1. **State the problem:** We are given a triangle ABC with sides AC = 7 cm, BC = 10 cm, and angle BAC = 65°. We need to find the size of angle ABC to the nearest 0.1°. 2. **Identify known values:** - Side AC = 7 cm - Side BC = 10 cm - Angle BAC = 65° 3. **Use the Law of Cosines to find side AB:** The Law of Cosines states: $$AB^2 = AC^2 + BC^2 - 2 \times AC \times BC \times \cos(\angle BAC)$$ Substitute the values: $$AB^2 = 7^2 + 10^2 - 2 \times 7 \times 10 \times \cos(65^\circ)$$ Calculate: $$AB^2 = 49 + 100 - 140 \times \cos(65^\circ)$$ $$AB^2 = 149 - 140 \times 0.4226$$ $$AB^2 = 149 - 59.164$$ $$AB^2 = 89.836$$ 4. **Find AB:** $$AB = \sqrt{89.836} \approx 9.48 \text{ cm}$$ 5. **Use the Law of Sines to find angle ABC:** The Law of Sines states: $$\frac{\sin(\angle ABC)}{AC} = \frac{\sin(\angle BAC)}{AB}$$ Rearranged to solve for $\sin(\angle ABC)$: $$\sin(\angle ABC) = \frac{AC \times \sin(\angle BAC)}{AB}$$ Substitute known values: $$\sin(\angle ABC) = \frac{7 \times \sin(65^\circ)}{9.48}$$ Calculate: $$\sin(\angle ABC) = \frac{7 \times 0.9063}{9.48} = \frac{6.3441}{9.48} \approx 0.669\$$ 6. **Find angle ABC:** $$\angle ABC = \sin^{-1}(0.669) \approx 42.0^\circ$$ **Final answer:** The size of angle ABC is approximately **42.0°** to the nearest 0.1°.