$解:過點(diǎn)?A?作?AF⊥ED,?垂足為?F?$
$由題意得:?ED⊥BD,??AB=FD=6.8m,?$
$?AF=BD,??AF//BD?$
$∴?∠FAC=∠ACB=22°?$
$在?Rt△ABC?中,?BC=\frac {AB}{tan{22}°}≈\frac {6.8}{\frac {2}{5}}=17(\mathrm {m})?$
$設(shè)?CD=x\ \mathrm {m}?$
$∴?AF=BD=BC+CD=(x+17)m?$
$在?Rt△ECD?中,?∠ECD=58°?$
$∴?ED=CD?tan{58}°≈\frac {8}{5}x(\mathrm {m})?$
$在?Rt△EAF?中,?∠EAF=37°?$
$∴?EF=AF?tan{37}°≈\frac {3}{4}(x+17)m?$
$∵?EF+DF=ED?$
$∴?\frac {3}{4}(x+17)+6.8=\frac {8}{5}x?$
$解得:?x=23?$
$∴?DE=\frac {8}{5}x=36.8(\mathrm {m})?$
$∴建筑物?DE?的高度約為?36.8m?$