原帖由 hww7 于 2007-11-25 14:45 发表
I suggest that:
//a,b,c,d,e,f,are known,x,y are unknown;
//d={d1,d2,d3,d4}are known;
(a,b),(c,d),(e,f),(x,y)
--------------------------------
sqrt((a-c)^2+(b-d)^2)=>D1
then match D1 with d ...
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