We could use the mathematical identity
[tex]\begin{gathered} \frac{2\sqrt[]{2}}{x}=\cos 45 \\ \frac{2\sqrt[]{2}}{x}=\frac{\sqrt[]{2}}{2} \\ \text{cross multiply} \\ 4\sqrt[]{2}=x\sqrt[]{2} \\ \text{divide both sides by }\sqrt[]{2} \\ x=4 \end{gathered}[/tex]To find y, we use
[tex]\begin{gathered} \frac{y}{2\sqrt[]{2}}=\tan 45 \\ \frac{y}{2\sqrt[]{2}}=1 \\ \text{cross multiply} \\ y=2\sqrt[]{2} \end{gathered}[/tex]This is to be expected since the triangle is isosceles.
Therefore, we have;
[tex]\begin{gathered} x=4 \\ y=2\sqrt[]{2} \end{gathered}[/tex]