Trigonometry (10th Edition)

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

Chapter 3 - Radian Measure and the Unit Circle - Section 3.3 The Unit Circle and Circular Functions - 3.3 Exercises - Page 119: 85f

Answer

$t=60^{\circ}$

Work Step by Step

Since October represents $x=9$, we substitute $x=9$ in the equation, $t=60−30\cos(\frac{\pi}{6}x)$ $t=60−30\cos(\frac{\pi}{6}\times9)$ $t=60−30\cos(\frac{9\pi}{6})$ $t=60−30(0)$ $t=60-0$ $t=60^{\circ}$
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