Draw an auxiliary line parallel to BC from point A and let it intersect CD at point F, then CF=AB=3.

At the same time AF=BC,

In addition, let's say CE=x, because E is on the 6th floor, 5 floors away from C, D is on 1 1 floor, and there is 10 floor away from C, so we can easily get CD=2CE=2x, thus DF=CD-CF=2x-3.

Let's enter the problem-solving stage. In fact, as long as AF and BC are expressed as functions of X, and then the two functions are equal, X can be obtained, as follows.

BC=EC*tan45=EC=x

AF=DF*cot60=DF* (root number 3)/3=(2x-3)* (root number 3)/3.

Bring BC=AF and get x=2+ (root number 3).

In this case, BC=x=6+3* (root number 3) is the final result (if you verify it yourself, maybe I was careless ...)

In fact, this kind of math problem can be easily solved as long as you set an unknown number closely related to the required value and then get the equation about this unknown number from two identical angles.