∴∠CED= 15，

∴∠CED=∠ECD.

∴DC=DE，

△ DCE is an isosceles triangle;

(2) In Rt△EDF,

From sin∠EDF=EFDE, EF=DE? sin∠EDF=24？ Sin 30 = 24× 12 = 12 (m)，

∫fg = ca = 1.5m，

Therefore, eg = ef+fg =12+1.5 =13.5 (m),

Answer: the height of flagpole EG is 13.5 meters.

(3) In Rt△EDF,

DF=DE from cos∠EDF=DFDE？ cos∠EDF=24？ Cos 30 = 24× 32 = 123 (m)，

Then cf = CD+df = 24+123 = 44.76 (m);

A: The length of line segment CF is 44.76 meters.