Answer
$17;331.9^{\circ}$
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 $\textbf{u}=\langle 15,-8 \rangle$, the magnitude is:
$|\textbf{u}|=\sqrt ((-15)^{2}+8^{2})=\sqrt (225+64)=\sqrt (289)=17$
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,
$\theta=\tan^{-1} (\frac{-8}{15})\approx-28.07^{\circ}$
The vector has a positive horizontal component and a negative vertical component which places it in the fourth quadrant. Since the direction angle is the positive angle between the x-axis and the position vector, we need to add $360^{\circ}$ to $-28.1^{\circ}$ to yield the direction angle $\theta$. Therefore, the direction angle $\theta=331.9^{\circ}$.
