A vertical tower stands on ground and is surmounted by a vertical flagpole of height 18 m. At a point on the ground, the angle of elevation of the bottom and the top of the flagpole are 30° and 60° respectively. What is the height of the tower?

Answer: Option A

Let DC be the vertical tower and AD be the vertical flagpole. Let B be the point of observation.
Given that AD = 18 m, ∠ ABC = 60°, ∠ DBC = 30°
Let DC be h
$$\begin{array}{rl}& \tan 30^{\circ }=\frac{DC}{BC}\\ & \frac{1}{\sqrt{3}}=\frac{h}{BC}\\ & h=\frac{BC}{\sqrt{3}}......\left(eq:1\right)\\ & \tan 60^{\circ }=\frac{AC}{BC}\\ & \sqrt{3}=\frac{18+h}{BC}\\ & 18+h=BC\times \sqrt{3}......\left(eq:2\right)\\ & \frac{eq:1}{eq:2}\Rightarrow \frac{h}{18+h}=\frac{\left(\frac{BC}{\sqrt{3}}\right)}{\left(BC\times \sqrt{3}\right)}\\ & =\frac{1}{3}\\ & \Rightarrow 3h=18+h\\ & \Rightarrow 2h=18\\ & \Rightarrow h=9\text{m}\end{array}$$
i.e., the height of the tower = 9 m