1. **Problem statement:** Find the measure of angle $\angle C$ in triangle $ABC$ where $\angle B = 55^\circ$, side $AB = 40$, and side $AC = 55$. 2. **Formula and rules:** We use the Law of Cosines to find the angle opposite a known side: $$\cos C = \frac{AB^2 + AC^2 - BC^2}{2 \times AB \times AC}$$ However, we do not have side $BC$ directly, so we need to find it or use the Law of Sines or another approach. Since only two sides and one angle are given, we can use the Law of Cosines to find side $BC$ first if we had $\angle A$, but we don't. Instead, we can use the Law of Sines or Law of Cosines with the given data. 3. Since $\angle B = 55^\circ$, and sides $AB$ and $AC$ are adjacent to $\angle B$, we can use the Law of Cosines to find side $BC$: $$BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos B$$ 4. Substitute values: $$BC^2 = 40^2 + 55^2 - 2 \times 40 \times 55 \times \cos 55^\circ$$ Calculate each term: $$40^2 = 1600$$ $$55^2 = 3025$$ $$2 \times 40 \times 55 = 4400$$ $$\cos 55^\circ \approx 0.5736$$ 5. Calculate: $$BC^2 = 1600 + 3025 - 4400 \times 0.5736$$ $$BC^2 = 4625 - 2523.84 = 2101.16$$ 6. Find $BC$: $$BC = \sqrt{2101.16} \approx 45.83$$ 7. Now use the Law of Cosines to find $\angle C$: $$\cos C = \frac{AB^2 + BC^2 - AC^2}{2 \times AB \times BC}$$ Substitute values: $$\cos C = \frac{40^2 + 45.83^2 - 55^2}{2 \times 40 \times 45.83}$$ Calculate numerator: $$1600 + 2101.16 - 3025 = 676.16$$ Calculate denominator: $$2 \times 40 \times 45.83 = 3666.4$$ 8. Calculate $\cos C$: $$\cos C = \frac{676.16}{3666.4} \approx 0.1844$$ 9. Find $\angle C$: $$C = \cos^{-1}(0.1844) \approx 79.4^\circ$$ **Final answer:** The measure of $\angle C$ is approximately $79.4^\circ$ to the nearest tenth.