1. **Problem statement:** We have a rhombus ABCD with side length 9 cm, angles $\angle D = 100^\circ$ and $\angle C = 80^\circ$. We need to find the length of diagonal AC to 1 decimal place. 2. **Properties of a rhombus:** All sides are equal in length, so $AB = BC = CD = DA = 9$ cm. 3. **Using the Law of Cosines in triangle ADC:** Diagonal AC is opposite angle $\angle D = 100^\circ$. The Law of Cosines formula is: $$c^2 = a^2 + b^2 - 2ab \cos(\theta)$$ where $c = AC$, $a = AD = 9$, $b = DC = 9$, and $\theta = 100^\circ$. 4. Substitute values: $$AC^2 = 9^2 + 9^2 - 2 \times 9 \times 9 \times \cos(100^\circ)$$ 5. Calculate step-by-step: $$AC^2 = 81 + 81 - 162 \cos(100^\circ)$$ 6. Calculate $\cos(100^\circ)$: $$\cos(100^\circ) \approx -0.1736$$ 7. Substitute: $$AC^2 = 162 - 162 \times (-0.1736) = 162 + 28.1232 = 190.1232$$ 8. Take the square root: $$AC = \sqrt{190.1232} \approx 13.79$$ 9. Round to 1 decimal place: $$AC \approx 13.8$$ cm **Final answer:** The length of diagonal AC is approximately **13.8 cm**.