Math, asked by Kushagg1454, 11 months ago

An aeroplane, when 3000m high, passes vertically above another aeroplane at an instant when the angles of elevation of aeroplanes at the same observation points are 60^0 and 45^0 respectively. How much high is the one plane than the other

Answers

Answered by bhartgarv
3

Answer:

1268 m

Step-by-step explanation:

In given figure Let AB=3000m and AD=x M then BD=3000-x

THEN THE SOLUTION IS SHOWN ABOVE IN PHOTO UPLOADED

Attachments:
Answered by sanjeevk28012
0

Answer:

The distance between both aeroplanes are 1267.899 meters .

Step-by-step explanation:

Given as :

The height of first aeroplane at position C = AC = 3000 m

The height of second aeroplane at position B = BA = 3000 - h

Let The distance between both plane = BC = h meter

The horizontal distance for both plane = OA = x meters

The angle made by first plane with ground = 60°

The angle made by second plane with ground = 45°

According to question

For Triangle OAC ,

Tan angle = \dfrac{perpendicular}{base}

or, Tan 60° = \dfrac{AC}{OA}

Or, Tan 60° = \dfrac{3000}{x}

Or, 1.732 = \dfrac{3000}{x}

∴   x = \dfrac{3000}{1.732}

i.e x = 1732.101           ............1

So, The horizontal distance of aeroplane = OA = x = 1732.101

Again

For Triangle OAB ,

Tan angle = \dfrac{perpendicular}{base}

Tan 45°  = \dfrac{AB}{OA}

1 = \dfrac{3000 - h}{x}

Or, x = 3000 - h

Put the value of x from 1

So, 1732.101 = 3000 - h

∴  h = 3000 - 1732.101

i.e h = 1267.899  meters

So, The distance between both plane = BC = h = 1267.899 meter

Hence, The distance between both aeroplanes are 1267.899 meters . Answer

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