# From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 100 m high, the distance of point P from the foot of the tower is.

A. 149 m B. 156 m C. 173 m D. 200 m Answer: Option C
$\begin{array}{l}\text{Let AB be the tower.}\\ \text{Then,}\ \angle\text{APB}=30^{\circ}\\ \text{and AB}=100\ \text{m}\\ \frac{\mathrm{\text{AB}}}{\mathrm{\text{AP}}}=\tan30^{\circ}=\frac{\text{1}}{\sqrt{\text{3}}}\\ \Rightarrow\mathrm{AP}=(\mathrm{AB}\times\sqrt{3})\mathrm{m}\\ =100\sqrt{3}\mathrm{~m}\\ =(100\times1.73)\mathrm{m}\\ =173\mathrm{~m}.\end{array}$