Find the missing side lengths. Answers should be simplified radicals (no decimals).

Given:
The right triangle is given.
Use the sin ratio to find the hypotenuse of the triangle.
[tex]\begin{gathered} \sin \theta=\frac{opposite\text{ side}}{\text{hypotenuse}} \\ \sin 45^{\circ}=\frac{2\sqrt[]{2}}{b} \\ \frac{1}{\sqrt[]{2}}=\frac{2\sqrt[]{2}}{b} \\ b=4 \end{gathered}[/tex]Now, use the cosine ratio to find the remaining side of the triangle,
[tex]\begin{gathered} \cos \theta=\frac{adjecent\text{ side}}{\text{hypotenuse}} \\ \cos 45^{\circ}=\frac{a}{4} \\ \frac{1}{\sqrt[]{2}}=\frac{a}{4} \\ a=2\sqrt[]{2} \end{gathered}[/tex]Answer:
[tex]\begin{gathered} a=2\sqrt[]{2} \\ b=4 \end{gathered}[/tex]