1. **State the problem:** Given a triangle with sides $a=3$, $b=7$, and included angle $C=20^\circ$, find the remaining angle $A$, angle $B$, and side $c$. 2. **Formula used:** Use the Law of Cosines to find side $c$: $$c^2 = a^2 + b^2 - 2ab\cos C$$ Then use the Law of Sines to find angles $A$ and $B$: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ 3. **Calculate side $c$:** $$c^2 = 3^2 + 7^2 - 2 \times 3 \times 7 \times \cos 20^\circ = 9 + 49 - 42 \cos 20^\circ$$ Calculate $\cos 20^\circ \approx 0.9397$: $$c^2 = 58 - 42 \times 0.9397 = 58 - 39.4674 = 18.5326$$ $$c = \sqrt{18.5326} \approx 4.304$$ 4. **Use Law of Sines to find angle $A$:** $$\frac{a}{\sin A} = \frac{c}{\sin C} \Rightarrow \sin A = \frac{a \sin C}{c} = \frac{3 \times \sin 20^\circ}{4.304}$$ Calculate $\sin 20^\circ \approx 0.3420$: $$\sin A = \frac{3 \times 0.3420}{4.304} = \frac{1.026}{4.304} \approx 0.2385$$ $$A = \arcsin(0.2385) \approx 13.8^\circ$$ 5. **Find angle $B$ using sum of angles:** $$B = 180^\circ - A - C = 180^\circ - 13.8^\circ - 20^\circ = 146.2^\circ$$ **Final answers:** $$C \approx 20^\circ$$ $$A \approx 14^\circ$$ $$B \approx 146^\circ$$