we have that
[tex]\sin (45^o)=\frac{4\sqrt[]{2}}{x}[/tex]by opposite side divided by the hypotenuse
Remember that
[tex]\sin (45^o)=\frac{\sqrt[]{2}}{2}[/tex]substitute
[tex]\frac{\sqrt[]{2}}{2}=\frac{4\sqrt[]{2}}{x}[/tex]solve for x
[tex]x=\frac{8\sqrt[]{2}}{\sqrt[]{2}}[/tex]simplify
