Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Section 5.4 - Product-to-Sum and Sum-to-Product Formulas - Exercise Set - Page 691: 63


The arc length is $\frac{20\pi }{5}\text{ inches or }20.94\,\text{inches}$

Work Step by Step

We know that the formula which connects the arc length $s$, the angle intercepted by the arc to the center of the circle $\theta $, and the radius of the circle or arc $r$ is: $s=r\theta $ Where $\theta $ is expressed in radians. To convert the angle mentioned in degrees to radians, multiply the angle in degree with $\frac{\pi }{180}.$ In the provided problem, r is 8 inches and the angle that is intercepted by the arc is 150 degrees. When converting the angle in degree to radians: $\begin{align} & \theta =150{}^\circ \times \frac{\pi }{180}\text{ radians} \\ & =\frac{5\pi }{6}\text{ radians} \end{align}$ So, the length of the arc is: $\begin{align} & s=r\theta \\ & =8\times \frac{5\pi }{6} \\ & =\frac{20\pi }{3}\text{ inches}\text{.} \end{align}$ Round off the length value to two decimal places; $\frac{20\pi }{3}\text{ inches}\approx \text{20}\text{.94 inches}\text{.}$
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