1. **State the problem:** Given $\cos \theta = \sqrt{\frac{11}{6}}$ and $\theta$ is in Quadrant I, find the exact value of $\sin \theta$ in simplest form. 2. **Recall the Pythagorean identity:** $$\sin^2 \theta + \cos^2 \theta = 1$$ 3. **Substitute the given value:** $$\sin^2 \theta + \left(\sqrt{\frac{11}{6}}\right)^2 = 1$$ 4. **Simplify the square of cosine:** $$\sin^2 \theta + \frac{11}{6} = 1$$ 5. **Isolate $\sin^2 \theta$:** $$\sin^2 \theta = 1 - \frac{11}{6} = \frac{6}{6} - \frac{11}{6} = -\frac{5}{6}$$ 6. **Interpretation:** The value $\sin^2 \theta = -\frac{5}{6}$ is negative, which is impossible for a square of a real number. This indicates an error in the given data because $\cos \theta = \sqrt{\frac{11}{6}}$ is greater than 1, which is not possible for cosine values. **Conclusion:** The given value $\cos \theta = \sqrt{\frac{11}{6}}$ is invalid since cosine values must be between $-1$ and $1$. Therefore, $\sin \theta$ cannot be determined from this data.