Question

An aeroplane when flying at a height of 4000 m from the ground passes vertically above another aeroplane at an instant when the angles of elevation of two planes from the same point on the ground at 60 degree and 45 degree respectively find the vertical distance between the aeroplanes at that instant

Answer 1

Answer:

Step-by-step explanation:

The answer is in the pics

Answer 2

ANSWER:

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Let P and Q be the position of two aeroplanes when Q is vertically below P and OP=4000m.

Let the angle of elevation of P and Q at point A on the ground be 60° and 45° respectively.

In trainagle AOP and AOQ ,

we have

tan 60°=OP/OA

and

tan 45°=OQ/OA

$\\ = > \sqrt{3 = \frac{4000}{OA} } \: \: \: and \: \: 1 = \frac{OQ}{OA} \\ \\ OA = \frac{4000}{ \sqrt{3 } } \: \: \: and \: OQ = OA \\ \\ = > OQ = \frac{4000}{ \sqrt{3} }m$

Therefore,

Vertical distance PQ between the aeroplane is given by,

PQ=OP-OQ

$= > PQ = (4000 - \frac{4000}{ \sqrt{3} } )m \\ \\ = 4000 \frac{( \sqrt{3} - 1 )}{ \sqrt{3} }m \\ \\ m = 1690.53m$

Therefore,

the vertical distance between the aeroplane is 1690.53m