已知三角形三个内角ABC满足A+C=2B,tanAtanC=2+根号3,顶点C对边上的高为4倍根号3

已知三角形三个内角ABC满足A+C=2B,tanAtanC=2+根号3,顶点C对边上的高为4倍根号3
求三角形的三边a,b,c

A+C=2B
A+C+B=180°
2B+B=180°
B=60°
A+C=2B=120°
tanB=-tan(A+C)=-(tanA+tanC)/(1-tanAtanC)
tanB=√3
tanAtanC=2+√3
√3 = -(tanA+tanC)/(1-2-√3) = (tanA+tanC)/(√3+1)
tanA+tanC = 3+√3
tanA、tanC相当于方程x^2-(3+√3）+（2+√3）的两个根
(x-1)(x-2-√3) = 0
tanA、tanC分别等于1,2+√3；或2+√3,1
即：A=45°,或C=45°
当A=45°时：
顶点C对边上的高为4√3
a = BC = 4√3/sinB = 4√3/sin60° = 4√3/(√3/2) = 8
b = AC = 4√3/sinA = 4√3/sin45° = 4√3/(√2/2) = 4√6
c = 4√3 /tanA + 4√3/tanB = 4√3 /tan45° + 4√3/tan60° = 4√3+4 = 4(√3+1)
当C=45°时：
tanC=1,tanA=2+√3,点C对边上的高为4√3
a = BC = 4√3/sinB = 4√3/sin60° = 4√3/(√3/2) = 8
c = 4√3 / tanA + 4√3 / tanB = 4√3/(2+√3) + 4√3/1 = 12√3-12 = 12(√3-1)
b = √(a^2+c^2-2accosB) = √{8^2+(12(√3-1))^2-2*8*12(√3-1)*cos60° } = 4√(46-24√3)