From the base of a pole of height 20 meter, the angle of elevation...

Plotting the diagram we get,

In $∆ ABD$

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

In $∆ AEC$

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

From equation $( i ) \text{ and } ( ii ) , \dfrac{( i )}{( ii )}$ we get, $\dfrac{\tan 60 °}{\tan 30 °} = \dfrac{20 + y}{y} ⇒ \dfrac{\sqrt{3}}{\sqrt{1 / 3}} = \dfrac{20 + y}{y}$

$⇒ \dfrac{20 + y}{y} = 3 ⇒ y = 10$

So, height of tower $= 20 + y = 30m$