Trigonometry
[tex]Sine\frac{Opposite}{Hypotenuse} Cos\frac{Adjacent}{Hypotenuse} Tan\frac{Adjacent}{Opposite} \\S\frac{O}{H} C\frac{A}{H} T\frac{A}{O}[/tex]
Depending on which angle you use, decides which formula you use.
We'll use the top right angle.
9 is the adjacent.
x is the opposite.
So you need to use
[tex]Tan\frac{Adjacent}{Opposite} \\Tan45\frac{9}{x} \\x = \frac{9}{Tan45} \\x = 9[/tex]
x = 9
If you solve y, you can check this is correct by doing pythagoras' theorem.
[tex]Cos\frac{Adjacent}{Hypotenuse} \\Cos45\frac{9}{y} \\y = \frac{9}{cos45} \\y = 9\sqrt{2}[/tex]
[tex]\sqrt{9^{2}+9^{2} } = 9\sqrt{2}[/tex]
