Solve for Radian Exactly $$\tan(A) = \frac{\sqrt{3}}{-3}$$ I've tried using special triangles but couldn't find a matching faction using sohcahtoa.

$\tan(A) = \frac{\sqrt{3}}{-3} = \tan(A) = \frac{-1}{\sqrt3}$ ... (1)

Note that $\tan(A) = \frac{\sqrt 3}{-3}$ is also $\frac { 1}{-\sqrt 3}$ ... (2)

That is, the angle A could be in the 2nd [as in case (1)]or the 4th quadrant [as in case (2)].