1. **Problem statement:** Given triangle ABC with sides AB = 7.2 cm, AC = 3.5 cm, and angle BAC = 28°, find the measure of angle ACB to the nearest degree. 2. **Formula used:** We will use the Law of Cosines to find side BC first, then use the Law of Cosines again or Law of Sines to find angle ACB. Law of Cosines: $$c^2 = a^2 + b^2 - 2ab\cos(C)$$ where $C$ is the angle opposite side $c$. 3. **Step 1: Find side BC** (opposite angle A): $$BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(\angle BAC)$$ $$BC^2 = 7.2^2 + 3.5^2 - 2 \times 7.2 \times 3.5 \times \cos(28^\circ)$$ 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, $$BC^2 = 51.84 + 12.25 - 50.4 \times 0.8829 = 64.09 - 44.48 = 19.61$$ 4. **Calculate BC:** $$BC = \sqrt{19.61} = 4.43 \text{ cm (approx)}$$ 5. **Step 2: Find angle ACB** (angle at C) using Law of Cosines again: $$\cos(\angle ACB) = \frac{AC^2 + BC^2 - AB^2}{2 \times AC \times BC}$$ Substitute values: $$\cos(\angle ACB) = \frac{3.5^2 + 4.43^2 - 7.2^2}{2 \times 3.5 \times 4.43} = \frac{12.25 + 19.61 - 51.84}{30.1} = \frac{-20.0}{30.1} = -0.664$$ 6. **Calculate angle ACB:** $$\angle ACB = \cos^{-1}(-0.664) \approx 131.6^\circ$$ 7. **Final answer:** The measure of angle ACB is approximately **132°** to the nearest degree.