Answer:
[tex]$\frac{5\sqrt{2} }{2}$[/tex]
Step-by-step explanation:
[tex]x: \text{opposite side of the angle of 45\º}[/tex]
[tex]5: \text{hypotenuse of the right triangle}[/tex]
[tex]$\sin(\theta)=\frac{\text{opp}}{\text{hyp}} \Rightarrow \sin(45\º)=\frac{x}{5} $[/tex]
[tex]$\text{Once }\sin(45\º)=\frac{\sqrt{2} }{2} $[/tex]
[tex]$\frac{\sqrt{2} }{2} =\frac{x}{5} \Rightarrow 2x=5\sqrt{2} \Rightarrow x=\frac{5\sqrt{2} }{2} $[/tex]
You can just remember that 5 is the diagonal of a square of side length x.
[tex]$5=x\sqrt{2} \Rightarrow x=\frac{5}{\sqrt{2} } \Rightarrow x=\frac{5}{\sqrt{2} } \cdot \frac{\sqrt{2} }{\sqrt{2} } = \frac{5\sqrt{2} }{2} $[/tex]
