The angle of elevation of the top of a building from the foot of the tower is $30^{\circ}$ and the angle of elevation of the top of the tower from the foot of the building is $60^{\circ}$. If the tower is $50\, m$ high,find the height of the building.

(N/A) Let $AB$ be the building and $CD$ be the tower.
In $\triangle CDB$,
$\frac{CD}{BD} = \tan 60^{\circ}$
$\frac{50}{BD} = \sqrt{3}$
$BD = \frac{50}{\sqrt{3}}$
In $\triangle ABD$,
$\frac{AB}{BD} = \tan 30^{\circ}$
$AB = BD \times \tan 30^{\circ} = \frac{50}{\sqrt{3}} \times \frac{1}{\sqrt{3}} = \frac{50}{3} = 16 \frac{2}{3} \, m$
Therefore,the height of the building is $16 \frac{2}{3} \, m$.