1. **Problem:** Given triangle $\triangle ABC$ with $|AB|=6$, $|BC|=8$, and $\angle A + \angle BC = 120^\circ$, find the length of side $|AC|$. 2. **Formula and rules:** We use the Law of Cosines to find the unknown side in a triangle: $$|AC|^2 = |AB|^2 + |BC|^2 - 2|AB||BC|\cos(\theta)$$ where $\theta$ is the angle between sides $AB$ and $BC$. 3. **Interpretation:** The problem states $\angle A + \angle BC = 120^\circ$. Usually, $\angle BC$ is not a standard notation for an angle; assuming it means $\angle B$, then $\angle A + \angle B = 120^\circ$. Since the sum of angles in a triangle is $180^\circ$, then $$\angle C = 180^\circ - 120^\circ = 60^\circ.$$ 4. **Apply Law of Cosines:** To find $|AC|$, we use the angle opposite to side $AC$, which is $\angle B$ or $\angle C$? Since $|AC|$ is opposite $\angle B$, but we only know $\angle C=60^\circ$, we need to clarify. Let's assume $\angle C=60^\circ$ is the angle between sides $AB$ and $BC$ (vertex $C$). Then, $$|AC|^2 = |AB|^2 + |BC|^2 - 2|AB||BC|\cos(60^\circ)$$ 5. **Calculate:** $$|AC|^2 = 6^2 + 8^2 - 2 \times 6 \times 8 \times \cos(60^\circ)$$ $$= 36 + 64 - 96 \times \frac{1}{2}$$ $$= 100 - 48 = 52$$ 6. **Final length:** $$|AC| = \sqrt{52} = 2\sqrt{13} \approx 7.211$$ **Answer:** The length of side $|AC|$ is $2\sqrt{13}$ or approximately 7.211.