From an aeroplane vertically above a horizontal road- the angles of depression of two friends standing on either side of the aeroplane are observed to be- - - - The distance between the two friends is 1 m- The height in metres of the aeroplane from the ground isA- Tan - Tan -TB- 1-Ta nC- 1-Tan-D- Tan - T an -

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Angle of Elevation and Angle of Depression

From an aerop...

Question

From an aeroplane vertically above a horizontal road, the angles of depression of two friends standing on either side of the aeroplane are observed to be $α & β$ . The distance between the two friends is 1 m. The height (in metres) of the aeroplane from the ground is

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

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

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

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

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Solution

The correct option is C
$\frac{T a n α . T a n β}{T a n α + T a n β}$

Let H be the height at which the aeroplane is flying.

In triangle ABD

$t a n β = \frac{A D}{B D} = \frac{H}{B D} \Rightarrow B D = \frac{H}{t a n β} (i)$

In $△$ ADC

$t a n α = \frac{A D}{D C} = \frac{H}{D C} D C = \frac{H}{t a n α} (i i)$

Now, BD + DC = BC = 1 m

$\frac{H}{t a n β} + \frac{H}{t a n α}$ = 1

$H [\frac{1}{t a n β} + \frac{1}{t a n α}] = 1 H [\frac{t a n α + t a n β}{t a n α . t a n β}] = 1 H = \frac{t a n α . t a n β}{t a n α + t a n β}$

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From an aeroplane vertically above a horizontal road, the angles of depression of two friends standing on either side of the aeroplane are observed to be α & β. The distance between the two friends is 1 m. The height (in metres) of the aeroplane from the ground is

From an aeroplane vertically above a horizontal road, the angles of depression of two friends standing on either side of the aeroplane are observed to be $α & β$ . The distance between the two friends is 1 m. The height (in metres) of the aeroplane from the ground is