1. **State the problem:** Given angles $A=40^\circ$, $B=80^\circ$, and side $a=15$ cm, find side $b$ 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 the sides and angles of a triangle. 3. **Calculate angle $C$:** $$C = 180^\circ - A - B = 180^\circ - 40^\circ - 80^\circ = 60^\circ$$ 4. **Apply the Law of Sines to find $b$:** $$\frac{a}{\sin A} = \frac{b}{\sin B} \implies b = a \times \frac{\sin B}{\sin A}$$ 5. **Substitute known values:** $$b = 15 \times \frac{\sin 80^\circ}{\sin 40^\circ}$$ 6. **Calculate sine values:** $$\sin 80^\circ \approx 0.9848, \quad \sin 40^\circ \approx 0.6428$$ 7. **Compute $b$:** $$b = 15 \times \frac{0.9848}{0.6428} = 15 \times 1.532 = 22.98$$ 8. **Final answer:** $$\boxed{b \approx 22.98 \text{ cm}}$$