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 .Find the distance travelled by the balloon during the interval.

Asked by Pragya Singh | 1 year ago |  173

Solution :-

Given,

Step 1: In right ΔBEC,

tan 30° = $$\dfrac{BE}{CE}$$

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

CE = $$87\sqrt{3}$$

Step 2:

tan 60° = $$\dfrac{AD}{CD}$$

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

CD = $$87\sqrt{3}$$ $$= 29\sqrt{3}$$

Step 3:

DE = CE – CD = ($$87\sqrt{3}$$ $$-29\sqrt{3}$$

$$29\sqrt{3}(3 – 1) = 58\sqrt{3}$$

Distance travelled by the balloon = $$58\sqrt{3}\;m$$.

Answered by Pragya Singh | 1 year ago

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