1. **State the problem:** Given side lengths $b=24$, $c=31$, and angle $\angle B=21^\circ$, use the Law of Sines to find all possible triangles (if any) with angles $\angle A_1$, $\angle A_2$, $\angle C_1$, $\angle C_2$ and sides $a_1$, $a_2$. 2. **Recall the Law of Sines:** $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ 3. **Find angle $C$ using the Law of Sines:** $$\frac{b}{\sin B} = \frac{c}{\sin C} \implies \sin C = \frac{c \sin B}{b} = \frac{31 \sin 21^\circ}{24}$$ Calculate $\sin 21^\circ \approx 0.3584$: $$\sin C = \frac{31 \times 0.3584}{24} = \frac{11.1104}{24} \approx 0.4629$$ 4. **Find possible values for $\angle C$:** Since $\sin C = 0.4629$, possible angles are: $$C_1 = \arcsin(0.4629) \approx 27.6^\circ$$ $$C_2 = 180^\circ - 27.6^\circ = 152.4^\circ$$ 5. **Check if triangles are possible:** - For $C_1 = 27.6^\circ$, sum of angles so far: $21^\circ + 27.6^\circ = 48.6^\circ$, so $A_1 = 180^\circ - 48.6^\circ = 131.4^\circ$ (valid) - For $C_2 = 152.4^\circ$, sum: $21^\circ + 152.4^\circ = 173.4^\circ$, so $A_2 = 180^\circ - 173.4^\circ = 6.6^\circ$ (valid) 6. **Find sides $a_1$ and $a_2$ using Law of Sines:** $$a = \frac{b \sin A}{\sin B}$$ Calculate $a_1$: $$a_1 = \frac{24 \sin 131.4^\circ}{\sin 21^\circ} = \frac{24 \times 0.7431}{0.3584} = \frac{17.8344}{0.3584} \approx 49.7$$ Calculate $a_2$: $$a_2 = \frac{24 \sin 6.6^\circ}{\sin 21^\circ} = \frac{24 \times 0.1150}{0.3584} = \frac{2.76}{0.3584} \approx 7.7$$ **Final answers:** $$\angle A_1 = 131.4^\circ, \quad \angle C_1 = 27.6^\circ, \quad a_1 = 49.7$$ $$\angle A_2 = 6.6^\circ, \quad \angle C_2 = 152.4^\circ, \quad a_2 = 7.7$$