An aeroplane at an altitude of 1500 m finds that two ships are sailing towards it in the same direction. The angles of depression as observed from the aeroplane are 45° and 30° respectively. Find the distance between the two ships.

From figure,

O is the position of aeroplane and P and Q are the position of ships.

OA = 1500 m

Let,

AQ = y

QP = x

In right angled ΔOAQ,

$\Rightarrow \tan \theta = \dfrac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 45^{\circ} = \dfrac{OA}{AQ} \\[1em] \Rightarrow 1 = \dfrac{1500}{y} \\[1em] \Rightarrow y = 1500 \text{ m.}$

In right angled ΔOAP,

$\Rightarrow \tan \theta = \dfrac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 30^{\circ} = \dfrac{OA}{AP} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{1500}{AQ + QP} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{1500}{y + x} \\[1em] \Rightarrow y + x = 1500\sqrt{3} \\[1em] \Rightarrow 1500 + x = 1500\sqrt{3} \\[1em] \Rightarrow x = 1500\sqrt{3} - 1500 \\[1em] \Rightarrow x = 1500(\sqrt{3} - 1) \\[1em] \Rightarrow x = 1500( 1.73 - 1) \\[1em] \Rightarrow x = 1500( 0.73) \\[1em] \Rightarrow x = 1095 \text{ m.}$

AQ = y = 1500 m

PQ = x = 1095 m

Hence, the distance between the two ships = 1095 m.