∵BE=AD,BE⊥AB,AD⊥AB，

The quadrangular bed is rectangular,

∴de∥ab,de=ab=80m,fh=ad= 1.6m，

∵CF⊥AB，

∴CF⊥DE，

Let CH=xm,

In Rt△CDH, DH = Chtan 30 = 3x ≈ 1.73x (m).

In Rt△CEH, eh = chtan 55 ≈ x 1.42 (m),

∫DH+EH = DE = 80m，

∴ 1.73x+x 1.42=80，

Solution: x≈32.9,

∴CH=32.9m，

∴cf=ch+fh=32.9+ 1.6=34.5(m).

Answer: The height CF of the flagpole in the school is 34.5 meters.