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From an airplane above a straight road the angles of depression of two positions at a distance 20 m apart on the road are observed to be $30 °$ and $45 °$ . The height of the airplane above the ground is

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10 \sqrt{3} m

10 (\sqrt{3} - 1) m

10 (\sqrt{3} + 1) m

20 m

answer is B.

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Detailed Solution

It is given that the angles of depression of two positions at a distance 20 m apart on the road are observed to be

30 °

and

45 °

from an airplane above a straight road .
According to the given data the figure has been drawn below.
Question Image

ΔABD,

tan θ = P e r p e n d i c u l a r b a s e

\frac{AD}{BD} = \tan 30 °

\frac{h}{20 - x} = \frac{1}{\sqrt{3}}

∵ tan 30 ° = 13

\sqrt{3} h = 20 - x

.......

(1)

and in

ΔADC,

tan θ = P e r p e n d i c u l a r b a s e

\frac{h}{x} = \tan 45 °

∵ tan 45 ° = 1

h = x

.......

(2)

From

(1)

and

(2)

, we get,

\sqrt{3} h = 20 - h

\Rightarrow \sqrt{3} h + h = 20

\Rightarrow h (\sqrt{3} + 1) = 20

\Rightarrow h = \frac{20}{\sqrt{3} + 1}

Rationalizing the denominator we get,

\Rightarrow h = 10 (\sqrt{3} - 1)

So height of the airplane is

10 (\sqrt{3} - 1)

m.
So the correct option is 2.

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