As observed from the top of a $100 \; m$ high light house 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 light house, find the distance between the two ships. [Given: $\sqrt{3} = 1.732$]
$OL$: Light house
$S_1, \; S_2$: Two ships, exactly one behind the other, on the same side of the light house
Height of light house $= OL = 100 \; m$
Distance between the two ships $= S_1S_2$
In $\triangle OLS_1$, $\;$ $\dfrac{OL}{OS_1} = \tan 45^\circ = 1$
$\implies$ $OS_1 = OL = 100 \; m$
In $\triangle OLS_2$, $\;$ $\dfrac{OL}{OS_2} = \tan30^\circ = \dfrac{1}{\sqrt{3}}$
$\implies$ $OS_2 = OL \times \sqrt{3} = 100 \times 1.732 = 173.2 \; m$
From the figure, $\;$ $S_1 S_2 = OS_2 - OS_1 = 173.2 - 100 = 73.2 \;m$
$\therefore \;$ Distance between the two ships $= 73.2 \; m$