How did you determine the angle in the components of $mg$ with mass on the incline? How do you know whether to use sine or cosine?

Here's a case where a force of interest is perpendicular to an incline (today's mass-on-incline problem had the force pointing down-- but geometric arguments are very similar).

By the alternate angle theorem from geometry, we have that the angle $\theta$ is the same as the angle $\alpha$. Since angles $\alpha$ and $\beta$ are complementary to the same angle, they are equal. So $\theta$, $\alpha$ and $\beta$ are all the same angle. Now look at the right triangle formed by the vector $\vec{F}$ and $F_y$, with angle $\beta$ between them. From this triangle you get that $F_x = F \sin \theta$ and $F_y= F\cos \theta$.

We'll be seeing lots of examples like this, so it will be worth your while to go through the geometry and trig carefully. Please ask again if it's not clear. This kind of thing will get much easier with practice.