1. **State the problem:** We need to find the measure of angle $Y$ in triangle $XYZ$ where the sides opposite to vertices $X$, $Y$, and $Z$ are 13, 7, and 12 respectively. 2. **Recall the Law of Cosines:** For any triangle with sides $a$, $b$, and $c$ opposite angles $A$, $B$, and $C$ respectively, the Law of Cosines states: $$c^2 = a^2 + b^2 - 2ab\cos(C)$$ 3. **Assign sides and angles:** Here, side opposite $Y$ is 7, so $a=13$ (opposite $X$), $b=12$ (opposite $Z$), and $c=7$ (opposite $Y$). We want to find $m\angle Y$, so: $$7^2 = 13^2 + 12^2 - 2 \times 13 \times 12 \times \cos(Y)$$ 4. **Calculate squares:** $$49 = 169 + 144 - 312 \cos(Y)$$ 5. **Simplify the right side:** $$49 = 313 - 312 \cos(Y)$$ 6. **Isolate the cosine term:** $$49 - 313 = -312 \cos(Y)$$ $$-264 = -312 \cos(Y)$$ 7. **Divide both sides by -312:** $$\frac{\cancel{-264}}{\cancel{-312}} = \cos(Y)$$ $$\cos(Y) = \frac{264}{312}$$ 8. **Simplify the fraction:** $$\cos(Y) = \frac{11}{13} \approx 0.8462$$ 9. **Find angle $Y$ using inverse cosine:** $$Y = \cos^{-1}(0.8462) \approx 32.2^\circ$$ **Final answer:** $$m\angle Y \approx 32.2^\circ$$