1. **State the problem:** Given an acute triangle $\triangle ABC$ with $\angle B = 58.8^\circ$, side $c = 10.3$ cm (opposite $\angle C$), and side $b = 10.5$ cm (opposite $\angle B$), find the measure of $\angle C$ to the nearest tenth of a degree. 2. **Formula used:** Use the Law of Sines, which states: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ 3. **Apply the Law of Sines to find $\angle C$:** $$\frac{b}{\sin B} = \frac{c}{\sin C} \implies \sin C = \frac{c \sin B}{b}$$ 4. **Substitute known values:** $$\sin C = \frac{10.3 \times \sin 58.8^\circ}{10.5}$$ Calculate $\sin 58.8^\circ$: $$\sin 58.8^\circ \approx 0.8526$$ 5. **Calculate $\sin C$:** $$\sin C = \frac{10.3 \times 0.8526}{10.5} = \frac{8.783}{10.5} \approx 0.8365$$ 6. **Find $\angle C$ by taking the inverse sine:** $$\angle C = \sin^{-1}(0.8365) \approx 56.3^\circ$$ 7. **Check if the triangle is acute:** Since $\angle B = 58.8^\circ$ and $\angle C \approx 56.3^\circ$, their sum is $115.1^\circ$. The remaining angle $\angle A = 180^\circ - 115.1^\circ = 64.9^\circ$, which is also acute. **Final answer:** $$\boxed{56.3^\circ}$$