From the top of a hill $h$ metres high the angles of depressions of the top and the bottom of a pillar are…
- $\frac{h(\tan \beta-\tan \alpha)}{\tan \beta}$
- $\frac{h(\tan \alpha-\tan \beta)}{\tan \alpha}$
- $\frac{h(\tan \beta+\tan \alpha)}{\tan \beta}$
- $\frac{h(\tan \beta+\tan \alpha)}{\tan \alpha}$
Solution
In $\triangle E D B$,
$\tan \alpha=\frac{h-h^{\prime}}{E D}$
and in $\triangle A C B$,

$\tan \beta=\frac{h}{A C}=\frac{h}{E D}$
Eliminate $E D$ from Eqs. (i) and (ii), we get
$\begin{aligned}
& \tan \alpha=\frac{h-h^{\prime}}{\frac{h}{\tan \beta}} \\
& \Rightarrow \quad h \frac{\tan \alpha}{\tan \beta}=h-h^{\prime} \\
& \Rightarrow \quad h^{\prime}=\frac{h(\tan \beta-\tan \alpha)}{\tan \beta}
\end{aligned}$
Asked in: AP EAMCET 2008