Question

An aeroplane at an altitude of 1200 meters finds that two ships are sailing towards it in the same direction . The angle of depression of the ships as observed from the planner are ${60}^{\circ }$ and ${30}^{\circ }$ respectively. Find the distance between the two ships.

Answer 1

Let the aeroplane be at B and let the two ships be at C and D such that their angles of depression from B are ${60}^{\circ }$ and ${30}^{\circ }$ respectively.
We have AB = 1200 metres. Let AC = x and CD = y.
In $\Delta CAB$ , we have.
$tan{60}^{\circ }=\frac{AB}{CA}$
$\Rightarrow \sqrt{3}=\frac{1200}{x}$
$\Rightarrow x=\frac{1200}{\sqrt{3}}=400\sqrt{3}$
In $\Delta BAD$ , we have
$tan{30}^{\circ }=\frac{AB}{AD}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{1200}{x+y}$
$\Rightarrow x+y=1200\sqrt{3}$
$\Rightarrow y=1200\sqrt{3}-x$
$\Rightarrow y=1200\sqrt{3}-400\sqrt{3}=800\sqrt{3}=800\times 1.732=1385.6$
Hence, the distance between the two ships is 1385.6 metres.