From the top of a light house, the angles of depression of two ships on

Question:

From the top of a light house, the angles of depression of two ships on the opposite sides of it are observed to be α and β. If the height of the light house be h metres and the line joining the ships passes through the foot of the light house, show that the distance between the ship is

$\frac{h(\tan \alpha+\tan \beta)}{\tan \alpha+\tan \beta}$ metres.

Solution:

Let  be the height of light house. And an angle of depression of the top of light house from two ships are and respectively. Let. And .

We have to find distance between the ships

We have the corresponding figure as follows

We use trigonometric ratios.

In $\triangle A B C$

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

$\Rightarrow \quad \tan \alpha=\frac{h}{x}$

$\Rightarrow \quad x=\frac{h}{\tan \alpha}$

Again in $\triangle A D C$

$\Rightarrow \quad \tan \beta=\frac{A C}{C D}$

$\Rightarrow \quad \tan \beta=\frac{h}{y}$

$\Rightarrow \quad y=\frac{h}{\tan \beta}$

Now,

$\Rightarrow \quad B D=x+y$

$\Rightarrow \quad B D=\frac{h}{\tan \alpha}+\frac{h}{\tan \beta}$

$\Rightarrow \quad B D=\frac{h(\tan \alpha+\tan \beta)}{\tan \alpha \tan \beta}$Hence the distance between ships is $\frac{h(\tan \alpha+\tan \beta)}{\tan \alpha \tan \beta}$

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