The angle made by the vector is [tex]\theta[/tex] such that
[tex]\tan(\theta) = \dfrac{18.4\,\rm m}{-28.7\,\rm m} \approx -0.6411[/tex]
Before taking the inverse tangent of both sides, recall that [tex]\arctan[/tex] returns a number between -π/2 and π/2 radians, or -90° and +90°. The vector in the diagram clearly makes an angle between 90° and 180°, however, so we use the fact that tangent has a period of π rad / 180° to write
[tex]\theta \approx \arctan(-0.6411) + 180^\circ n[/tex]
where [tex]n\in\Bbb Z[/tex].
Now,
[tex]\arctan(-0.6411) \approx -0.5701 \,\mathrm{rad} \approx -32.6639^\circ[/tex]
Add 180° to this to recover the correct angle.
[tex]\theta = \arctan(-0.6411) + 180^\circ \approx \boxed{147.34^\circ}[/tex]
