Question

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

Answer 1

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

Answer 2

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