Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 267: 109


44.7% of the equator can be seen from the satellite.

Work Step by Step

Let $S$ be the satellite's position. Let $C$ be the center of the Earth. Let $P$ be the point on the Earth's surface where the tangent line from the satellite meets the Earth's surface. Let $\theta = \angle SCP$ Note that $\angle SPC = 90^{\circ}$ From the diagram, we can see that a little less than 50% of the equator can be seen from the satellite. Note that the half of the equator on the opposite side of the Earth is not visible from the satellite. To find the exact percentage, we can find the angle $\theta$ Note that the Earth's radius is $\frac{7927}{2} = 3963.5~miles$ Also, the distance from the center of the Earth to the satellite is $(20,000+3963.5)~miles$ $cos~\theta = \frac{\overline{CP}}{\overline{SC}}$ $cos~\theta = \frac{3963.5}{20,000+3963.5}$ $\theta = arccos(\frac{3963.5}{23,693.5})$ $\theta = 80.48^{\circ}$ We can calculate the percentage of the equator that can be seen from the satellite: $\frac{2\theta}{360^{\circ}} = \frac{(2)(80.48^{\circ})}{360^{\circ}} = 0.447$ 44.7% of the equator can be seen from the satellite.
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