Question

An aeroplane at an altitude of $1500$ metres finds that two ships are sailing towards it in the same direction. The angles of depression as observed from the aeroplane are ${45}^{\circ }$ and ${30}^{\circ }$ respectively. Find thee distance (in $m$) between the two ships

Answer 1

Let $D$ is the aeroplane $A$ and $B$ are two ships.
Given: $CD=1500m,\angle EDA={30}^{\circ },\angle EDB={45}^{\circ }$
In $\triangle DBC\ ,

$\tan {45}^{\circ }=\frac{DC}{BC}$

$=\frac{1500}{BC}$
$1=\frac{1500}{BC}$

$\Rightarrow BC=1500m$
In $△ADC$,

$\tan {30}^{\circ }=\frac{DC}{AC}$
$\frac{1}{\sqrt{3}}=\frac{1500}{AB+BC}$
$AB+BC=1500\sqrt{3}$
$AB=1500\sqrt{3}-1500$
$AB=1500(\sqrt{3}-1)$
$=1500(1.73-1)=1500\times 0.73$
$=1095m$.
Hence, distance between two ships $=1095m$.