# If the angle of elevation of a tower from a distance of 100 metres from its foot is 60?, the height of the tower is.

A. $100 \sqrt{3} m$ B. $\frac{100}{\sqrt{3}} m$ C. $50 \sqrt{3} m$ D. $\frac{200}{\sqrt{3}} m$ Answer: Option A
Let AB be the tower and a point P at a distance of 100 m from its foot, angle of elevation of the top of the tower is 60° Let height of the tower = h $\begin{array}{l}\text{Then in }\text{right }\triangle\text{ABP}\\ \begin{array}{l}\tan\theta=\frac{\text{ Perpendicular }}{\text{ Base }}=\frac{\text{AB}}{\text{PB}}\\ \Rightarrow\tan60^{\circ}=\frac{\text{h}}{100}\Rightarrow\sqrt{3}=\frac{\text{h}}{100}\\ \Rightarrow h=100\sqrt{3}\\ \therefore\text{ Height of tower }=100\sqrt{3}\end{array}\end{array}$