Answer :
The figure demonstrates the condition given in the question.
We have to find the distance between D and C.
In Δ BCD,
( tan θ = Height/Base)
⇒ 
⇒ y = x√3 m ………….(1)
In Δ ABC,
( tan θ = Height/Base)
⇒ 
⇒ 5 + x = y√3
⇒ 5 + x = (x√3)(√3) (From eq 1)
⇒ 5 + x = 3x
⇒ 2x = 5
⇒ x = 2.5m
∴ Height of tower = x = 2.5m
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