Question

The top of a $15m$ high tower makes an angle of depression of ${60}^{\circ }$ with the bottom of an electronic pole and angle of depression of ${30}^{\circ }$ with the top of the pole. What is the height of the electric pole?

• A. $10m$
• B. $5m$
• C. $15m$
• D. $20m$

Answer 1

The correct option is A $10m$


Consider the diagram shown above. AC represents the tower and DE represents the pole.
Consider a point B on AC such that, $BC=DE=h$ (height of the electric pole).

Given:
Height of the tower = $AC=15m$
Angle of depression to the top of the electric pole = $\angle ADB={30}^{\circ }$
Angle of depression to the bottom of the electric pole = $\angle AEC={60}^{\circ }$.

Let $DE=h$
Then, $BC=DE=h$,
$AB=(15-h)[\because$ $AC=15$ and $BC=h$]
Since distance between the elctric pole and tower remains constant, $BD=CE$.

In triangle ACE,
$\tan {60}^{\circ }=\frac{AC}{CE}$
We know that, $tan{60}^o=\sqrt{3}$
$\Rightarrow \sqrt{3}=\frac{15}{CE}$
$\Rightarrow CE=\frac{15}{\sqrt{3}}$ ------(i)

$\tan {30}^{\circ }=\frac{AB}{BD}$
We know that, $tan{30}^o=\frac{1}{\sqrt{3}}$
$\frac{1}{\sqrt{3}}=\frac{15-h}{BD}$
$\frac{1}{\sqrt{3}}=\frac{15-h}{\frac{15}{\sqrt{3}}}$ [$\because$ $BD=CE$ and substituted the value of CE from (1)]

$\Rightarrow (15-h)=\frac{1}{\sqrt{3}}\times \frac{15}{\sqrt{3}}=\frac{15}{3}=5$
$\Rightarrow h=15-5=10m$
Height of the electric pole = $10m$