Answer:
C) 113°
Step-by-step explanation:
The direction angle for a position vector [tex]v=\langle x,y\rangle[/tex] is [tex]\alpha=tan^{-1}(\frac{y}{x})[/tex] plus a correction based on the quadrant (more on that later).
Hence:
[tex]\theta=tan^{-1}(\frac{y}{x})\\\\\theta=tan^{-1}(\frac{12}{-5})\\\\\theta\approx-67.38^\circ[/tex]
We are not done here, however, as we need to account for which quadrant the vector is located in. Since v=⟨-5, 12⟩ is located in Quadrant II, then the direction angle must also be in Quadrant II.
Therefore, we use the formula [tex]\theta=180+\alpha[/tex] to account for this:
[tex]\theta=180^\circ+\alpha\\\theta=180^\circ+(-67.38^\circ)\\\theta=180^\circ-67.38^\circ\\\theta=112.62^\circ\\\theta\approx113^\circ[/tex]
