The angle of elevation of the top of a building from the foot of a 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 $60 \; m$ high, find the height of the building.
$T_1T_2 =$ Tower of height $60 \; m$
$B_1B_2 =$ Building
In $\triangle T_1T_2B_1$, $\;$ $\dfrac{T_1T_2}{T_1B_1} = \tan 60^\circ = \sqrt{3}$
$\implies$ $T_1B_1 = \dfrac{T_1T_2}{\sqrt{3}} = \dfrac{60}{\sqrt{3}}$
In $\triangle B_1B_2T_1$, $\;$ $\dfrac{B_1B_2}{T_1B_1} = \tan 30^\circ = \dfrac{1}{\sqrt{3}}$
$\implies$ $B_1B_2 = \dfrac{T_1B_1}{\sqrt{3}} = \dfrac{60}{\sqrt{3}} \times \dfrac{1}{\sqrt{3}} = 20$
$\therefore \;$ Height of building $= 20 \; m$