所以 BA2=L2-C'A2=2-R2-r2+2Rc0S0 故： BA2-BA"2Rr(1-cos0) h= (2) BA+BA BA+BA BA+BA所以 故： 2 2 2 2 2 2 BA L C A L R r Rr    = − = − − + 2 cos  2 2 2 ' ' ' ' 4 sin ( ) 2 (1 cos ) 2 Rr BA BA Rr h BA BA BA BA BA BA − −  = = = + + + (2)