Answer :
[tex]P=(\stackrel{\stackrel{x}{adjacent}}{2\sqrt{2}}~~,~~\stackrel{\stackrel{y}{opposite}}{2\sqrt{2}}) \\\\\\ tan(\theta )=\cfrac{\stackrel{opposite}{2\sqrt{2}}}{\underset{adjacent}{2\sqrt{2}}}\implies \theta =tan^{-1}\left( \cfrac{2\sqrt{2}}{2\sqrt{2}} \right)\implies \theta =tan^{-1}(1)\implies \theta =45^o[/tex]
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