Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 7 - Test - Page 353: 9


$|\textbf{v}|=10$ $\theta=126.9^{\circ}$

Work Step by Step

The magnitude of a vector $\textbf{v}=\langle a,b \rangle$ is given as $|\textbf{v}|=\sqrt (a^{2}+b^{2})$. Since $\textbf{v}=\langle -6,8 \rangle$, the magnitude is: $|\textbf{v}|=\sqrt ((-6)^{2}+(8)^{2})=\sqrt (36+64)=\sqrt (100)=10$ 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, $\theta=\tan^{-1} (\frac{8}{-6})=\tan^{} (\frac{4}{-3})=-53.1^{\circ}$ 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 $-53.1^{\circ}$ to yield the direction angle $\theta$. Therefore, the direction angle $\theta=126.9^{\circ}$.
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