## Fundamentals of Physics Extended (10th Edition)

$\theta\approx122°$
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. $\theta=180°-57.0^{\circ}\approx122°$