Answer
$\theta\approx122°$
Work Step by Step
We use the following formula to find the angle:
$\theta=arctan(\frac{A_{y}}{A_{x}})$
We plug in the known values to obtain:
$=arctan(\frac{40.0m}{-25.0m})\approx-57.0^{\circ}$
The angle is in quadrant 2 because the vector has a negative $\hat{x}$ component and a positive $\hat{y}$ component.
$\theta=180°-57.0^{\circ}\approx122°$
