Answer
\[
150^{\circ}=\theta, 4=|W|
\]
Work Step by Step
Given
\[
-2 \sqrt{3} i+2 j=F
\]
The magnitude of the vector $\mathrm{W}$ is
\[
\sqrt{F_{x}^{2}+F_{y}^{2}}=|F|
\]
Replace $F_{x}$ with $F_{y}$ values
\[
\begin{aligned}
&\sqrt{(2)^{2}+(-2 \sqrt{3})^{2}}=|F| \\
&=\sqrt{4+4 \cdot 3} \\
&=\sqrt{16} \\
&=4
\end{aligned}
\]
$\theta$ is the angle made with the positive $x-$axis, the vector.
\[
\begin{aligned}
&\frac{F_{x}}{F_{y}}=\tan \theta \\
&=\frac{2}{-2 \sqrt{3}} \\
&-\frac{1}{\sqrt{3}}=\tan \theta \\
&\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)=\theta \\
&=150^{\circ}
\end{aligned}
\]
The angle is $150^{\circ}$ but not $330^{\circ}$ because the $x$ component is negative, and the $y$ component is positive. Thus the angle must be in Quadrant II.
