From the top of a light house, the angles of depression of two ships on the opposite sides of it are observed to be $ \alpha $ and $ \beta $. 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.
Given:
From the top of a light house, the angles of depression of two ships on the opposite sides of it are observed to be \( \alpha \) and \( \beta \).
The height of the light house be \( h \) metres and the line joining the ships passes through the foot of the light house.
To do:
We have to show that the distance between the ship is \( \frac{h(\tan \alpha+\tan \beta)}{\tan \alpha \tan \beta} \) metres.
Solution:
Let $AB$ be the light house and $C$ and $D$ be the two ships which make angles of elevation with $B$ are $\alpha$ and $\beta$ respectively.
From the figure,
$AB=h \mathrm{~m}$
Let $\mathrm{AC}=x\ m$ and $\mathrm{AD}=y\ m$.
In $\Delta \mathrm{BCA}$,
$\tan \alpha=\frac{\mathrm{AB}}{\mathrm{AC}}$
$=\frac{h}{x}$
$\Rightarrow x=\frac{h}{\tan \alpha}$..........(i)
Similarly,
In $\triangle \mathrm{BDA}$,
$\tan \beta=\frac{\mathrm{AB}}{\mathrm{AD}}$
$=\frac{h}{y}$
$\Rightarrow y=\frac{h}{\tan \beta}$.............(ii)
$\mathrm{CD}=x+y$
$=\frac{h}{\tan \alpha}+\frac{h}{\tan \beta}$
$=h(\frac{1}{\tan \alpha}+\frac{1}{\tan \beta}) \mathrm{m}$
$=h \frac{(\tan \beta+\tan \alpha)}{\tan \alpha \tan \beta} \mathrm{m}$
$=\frac{h(\tan \alpha+\tan \beta)}{\tan \alpha \tan \beta} \mathrm{m}$
Hence proved.
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