Question

From the top of a lighthouse $120m$ high two ships on the same side of the lighthouse are observed. The angles of depression of the ships as seen from the lighthouse are found to be ${30}^o$ and ${60}^o$. Find the distance between the two ships (Assume that the two ships and bottom of the lighthouse are in a line).

Answer 1

Let say the height of the lighthouse $AB=120$ metres. The observer is at $A$.
The ship at $D$ makes an angle of depression ${30}^o$ and the ship at $C$ makes an angles of depression ${60}^o$.
Taking $DC=y$ and $CB=x$
In right angled $\Delta ACB$
$tan{60}^o=\frac{AB}{CB}$
$\therefore \sqrt{3}=\frac{120}{x}$
$\therefore x=\frac{120}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$ (Rationalizing)
$\therefore$ $=\frac{120\sqrt{3}}{3}=40\sqrt{3}m$
In right angled $\Delta ADB$
$\tan {30}^o=\frac{AB}{BD}$
$\frac{1}{\sqrt{3}}=\frac{120}{x+y}$
$x+y=120\sqrt{3}m$

Therefore, $Y=120\sqrt{3}-40\sqrt{3}=80\sqrt{3}m$