1. **Problem statement:** Given a triangle ABC with angles $\alpha = 32^\circ$, $\beta = 58^\circ$, and side $c = 15$ cm opposite angle $C$, find side $b$ opposite angle $B$ using the sine function. 2. **Formula used:** The Law of Sines states: $$\frac{a}{\sin \alpha} = \frac{b}{\sin \beta} = \frac{c}{\sin \gamma}$$ where $a,b,c$ are sides opposite angles $\alpha, \beta, \gamma$ respectively. 3. **Given:** $c = 15$ cm, $\beta = 58^\circ$, and $\alpha = 32^\circ$. Since $\alpha + \beta + \gamma = 180^\circ$, angle $\gamma = 180^\circ - 32^\circ - 58^\circ = 90^\circ$. 4. **Calculate side $b$:** Using the Law of Sines, $$\frac{b}{\sin 58^\circ} = \frac{15}{\sin 90^\circ}$$ 5. **Simplify:** Since $\sin 90^\circ = 1$, $$b = 15 \times \sin 58^\circ$$ 6. **Evaluate $\sin 58^\circ$:** $$\sin 58^\circ \approx 0.8480$$ 7. **Calculate $b$:** $$b = 15 \times 0.8480 = 12.72$$ 8. **Conclusion:** The calculation $b = 12.72$ cm is correct. Therefore, the answer is $b = 12.72$ cm.