1. **State the problem:** Given triangle ABC with sides $b=28$, $c=33$, and angle $B=30^\circ$, solve for angle $C$, angle $A$, and side $a$ using the Law of Sines. 2. **Recall the Law of Sines formula:** $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ This relates sides and their opposite angles in any triangle. 3. **Find angle $C$ using the Law of Sines:** $$\frac{b}{\sin B} = \frac{c}{\sin C} \implies \sin C = \frac{c \sin B}{b} = \frac{33 \times \sin 30^\circ}{28}$$ Calculate $\sin 30^\circ = 0.5$: $$\sin C = \frac{33 \times 0.5}{28} = \frac{16.5}{28} = 0.5893$$ 4. **Find angle $C$:** $$C = \sin^{-1}(0.5893) \approx 36.1^\circ$$ 5. **Find angle $A$ using the triangle angle sum:** $$A = 180^\circ - B - C = 180^\circ - 30^\circ - 36.1^\circ = 113.9^\circ$$ 6. **Find side $a$ using Law of Sines:** $$\frac{a}{\sin A} = \frac{b}{\sin B} \implies a = \frac{b \sin A}{\sin B} = \frac{28 \times \sin 113.9^\circ}{\sin 30^\circ}$$ Calculate $\sin 113.9^\circ \approx 0.9171$ and $\sin 30^\circ = 0.5$: $$a = \frac{28 \times 0.9171}{0.5} = \frac{25.679}{0.5} = 51.4$$ **Final answers rounded to nearest tenth:** $$C = 36.1^\circ, \quad A = 113.9^\circ, \quad a = 51.4$$