Compute $\tan 120^\circ$.
I know that $\tan 120^\circ = \frac{\sin 120^\circ}{\cos 120^\circ}$ by the definition of tangent. I also know that $\sin 120^\circ = \frac{\sqrt{3}}{2}$ and $\cos 120^\circ = -\frac{1}{2}$ by the 30-60-90 triangle. So, $\tan 120^\circ = \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = -\sqrt{3}$. 