1. **State the problem:** We have triangle ABC with sides AC = 8 cm, BC = 15 cm, and angle ACB = 70°. (a) Calculate the length of side AB. (b) Calculate the size of angle BAC. --- 2. **Calculate AB using the Law of Cosines:** The Law of Cosines states: $$AB^2 = AC^2 + BC^2 - 2 \times AC \times BC \times \cos(\angle ACB)$$ Substitute the values: $$AB^2 = 8^2 + 15^2 - 2 \times 8 \times 15 \times \cos(70^\circ)$$ Calculate each term: $$8^2 = 64$$ $$15^2 = 225$$ $$2 \times 8 \times 15 = 240$$ $$\cos(70^\circ) \approx 0.3420$$ So: $$AB^2 = 64 + 225 - 240 \times 0.3420 = 289 - 82.08 = 206.92$$ Take the square root: $$AB = \sqrt{206.92} \approx 14.39$$ Rounded to 3 significant figures: $$AB = 14.4\text{ cm}$$ --- 3. **Calculate angle BAC using the Law of Sines:** The Law of Sines states: $$\frac{\sin(\angle BAC)}{BC} = \frac{\sin(\angle ACB)}{AB}$$ Rearranged to find $\sin(\angle BAC)$: $$\sin(\angle BAC) = \frac{BC \times \sin(\angle ACB)}{AB}$$ Substitute known values: $$\sin(\angle BAC) = \frac{15 \times \sin(70^\circ)}{14.39}$$ Calculate $\sin(70^\circ)$: $$\sin(70^\circ) \approx 0.9397$$ So: $$\sin(\angle BAC) = \frac{15 \times 0.9397}{14.39} = \frac{14.0955}{14.39} \approx 0.9799$$ Find $\angle BAC$: $$\angle BAC = \sin^{-1}(0.9799) \approx 78.7^\circ$$ Rounded to 1 decimal place: $$\angle BAC = 78.7^\circ$$