Question

an aeroplane when 3000m high passes vertically above another aeroplane at an instant when the angles of elevation of aeroplane at the same observation points are 60° and 45°.how much high is the one plane than the other?

Answer 1

i hope this will usful to u

Answer 2

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 = 3000m.

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

$\tan 60=\frac{3000}{A B}$

The angle for the second plane is 45, which means

$\tan 45=\frac{h}{A B}$

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

$\begin{array}{l}{A B=\frac{3000}{\tan 60} ; \quad A B=\frac{h}{\tan 45}} \\ {A B=\frac{3000}{\sqrt{3}} ; \quad A B=\frac{h}{1}}\end{array}$

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

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

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

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.