Question

A vertical pole subtends an angle ${\tan }^{-1}\left(\frac{1}{2}\right)$ at a point $P$ on the ground. If the angles substended by the upper half and the lower half of the pole at $P$ are respectively $\alpha$ and $\beta$, then $(\tan \alpha ,\tan \beta )$ is equal to

• A. $\left(\frac{1}{4},\frac{1}{5}\right)$
• B. $\left(\frac{1}{5},\frac{2}{9}\right)$
• C. $\left(\frac{2}{9},\frac{1}{4}\right)$
• D. $\left(\frac{1}{4},\frac{2}{9}\right)$

Answer 1

Answer: (C)

Let $AC$ be a pole and point $P$ be the position on of the ground.

Given, $\theta ={\tan }^{-1}\frac{1}{2}$
$\Rightarrow \tan \theta =\frac{1}{2}$
Also, $\theta =\alpha +\beta$
$\Rightarrow \tan \theta =\tan (\alpha +\beta )$
$\Rightarrow \frac{1}{2}=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$
(a) When $(\tan \alpha ,\tan \beta )=\left(\frac{1}{4},\frac{1}{5}\right)$
$\therefore RHS=\frac{\frac{1}{4}+\frac{1}{5}}{1-\frac{1}{4}\times \frac{1}{5}}=\frac{\frac{9}{20}}{\frac{19}{20}}$
$=\frac{9}{19}\ne \frac{1}{2}$, not true
(b) When $(\tan \alpha ,\tan \beta )=\left(\frac{1}{5},\frac{2}{9}\right)$
$RHS=\frac{\frac{1}{5}+\frac{2}{9}}{1-\frac{1}{5}\times \frac{2}{9}}=\frac{\frac{19}{45}}{\frac{43}{45}}$
$=\frac{19}{43}\ne \frac{1}{2},$ not true
(c) When $(\tan \alpha ,\tan \beta )=\left(\frac{2}{9},\frac{1}{4}\right)$
$RHS=\frac{\frac{2}{9}+\frac{1}{4}}{1-\frac{2}{9}\times \frac{1}{4}}=\frac{\frac{17}{36}}{\frac{34}{36}}=\frac{1}{2}$, true