The angle of depression of the top and bottom of a $20 \; m$ tall building from the top of a multi-storied building are $30^\circ$ and $60^\circ$ respectively. Find the height of the multi-storied building and the distance between two buildings.
$AB =$ Multi-storied building
$CD =$ Building of height $20 \; m$
$AC =$ Distance between the two buildings
Draw $DP \perp AB$
Then $\;$ $AP = CD = 20$ and $PD = AC$
In $\triangle BPD$, $\;$ $\dfrac{PB}{PD} = \tan 30^\circ$
$\implies$ $PB = PD \; \tan 30^\circ = \dfrac{PD}{\sqrt{3}} = \dfrac{AC}{\sqrt{3}}$ $\;\;\; \cdots \; (1)$
In $\triangle BAC$, $\;$ $\dfrac{AB}{AC} = \tan 60^\circ$
$\implies$ $AB = AC \; \tan 60^\circ = \sqrt{3} \; AC$ $\;\;\; \cdots \; (2)$
But $\;$ $AB = AP + PB$ $\;\;\; \cdots \; (3)$
$\therefore \;$ In view of equations $(1)$ and $(2)$ equation $(3)$ becomes,
$\sqrt{3} \; AC = 20 + \dfrac{AC}{\sqrt{3}}$
i.e. $\;$ $AC \left(\sqrt{3} - \dfrac{1}{\sqrt{3}}\right) = 20$
i.e. $\;$ $\dfrac{2 \; AC}{\sqrt{3}} = 20$ $\implies$ $AC = 10 \sqrt{3}$
Substituting the value of $AC$ in equation $(2)$ we have,
$AB = 10 \sqrt{3} \times \sqrt{3} = 30$
$\therefore \;$ Height of the multi-storied building $= 30 \; m$
Distance between the two buildings $= 10 \sqrt{3} \; m$