From the top a 7

Let the distance between the foots of building and cable tower is (m).

The height of cable tower = AB = AE+EB (h+7)m.

In ∆AED,

tan 60° =

√3 =

h = √3 ---------(1)

The height of cable tower = AB = AE+EB (h+7)m.

In ∆DEB,

tan 45° =

1 =

----------(2)

On substituting value of in eqn. (1)

h = 7√3

Height of cable tower is (h+7)m.

7√3+7

7(√3+1)m.

Therefore height of cable tower is 7(√3+1)m.

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