1. **State the problem:** We have triangle ABC with side $AB = 3$ cm, angle $\angle ABC = 40^\circ$, and angle $\angle BAC = 70^\circ$. We want to find the missing side lengths and the remaining angle. 2. **Find the missing angle:** The sum of angles in a triangle is $180^\circ$. $$\angle ACB = 180^\circ - 40^\circ - 70^\circ = 70^\circ$$ 3. **Use the Law of Sines:** The Law of Sines 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. 4. **Assign sides:** Given $AB = 3$ cm is opposite $\angle C = 70^\circ$, so side $c = 3$ cm. 5. **Find side $a$ opposite $\angle A = 70^\circ$:** $$\frac{a}{\sin 70^\circ} = \frac{3}{\sin 70^\circ}$$ Since $\sin 70^\circ$ cancels out, $a = 3$ cm. 6. **Find side $b$ opposite $\angle B = 40^\circ$:** $$\frac{b}{\sin 40^\circ} = \frac{3}{\sin 70^\circ}$$ Multiply both sides by $\sin 40^\circ$: $$b = \frac{3 \sin 40^\circ}{\sin 70^\circ}$$ 7. **Calculate $b$ numerically:** $$b = \frac{3 \times 0.6428}{0.9397} \approx \frac{1.9284}{0.9397} \approx 2.05 \text{ cm}$$ **Final answers:** - $\angle ACB = 70^\circ$ - Side $a = 3$ cm - Side $b \approx 2.05$ cm