The angle of elevation of the top of a tower from a certain point is $30^{\circ}$. If the observer moves $20 \, m$ towards the tower,the angle of elevation of the top increases by $15^{\circ}$. Find the height of the tower.

(C) Let the height of the tower be $h \, m$.
Let the tower be $PR$ where $R$ is the base and $P$ is the top.
Let the initial position of the observer be $Q$ and the new position be $S$.
Given $QS = 20 \, m$,$\angle PQR = 30^{\circ}$,and the angle of elevation increases by $15^{\circ}$ at $S$,so $\angle PSR = 30^{\circ} + 15^{\circ} = 45^{\circ}$.
In $\triangle PSR$,$\tan 45^{\circ} = \frac{PR}{SR} \implies 1 = \frac{h}{SR} \implies SR = h$.
In $\triangle PQR$,$\tan 30^{\circ} = \frac{PR}{QR} = \frac{h}{QS + SR} = \frac{h}{20 + h}$.
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$,we have $\frac{1}{\sqrt{3}} = \frac{h}{20 + h}$.
$20 + h = h\sqrt{3}$.
$20 = h(\sqrt{3} - 1)$.
$h = \frac{20}{\sqrt{3} - 1}$.
Rationalizing the denominator: $h = \frac{20(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{20(\sqrt{3} + 1)}{3 - 1} = \frac{20(\sqrt{3} + 1)}{2} = 10(\sqrt{3} + 1) \, m$.
Thus,the height of the tower is $10(\sqrt{3} + 1) \, m$.