Pic related. I need the length of that cathetus. We know R, alfa and that the second triangle is right angled. The problem is that I don't know a formula for the length of that side that unites the triangles. I know pythagoras theorem but first I need that length.
sin α * R = x (the side shared by both triangles)
if you haven't done trig functions yet, it should be possible to solve with pythagoras though, since it's isoceles. just imagine the hight above the unknown side which splits the upper triangle it into two mirrored right-angled triangles with angles 90°, α/2 and 180-90-α/2 and with hypothenuse R, height h and x/2 as sides.
but something's fishy, i think you can't really solve the problem. you would need two sides to apply pythagoras on the lower triangle. otherwise there are infinite variants of right-angled triangles. every point on a half-circle above a line (center = mid point, radius = half the length) completes into a right-angled triangle. some more angles would also help, but you need *something* more...
Damn, I got your/you're mixed up. Here's a cutie to fix my mistake
Let M be the center of the circle with radius R.
Draw a perpendicular line from B to AM and you see that the radius R is subdivided into two parts: The height h and a segment which has length R cos alpha .
=> h = R - R cos alpha = R(1 - cos alpha)