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

Let the distance between the buildings be $AC$ and let the upper excess height above the shorter building be $PD$. From the $45^{\circ}$ depression to the bottom, $\tan 45^{\circ}=\frac{PC}{AC}=1$, so $PC=AC$. From the $30^{\circ}$ depression to the top, $\tan 30^{\circ}=\frac{PD}{BD}=\frac{1}{\sqrt{3}}$, hence $BD=PD\sqrt{3}$. Since $BD=AC$ and $PC=PD+8$, we get $PD+8=PD\sqrt{3}$, so $PD=4(\sqrt{3}+1)$. Therefore $AC=PC=PD+8=4(3+\sqrt{3})\mathrm{m}$.