1. **State the problem:** Given triangle ABC with angles $\angle A = 32^\circ$, $\angle C = 76^\circ$, and side $a = 9$ opposite angle A, find the lengths of sides $b$ and $c$. 2. **Find the missing angle:** The sum of angles in a triangle is $180^\circ$. $$\angle B = 180^\circ - 32^\circ - 76^\circ = 72^\circ$$ 3. **Use the Law of Sines:** $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ 4. **Calculate side $b$:** $$b = \frac{a \sin B}{\sin A} = \frac{9 \sin 72^\circ}{\sin 32^\circ}$$ Calculate intermediate values: $$\sin 72^\circ \approx 0.9511, \quad \sin 32^\circ \approx 0.5299$$ $$b = \frac{9 \times 0.9511}{0.5299} = \frac{8.5599}{0.5299}$$ Show cancellation: $$b = \frac{\cancel{9} \times 0.9511}{\cancel{0.5299}} = 16.15$$ 5. **Calculate side $c$:** $$c = \frac{a \sin C}{\sin A} = \frac{9 \sin 76^\circ}{\sin 32^\circ}$$ Calculate intermediate values: $$\sin 76^\circ \approx 0.9703$$ $$c = \frac{9 \times 0.9703}{0.5299} = \frac{8.7327}{0.5299}$$ Show cancellation: $$c = \frac{\cancel{9} \times 0.9703}{\cancel{0.5299}} = 16.48$$ 6. **Final answers:** $$b \approx 16.15, \quad c \approx 16.48$$