The shadow of a tower is $\sqrt{3}$ times its height. Then the angle of elevation of the top of the tower is

AB = Tower = x units

$\mathrm{BC}=$ Shadow $=\sqrt{3} x$ units

$\quad \tan \mathrm{ACB}=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{x}{\sqrt{3} x}=\frac{1}{\sqrt{3}}=\tan 30^{\circ}$

$\therefore \angle \mathrm{ACB}=30^{\circ}$