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An airplane when flying at a height of 5000 m from the ground passes vertically above another airplane at an instant when the angles of the elevation of the two planes from the same point on the ground are 60∘ and 45∘respectively. ____ is the vertical distance between the airplanes at the instant.

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Detailed Solution

Let the height of first plane be AB=4000m and the height of second plane be BC=xm
∠BDC=45∘ and ∠ADB=60∘
In △CBD,

\frac{x}{y} = \tan 45 ° = 1

Or, x = y
In triangle ABD,

\frac{4000}{y} = \tan 60 ° = \sqrt{3}

Or, x = y =

\frac{4000 \sqrt{3}}{3}

= 2306.67m
Then, the vertical distance between two = 4000 − y
= 4000 − 2306.67
=1693.33m

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