From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of a 20m high building are $45^{\circ}$ and $60^{\circ}$ respectively. Find the height of the tower.
OB: Building of height 20m
BT: Transmission tower of height 'h'
C: Point of observation
From the figure, in triangle BOC,
$\dfrac{BO}{OC} = \tan 45^{\circ} = 1$ $\implies$ $BO = OC = 20$
In triangle TOC,
$\dfrac{TO}{OC} = \tan 60^{\circ} = \sqrt{3}$
$\implies$ $TO = OC\sqrt{3} = 20\sqrt{3}$
i.e. $h+20 = 20\sqrt{3}$
i.e. $h=14.64$
Therefore, height of the tower is 14.64m