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An airplane flying horizontally $1$ km above the ground is observed at an elevation of $60 °$ . If after $10$ secs the elevation is observed to be $30 °$ . Then the uniform speed per hour the airplane is

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20 \sqrt{3} km / h

240 \sqrt{3} km / h

256 \sqrt{3} km / h

250 \sqrt{3} km / h

answer is B.

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Detailed Solution

It is given an airplane flying horizontally

1

km above the ground is observed at an elevation of

60 °

.
According to the given data, the figure has been drawn below.
Question Image

△ AED

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

∠AED

= 60 °

\tan 60^{o} = \frac{AD}{ED}

\Rightarrow

\sqrt{3} = \frac{1}{ED}

∵ tan 60 ° = 3

\Rightarrow ED = \frac{1}{\sqrt{3}}

Similarly, in

△ BEC

,
∠BEC= 30

°

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

\tan 30 ° = \frac{BC}{EC}

\Rightarrow \frac{1}{\sqrt{3}}

= \frac{1}{EC}

∵ tan 30 ° = 13

\Rightarrow EC = \sqrt{3}

Distance travelled by airplane in 10 seconds

\Rightarrow EC - ED

\Rightarrow \sqrt{3} - \frac{1}{\sqrt{3}}

\Rightarrow \frac{3 - 1}{\sqrt{3}}

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

We know,

Speed = \frac{Distance}{Time}

.
Here time is 10 sec, converting in hr we get
3600sec=1hr

\Rightarrow

10sec=

\frac{10}{3600}

hr
Substituting values in speed formula,

\Rightarrow Speed = \frac{\frac{2}{\sqrt{3}}}{\frac{10}{3600}}

\Rightarrow Speed = \frac{\frac{2}{\sqrt{3}}}{\frac{1}{360}}

\Rightarrow Speed = 240 \sqrt{3}

Therefore, speed is

240 \sqrt{3}

km/h.
So the correct option is 2.

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