Respuesta :

Answer:

Part 1) [tex]y=6\ units[/tex]

Part 2) [tex]x=3\ units[/tex]

Step-by-step explanation:

see the attached figure with letters to better understand the problem

In the right triangle ABE

we have that

AE=BE

so

Is a 45°-90°-45° triangle

[tex]cos(45\°)=\frac{\sqrt{2}}{2}[/tex]

[tex]cos(45\°)=\frac{3\sqrt{2}}{y}[/tex]

so

[tex]\frac{\sqrt{2}}{2}=\frac{3\sqrt{2}}{y}[/tex]

[tex]y=6\ units[/tex]

In the right triangle EBC

we have that

Is a 45°-90°-45° triangle

[tex]sin(45\°)=\frac{\sqrt{2}}{2}[/tex]

[tex]sin(45\°)=\frac{x}{3\sqrt{2}}[/tex]

[tex]\frac{x}{3\sqrt{2}}=\frac{\sqrt{2}}{2}[/tex]

[tex]x=3\ units[/tex]

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