1. **State the problem:** We are given a triangle with vertices A, B, and C. Side AB = 10 cm, side AC = 8 cm, and angle B = 40°. 2. **Identify what is asked:** Find the size of angle A. 3. **Use the Law of Cosines:** To find angle A, we first need the length of side BC opposite angle A. However, we only have two sides and the included angle B, so we can use the Law of Cosines to find side BC. 4. **Apply Law of Cosines to find BC:** $$BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(B)$$ $$BC^2 = 10^2 + 8^2 - 2 \times 10 \times 8 \times \cos(40^\circ)$$ $$BC^2 = 100 + 64 - 160 \times \cos(40^\circ)$$ 5. **Calculate numeric value:** $$\cos(40^\circ) \approx 0.7660$$ $$BC^2 = 164 - 160 \times 0.7660 = 164 - 122.56 = 41.44$$ $$BC = \sqrt{41.44} \approx 6.44 \text{ cm}$$ 6. **Use Law of Cosines again to find angle A:** $$\cos(A) = \frac{AB^2 + BC^2 - AC^2}{2 \times AB \times BC}$$ $$\cos(A) = \frac{10^2 + 6.44^2 - 8^2}{2 \times 10 \times 6.44}$$ $$\cos(A) = \frac{100 + 41.44 - 64}{128.8} = \frac{77.44}{128.8} \approx 0.6017$$ 7. **Calculate angle A:** $$A = \cos^{-1}(0.6017) \approx 53.1^\circ$$ **Final answer:** Angle A is approximately $53.1^\circ$.