Question

A 1.2m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2m from the ground. The angle of elevation of the balloon from the eyes of the girl at that instant is 60°. After sometime, the angle of elevation reduces to 30° find the distance traveled by the balloon during the interval

Answer 1

Given that 1.2 m tall girl sees a balloon
So,AG=1.2 m
Also,AG&BF are parallel
Bf=AG=1.2 m
And the balloon is at a height of 88.2 m
So,EF=88.2

Girl sees balloon first at 60°
So,angle DAC=30°

Distance travelled by the balloon=BC
Now,
BF=EF-BF
BF=88.2-1.2m
BE=87m
Also,BE=DC=87m

here,
angleABE=90°&angle=90°

in right angle triangle EBA

tanA=BF/AB
√3=87/AB
AB=87/√3 m
in right angle triangle DAC
tanA=CD/AC
tan30°=CD/AC
1/√3=87/AC
AC=87√3m

Now,
AC=AB+BC
87√3=87/√3+BC
87√3-87/√3=BC
BC=87√3-87/√3
BC=87(√3-1/√3)
BC=87(√3√3-1/√3)
BC=87(2/√3)
BC=87×2/√3
BC=58√3