As observed from the top of a $80 \; m$ tall light house, the angles of depression of two ships on the same side of the lighthouse, in horizontal line with its base, are $30^\circ$ and $40^\circ$
Find the distance between the two ships, giving the answer correct to the nearest meter.
$OL: $ Light house of height $= 80 \; m$
$S_1, \; S_2: $ Two ships
$S_1 S_2 = $ Distance between the ships
In $\triangle OLS_1$, $\;$ $\dfrac{OL}{OS_1} = \tan 40^\circ$
i.e. $\;$ $OS_1 = \dfrac{OL}{\tan 40^\circ}$ $\;\;\; \cdots \; (1)$
In $\triangle PLS_2$, $\;$ $\dfrac{OL}{OS_2} = \tan 30^\circ$
i.e. $\;$ $OS_2 = \dfrac{OL}{\tan 30^\circ}$
i.e. $\;$ $OS_1 + S_1 S_2 = \dfrac{OL}{\tan 30^\circ}$ $\;\;\; \cdots \; (2)$
In view of equation $(1)$, equation $(2)$ becomes
$\dfrac{OL}{\tan 40^\circ} + S_1 S_2 = \dfrac{OL}{\tan 30^\circ}$
i.e. $\;$ $S_1 S_2 = OL \left[\dfrac{1}{\tan 30^\circ} - \dfrac{1}{\tan 40^\circ}\right]$
i.e. $\;$ $S_1 S_2 = 80 \left[\dfrac{1}{0.5774} - \dfrac{1}{0.8391}\right] = 43.21$
$\therefore \;$ Distance between the ships $= S_1 S_2 = 43 \; m$ (to the nearest meter)