Solution: Make FH⊥CD in H and cross AB in G, then HD = GB = EF = 1.6m, EG = FB = 3.2m, GH = BD = 23.6m, AG = AB-GB = 2m-1.6m = 0.4m.

∫AB∨CD

∴AG/CH=EG/EH.

Ch = ag eh/eg = 0.4× 26.8/3.2 = 3.35 (m)

∴ Tree height CD = DH+HC =1.6m+3.35m = 4.95m.