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: 42

Answer

$$29;316.4^\circ $$

Work Step by Step

$$\eqalign{ & {\text{Let }}{\bf{u}} = \left\langle {21, - 20} \right\rangle \cr & \cr & {\text{Calculate the magnitude, recall that }}\left| {\left\langle {a,b} \right\rangle } \right| = \sqrt {{a^2} + {b^2}} \cr & \left| {\bf{u}} \right| = \left| {\left\langle {21, - 202} \right\rangle } \right| = \sqrt {{{\left( {21} \right)}^2} + {{\left( { - 20} \right)}^2}} \cr & \left| {\bf{u}} \right| = \sqrt {441 + 400} = \sqrt {841} \cr & \left| {\bf{u}} \right| = 29 \cr & \cr & {\text{The direction of the angle is given by }}\theta = {\tan ^{ - 1}}\left( {\frac{b}{a}} \right) \cr & \theta = {\tan ^{ - 1}}\left( {\frac{{ - 20}}{{21}}} \right) + 360^\circ \cr & \theta \approx 316.4^\circ \cr} $$
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