Let AB be the 1st pole and CE = 24 m be the 2nd pole
In Δ ADE,
tan 30° = DE/AD
⇒ 1/√3 = DE/15
∴ DE = 15/√3 = 5√3 = 5× 1.732 = 8.66 m
Thus, Height of the first pole = AB = CD = CE-DE = 24-8.66
= 15.34 m
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