From the top of a lighthouse, the angle of depression of two ships on the opposite sides of it are observed to be $30^\circ$ and $60^\circ$. If the height of the lighthouse is $h$ meter and the line joining the ships passes through the foot of the lighthouse, show that the distance between the ships is $\dfrac{4h}{\sqrt{3}} \; m$
$OL =$ Lighthouse of height $h$ meter
$S_1, \; S_2 =$ Position of the two ships on either side of the lighthouse
$S_1S_2 =$ Distance between the two ships
In $\triangle S_1OL$, $\;$ $\dfrac{OL}{S_1O} = \tan 30^\circ$
i.e. $\;$ $\dfrac{h}{S_1O} = \dfrac{1}{\sqrt{3}}$
$\implies$ $S_1O = h \sqrt{3}$
In $\triangle S_2OL$, $\;$ $\dfrac{OL}{OS_2} = \tan 60^\circ$
i.e. $\;$ $\dfrac{h}{OS_2} = \sqrt{3}$
$\implies$ $OS_2 = \dfrac{h}{\sqrt{3}}$
Now, $\;$ $S_1S_2 = S_O + OS_2 = h \sqrt{3} + \dfrac{h}{\sqrt{3}} = \dfrac{4h}{\sqrt{3}}$
$\therefore \;$ Distance between the two ships $= \dfrac{4h}{\sqrt{3}}$ meter