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

Question:

The angles of depression of two ships from the top of a light house are 45° and 30° towards east. If the ships are 100 m apart. the height of the light house is

(a) $\frac{50}{\sqrt{3+1}} m$

 

(b) $\frac{50}{\sqrt{3-1}} m$

(c) $50(\sqrt{3}-1) m$

 

(d) $50(\sqrt{3}+1) m$

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 $d$.

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