As observed from the top of a $75 \, m$ high lighthouse from the sea-level,the angles of depression of two ships are $30^{\circ}$ and $45^{\circ}$. If one ship is exactly behind the other on the same side of the lighthouse,find the distance between the two ships.

(N/A) Let $AB$ be the lighthouse and the two ships be at points $C$ and $D$ respectively.
In $\triangle ABC$,
$\frac{AB}{BC} = \tan 45^{\circ}$
$\frac{75}{BC} = 1$
$BC = 75 \, m$
In $\triangle ABD$,
$\frac{AB}{BD} = \tan 30^{\circ}$
$\frac{75}{BC + CD} = \frac{1}{\sqrt{3}}$
$\frac{75}{75 + CD} = \frac{1}{\sqrt{3}}$
$75\sqrt{3} = 75 + CD$
$CD = 75(\sqrt{3} - 1) \, m$
Therefore,the distance between the two ships is $75(\sqrt{3} - 1) \, m$.