From the given triangle, the dimensions are
[tex]\begin{gathered} XY=110\operatorname{mm}=z \\ YZ=141\operatorname{mm}=x \\ XZ=203\operatorname{mm}=y \end{gathered}[/tex]
To find the unknown angle, we find angle m∠Y
Using the cosine rule formula which is shown below
[tex]\cos Y=\frac{x^2+z^2-y^2}{2xz}[/tex]
Where
[tex]\begin{gathered} XY=z \\ YZ=x\text{ and} \\ XZ=y \end{gathered}[/tex]
Substitute the values of x, y and z into the cosine rule formula
[tex]\begin{gathered} \cos Y=\frac{x^2+z^2-y^2}{2xz} \\ \cos Y=\frac{141^2+110^2-203^2}{2(141)(110)} \\ \cos Y=\frac{-9228}{31020}=-0.2975 \\ Y=\cos ^{-1}(-0.2975) \\ Y=107.3^0 \end{gathered}[/tex]
Let the unknown angle be k, since m∠Y is known,
[tex]k+m\angle Y=180^0\text{ (angle on a straight line)}[/tex]
Substitute 107.3° of m∠Y into the formula above
[tex]\begin{gathered} k+m\angle Y=180^0 \\ k+107.3^o=180^o_{} \\ \text{Collect like terms} \\ k=180^o-107.3^{} \\ k=72.7^0 \\ k=73^o\text{ (nearest degree)} \end{gathered}[/tex]
Hence, the unknown angle 73^