From the top of a hill $h$ metres high the angles of depressions of the top and the bottom of a pillar are $\alpha$ and $\beta$ respectively. The height (in metres) of the pillar is

• Correct answer
  A

  $\frac{h(\tan \beta-\tan \alpha)}{\tan \beta}$
• B

  $\frac{h(\tan \alpha-\tan \beta)}{\tan \alpha}$
• C

  $\frac{h(\tan \beta+\tan \alpha)}{\tan \beta}$
• D

  $\frac{h(\tan \beta+\tan \alpha)}{\tan \alpha}$

Solution

Let $A B$ be a hill whose height is $h$ metres and $C D$ be a pillar of height $h^{\prime}$ metres.
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
$\begin{aligned}<br />& \tan \alpha=\frac{h-h^{\prime}}{\frac{h}{\tan \beta}} \\<br />& \Rightarrow \quad h \frac{\tan \alpha}{\tan \beta}=h-h^{\prime} \\<br />& \Rightarrow \quad h^{\prime}=\frac{h(\tan \beta-\tan \alpha)}{\tan \beta}<br />\end{aligned}$

Hence, the correct answer is option: 'A'.