1. **State the problem:** Given that $\sin A = -\frac{20}{29}$ and angle $A$ is in Quadrant III, find the exact value of $\tan A$ in simplest radical form with a rational denominator. 2. **Recall the formula:** $$\tan A = \frac{\sin A}{\cos A}$$ We know $\sin A$, but we need $\cos A$. Use the Pythagorean identity: $$\sin^2 A + \cos^2 A = 1$$ 3. **Calculate $\cos A$:** $$\cos^2 A = 1 - \sin^2 A = 1 - \left(-\frac{20}{29}\right)^2 = 1 - \frac{400}{841} = \frac{841 - 400}{841} = \frac{441}{841}$$ 4. **Simplify $\cos A$:** $$\cos A = \pm \frac{21}{29}$$ Since $A$ is in Quadrant III, both sine and cosine are negative, so: $$\cos A = -\frac{21}{29}$$ 5. **Find $\tan A$:** $$\tan A = \frac{\sin A}{\cos A} = \frac{-\frac{20}{29}}{-\frac{21}{29}} = \frac{-20}{29} \times \frac{29}{-21} = \frac{20}{21}$$ 6. **Final answer:** $$\boxed{\tan A = \frac{20}{21}}$$