In $\triangle \mathrm{XYZ}, \mathrm{z}=46 \mathrm{~cm}, x=45 \mathrm{~cm}$ and $\angle \mathrm{Y}=81^{\circ}$. Find the length of $y$, to the nearest centimeter.

Step 1 ：Use the Law of Cosines: \(y^2 = x^2 + z^2 - 2xz \cdot \cos(Y)\)

Step 2 ：Plug in the values: \(y^2 = 45^2 + 46^2 - 2(45)(46) \cdot \cos(81^\circ)\)

Step 3 ：Calculate: \(y^2 \approx 3493.36\)

Step 4 ：Take the square root: \(y \approx 59.10\)

Step 5 ：\(\boxed{y \approx 59}\) cm