1. **State the problem:** We are given a triangle with vertices Y, W, and X, and side lengths $YW=7$, $WX=9$, and $YX=12$. We need to find the measures of angles $\angle W$, $\angle X$, and $\angle Y$. 2. **Formula used:** To find the angles of a triangle when all three sides are known, we use the Law of Cosines: $$\cos(\theta) = \frac{a^2 + b^2 - c^2}{2ab}$$ where $\theta$ is the angle opposite side $c$, and $a$ and $b$ are the other two sides. 3. **Find $\angle W$ (opposite side $YX=12$):** $$\cos(\angle W) = \frac{YW^2 + WX^2 - YX^2}{2 \times YW \times WX} = \frac{7^2 + 9^2 - 12^2}{2 \times 7 \times 9} = \frac{49 + 81 - 144}{126} = \frac{-14}{126} = -\frac{1}{9}$$ 4. Calculate $\angle W$: $$\angle W = \cos^{-1}(-\frac{1}{9}) \approx 96.4^\circ$$ 5. **Find $\angle X$ (opposite side $YW=7$):** $$\cos(\angle X) = \frac{WX^2 + YX^2 - YW^2}{2 \times WX \times YX} = \frac{9^2 + 12^2 - 7^2}{2 \times 9 \times 12} = \frac{81 + 144 - 49}{216} = \frac{176}{216} = \frac{22}{27}$$ 6. Calculate $\angle X$: $$\angle X = \cos^{-1}(\frac{22}{27}) \approx 35.1^\circ$$ 7. **Find $\angle Y$ (opposite side $WX=9$):** Since the sum of angles in a triangle is $180^\circ$: $$\angle Y = 180^\circ - \angle W - \angle X = 180^\circ - 96.4^\circ - 35.1^\circ = 48.5^\circ$$ **Final answers:** $$m\angle W \approx 96.4^\circ$$ $$m\angle X \approx 35.1^\circ$$ $$m\angle Y \approx 48.5^\circ$$