-1}\right(\frac{1}{2})-\varphi )=\frac{h}{b}$ .......................(2)

(1) divided by (2) we get

$2=\frac{\frac{1}{2}}{\frac{1-\frac{tan\varphi }{2}}{1+\frac{tan\varphi }{2}}}$

solving the above equation we get

$tan\varphi =\frac{2}{9}$

$\varphi =ta{n}^{-1}\frac{2}{9}$