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The angles of elevation of an artificial earth satellite is measured from two earth stations, situated on the same side of the satellite, are found to be $30 °$ and $60 °$ . The two earth stations and the satellite are in the same vertical plane. If the distance between the earth stations is $4000$ km, find the distance between the satellite and earth. $(\sqrt{3} = 1.732)$

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3446

3464

3488

3400

answer is B.

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Detailed Solution

It is given that the angles of elevation of an artificial earth satellite is measured from two earth stations which are situated on the same side of the satellite are found to be

30 °

and 60

°

. The two earth stations and the satellite are in the same vertical plane and the distance between the stations is

4000

km.
Consider A and B are two earth stations.
Let AB

= 4000

km.
Let BC

= y

km and CD

= x

km, where CD is the distance of the satellite from the earth.
According to the given data the figure has been drawn below.
Question Image

From

△ ACD

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

\tan 30 ° = \frac{DC}{AC}

\Rightarrow \frac{x}{4000 + y} = \frac{1}{\sqrt{3}}

∵ tan 30 ° = 13

\Rightarrow \sqrt{3} x = 4000 + y

........(1)
From

△ BCD

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

\tan 60 ° = \frac{DC}{BC}

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

∵ tan 60 ° = 3

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

.......(2)
From (1) and (2)

\sqrt{3} x = 4000 + \frac{x}{\sqrt{3}}

\Rightarrow 3 x = 4000 \sqrt{3} + x

\Rightarrow 2 x = 4000 \sqrt{3}

\Rightarrow x = 2000 \sqrt{3} km

\Rightarrow x = 2000 \times 1.732

(∵ \sqrt{3} = 1.732)

∴

x = 3464 km

Hence, the distance of satellite from the earth is

3464 km .

So the correct option is 2.

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