For 2, we have the following triangle:
Using sine and cosine we have the following:
[tex]\begin{gathered} \cos (45)=\frac{x}{10\sqrt[]{2}} \\ \Rightarrow x=\cos (45)\cdot10\sqrt[]{2}=(\frac{1}{\sqrt[]{2}})\cdot10\sqrt[]{2}=10 \\ \sin (45)=\frac{y}{10\sqrt[]{2}} \\ \Rightarrow y=\sin (45)\cdot10\sqrt[]{2}=(\frac{1}{\sqrt[]{2}})10\sqrt[]{2}=10 \end{gathered}[/tex]
Therefore, the remaining sides are y= 10 and x = 10
