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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.

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It appears that you might mean for the point to be (-25,40) although you stated (25,40). The atan2(y,x) function using (-25,40) will yield ~122 deg.

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