答案
解:延长AF交BH于点N,过点F作FM⊥BH于点M,
∵∠FEH=30°,EF=1.7m,
∴FM=0.85m,
∴EM=
×1.7
≈1.47,

由题意可得出:AB∥FM,
∴
=
,
∵CD=1.6m,DG=0.8m,
∴MN=0.68m,
∵BE=2.1m,
∴BN=2.1+1.47+0.68=4.25(m),
∵
=
,
∴
=
,
解得:AB≈5.3(m).
答:电杆的高约为5.3m.
解:延长AF交BH于点N,过点F作FM⊥BH于点M,
∵∠FEH=30°,EF=1.7m,
∴FM=0.85m,
∴EM=
×1.7
≈1.47,

由题意可得出:AB∥FM,
∴
=
,
∵CD=1.6m,DG=0.8m,
∴MN=0.68m,
∵BE=2.1m,
∴BN=2.1+1.47+0.68=4.25(m),
∵
=
,
∴
=
,
解得:AB≈5.3(m).
答:电杆的高约为5.3m.