解:過點?$A$?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?$36.8m$?