Question

14. A 1.2 m tall girl spots a balloon moving
with the wind in a horizontal line at a
height of 88.2 m from the ground. The
angle of elevation of the balloon from
the eyes of the girl at any instant is
60°. After some time, the angle of
elevation reduces to 30° (see Fig. 9.13).
Find the distance travelled by the
balloon during the interval.
88.2 m
30°
Fig. 9.13​

Answer 1

Given :-

A 1.2 m tall girl spots a balloon moving  with the wind in a horizontal line at a  height of 88.2 m from the ground.

The  angle of elevation of the balloon from  the eyes of the girl at any instant is  60°.

After some time, the angle of  elevation reduces to 30°.

To Find :-

The distance travelled by the  balloon during the interval.

Solution :-

Let the initial position of the balloon be A and final position be B.

According to the question,

Height of balloon above the girl height = 88.2 – 1.2

= 87 m

Now, finding

Distance travelled by the balloon = DE = CE – CD

In right ΔBEC,

$\sf tan \ 30^{o}=\dfrac{BE}{CE}$

$\sf \dfrac{1}{\sqrt{3} } =\dfrac{87}{CE}$

$\sf CE=87\sqrt{3}$

In right ΔADC,

$\sf tan \ 60^{o}=\dfrac{AD}{CD}$

$\sf \sqrt{3} =\dfrac{87}{CD}$

$\sf CD=\dfrac{87}{\sqrt{3} }=29\sqrt{3}$

Next,

DE = CE – CD

$\sf (87\sqrt{3} -29\sqrt{3} )=29\sqrt{3} (3-1)$

$\sf =58\sqrt{3}$

Distance travelled by the balloon is 58√3 m