Why does a=g*sin(theta) on an inclined plane

[Jow]

I understand the derivation that on an inclined plane of angle θ, the acceleration of the object on the plane, parallel to the plane, is a=g*sinθ. However, I was just thinking about it, and should it not be a=g/sinθ ?

I got this because sinθ = g/a. θ is the angle between the ramp and the earth, g is the acceleration straight downwards (the opposite side from θ) and a is the hypotenuse of the triangle.
Rearranging, a=g/sinθ.

I feel like something must be wrong, but I can't see what it is.

 

[Chestermiller]

When g is resolved into components, it represents the hypotenuse of the triangle.

 

[TysonM8]

The problem is the two formulas you have created (a=g/sinx and a=gsinx), use different angles. The first formula you created uses the angle between the ramp and the earth. The second one uses the angle between the vertical axis and the direction of acceleration. So both formulas are correct but they should be written as a=g/sinx and a=gsiny where y=90-x

 

[Chestermiller]

The problem is the two formulas you have created (a=g/sinx and a=gsinx), use different angles. The first formula you created uses the angle between the ramp and the earth. The second one uses the angle between the vertical axis and the direction of acceleration. So both formulas are correct but they should be written as a=g/sinx and a=gsiny where y=90-x

This is clearly incorrect. If you substitute y = 90-x into the second equation, you get a=g cosx. The acceleration cannot simultaneously be equal to g/sinx and gcos x, since cosx≠1/sinx. In fact, both these equations are wrong. The acceleration (in the absence of friction) is a=gsinx.

Chet

 

[nasu]

I got this because sinθ = g/a. θ is the angle between the ramp and the earth, g is the acceleration straight downwards (the opposite side from θ) and a is the hypotenuse of the triangle.
Rearranging, a=g/sinθ.

I feel like something must be wrong, but I can't see what it is.

g is not opposite to the angle theta.
It may be that you are mixing triangles. The triangle made by the components of the accelerations with the triangle made by the inclined planer itself.
Draw a diagram of the forces (or accelerations) and you will see.

 

[srijag]

http://thecraftycanvas.com/library/...clined-plane-or-ramp-with-free-body-diagrams/
I hope that helps.

 

[Chestermiller]

Here's a trick: Whenever a vector is resolved into orthogonal components, the vector is always the hypotenuse of the triangle.