Hint: (I'm just hinting, look at the picture as an auxiliary line)

(1) As shown in the figure,

Extend the intersection CB of AD to point G, and there is ∠DGF(B)= elevation angle at point D = 60 to get GF;

② Obtaining CG from the CF-GF you obtained is the key;

(3) Let the building height AB=x meters, which is obtained from the tangent of ∠ACB and ∠AGB:

CF=x/tan∠ACB，GF=x/tan∠AGB

∴? CF-GF=x/tan∠ACB-x/tan∠AGB=CG→ x;

④ Find CB from X and ③, and from the tangent of similarity or ∠ ECB;

⑤ finally x-BE? =AE .

Perhaps, you can think of a better way, I hope it will help you!