Question

An aero plane, when flying at a height of 4000m from the ground passes
vertically above another aero plane at an instant when the angle of elevation of
the two planes from the same point on the ground are 60° and 45° respectively.
Find the vertical distance between the aero planes at that instant.​

Answer 1

Answer:

$\boxed{\sf \: Vertical\:distance\:between\:two\:planes= \: \dfrac{4000 (3- \sqrt{3} )}{3} \: m \: }$

Step-by-step explanation:

Let A and B be the position of two aeroplanes, when B is vertically below the A and height of plane A from ground is 4000 m.

Let C be some point on plane such that the angles of elevation of the two aeroplanes from the point C are 60° and 45° respectively.

Now, In right-angle triangle ADC

$\sf \: tan {60}^{ \circ} = \dfrac{AD}{CD}$

$\sf \: \sqrt{3} = \dfrac{4000}{CD}$

$\sf \: CD = \dfrac{4000}{ \sqrt{3} }$

$\sf \: CD = \dfrac{4000}{ \sqrt{3} } \times \dfrac{ \sqrt{3} }{ \sqrt{3} }$

$\implies\sf \:\boxed{\sf \: \sf \: CD = \dfrac{4000 \sqrt{3} }{3} \: m \: } \: - - - (1)$

Now, In right-angle triangle BCD

$\sf \: tan {45}^{ \circ} = \dfrac{BD}{CD}$

$\sf \: 1 = \dfrac{BD}{CD}$

$\implies\sf \: \sf \: {BD} = {CD}$

So, using equation (1), we get

$\implies\sf \:\boxed{\sf \: \sf \: BD = \dfrac{4000 \sqrt{3} }{3} \: m \: } \:$

Now,

$\sf \: Vertical\:distance\:between\:two\:planes$

$\sf \: = \: AB$

$\sf \: = \: AD - BD$

$\sf \: = \: 4000 - \dfrac{4000 \sqrt{3} }{3}$

$\sf \: = \: \dfrac{12000 - 4000 \sqrt{3} }{3}$

$\sf \: = \: \dfrac{4000 (3- \sqrt{3} )}{3} \: m$

Hence,

$\implies\sf \: Vertical\:distance\:between\:two\:planes= \: \dfrac{4000 (3- \sqrt{3} )}{3} \: m$