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Q.

An aero plane flying horizontally 1 km above the ground is observed at an angle of $60^{\circ}$ . If after 10 seconds the elevation is observed to be $30^{\circ}$ , find the uniform speed of the aeroplane.

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a

415.68 km/hr.

b

315.1 km/hr.

c

550.68 km/hr.

d

640.68 km/hr.

answer is A.

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Detailed Solution

Suppose observation point = O
Initial location of aero plane= C
Location after 10 seconds of aero plane= E
F:New DownloadQ.31.jpg

F:New DownloadQ.31.jpg

Total height =1 km.

∠ AOC = 60^{\circ}

AC = BD = 1 km

∠ BOD = 30^{\circ}

In right angled triangle,

△ ABC

\tan 60^{\circ} = \frac{BC}{AB}

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

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

In right angled triangle,

△ ADE

,

\tan 30^{\circ} = \frac{ED}{AD}

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

\Rightarrow AD = \sqrt{3}

Now,

BD = AD - AB

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

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

BD = \frac{2}{\sqrt{3}}

So, in 10 seconds

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

distance covered by aero plane.
Time

t = 10 secs = \frac{10}{3600}

t = \frac{1}{360} hrs

Speed,

s = \frac{Distance}{Time}

s = \frac{\frac{2}{\sqrt{3}}}{\frac{1}{360}}

s = \frac{720 \sqrt{3}}{3}

s = 240 \times 1.732

s = 415.68 km / hr

Correct option is 1.

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