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 60m high, find the height of the building.
AB: Building of height 'h'
OT: Tower of height 60m
From the figure, in triangle ATO,
$\dfrac{TO}{AO} = \tan 60^{\circ} = \sqrt{3}$
$\therefore$ $AO = \dfrac{TO}{\sqrt{3}} = \dfrac{60}{\sqrt{3}} = 20\sqrt{3}$
In triangle BAO,
$\dfrac{BA}{OA} = \tan 30^{\circ} = \dfrac{1}{\sqrt{3}}$
$\therefore$ $BA = h = \dfrac{OA}{\sqrt{3}} = \dfrac{20\sqrt{3}}{\sqrt{3}} = 20$
$\therefore$ The height of the building is 20m.