Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 5 - Trigonometric Functions - Section 5.1 Angles and Their Measures - 5.1 Assess Your Understanding - Page 386: 82


$6.928 \text{ cm}$

Work Step by Step

$\because A = \dfrac{1}{2}r^2 \theta$ where $r$: Radius of the circle $\theta$: Central angle that subtends the arc (in radians) $A$: Area of the sector of the circle formed by the central angle $\theta$ Derive the formula for $r$: \begin{align*} A&=\frac{1}{2}r^2\theta\\\\ 2A&=r^2\theta\\\\ \dfrac{2A}{\theta}&=r^2\\\\ \pm\sqrt{\dfrac{2A}{\theta}}&=r\\ \end{align*} Since $r$ is a length, it has to be non-negative. So, $r= \sqrt{\dfrac{2 A}{\theta}}$ Using this formula gives: $r= \sqrt{\dfrac{2 A}{\theta}}$ $r = \sqrt{\dfrac{2 \times 6}{\frac{1}{4}}} =\sqrt{12 \cdot 4}= \sqrt{24} \approx \boxed{6.928 \text{ cm}}$
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