The top of a 15 metre high tower makes an angle of elevation of 60° with the bottom of an electric pole and angle of elevation of 30° with the top of the pole. What is the height of electric pole ?
A. 5 meters B. 8 meters C. 10 meters D. 12 meters Answer: Option CShow Answer
Solution(By Apex Team)
$\begin{array}{l}\text{Let AB be the tower and}\\
\text{CD be the electric pole.}\\
\text{Then,}\\
\angle\mathrm{ACB}=60^{\circ},\angle\mathrm{EDB}=30^{\circ}\\
\text{and AB }=15\ \text{m}\\
\text{Let CD}=\text{h}\\
\text{Then BE}=\left(\text{AB}-\text{AE}\right)\\
=(\mathrm{AB}-\mathrm{CD})=(15-\mathrm{h}) \\
\text { We have } \frac{\mathrm{AB}}{\mathrm{AC}}=\tan 60^{\circ}=\sqrt{3} \\
\Rightarrow \mathrm{AC}=\frac{\mathrm{AB}}{\sqrt{3}}=\frac{15}{\sqrt{3}} \\
\text { And } \frac{\mathrm{BE}}{\mathrm{DE}}=\tan 30^{\circ}=\frac{1}{\sqrt{3}} \\
\Rightarrow \mathrm{DE}=(\mathrm{BE} \times \sqrt{3})=\sqrt{3}(15-\mathrm{h}) \\
\Rightarrow 3 \mathrm{~h}=(45-15) \\
\Rightarrow \mathrm{h}=10 \mathrm{~m}
\end{array}$
