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

Question:

From a light house the angles of depression of two ships on opposite sides of the light house are observed to be 30° and 45°. If the height of the light house is h metres, the distance between the ships is

(a) $(\sqrt{3}+1)$ h metres

 

(b) $(\sqrt{3}-1)$ hetres

(c) $\sqrt{3} h$ metres

(d) $1+\left(1+\frac{1}{\sqrt{3}}\right)$ h metres

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 $a$.

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