Question

An aeroplane is flying at a height of 3000 metres above the ground -------------- .​

Answer 1

Answer:

Let height of first aeroplane is BD which is flying at 3000 m height and height of second aeroplane BC = h m. 

Let angles of elevation from point A are 60° and 45° respectively. 

So, ∠BAD = 60° and ∠CAB = 45°and 

Let AB = x m.

From right angled ΔABD,

tan 60° = BD/AB ⇒ √3 = 3000/x

Hence, height of second plane from first plane CD = BD – BC  = 3000 – 1732 = 1268

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