Discussion

Explanation:

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Given, height of plane initially (BC) = 1000 meters.

Let CF = x

In ∆CFE, cos30° = √3/2 = CE/CF

⇒ CE = √3x/2 = BD   …(1)

Also, sin30° = 1/2 = FE/CF

⇒ FE = x/2   …(2)

In ∆ABC, tan60° = √3 = CB/AB

⇒ AB = CB/√3 = 1000/√3   …(3)

In ∆AFD, tan45° = 1 = FD/AD

⇒ AD = FD   …(4)

We know, BD = AD – AB = FD – AB

⇒ BD = (DE + EF) - AB

From (1), (2) and (3)

√3x/2 = (1000 + x/2) – 1000/√3

Solving this we get x = 2000/√3

∴ Plane travels 2000/√3 meters in 5 seconds.

⇒ Speed of the plane = (2000/√3) ÷ 5 = 400/√3.

Hence, option (d).

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