I would think in terms of vectors. If the slopes in the [MATH]x[/MATH] and [MATH]y[/MATH] directions are [MATH]a[/MATH] and [MATH]b[/MATH], respectively, then the equation of the plane can be written as [MATH]z = ax +by + c[/MATH], or [MATH]ax +by -z = c[/MATH]. So the normal to the plane is [MATH]\vec N = \langle a, b, -1\rangle[/MATH]. A vector along the [MATH]z[/MATH] axis is [MATH]\langle 0,0,1\rangle[/MATH]. Now the cross product of those two involves the sine of the angle you seek. Do you know that formula? Can you take it from there?