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. 5 metres
• B. 8 metres
• C. 10 metres
• D. 12 metres

Answer 1

The correct option is C

10 metres

Consider the diagram shown above. AC represents the tower and DE represents the pole.
Given that $AC=15m$, $\angle ADB={30}^{\circ }$, $\angle AEC={60}^{\circ }$.
Let $DE=h$
Then, $BC=DE=h$,
$AB=(15-h)$ [$\because$ $AC=15$ and $BC=h$),
$BD=CE$
$\tan {60}^{\circ }=\frac{AC}{CE}$
$\Rightarrow \sqrt{3}=\frac{15}{CE}$
$\Rightarrow CE=\frac{15}{\sqrt{3}}$ $\dots \left(1\right)$
$\tan {30}^{\circ }=\frac{AB}{BD}$
$\frac{1}{\sqrt{3}}=\frac{15-h}{BD}$
$\frac{1}{\sqrt{3}}=\frac{15-h}{\frac{15}{\sqrt{3}}}$[$\because$ $BD=CE$ and substituted the value of CE from (1)]
$\Rightarrow (15-h)=\frac{1}{\sqrt{3}}\times \frac{15}{\sqrt{3}}=\frac{15}{3}=5$
$\Rightarrow h=15-5=10m$
i.e., height of the electric pole = $10m$