The angles of depression of two ships from the top of a light house are 45°

Solution:

Let =  be the light house.

The given situation can be represented as,

It is clear that  and 

Again, let  and m is given.

Here, we have to find the height of light house.

So we use trigonometric ratios.

In a triangle,

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

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

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

$\Rightarrow h=x$

Again in a triangle ABD,,

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

$\Rightarrow \tan 30^{\circ}=\frac{h}{x+100}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+100}$

$\Rightarrow \sqrt{3} h=x+100$

Put $x=h$

$\Rightarrow \sqrt{3} h=h+100$

$\Rightarrow h(\sqrt{3}-1)=100$

$\Rightarrow h=\frac{100}{\sqrt{3}-1}$

$\Rightarrow h=\frac{100}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$

$\Rightarrow h=50(\sqrt{3}+1)$

Hence the correct option is $d$.