Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 7 - Review Exercises - Page 345: 42

Answer

$29;316.4^{\circ}$

Work Step by Step

The magnitude of a vector $\textbf{u}=\langle a,b \rangle$ is given as $|\textbf{u}|=\sqrt (a^{2}+b^{2})$. Since $\textbf{u}=\langle 21,-20 \rangle$, the magnitude is: $|\textbf{u}|=\sqrt ((21)^{2}+(-20)^{2})=\sqrt (441+400)=\sqrt (841)=29$ The direction angle $\theta$ can be found through the equation $\tan\theta=\frac{b}{a}$. Substituting the values of $a$ and $b$ in the formula and solving using a calculator, $\theta=\tan^{-1} (\frac{-20}{21})=-43.6^{\circ}$ The vector has a positive horizontal component and a negative vertical component which places it in the fourth quadrant. Since the direction angle is supposed to be the positive angle between the x-axis and the position vector, we need to add $360^{\circ}$ to $-43.6^{\circ}$ to yield the direction angle $\theta$. Therefore, the direction angle $\theta=316.4^{\circ}$.
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