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 $\theta=120^{\circ}$.