motion on an inclined plane

Assume we have a plane, inclined at a degree \[\theta\]. We then decompose the gravitational force into two components, \[F_{H}\], the force parallel to the plane and \[F_{V}\], the force perpendicular to the plane.

We can decompose it as such:

To find the acceleration of the particle going down the inclined plane, only \[F_{H}\] is essential.
From the free-body diagram, it is shown that \[\sin \theta = \frac{F_{H}}{mg}\], rearranging it, we get \[F_{H}=mg\sin \theta\].
Since \[F=ma\], we can substitute it into the equation above to get \[ma=mg\sin \theta\], simplifying it we get \[a=g\sin \theta\].