Please... Slove this question with figure

Dear Student,

let the boy spot the balloon at point A. 

and the angle of elevation from his eye point P to the balloon is 60 deg.

PQ=1.3 m.

let after 2 seconds the balloon reaches to the point B. and  now its angle of elevation from the eye is 30 deg. 

QCF is the horizontal line at the ground level.

since speed of the wind =  29√3 m/s

the distance covered by the balloon in 2 sec be AB=29√3 x 2 = 58√3m.........(a)

in the right angled triangle APD,
tan 60=AD/PD
√3=AD/PD
PD=AD/√3 m------(1)

in the right angled triangle BEP,
tan 30=BE/PE
1/√3=AD/PE    [since BE=AD]
PE=√3AD m----(2)

PE=PD+DE  

PE=PD+AB  [since DE=AB]
PE=PD+58√3  [using (a)]

from (1) and (2), substitute the values in terms of AD.
√3AD=(1/√3)AD+58√3
3AD=AD+174
2AD=174
AD=87m.

therefore the height of the balloon from the ground level is AC=AD+DC

AC=AD+PQ

AC=87+1.3=88.3 m.

thus the height of the balloon from the ground is 88.3 m.

Regards!