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.
\(\boxed{y \approx 59}\) cm
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