在三角形ABC中,角A.B.C的对边分别为a.b.c.已知a+b=51c=根号7.且4sin平方2分之A+B减去cos2C=2分之7（1)求

a+b=5c=√74sin^2[(A+B)/2] - cos2C=7/24sin^2(90°-C/2) - (2cos^2C-1)=7/24cos^2(C/2) - (2cos^2C-1) = 7/22(cosC+1) - (2cos^2C-1) = 7/22cosC+2 - 2cos^2C +1 = 7/2cos^2C-cosC +1/4 = 0(cosc-1/2) = 0cosC = 1/...