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The height of a hill is $3, 300$ meters. From the point $D$ on the ground the angle of elevation of the top of the hill is 60 $°$ . A balloon is moving with constant speed vertically upwards from $D .$ After $5$ minutes of its movement a person sitting in it observes the angle of elevation of the top of the hill as $30 °$ . The speed of the balloon is

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2.64

km/hr

26.4

km/hr

22.4

km/hr

2.24

km/hr

answer is B.

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Detailed Solution

It is given that the height of a hill =

3, 300

meters. The point

D

on the ground makes the angle of elevation

60 °

with the top of the hill. A man in the balloon, which is moving with a constant speed vertically upwards from D observed the angle of elevation of the top of the hill as

30 °

after 5 minutes.
According to the given data the figure has been drawn below.
Question Image

We can see that,

BCDE

is a rectangle.
So,

BE = CD

and

C = ED

.
In

△ AED

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

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

\Rightarrow \sqrt{3} = \frac{3300}{ED}

∵ tan 60 ° = 3

\Rightarrow ED = \frac{3300}{\sqrt{3}}

\Rightarrow ED = 1100 \sqrt{3}

ED = BC = 1100 \sqrt{3}

….(1)
In

△ ABC

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

\tan 30^{o} = \frac{AB}{BC}

\Rightarrow AB = BCtan 30^{o}

Substitute the value of BC from equation 1 and

tan 30 ° = 13

\Rightarrow AB = 1100 \sqrt{3} \times \frac{1}{\sqrt{3}}

\Rightarrow AB = 1100

Distance covered by balloon is

\Rightarrow AE - AB

\Rightarrow 3300 - 1100

\Rightarrow 2200 m

\Rightarrow 2.2 km

So, speed will be

\Rightarrow \frac{Distance}{Time}

\Rightarrow \frac{2.2 km}{\frac{5}{60} hr}

\Rightarrow 26.4 km / hr

So the correct option is 2.

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