1. **State the problem:** We are given a triangle ABC with angle ACB = 38°, side AB = 29 mm, and side AC = 40 mm. We need to find the two possible sizes of angle BAC to 3 significant figures. 2. **Relevant formula:** Use the Law of Sines, which states: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ where $a$, $b$, and $c$ are sides opposite angles $A$, $B$, and $C$ respectively. 3. **Assign sides and angles:** - Angle $C = 38^\circ$ - Side $AB = c = 29$ mm (opposite angle $C$) - Side $AC = b = 40$ mm (opposite angle $B$) - Angle $A$ is unknown 4. **Apply Law of Sines to find angle $B$:** $$\frac{c}{\sin C} = \frac{b}{\sin B} \implies \sin B = \frac{b \sin C}{c} = \frac{40 \times \sin 38^\circ}{29}$$ Calculate: $$\sin 38^\circ \approx 0.6157$$ $$\sin B = \frac{40 \times 0.6157}{29} \approx 0.849$$ 5. **Find possible values of angle $B$:** Since $\sin B = 0.849$, angle $B$ can be: - $B_1 = \sin^{-1}(0.849) \approx 58.1^\circ$ - $B_2 = 180^\circ - 58.1^\circ = 121.9^\circ$ 6. **Find corresponding angle $A$ for each case:** Using the triangle angle sum: $$A = 180^\circ - B - C$$ - For $B_1 = 58.1^\circ$: $$A_1 = 180^\circ - 58.1^\circ - 38^\circ = 83.9^\circ$$ - For $B_2 = 121.9^\circ$: $$A_2 = 180^\circ - 121.9^\circ - 38^\circ = 20.1^\circ$$ 7. **Check validity:** Both $A_1$ and $A_2$ are positive and less than 180°, so both are valid. **Final answers:** $$\boxed{A_1 = 83.9^\circ, \quad A_2 = 20.1^\circ}$$ Both answers are to 3 significant figures.