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: 70



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 $\alpha=tan^{-1}(\frac{-4}{6})=tan^{-1}(\frac{-2}{3})=326.30^\circ.$ (because it must be in Quadrant IV).
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.