在△ABC中,角A,B,C所对的边分别为a,b,c,已知△ABC的外接圆的半径R=1...答:先用正玄定理,a/sinA=b/sinB=c/sinC=2R=2,cos2C-cos2A=(6/5a-b)sinB,1-2sinC^2-1+2sinA^2=(6/5a-b)sinB,式子两边乘以2R,得到a^2-c^2=(6/5a-b)b 也就是c^2=a^2+b^2-6/5*ab 有余弦定理的,cosC=3/5 △ABC的面积=1/2 absinC cosC=3/5,那么sinC=4/5,c=2...