Answer :

In the fig CE is the height of the pole and x be the distance between tower and pole.


In ∆ADE


tan 45° =


1 =


x = 50-h ----- (1)


In ∆ABC


tan 60° =


√3 =


x = -----(2)


On substituting value of x in eqn (1), we get


50 – h =


h = 50 -


21.13m



Therefore the height of the pole is 21.13m


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