An observed from the top of a 150 m tall light house, the angles of depression of two ships approaching it are 30° and 45°. If one ship is directly behind the other, find the distance between the two ships.
Let be the light house of m. and angle of depression of two ship C and D are and respectively.
Let, $B C=x, C D=y$ and $\angle A D B=30^{\circ}, \angle A C B=45^{\circ}$.
We use trigonometric ratios.
In a triangle,
$\Rightarrow \tan 45^{\circ}=\frac{A B}{B C}$
$\Rightarrow 1=\frac{150}{x}$
$\Rightarrow x=150$
Again in a triangle,
$\Rightarrow \tan 30^{\circ}=\frac{A B}{B D}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{150}{x+y}$
$\Rightarrow x+y=150 \sqrt{3}$
$\Rightarrow 150+y=150 \sqrt{3}$
$\Rightarrow y=150 \sqrt{3}-150$
$\Rightarrow y=150(\sqrt{3}-1)$
$\Rightarrow y=150 \times 0.732$
Hence distance between the ships is $109.8 \mathrm{~m}$.