A plane flies above a man at a height of 1000m. The angle of elevation changes from 60° to 30° in 5 sec. Find speed of plane.
Explanation:
Given, BC = DE = 1000 m
In ∆ABC, tan60° = √3 = BC/AB
⇒ AB = BC/√3 = 1000/√3
In ∆ABC, tan30° = 1/√3 = DE/AD
⇒ AD = DE × √3 = 1000√3
∴ BD = AD - AB = 1000√3 – 1000/√3 = 2000/√3.
∴ Plane travels 2000/√3 meters in 5 seconds.
⇒ Speed of the plane = (2000/√3) ÷ 5 = 400/√3 m/s.
Hence, option (c).
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