Question

An observer flying in an aeroplane, at an altitude of 900 m, observes two ships in front of him, which are in the same direction at an angles of depression of ${60}^o$ and ${30}^o$ respectively. Find the distance between the two ships.

Answer 1

Based upon the given information, we can draw the above shown diagram.

Here,

$\angle OBP={30}^{\circ },\angle OAP={60}^{\circ },OP=900m$

To find: $AB$

Solution:

We know that, $\tan \theta =\frac{P}{B}$

$\therefore$In $△OPB,$

$\tan {30}^{\circ }=\frac{900m}{PB}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{900}{PB}$

$\Rightarrow PB=900\times \sqrt{3}m$

$\therefore PB=900\sqrt{3}m$

Now, in $△OAB,$

$\tan {60}^{\circ }=\frac{900m}{PA}$

$\Rightarrow \sqrt{3}=\frac{900}{PA}$

$\Rightarrow PA=\frac{900}{\sqrt{3}}m=\frac{900\sqrt{3}}{3}m$

$\therefore PA=300\sqrt{3}m$

$AB=PB-PA=900\sqrt{3}-300\sqrt{3}=600\sqrt{3}m$

$\therefore AB=600\sqrt{3}m$

Hence, the distance between the two ships is $600\sqrt{3}m.$