1. **Problem statement:** Calculate the measure of angle $\angle XYZ$ in triangle $XYZ$ with sides $XY=4$ cm, $XZ=6$ cm, and $YZ=3$ cm. 2. **Formula used:** To find an angle in a triangle when all sides are known, use the Law of Cosines: $$\cos(\theta) = \frac{a^2 + b^2 - c^2}{2ab}$$ where $\theta$ is the angle opposite side $c$, and $a$, $b$ are the other two sides. 3. **Identify sides:** We want $\angle XYZ$, which is the angle at vertex $Y$. The side opposite $Y$ is $XZ=6$ cm. The other two sides forming angle $Y$ are $XY=4$ cm and $YZ=3$ cm. 4. **Apply Law of Cosines:** $$\cos(\angle XYZ) = \frac{XY^2 + YZ^2 - XZ^2}{2 \times XY \times YZ} = \frac{4^2 + 3^2 - 6^2}{2 \times 4 \times 3}$$ 5. **Calculate numerator:** $$4^2 + 3^2 - 6^2 = 16 + 9 - 36 = -11$$ 6. **Calculate denominator:** $$2 \times 4 \times 3 = 24$$ 7. **Calculate cosine:** $$\cos(\angle XYZ) = \frac{-11}{24}$$ 8. **Find angle:** $$\angle XYZ = \cos^{-1}\left(\frac{-11}{24}\right)$$ 9. **Evaluate angle to nearest tenth:** $$\angle XYZ \approx 117.3^\circ$$ **Final answer:** The measure of $\angle XYZ$ is approximately $117.3^\circ$.