1. We are given that $\cos \theta = -\frac{1}{6}$ and that $\theta$ is in Quadrant III. 2. In Quadrant III, both sine and cosine are negative. 3. Use the Pythagorean identity: $$\sin^2 \theta + \cos^2 \theta = 1$$ 4. Substitute $\cos \theta = -\frac{1}{6}$: $$\sin^2 \theta + \left(-\frac{1}{6}\right)^2 = 1$$ $$\sin^2 \theta + \frac{1}{36} = 1$$ 5. Solve for $\sin^2 \theta$: $$\sin^2 \theta = 1 - \frac{1}{36} = \frac{36}{36} - \frac{1}{36} = \frac{35}{36}$$ 6. Take the square root to find $\sin \theta$: $$\sin \theta = \pm \sqrt{\frac{35}{36}} = \pm \frac{\sqrt{35}}{6}$$ 7. Since $\theta$ is in Quadrant III, $\sin \theta$ is negative. **Final answer:** $$\sin \theta = -\frac{\sqrt{35}}{6}$$