Answer
The angle of depression from the ballon is $\approx 8^{\circ}$.
Work Step by Step
Refer to the attachment:
$B =$ Hot air Balloon
$D = $ Stadium
$α =$ Angle of depression from the hot air balloon
$AB = 300$ ft
$BD = 2200$ ft
1. Find $\angle x$ in the triangle $\triangle ABD$ using the sine law
$\frac{sin(x)}{AB} = \frac{sin90}{BD}$
$\frac{sin(x)}{300} = \frac{sin90}{2200}$
$sin(x) = \frac{300sin90}{2200}$
$sin(x) = \frac{300}{2200} $
$sin(x) = \frac{300}{2200} $
$sin(x) = 0.1364... $
$x = sin^{-1}(0.1364...)$
$x = 7.84^{\circ}$
$x \approx 8^{\circ}$
2. Apply the concept of alternate angles
$x = α$ (Because of alternate angles)
Therefore, $α \approx 8^{\circ}$.
