Let EH=x,

In Rt△CEH ∠ ech = 45,

∴∠CEH=45，

∴CH=EH=x，

At Rt△DEH,

∠EDH=36，tan∠DEH=HDEH，

∠DEH=90 -36 =54，

∴tan54 =HDEH，

∴HD=tan54 x,

∫CH+DH = AB = 20，

∴x+tan54 x=20，

Solution: x≈8.40,

∴EF=EH+HF≈8.40+ 1.5=9.9(m)，

That is, the height of the telephone pole is 9.9 m.