Question

An aeroplain when 'x' metres high passes vertcally above another aeroplane at instant when angle of elevation of two aeroplane from the same point on the ground is 60 and 45 respectively . find vertical distance between two aerplane is​

Answer 1

Question :-

An aeroplane when flying at a height of 3000 m above 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 are 60 degree and 45 degree respectively. Find the vertical distance between the aeroplanes at that instant.  

Answer :-

1268 m is the height of one plane than the other.  

Given :-  

The aeroplane when 3000 m high passes vertically above another aeroplane.

Angle of elevation of the aeroplane’s at the same observation = 60°and 45°

To find :-

How much high is the one plane to other = ?

Solution :-  

The height at which the aeroplane (1) is from the ground is of height = 3000 m.  

The observer is standing at point A from where he observe both the planes one at D and another at C  

The height at which the aeroplane (2) is from the ground is of height = h meters.

To find the height difference between the two planes, we first have to find the height of the second plane  

$tan 60 = \frac{3000}{AB}$

The angle for the second plane is 45, which means  

$tan 45 = \frac{h}{AB}$

Now equating the value of the length between the planes and the observer  

$AB = \frac{3000}{tan60} ; AB = \frac{h}{tan45}$

$AB = \frac{3000}{\sqrt{3} } ; AB = \frac{h}{1}$

$\frac{3000}{\sqrt{3} } X \frac{\sqrt{3} }{\sqrt{3} } = h$

$h = \frac{3000\sqrt{3} }{3} = 1000 X 1.732 = 1732m$

The height of the second plane from the ground is 1732 m  

Therefore, the Distance between the planes or the second plane is (3000 – 1732) = 1268 m lower than the first plane,  

The height that the first plane is from second plane is 1268 m.

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