1. **State the problem:** Find the angle $C$ in a triangle with sides $AC=12$, $CB=10$, and $AB=9$. 2. **Formula used:** Use the Law of Cosines to find angle $C$: $$\cos(C) = \frac{AC^2 + CB^2 - AB^2}{2 \cdot AC \cdot CB}$$ 3. **Apply the values:** $$\cos(C) = \frac{12^2 + 10^2 - 9^2}{2 \cdot 12 \cdot 10} = \frac{144 + 100 - 81}{240} = \frac{163}{240}$$ 4. **Calculate the cosine value:** $$\cos(C) = 0.6791667$$ 5. **Find angle $C$ by taking the inverse cosine:** $$C = \cos^{-1}(0.6791667) \approx 47.3^\circ$$ 6. **Conclusion:** The angle $C$ in the triangle is approximately $47.3^\circ$.