From a light house the angles of depression of two ships on opposite sides

Solution:

Let the height of the light house AB be  meters

Given that: angle of depression of ship are  and.

Distance of the ship C =  and distance of the ship D =

Here, we have to find distance between the ships.

So we use trigonometric ratios.

In a triangle,

$\Rightarrow \tan C=\frac{A B}{B C}$

$\Rightarrow \tan 30^{\circ}=\frac{h}{x}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x}$

$\Rightarrow x=\sqrt{3} h$

Again in a triangle ABD,

$\tan D=\frac{A B}{B D}$

$\Rightarrow \tan 45^{\circ}=\frac{h}{y}$

$\Rightarrow \mathrm{I}=\frac{h}{y}$

$\Rightarrow y=h$

Now, distance between the ships $=x+y=\sqrt{3} h+h=(\sqrt{3}+1) h$

Hence the correct option is $a$.