An aeroplane at an altitude of 900 m finds that two ships are sailing towards it in the same direction. The angles of depression of the ships, as observed from the plane, are 60° and 30° respectively. Find the distance between the ships.

Let A be the position of the aeroplane and B be the point on the sea surface vertically below the plane.

Let C and D be the positions of the two ships.

Let AB be the altitude of the aeroplane = 900 m

BC = y meters and CD = x meters

In ∆ABC,

$\Rightarrow \tan 60^{\circ} = \dfrac{\text{perpendicular}}{\text{base}} = \dfrac{AB}{BC} \\[1em] \Rightarrow \sqrt3 = \dfrac{900}{y} \\[1em] \Rightarrow y = \dfrac{900}{\sqrt3} \\[1em] \Rightarrow y = \dfrac{900}{\sqrt3} \times \dfrac{\sqrt3}{\sqrt3} \\[1em] \Rightarrow y = 300\sqrt3 \text{ m …….(1)}$

In ∆ABD,

$\Rightarrow \tan 30^{\circ} = \dfrac{\text{perpendicular}}{\text{base}} = \dfrac{AB}{BD} \\[1em] \Rightarrow \dfrac{1}{\sqrt3} = \dfrac{900}{x + y} \\[1em] \Rightarrow x + y = 900\sqrt3 \text{ m ………….(2)}$

Subtracting equation (1) from (2), we get :

(x + y) - y = $900\sqrt3 - 300\sqrt3$

x = $600\sqrt3$

x = 600(1.732)

x = 1039.20 m.

Hence, the distance between the ships = 1039.20 m