What is the angle between the direction of $A$ and the positive direction of x?

The x component of vector $$A$$ is 25.0 m and the y component is 40.0 m. (a) What is the magnitude of $$A$$? (b) What is the angle between the direction of $$A$$ and the positive direction of x?

I solved (a). When attempting (b) I tried using the formula $$\tan\theta=\frac{a_y}{a_x}=\frac{40}{-25}=-1.6$$, thus $$arctan(-1.6)=-58º$$ but the answer key is $$122º$$.

Any help is appreciated.