From the top of a building $60 \; m$ high, angles of elevation and depression were observed to the top and base of a tower to be $30^\circ$ and $60^\circ$ respectively. Find the height of the tower, given both the structures are on the same level ground and opposite each other.
$AB = $ Building of height $60 \; m$
$OT = $ Tower
$AO = $ Distance between the building and the tower
From $B$ draw $BP \perp OT$
Then, $\;$ $AB = OP = 60 \; m$ $\;$ and $\;$ $AO= BP$
In $\triangle ABO$, $\;$ $\dfrac{AB}{AO} = \tan 60^\circ$
$\implies$ $AO = \dfrac{AB}{\tan 60^\circ} = \dfrac{60}{\sqrt{3}} = 20 \sqrt{3} \; m$
$\therefore \;$ $BP = AO = 20 \sqrt{3} \; m$
In $\triangle PTB$, $\;$ $\dfrac{PT}{BP} = \tan 30^\circ$
$\implies$ $PT = BP \times \tan 30^\circ = 20 \sqrt{3} \times \dfrac{1}{\sqrt{3}} = 20 \; m$
Height of tower $= OP + PT = 60 + 20 = 80 \; m$