三角答案

for simplicity, assume radius as 1

DEFINE: <EAO = a,     <FBO = b
draw HX perpendicular to BF, where x is on BF; easy to show <BHX = a  (as BH perpend. AO, and HX perpend. AE)

then, <HBO = 90 - a - b

BH=BO * COS(90 -a-b)=sin(a+b)
BX=BH*sin a=sin(a+b)*sin a
HX=BH*cos a=sin(a+b)*cos a
FX=BF-BX = cos b - sin(a+b)*sin a

now it's straight forward:
AE = cos a

HF=SQRT(HX^2+FX^2)=SQRT(sin(a+b)^2*cosa^2+(cos b - sin(a+b)sina)^2), which can be simplified all the way to cos a.