we know that
In the right triangle of the figure
For the angle x
[tex]cos(x)=\frac{a}{b}[/tex]
[tex]x=cos^{-1}(\frac{a}{b})[/tex]
[tex]sin(x)=\frac{c}{b}[/tex]
[tex]x=sin^{-1}(\frac{c}{b})[/tex]
[tex]tan(x)=\frac{c}{a}[/tex]
[tex]x=tan^{-1}(\frac{c}{a})[/tex]
For the angle y
[tex]cos(y)=\frac{c}{b}[/tex]
[tex]y=cos^{-1}(\frac{c}{b})[/tex]
[tex]sin(y)=\frac{a}{b}[/tex]
[tex]y=sin^{-1}(\frac{a}{b})[/tex]
[tex]tan(y)=\frac{a}{c}[/tex]
[tex]y=tan^{-1}(\frac{a}{c})[/tex]
therefore
the answer is
[tex]x=sin^{-1}(\frac{c}{b})[/tex]
[tex]x=tan^{-1}(\frac{c}{a})[/tex]
[tex]y=cos^{-1}(\frac{c}{b})[/tex]