Solution: solution: as shown in the figure: D is DH⊥AB, and the vertical foot is H.

Let AC=x meters,

In Rt△ACD, ∠ ACD = 90, ∠ DAC = 25,

∴CD=AC？ tan∠DAC=xtan25。

When Rt△BDH, ∠ BHD = 90,

∠BDH =∠BDE = 15 30′，

∴BH=DH？ tan 15 30′= AC？ tan 15 30′= x？ tan 15 30’。

CD = AH，AH+HB=AB，

∴x(tan25+tan 15 30 ')= 30。

∴x=30tan25+ tan 15 30'≈40.3.

A: The horizontal distance AC between the two buildings is 40.3 meters.