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I was trying to solve frictionless inclined plane problem using a diff. frame as shown in the figure, and can't figure out that acceleration along the plane = g.sin(Θ), but I think it should be = g/sin(Θ) as according to the figure.

I have gone through numerous examples and theory regarding such cases, but all of them use the frame with x axis along the incline plane and y axis perpendicular to it.

Frictionless inclined plane

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It does not matter what frame you resolve the components at, the answer should be the same. –  ja72 Nov 18 '12 at 0:42
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I think $\cos(90^\circ-\theta)=\frac{g}{a}$ is wrong. Angle between $\vec{g}$ and $\vec{a}$ is that, but that is not the same thing. Think about what happens when $\theta=0$ in the acceleration infinite?

To solve this problem you need to find the projection of $\vec{g}$ along the incline. See picture below that might help you.

Incline

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