Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 13, Trigonometric Ratios and Functions - 13.4 Evaluate Inverse Trigonometric Functions - 13.4 Exercises - Quiz for Lessons 13.3-13.4 - Page 880: 15



Work Step by Step

Here, we have $\cos \theta =-0.72; 90^{\circ} \lt \theta \lt 180^{\circ}$ $\theta'=\cos^{-1} (0.72) \approx 136.05^{\circ}$ This angle $\theta'$ is in the $II$ quadrant. Then we have $ \cos 136.05^{\circ} \approx -0.719945 \approx -0.72^{\circ}$
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