Trigonometry (10th Edition)

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

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.4 Vectors, Operations, and the Dot Product - 7.4 Exercises - Page 329: 35

Answer

$8;120^{\circ}$

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 $\textbf{u}=\langle -4,4\sqrt 3 \rangle$, the magnitude is: $|\textbf{u}|=\sqrt ((-4)^{2}+(4\sqrt 3)^{2})=\sqrt (16+48)=\sqrt (64)=8$ 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, $\theta=\tan^{-1} (\frac{4\sqrt 3}{-4})=-60^{\circ}$ The vector has a negative horizontal component and a positive vertical component which places it in the second quadrant. Since the direction angle is supposed to be the positive angle between the x-axis and the position vector, we need to add $180^{\circ}$ to $-60^{\circ}$ to yield the direction angle $\theta$. Therefore, the direction angle $\theta=120^{\circ}$.
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