Elementary Geometry for College Students (5th Edition)

Published by Brooks Cole
ISBN 10: 1439047901
ISBN 13: 978-1-43904-790-3

Chapter 11 - Review Exercises: 24

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, $α \approx 8^{\circ}$.
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