Question

A child on a toboggan slides down a hill with an acceleration of magnitude 1.5 m/s^2. If friction is negligible, what is the angle between the hill and the horizontal?

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I don't even know how to get started on this question. Well, I first drew the free body diagram of the toboggan on an incline; however, the only variable I have is acceleration... Don't I also need at least one other variable (force, mass?) to calculate the angle?
Thanks.
a = 1.5 m/s^2 = g sin A

Solve for the angle A. g is the acceleration of gravity
Since g = 9.8 m/s^2, sin A = 1.5/9.81 = 0.1529 and A = 8.8 degrees

It is the component of the weight force down the hill, M g sin A, that causes (and equals) the acceleration, M a.

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