Answer :

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Let AB = DE = y m be the two poles such that BD = 80 m.


Let CD = x m


BC = (80 – x) m


In Δ CDB,


tan 30° = BD/CD


1/√3 = y/x


x = √3y ....(1)


In Δ CBA,


tan 60° = AB/BC


√3 = y/(80-x)


√3(80-x) = y


√3(80-√3y) = y [from(1)]


80√3-3y = y


4y = 80√3


y = 20√3 m


Substituting the value of y in equation (1), we get-


x = 60 m


Hence, Height of each pole = 20√3 m


and, Distance between the poles = 60 m.


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