1. **State the problem:** We need to find the measure of angle $\angle C$ in triangle $ABC$ where sides $AB=5$, $BC=10$, and $AC=11$. 2. **Formula used:** Use the Law of Cosines to find an angle when all three sides are known: $$\cos(\angle C) = \frac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC}$$ 3. **Substitute the known values:** $$\cos(\angle C) = \frac{5^2 + 10^2 - 11^2}{2 \cdot 5 \cdot 10} = \frac{25 + 100 - 121}{100} = \frac{4}{100}$$ 4. **Simplify the fraction:** $$\cos(\angle C) = \frac{\cancel{4}}{\cancel{100}} = 0.04$$ 5. **Find the angle:** $$\angle C = \cos^{-1}(0.04)$$ Using a calculator, $$\angle C \approx 87.7^\circ$$ 6. **Round to the nearest degree:** $$\angle C \approx 88^\circ$$ **Final answer:** $\boxed{88^\circ}$