The angle of elevation of the top of a tower from a point on the ground,which is $30\, m$ away from the foot of the tower,is $30^{\circ}$. Find the height of the tower.

(N/A) Let $AB$ be the tower and the angle of elevation from point $C$ (on the ground) is $30^{\circ}$.
In $\triangle ABC$,
$\frac{AB}{BC} = \tan 30^{\circ}$
$\frac{AB}{30} = \frac{1}{\sqrt{3}}$
$AB = \frac{30}{\sqrt{3}} = \frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{30\sqrt{3}}{3} = 10\sqrt{3}\, m$
Therefore,the height of the tower is $10\sqrt{3}\, m$.