#### Answer

5 units

#### Work Step by Step

We can use the law of cosines here because we know the lengths of two sides of the triangle and the measure of the included angle.
The law of cosines is:
$a^{2}=b^{2}+c^{2}-2bc\cos A$
where $b,c$ are the two known sides of the triangle while $A$ is the known angle. The unknown side opposite the known angle is $a$.
Substituting the values in the formula and solving:
$a^{2}=b^{2}+c^{2}-2bc\cos A$
$a^{2}=1^{2}+(4\sqrt 2)^{2}-2(1)(4\sqrt 2)\cos 45^{\circ}$
$a^{2}=1+[4^{2}\times(\sqrt 2)^{2}]-8\sqrt 2\cos 45^{\circ}$
$a^{2}=1+[16\times 2]-8\sqrt 2\cos 45^{\circ}$
$a^{2}=1+[32]-8\sqrt 2\cos 45^{\circ}$
$a^{2}=33-8\sqrt 2\cos 45^{\circ}$
We know that $\cos 45^{\circ}=\frac{\sqrt 2}{2}$. Therefore,
$a^{2}=33-8\sqrt 2\cos 45^{\circ}$
$a^{2}=33-8\sqrt 2(\frac{\sqrt 2}{2})$
$a^{2}=33-(\frac{8\times\sqrt 2\times\sqrt 2}{2})$
$a^{2}=33-(4\times\sqrt 2\times\sqrt 2)$
$a^{2}=33-(4\times2)$
$a^{2}=33-(8)$
$a^{2}=25$
$a=\sqrt {25}$
$a=5$
Therefore, the length of the unknown side of the triangle is 5 units.