From the top of a $7 \; m$ high building, the angle of elevation of the top of a tower is $60^\circ$ and the angle of depression of the foot of the tower is $30^\circ$. Find the height of the tower.
$O_1B =$ Building of height $7 \; m$
$O_2T =$ Tower
Draw $BP \perp O_2T$
From the figure, $\;$ $O_2P = O_1B = 7 \; m$ and $BP = O_1 O_2$
In $\triangle O_1BO_2$, $\;$ $\dfrac{O_1 B}{O_1 O_2} = \tan 30^\circ = \dfrac{1}{\sqrt{3}}$
$\implies$ $O_1 O_2 = \sqrt{3} \times O_1 B = 7 \sqrt{3} \; m$
$\therefore \;$ $BP = 7 \sqrt{3} \; m$
In $\triangle BPT$, $\;$ $\dfrac{PT}{BP} = \tan 60^\circ = \sqrt{3}$
$\implies$ $PT = BP \sqrt{3} = 7 \sqrt{3} \times \sqrt{3} = 21 \; m$
$\therefore \;$ Height of tower $= O_2P + PT = 7 + 21 = 28 \; m$