1. **State the problem:** We are given a triangle with sides $a=8.9$ cm, $b=11.2$ cm, and $c=13.0$ cm, and we need to find the measure of angle $\angle C$ using the cosine law. 2. **Recall the cosine law formula:** $$c^2 = a^2 + b^2 - 2ab \cos(C)$$ This formula relates the lengths of the sides of a triangle to the cosine of one of its angles. 3. **Rearrange the formula to solve for $\cos(C)$:** $$\cos(C) = \frac{a^2 + b^2 - c^2}{2ab}$$ 4. **Substitute the given values:** $$\cos(C) = \frac{8.9^2 + 11.2^2 - 13.0^2}{2 \times 8.9 \times 11.2}$$ 5. **Calculate the squares:** $$8.9^2 = 79.21, \quad 11.2^2 = 125.44, \quad 13.0^2 = 169.00$$ 6. **Plug these into the numerator:** $$79.21 + 125.44 - 169.00 = 204.65 - 169.00 = 35.65$$ 7. **Calculate the denominator:** $$2 \times 8.9 \times 11.2 = 199.36$$ 8. **Calculate $\cos(C)$:** $$\cos(C) = \frac{35.65}{199.36}$$ 9. **Simplify the fraction:** $$\cos(C) \approx 0.1788$$ 10. **Find angle $C$ by taking the inverse cosine:** $$C = \cos^{-1}(0.1788)$$ 11. **Calculate $C$ (in degrees):** $$C \approx 79.7^\circ$$ 12. **Round to the nearest degree:** $$C \approx 80^\circ$$ **Final answer:** $\boxed{80^\circ}$