1. **State the problem:** Find the measure of angle B in a triangle where side $b=42.8$ cm, side $c=30.6$ cm, and angle $C=41^\circ$. 2. **Recall the Law of Cosines formula:** $$b^2 = a^2 + c^2 - 2ac \cos B$$ But since we don't have side $a$, we use the Law of Sines instead: $$\frac{b}{\sin B} = \frac{c}{\sin C}$$ 3. **Apply the Law of Sines:** $$\frac{42.8}{\sin B} = \frac{30.6}{\sin 41^\circ}$$ 4. **Calculate $\sin 41^\circ$:** $$\sin 41^\circ \approx 0.6561$$ 5. **Set up the equation for $\sin B$:** $$\sin B = \frac{42.8 \times 0.6561}{30.6}$$ 6. **Calculate the numerator:** $$42.8 \times 0.6561 = 28.08$$ 7. **Calculate $\sin B$:** $$\sin B = \frac{28.08}{30.6} \approx 0.9176$$ 8. **Find angle $B$ by taking the inverse sine:** $$B = \sin^{-1}(0.9176) \approx 66.8^\circ$$ 9. **Round to three significant figures:** $$B \approx 66.8^\circ$$ **Final answer:** $$\boxed{66.8^\circ}$$