Question

A tower subtends an angle $\alpha$ at a point A in the plane of its base and the angle of depression of the foot of the tower at a point b feet just above A is $\beta$. Then, the height of the tower is

• A. $b\tan \left(\alpha \right)cot\left(\beta \right)$
• B. $bcot\left(\alpha \right)\tan \left(\beta \right)$
• C. $bcot\left(\alpha \right)cot\left(\beta \right)$
• D. $b{\tan }^2\left(\alpha \right)cot\left(\beta \right)$

Answer 1

The correct option is A

$b\tan \left(\alpha \right)cot\left(\beta \right)$

Finding the height of the tower:

Given, A tower subtends an angle $\alpha$ at a point A in the plane of its base and the angle of depression of the foot of the tower at a point b feet just above A is $\beta$.

Assume ‘X’ is the distance from point ‘A’ to foot of the tower.

Let ‘h’ be the height of the tower $\mathrm{h}=\mathrm{PQ}$

In $△\mathrm{APQ},$

$\Rightarrow \mathrm{h}=\mathrm{xtan}\left(\mathrm{\alpha }\right).....\left(\mathrm{i}\right)$

In $△\mathrm{PRB}$

$\begin{array}{rcl}& \Rightarrow & \mathrm{b}=\mathrm{xtan}\left(\mathrm{\beta }\right)\\ & \Rightarrow & \mathrm{x}=\frac{\mathrm{b}}{\tan \left(\mathrm{\beta }\right)}\\ & \Rightarrow & x=bcot\left(\beta \right)...\left(\mathrm{ii}\right)\end{array}$

From equation $\left(i\right)\&\left(ii\right)$

$\Rightarrow \mathrm{h}=\mathrm{btan}\left(\mathrm{\alpha }\right)\cot \left(\mathrm{\beta }\right)$

Hence, correct option is $\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{.}$