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.

Let CD is 1.2m high girl and FG is the distance travelled by balloon.
In ΔACE,
AE/CE = tan 60°
⇒ (AF – EF)/CE = √3
⇒ (88.2 – 1.2)/CE = √3
⇒ 87/CE = √3 ⇒ CE = 87/√3 = 87√3/3 = 29√3
In ΔBCG,
BG/GC = tan 30° ⇒ 88.2 – 1.2/CG = 1/√3
⇒ 87/CG = 1/√3 ⇒ CG = 87√3
Distance travelled by balloon = EG = CG – CE = 87√3 – 29√3 = 58√3
Hence, the distance travelled by balloon is 58√3m.