From an aeroplane vertically above a straight horizontal road, the angles of depression of two stones on opposite sides of the aeroplane are observed to be $α$ and $β$ . The distance between the stones is 1m.Then the height in meters of the aeroplane above the road is

$\frac{t a n α . t a n β}{t a n β + t a n α}$

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$\frac{c o t α . t a n β}{c o t β + c o t α}$

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$\frac{t a n α . t a n β}{t a n β - t a n α}$

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$\frac{c o t α . t a n β}{c o t β - c o t α}$

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Solution

The correct option is A
$\frac{t a n α . t a n β}{t a n β + t a n α}$

Let h be the height of aeroplane.

In $Δ A B D, t a n α = \frac{h}{x}$
$\Rightarrow x = \frac{h}{t a n α} \dots (1)$

In $Δ A D C, t a n β = \frac{h}{1 - x}$
$\Rightarrow 1 - x = \frac{h}{t a n β} \Rightarrow x = 1 - \frac{h}{t a n β} \dots (2)$

From (1) and (2), we get
$\frac{h}{t a n α} = 1 - \frac{h}{t a n β} \Rightarrow h [\frac{1}{t a n α} + \frac{1}{t a n β}] = 1 \Rightarrow h [\frac{t a n β + t a n α}{t a n α . t a n β}] = 1 \Rightarrow h = \frac{t a n α . t a n β}{t a n α + t a n β}$

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From an aeroplane vertically above a straight horizontal road, the angles of depression of two stones on opposite sides of the aeroplane are observed to be α and β. The distance between the stones is 1m.Then the height in meters of the aeroplane above the road is

From an aeroplane vertically above a straight horizontal road, the angles of depression of two stones on opposite sides of the aeroplane are observed to be $α$ and $β$ . The distance between the stones is 1m.Then the height in meters of the aeroplane above the road is