Cut off cycloid at an angle for Mountain Bike Jump / Ramp So I'm making an MTB jump to double as a ramp for my 6-year-old sister to carry her bike up this step in our front yard. I want to totally over engineer it and decided I wanted it to have a lip angle of $42.492°$ (golden angle-$180$) then I want the jump to be an upside-down cycloid/ Brachistochrone/ tautochrone/ isochrone with the curve going from $0°$ relative to the ground plane to $42.492°$ from the ground plane. I'm trying to figure out the size of circle I need to trace out the cycloid if the step/ jump is $7$ inches tall, and the lip angle is $42.492°$ or at least some way to calculate it. Essentially I want to find out the size of the circle or a way to calculate the size of the circle to be able to cut off the cycloid drawn by said circle at the right angle. I'm only a Freshman in HS and thus only have limited knowledge in Calculus.
 A: This was my original interpretation of the question:

An arch of a cycloid can be parameterised as $x=r(t-\sin(t)), \ y=r(1-\cos(t)$, where $r$ is the radius of the circle used to draw the cycloid.
From this we can compute $\frac{\mathrm{d}x}{\mathrm{d}t}=r(1-\cos(t))$, $\frac{\mathrm{d}y}{\mathrm{d}t}=r\sin(t)$ and $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\sin(t)}{1-\cos(t)}$.
For the lip angle to be $42^\circ$,we require that $\frac{\mathrm{d}y}{\mathrm{d}x}=\tan(48^\circ)$ at the point $(x,y)$ where the cycloid is to be cut. Using the half-angle formula $\cot(\frac{\theta}{2})=\frac{\sin(\theta)}{1-\cos(\theta)}$ we can deduce that $t=2\cot^{-1}(\tan(48^\circ))$. 
Plugging $x=7$ and $t=2\cot^{-1}(\tan(48^\circ))$ into $x=r(t-\sin(t)$ will give the required value of $r$.

Edit:
As it turns out, the cycloid section is to start at the middle of the arch (where $x=\pi r$) rather than from the origin and the lip angle is now $42.492$. So we now require that $x=\pi r - 7$ when $t=2\cot^{-1}(\tan(42.492^\circ))$
