Answer
\[
2=|W|, 240^{\circ}=\theta
\]
Work Step by Step
Given
\[
-i-\sqrt{3} j=W
\]
Here the $x$ component of the vector is $-1=W_{x}$, and the $y$ component of the vector is $-\sqrt{3}=W_{y}$
The magnitude of the vector W is
\[
|\sqrt{W_{x}^{2}+W_{y}^{2}}=W|
\]
Substitute $W_{x}$ and $W_{y}$ values
\[
\begin{aligned}
&\sqrt{(-1)^{2}+(-\sqrt{3})^{2}}=|W| \\
&=\sqrt{3+1} \\
&=\sqrt{4} \\
&=2
\end{aligned}
\]
$\theta$ is the angle made with the positive $x-$ axis and the vector.
\[
\frac{W_{x}}{W_{y}}=\tan \theta
\]
\[
\begin{aligned}
&=\frac{-\sqrt{3}}{-1} \\
\tan \theta &=\sqrt{3} \\
\theta &=\tan ^{-1} \sqrt{3} \\
&=240^{\circ}
\end{aligned}
\]
The angle is $240^{\circ}$ but not $60^{\circ}$ because the $x$ component and $y$ component of the given vector are negative. Thus the angle must be in Quadrant III.
