1. **State the problem:** Given triangle ABC with angles $A=72^\circ$, $B=43^\circ$, and side $a=23$ opposite angle $A$, find angle $C$ and sides $b$ and $c$. 2. **Find angle $C$:** The sum of angles in a triangle is $180^\circ$. $$C = 180^\circ - A - B = 180^\circ - 72^\circ - 43^\circ = 65^\circ$$ 3. **Use the Law of Sines:** $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ 4. **Calculate side $b$:** $$b = \frac{a \sin B}{\sin A} = \frac{23 \times \sin 43^\circ}{\sin 72^\circ}$$ Calculate intermediate values: $$\sin 43^\circ \approx 0.6820, \quad \sin 72^\circ \approx 0.9511$$ $$b = \frac{23 \times 0.6820}{0.9511} = \frac{15.686}{0.9511}$$ Show cancellation: $$b = \frac{\cancel{15.686}}{\cancel{0.9511}} = 16.5$$ 5. **Calculate side $c$:** $$c = \frac{a \sin C}{\sin A} = \frac{23 \times \sin 65^\circ}{\sin 72^\circ}$$ Calculate intermediate values: $$\sin 65^\circ \approx 0.9063$$ $$c = \frac{23 \times 0.9063}{0.9511} = \frac{20.8469}{0.9511}$$ Show cancellation: $$c = \frac{\cancel{20.8469}}{\cancel{0.9511}} = 21.9$$ 6. **Final answers rounded to nearest tenth:** $$C = 65.0^\circ, \quad b = 16.5, \quad c = 21.9$$