A vertical tower stands on ground and is surrounded by a vertical flag pole of height  $18m$.  At a point on the ground, the angle of elevation of the bottom and the top of the flagpole are  $30°\text{\hspace{0.17em}and\hspace{0.17em}}60°$ respectively.  What is the height of the tower?  (in meters)

Let  $DC$ be the vertical tower and  $AD$ be the vertical flagpole

Let $B$ be the point of observation

Given that  $AD=18m,\angle ABC=60°,\angle DBC=30°$

Let  $DC$ be $\text{'}h\text{'}$,  $BC=x$

$\tan 30°=\frac{DC}{BC}$

$\frac{1}{\sqrt{3}}=\frac{h}{BC}$

$\frac{1}{\sqrt{3}}=\frac{h}{x}$

$x=\sqrt{3}h\to \left(1\right)$

$\tan 60°=\frac{AC}{BC}$

$\sqrt{3}=\frac{AD+DC}{BC}$

$\sqrt{3}=\frac{18+h}{x}\to \left(2\right)$

From (1) and (2)

$\sqrt{3}x=18+h$

$\Rightarrow \sqrt{3}\left(\sqrt{3}h\right)=18+h$

$\Rightarrow 3h=18+h$

$\Rightarrow 3h-h=18$

$\Rightarrow 2h=18$

$\Rightarrow h=\frac{18}{2}$

$h=9$

$\therefore$ height of the tower is 9 meters.