1. **Stating the problem:** We have triangle ABC with sides AB = 7.2 cm, AC = 3.5 cm, and angle BAC = 28°. We need to find the measure of angle ACB. 2. **Formula used:** We can use the Law of Cosines or Law of Sines to find the unknown angles. Since we know two sides and the included angle, the Law of Cosines is suitable to find side BC first: $$BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(\angle BAC)$$ 3. **Calculate side BC:** $$BC^2 = 7.2^2 + 3.5^2 - 2 \times 7.2 \times 3.5 \times \cos(28^\circ)$$ Calculate each term: $$7.2^2 = 51.84$$ $$3.5^2 = 12.25$$ $$2 \times 7.2 \times 3.5 = 50.4$$ $$\cos(28^\circ) \approx 0.8829$$ So: $$BC^2 = 51.84 + 12.25 - 50.4 \times 0.8829 = 64.09 - 44.48 = 19.61$$ Therefore: $$BC = \sqrt{19.61} = 4.43 \text{ cm}$$ 4. **Find angle ACB:** Using Law of Sines: $$\frac{\sin(\angle ACB)}{AB} = \frac{\sin(\angle BAC)}{BC}$$ Rearranged: $$\sin(\angle ACB) = \frac{AB \times \sin(\angle BAC)}{BC}$$ Calculate: $$\sin(28^\circ) \approx 0.4695$$ $$\sin(\angle ACB) = \frac{7.2 \times 0.4695}{4.43} = \frac{3.38}{4.43} = 0.763$$ 5. **Calculate angle ACB:** $$\angle ACB = \arcsin(0.763) \approx 49.8^\circ$$ Rounded to the nearest degree: $$\angle ACB \approx 50^\circ$$ **Final answer:** Angle ACB is approximately 50 degrees.