From the base of a pole of height 20 meter, the angle of elevation of the top of a tower is 60 ° . The…

Plotting the diagram we get,

In $∆ABD$

$\tan 60°=\frac{20+y}{x} ...\left(\mathrm{i}\right)$

In $∆AEC$

$\tan 30°=\frac{y}{x} ...\left(\mathrm{ii}\right)$

From equation $\left(\mathrm{i}\right) \& \left(\mathrm{ii}\right),\frac{\left(\mathrm{i}\right) }{\left(\mathrm{ii}\right) }$ we get, $\frac{\tan 60°}{\tan 30°}=\frac{20+y}{y}\Rightarrow \frac{\sqrt{3}}{\sqrt{1/3}}=\frac{20+y}{y}$

$\Rightarrow \frac{20+y}{y}=3\Rightarrow y=10$

So, height of tower $=20+y=30 \mathrm{m}$