Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 8 - Application of Trigonometry - Chapter 8 Test Prep - Review Exercises - Page 840: 41

Answer

$$15;126.9^\circ $$

Work Step by Step

$$\eqalign{ & {\text{Let }}{\bf{u}} = \left\langle { - 9,12} \right\rangle \cr & \cr & {\text{Calculate the magnitude, recall that }}\left| {\left\langle {a,b} \right\rangle } \right| = \sqrt {{a^2} + {b^2}} \cr & \left| {\bf{u}} \right| = \left| {\left\langle { - 9,12} \right\rangle } \right| = \sqrt {{{\left( { - 9} \right)}^2} + {{\left( {12} \right)}^2}} \cr & \left| {\bf{u}} \right| = \sqrt {81 + 144} = \sqrt {225} \cr & \left| {\bf{u}} \right| = 15 \cr & \cr & {\text{The direction of the angle is given by }}\theta = {\tan ^{ - 1}}\left( {\frac{b}{a}} \right) \cr & \theta = {\tan ^{ - 1}}\left( {\frac{{12}}{{ - 9}}} \right) + 180^\circ \cr & \theta \approx 126.9^\circ \cr} $$
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.