1. **Problem statement:** We have a rhombus ABCD with side length 7 cm, angle $\angle D = 120^\circ$, and $\angle C = 60^\circ$. 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 $AB = BC = CD = DA = 7$ cm. 3. **Using the law of cosines in triangle ADC:** Since $\angle D = 120^\circ$, triangle ADC has sides $AD = 7$, $DC = 7$, and diagonal AC opposite $\angle D$. The law of cosines states: $$AC^2 = AD^2 + DC^2 - 2 \times AD \times DC \times \cos(\angle D)$$ 4. **Substitute values:** $$AC^2 = 7^2 + 7^2 - 2 \times 7 \times 7 \times \cos(120^\circ)$$ 5. **Calculate cosine:** $$\cos(120^\circ) = -\frac{1}{2}$$ 6. **Simplify:** $$AC^2 = 49 + 49 - 2 \times 7 \times 7 \times \left(-\frac{1}{2}\right)$$ $$AC^2 = 98 + 49 = 147$$ 7. **Find AC:** $$AC = \sqrt{147} = \sqrt{49 \times 3} = 7\sqrt{3} \approx 12.1$$ **Final answer:** The length of diagonal AC is approximately **12.1 cm**.