From the top of a $\displaystyle 45$ m high light house, the angles of depression of two ships, on the opposite side of it, are observed to be $\displaystyle 30$° and $\displaystyle 60$°. If the line joining the ships passes through the foot of the light house, find the distance between the ships. (Use $\displaystyle \sqrt{3}=1.73$ )

Let AB be the light house and C and D be positions of ships. \[\begin{aligned} & \tan 60^{\circ}=\sqrt{3}=\frac{45}{y} \\ & \Rightarrow y=15 \sqrt{3} \\ & \tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{45}{x} \\ & \Rightarrow x=45 \sqrt{3} \end{aligned} \] Distance between two ships \(\displaystyle =\mathrm{x}+\mathrm{y}=60 \sqrt{3}\) \[=60 \times 1.73=103.8 \mathrm{~m} \]