Angels don't help with math homework.
We have x=-2 and y=3 and magnitude aka hypotenuse
[tex]h = \sqrt{x^2 + y^2} =\sqrt{(-2)^2+(3)^2}=\sqrt{13}[/tex]
Now the trig functions are straightforward;
[tex]\cos \theta = \dfrac x h = - \dfrac{2}{\sqrt{13}}[/tex]
[tex]\sin \theta = \dfrac y h = \dfrac{3}{\sqrt{13}}[/tex]
[tex]\tan \theta = \dfrac y x = - \dfrac 3 2 [/tex]
I won't bother to move the radicals to the numerator because I think that's a waste of time, but your teacher may feel differently.
