Answer:
Solution given;
∆ABD & ∆ABC & ∆ACD are right angled triangle.
AC=8ft
BC=4ft.
BD=?
In ∆ ABC
Tan A=P/b
<BAC=Tan-¹ (4/8)
=26.565°
again
<CAD=90°-26.564=63.436°
In ∆ ACD
Tan <CAD=P/b
Tan (63.436°)=CD/8
CD=2×8=16
So
BD=4+16=20ft.
Therefore Approximate the distance from
Approximate the distance fromthe ground to the balloon is 20ft.