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An aeroplane flying horizontally at a height of $1.5$ km above the ground is observed at a certain point on earth to subtend an angle of 60 $° .$ After $15$ seconds, its angle of elevation is observed to be 30 $° .$ Calculate the speed of the, aeroplane in km/h.

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240 \sqrt{3}

km/h

230 \sqrt{3}

km/h

210

km/h

220

km/h

answer is A.

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Detailed Solution

Given, the height of the aero plane flying from the ground is 1.5km.
Angle subtended is 60

° .

We have to find the speed of the airplane in km/h.
Question Image

The angles of elevation of the plane in two positions

P

and

Q

∠ PAB = 60^{0}

and

∠ QAC = 30^{0}

PB = QC =

1.5

km.
We know that tan

θ = \frac{perpendicular}{base}

.
Now in right angled

△ ABP,

\tan 60^{0} = \frac{BP}{AB}

\Rightarrow \sqrt{3} = \frac{1.5}{AB}

\Rightarrow AB = \frac{1.5}{\sqrt{3}}

\Rightarrow (0.5) \sqrt{3}

km.
Again, in right angled triangle

△ ACQ

\tan 30^{0} = \frac{QC}{AC}

\Rightarrow \frac{1}{\sqrt{3}} = \frac{1.5}{AB}

\Rightarrow AC = (1.5) \sqrt{3}

PQ = BC = AC - AB

\Rightarrow PQ = (1.5) \sqrt{3} - (0.5) \sqrt{3}

\Rightarrow PQ = (1.5 - 0.5) \sqrt{3}

\Rightarrow PQ = \sqrt{3}

km
The plane travels

PQ =

\sqrt{3}

km in

15

secs
We know,
Speed of the airplane

=

\frac{distance}{time}

\Rightarrow speed = \frac{\sqrt{3}}{\frac{15}{3600}}

\Rightarrow speed = 240 \sqrt{3}

km/h.
Hence, the speed of the airplane is

240 \sqrt{3}

km/h.
Therefore, the correct option is 1.

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An aeroplane flying horizontally at a height of 1.5km above the ground is observed at a certain point on earth to subtend an angle of 60°. After 15seconds, its angle of elevation is observed to be 30°. Calculate the speed of the, aeroplane in km/h.

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Detailed Solution

Get Expert Academic Guidance – Connect with a Counselor Today!

An aeroplane flying horizontally at a height of $1.5$ km above the ground is observed at a certain point on earth to subtend an angle of 60 $° .$ After $15$ seconds, its angle of elevation is observed to be 30 $° .$ Calculate the speed of the, aeroplane in km/h.