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From an airplane flying vertically above a horizontal road, the angle of depression of two consecutive kilometer stones on the same side of the airplane are observed to be $30 °$ and $60 °$ , respectively. The height at which the airplane is flying (in km) is

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\frac{4}{\sqrt{3}}

\frac{\sqrt{3}}{2}

\frac{2}{\sqrt{3}}

2

answer is B.

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Detailed Solution

It is given that the angles of depression of two consecutive kilometer stones from an airplane which is flying vertically above a horizontal road, are observed to be

30 °

and

60 °

and the stones are on the same side of the airplane.
According to the given data the figure has been drawn below.
Question Image

Let the distance of two consecutive stones are

x

x + 1

and height of airplane be h.
In

ΔBCD

, we have

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

\tan 60 ° = \frac{BC}{DC}

\Rightarrow \sqrt{3} = \frac{h}{x}

(\tan 60 ° = \sqrt{3})

\Rightarrow x = \frac{h}{\sqrt{3}}

.... (i)
In

ΔABC

, we have

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

\tan 30 ° = \frac{h}{x + 1}

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

(\tan 30 ° = \frac{1}{\sqrt{3}})

\Rightarrow x + 1 = h \sqrt{3}

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

......[from equation (i)]

\Rightarrow \frac{2 h}{\sqrt{3}} = 1

\Rightarrow h = \frac{\sqrt{3}}{2} km

So the height at which the airplane is flying (in km) is

\frac{\sqrt{3}}{2} km

. So the correct option is 2.

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