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  4. A vertical pole subtends an angle tan -1((1/2)) 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 α and β, then ( tan α, tan β) is equal to

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Q. 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

EAMCETEAMCET 2012

Solution:

Let $A C$ be a pole and point $P$ be the position on of the ground.
image
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 $(\tan \alpha, \tan \beta)=\left(\frac{1}{4}, \frac{1}{5}\right)$
$\therefore RHS =\frac{\frac{1}{4}+\frac{1}{5}}{1-\frac{1}{4} \times \frac{1}{5}}=\frac{\frac{9}{20}}{\frac{19}{20}}$
$=\frac{9}{19} \neq \frac{1}{2}$, 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}} $
$=\frac{19}{43} \neq \frac{1}{2},$ 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