由正弦定理a/sinA=b/sinB=c/sinC=2R
得a=2R*sinA,b=2R*sinB,c=2R*sinC
又因 a+b+10c=2(sinA+sinB+10sinC)
得 2R*sinA + 2R*sinB + 10*2R*sinC =2(sinA+sinB+10sinC),
整理得 (R-1)*(2sinA+2sinB+20sinC)=0
因为三角形中,sinA>0 ,sinB>0,sinC>0
所以(sinA+sinB+10sinC)>0
所以只能R-1=O,即R=1
又A=60°
所以a=2R*sinA=2*1*sin60°=√3