高H=1米,两端距离L=12米.求它的弧度A?
弧半径为R.
R^2=(R-H)^2+(L/2)^2
R^2=R^2-2*R*H+H^2+L^2/4
2*R*H=H^2+L^2/4
R=H/2+L^2/(8*H)
=1/2+12^2/(8*1)
=18.5米
A=2*ARC SIN((L/2)/R)
=2*ARC SIN((12/2)/18.5)
=37.85度
=37.85*PI/180
=0.660595弧度