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

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
$$\begin{array}{rl}& \text{Then in right }\Delta ABP\\ & \tan \theta =\frac{\text{Perpendicular}}{\text{Base}}=\frac{AB}{PB}\\ & \Rightarrow \tan 60^{\circ }=\frac{h}{100}\Rightarrow \sqrt{3}=\frac{h}{100}\\ & \Rightarrow h=100\sqrt{3}\\ & \therefore \text{Height}\text{of}\text{tower}=100\sqrt{3}\end{array}$$