The angle of elevation of the top of a tower $30 \, m$ high from the foot of another tower in the same plane is $60^{\circ}$ and the angle of elevation of the top of the second tower from the foot of the first tower is $30^{\circ}$. Find the distance between the two towers and also the height of the other tower.

(A) Let the distance between the two towers be $AB = x \, m$ and the height of the second tower be $PA = h \, m$.
Given: Height of the first tower $QB = 30 \, m$,$\angle QAB = 60^{\circ}$,and $\angle PBA = 30^{\circ}$.
In $\triangle QAB$,$\tan 60^{\circ} = \frac{QB}{AB} = \frac{30}{x}$.
Since $\tan 60^{\circ} = \sqrt{3}$,we have $\sqrt{3} = \frac{30}{x}$,which gives $x = \frac{30}{\sqrt{3}} = 10\sqrt{3} \, m$.
In $\triangle PBA$,$\tan 30^{\circ} = \frac{PA}{AB} = \frac{h}{x}$.
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$,we have $\frac{1}{\sqrt{3}} = \frac{h}{10\sqrt{3}}$.
Solving for $h$,we get $h = \frac{10\sqrt{3}}{\sqrt{3}} = 10 \, m$.
Thus,the distance between the towers is $10\sqrt{3} \, m$ and the height of the second tower is $10 \, m$.