已知有一弧形弦长L1=44.6高H1=15,半径为R1,这弧所对的圆心角为A1；另一弧形弦长L2=28.7且两弧形弧长C1=C2相等.求弦长28.7弧形的半径R2?
R1^2=(R1-H1)^2+(L1/2)^2
R1^2=R1^2-2*R1*H1+H1^2+L1^2/4
2*R1*H1=H1^2+L1^2/4
R1=H1/2+L1^2/(8*H1)
=15/2+44.6^2/(8*15)
=24.0763
A1=2*ARC SIN((L/2)/R)
=2*ARC SIN((44.6/2)/24.0763)
=135.706度
C1=PI*R*A/180=PI*24.0763*135.706/180=128.307
C2=C1=128.307
L2=28.7
Rn+1=(1+(L2-2*Rn*SIN(C2/(2*Rn)))/(L2-C2*COS(C2/(2*Rn))))*Rn
R0=29
R1=24.113
R2=25.194
R3=25.278
R4=25.278
R2=25.278