∴∠B=60，tanB=√3

∫BC = 1

∴AC=√3

Tan∠ADC is a root of equation 3 (x 2+1/x 2)-5 (x+1/x) = 2, but this equation contains fractions, so it is difficult to solve and needs to be transformed.

The equation is transformed as follows:

3(x^2+2+ 1/x^2-2)-5(x+ 1/x)=2

3(x+ 1/x)^2-5(x+ 1/x)=8

Let y=x+ 1/x, and we get

3y^2-5y=8

seek

y 1= 16/6=8/3，y2=- 1

That is, x+ 1/x=8/3 and x+ 1/x=- 1.

X+ 1/x=8/3 has a real number solution and can be obtained.

x 1=(8+2√7)/6，x2=(8-2√7)/6

∠ADC & gt； ∠B, and tan ∠ b = ∠ 3, obviously x 1 = (8+2 ∠ 7)/6 > 3 is meaningful.

∴CD=AC/tan∠ADC

=√3/[(8+2√7)/6]

=6√3/(8+2√7)

Compared with high school students, this problem is obviously more difficult, involving many knowledge points and several mathematical skills, which is difficult for ordinary high school students to master.