Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 2 - Acute Angles and Right Triangles - Section 2.4 Solving Right Triangles - 2.4 Exercises: 44

Answer

The sun's diameter is approximately $864,900~miles$

Work Step by Step

We can convert the angle to radians: $\theta = 32' = (\frac{32}{60})^{\circ}(\frac{\pi~rad}{180^{\circ}}) = 0.0093084~rad$ The arc length is equal to $\theta~d$. Since $\theta$ is a small angle, the diameter of the sun is almost equal to the arc length. Therefore, we can use the arc length to approximate the sun's diameter. We can find the arc length $S$ to approximate the sun's diameter: $S = \theta~d$ $S = (0.0093084~rad)(92,919,800~mi)$ $S = 864,900~mi$ The sun's diameter is approximately $864,900~miles$
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