Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 9 - Polar Coordinates; Vectors - 9.4 Vectors - 9.4 Assess Your Understanding - Page 607: 67



Work Step by Step

If $v=ai+bj$ then by definition $tan(\alpha)=\frac{b}{a}$, hence $\alpha=tan^{-1}(\frac{b}{a})$. Also $0^\circ\lt\alpha\lt360^\circ.$ Therefore here $\alpha=tan^{-1}(\frac{3}{-3\sqrt3})=tan^{-1}(-\frac{\sqrt{3}}{3})=150^\circ.$
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