Question

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 60º. After some time, the angle of elevation reduces to 30º. Find the distance travelled by the balloon during the interval.

Answer 1

Answer:

1905.84 is the answer i can think

Answer 2

Answer:

58√3 m.

Step-by-step explanation:

If the position of the first balloon is A and of the second or the final balloon is B then we will get that.

Again we get that the height of the balloon which is above the height of the girl = 88.2 m - 1.2 m = 87 m .

So, the distance which is traveled by the balloon can be written by:-

If we take the two right angled triangles ACD of 60 degree and BCE of angle 30 degree.

So, DE = CE - CD  .

The value of tan 30° = BE/CE .

1/√3= 87/CE .

CE = 87√3 m .

Also, the value of tan 60° = AD/CD .

√3= 87/CD

CD = 87/√3 m = 29√3 m .

So, finally the distance which is traveled by balloon will be =  DE = CE - CD

= (87√3 - 29√3) m = 29√3(3 - 1) m which on solving we will get 58√3 m.