[tex]\qquad\qquad\huge\underline{{\sf Answer}}[/tex]
Let's solve for x and y ~
Since the given triangle is a right angled triangle, we can use Trigonometric ratios here to get the required values
[tex]\qquad \sf \dashrightarrow \: \tan(45 \degree) = \dfrac{y}{2 \sqrt{2} } [/tex]
[tex]\qquad \sf \dashrightarrow \: 1= \dfrac{y}{2 \sqrt{2} } [/tex]
[tex]\qquad \sf \dashrightarrow \:y = 2 \sqrt{2} \: \: units[/tex]
Now, let's get to x ~
[tex]\qquad \sf \dashrightarrow \: \sin(45 \degree) = \dfrac{y}{x} [/tex]
[tex]\qquad \sf \dashrightarrow \: \dfrac{1}{ \sqrt{2} } = \dfrac{2 \sqrt{2} }{x} [/tex]
[tex]\qquad \sf \dashrightarrow \:x = 2 \sqrt{2}{} \times \sqrt{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:x = 2 \times 2[/tex]
[tex]\qquad \sf \dashrightarrow \:x = 4 \: \: units[/tex]
I hope you understood the whole procedure ~
