A balloon rises to the ceiling of a gymnasium. You want to find the distance from
the ground to the balloon. You use a cardboard square to line up the balloon and
the ground. Your friend measures the vertical distance from the ground to your
eye and the distance from you to the gym wall. Approximate the distance from
the ground to the balloon.

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.

lockers 1, 4, 9, 16, 25, 40, and 49 are open

i ran through it by the other answer is wrong