1. **State the problem:** We have a triangle with sides AC = 12 cm, AB = 19 cm, and angle C = 80° opposite side AB. We want to find side BC = x. 2. **Identify the known elements:** - Side AC = 12 cm - Side AB = 19 cm - Angle C = 80° 3. **Use the Law of Cosines to find side BC (x):** The Law of Cosines states: $$c^2 = a^2 + b^2 - 2ab \cos(C)$$ where $c$ is the side opposite angle $C$. Here, $c = x$, $a = 12$, $b = 19$, and $C = 80^\circ$. 4. **Apply the formula:** $$x^2 = 12^2 + 19^2 - 2 \times 12 \times 19 \times \cos(80^\circ)$$ Calculate each term: $$12^2 = 144$$ $$19^2 = 361$$ 5. **Calculate the cosine term:** $$\cos(80^\circ) \approx 0.1736$$ 6. **Substitute values:** $$x^2 = 144 + 361 - 2 \times 12 \times 19 \times 0.1736$$ $$x^2 = 505 - 2 \times 12 \times 19 \times 0.1736$$ 7. **Calculate the product:** $$2 \times 12 \times 19 = 456$$ $$456 \times 0.1736 \approx 79.1616$$ 8. **Simplify:** $$x^2 = 505 - 79.1616 = 425.8384$$ 9. **Find $x$ by taking the square root:** $$x = \sqrt{425.8384} \approx 20.64$$ 10. **Round to 1 decimal place:** $$x \approx 20.6$$ **Final answer:** $$\boxed{20.6 \text{ cm}}$$