The angles of depression of the top and the bottom of an $8 \, m$ tall building from the top of a multi-storeyed building are $30^{\circ}$ and $45^{\circ}$,respectively. Find the height of the multi-storeyed building and the distance between the two buildings.

(N/A) Let $PC$ be the multi-storeyed building and $AB$ be the $8 \, m$ tall building. We need to find the height of the multi-storeyed building $(PC)$ and the distance between the two buildings $(AC)$.
From the figure,$PQ$ is the horizontal line from the top of the building $P$. Since $PQ \parallel BD$,the alternate interior angles are equal.
Therefore,$\angle PBD = 30^{\circ}$ and $\angle PAC = 45^{\circ}$.
In right-angled $\triangle PBD$:
$\tan 30^{\circ} = \frac{PD}{BD} \implies \frac{1}{\sqrt{3}} = \frac{PD}{BD} \implies BD = PD\sqrt{3}$.
In right-angled $\triangle PAC$:
$\tan 45^{\circ} = \frac{PC}{AC} \implies 1 = \frac{PC}{AC} \implies PC = AC$.
Since $AC = BD$ and $PC = PD + DC$,where $DC = AB = 8 \, m$:
$PD + 8 = AC = BD = PD\sqrt{3}$.
Solving for $PD$:
$PD\sqrt{3} - PD = 8 \implies PD(\sqrt{3} - 1) = 8 \implies PD = \frac{8}{\sqrt{3} - 1}$.
Rationalizing the denominator:
$PD = \frac{8(\sqrt{3} + 1)}{3 - 1} = \frac{8(\sqrt{3} + 1)}{2} = 4(\sqrt{3} + 1) \, m$.
Height of the multi-storeyed building $PC = PD + DC = 4\sqrt{3} + 4 + 8 = 4\sqrt{3} + 12 = 4(3 + \sqrt{3}) \, m$.
Distance between the buildings $AC = PC = 4(3 + \sqrt{3}) \, m$.