From an aeroplane vertically above a straight horizontal road, the angles of depression of two consecutive mile stones on opposite sides of the aeroplane are observed to be $ \alpha $ and $ \beta $. Show that the height in miles of aeroplane above the road is given by $ \frac{\tan \alpha \tan \beta}{\tan \alpha+\tan \beta} $.

Given:

From an aeroplane vertically above a straight horizontal road, the angles of depression of two consecutive mile stones on opposite sides of the aeroplane are observed to be \( \alpha \) and \( \beta \).

To do:

We have to show that the height in miles of aeroplane above the road is given by \( \frac{\tan \alpha \tan \beta}{\tan \alpha+\tan \beta} \).

Solution:

Let $B$ be the aeroplane and $C$ and $D$ be two points such that the angles of depression from $B$ are $\alpha$ and $\beta$ respectively.

$CD=1\ km$

Let the height of the plane be $h$, the distance between $C$ and $A$ be $x\ km$.

This implies,

$CA=x\ km$ and $AD=1-x\ km$

$\angle \mathrm{C}=\alpha, \angle \mathrm{D}=\beta$              (Alternate angles are equal)

In $\Delta \mathrm{ACB}$,

$\tan \alpha=\frac{\mathrm{AB}}{\mathrm{AC}}$

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

$\Rightarrow x=\frac{h}{\tan \alpha}$...........(i)

Similarly,

In $\triangle \mathrm{ABD}$,

$\tan \beta=\frac{\mathrm{AB}}{\mathrm{AD}}$

$\Rightarrow \tan \beta=\frac{h}{1-x}$

$\Rightarrow h=\tan \beta(1-x)$

$\Rightarrow h=\tan \beta-x \tan \beta$

$\Rightarrow h=\tan \beta-\frac{h}{\tan \alpha} \tan \beta$         [From (i)]

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

$\Rightarrow h(1+\frac{\tan \beta}{\tan \alpha})=\tan \beta$

$\Rightarrow h\frac{(\tan \alpha+\tan \beta)}{\tan \alpha}=\tan \beta$

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

Hence proved. 

Tutorialspoint

Updated on: 10-Oct-2022

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