1. **State the problem:** We are given a triangle with sides $a=3.6$ cm, $b=7.5$ cm, and $c=5.1$ cm. We need to find the measures of angles $A$, $B$, and $C$. 2. **Formula used:** Use the Law of Cosines to find each angle. The Law of Cosines states: $$\cos A = \frac{b^2 + c^2 - a^2}{2bc}, \quad \cos B = \frac{a^2 + c^2 - b^2}{2ac}, \quad \cos C = \frac{a^2 + b^2 - c^2}{2ab}$$ 3. **Calculate angle A:** $$\cos A = \frac{7.5^2 + 5.1^2 - 3.6^2}{2 \times 7.5 \times 5.1} = \frac{56.25 + 26.01 - 12.96}{76.5} = \frac{69.3}{76.5}$$ 4. Simplify fraction with cancellation: $$\cos A = \frac{\cancel{69.3}}{\cancel{76.5}} \approx 0.9059$$ 5. Find angle $A$ by taking inverse cosine: $$A = \cos^{-1}(0.9059) \approx 24.8^\circ$$ 6. **Calculate angle B:** $$\cos B = \frac{3.6^2 + 5.1^2 - 7.5^2}{2 \times 3.6 \times 5.1} = \frac{12.96 + 26.01 - 56.25}{36.72} = \frac{-17.28}{36.72}$$ 7. Simplify fraction with cancellation: $$\cos B = \frac{\cancel{-17.28}}{\cancel{36.72}} \approx -0.4705$$ 8. Find angle $B$ by inverse cosine: $$B = \cos^{-1}(-0.4705) \approx 118.1^\circ$$ 9. **Calculate angle C:** Use the fact that the sum of angles in a triangle is $180^\circ$: $$C = 180^\circ - A - B = 180^\circ - 24.8^\circ - 118.1^\circ = 37.1^\circ$$ 10. **Final answer:** - Angle $A \approx 24.8^\circ$ - Angle $B \approx 118.1^\circ$ - Angle $C \approx 37.1^\circ$