To solve this problem, we will use the following formula for the direction angle of a vector x=ai+bj:
[tex]\tan \theta=(\frac{b}{a}).[/tex]
In the given problem:
[tex]\begin{gathered} a=-8\sqrt[]{3}, \\ b=8. \end{gathered}[/tex]
Therefore:
[tex]tan\theta=(\frac{8}{-8\sqrt[]{3}})\text{.}[/tex]
Simplifying the above result, we get:
[tex]\tan \theta=(-\frac{1}{\sqrt[]{3}}).[/tex]
Therefore, since the vector is in the second quadrant, we get:
[tex]\theta=150^{\circ}.[/tex]
Answer:
[tex]150^{\circ}.[/tex]
