三角形ABC中,若a^2+c^2=b^2+ac,且a除c等于1+根号3除2,求角C的大小

a²+c²=b²+ac
即a²+c²-b²=ac
因为a²+c²-b²=2ac cosB
则cosB=1/2
B=π/3
即A+C=2π/3
a/c = sinA/sinC = sin(2π/3 - C)/sinC
=1/2 + √3/2 cotC = (1 + √3)/2
故cotC = 1
C=π/4