Question

an aeroplane when flying at a height of 4000m from the ground passes vertically above another aeroplane at an instant when the angles of the elevation of the two planes from the same point on the ground are 60 and 45 respectively. find the vertical distance between the aeroplane 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,

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