∴∠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.