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Answer 1

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Let the height of tower be h

Let BC = x

AB = 30 m

In ∆ ABC,

Tan 60°= AB / BC

Tan60° = 30/x

√3 = 30/x

x = 30/√3

x = 10√3

In ∆CBD,

Tan 30° = CD / BC

1 / √3 = h / x

1 / √3 = h / 10√3

h = 10 m

so, height of the building is 10 m