tan30∘=ECAD=13

tan 30∘ = ECAD = 13(ED=h)(ED=h)
=> AD==3

AD = h 3 ---- (1)
tan60∘=FDAD

tan 60∘ = FDAD = 3√3
FD=h+40FD = h + 40
=h+403√AD = h + 403 ---- (2)
h3

h+403√h3 = h + 403
3h=h+403h = h + 40
2h=40=>h=20m2h = 40 => h = 20 m
EC=h+20=20+20EC = h + 20 = 20+20
=40m= 40 m

40m