The angles of depression of the top and bottom of a tower as seen from the top of a $60 \sqrt{3} \; m$ high cliff are $45^\circ$ and $60^\circ$ respectively. Find the height of the tower.
$OC =$ Cliff of height $60 \sqrt{3} \; m$
$TA =$ Tower
Draw $TP \perp OC$
Then, $\;$ $PO = TA$, $\;$ $PT = OA$
In $\triangle COA$, $\;$ $\dfrac{OC}{OA} = \tan 60^\circ = \sqrt{3}$
$\implies$ $OA = \dfrac{OC}{\sqrt{3}} = \dfrac{60 \sqrt{3}}{\sqrt{3}} = 60 \; m$
$\therefore \;$ $PT = 60 \; m$
In $\triangle CPT$, $\;$ $\dfrac{CP}{PT} = \tan 45^\circ = 1$
$\implies$ $CP = PT = 60 \; m$
Now, $\;$ $PO = OC - CP = 60 \sqrt{3} - 60 = \left(\sqrt{3} - 1\right) \times 60 = 0.732 \times 60 = 43.92$
$\implies$ $TA = 43.92$
$\therefore \;$ Height of tower $= 43.92 \; m$