Question

Q 5 An aero plane when flying at a height of 5000m from the ground passes vertically above
another aero plane 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
aero planes at the instant.​

Answer 1

Answer:

2113.239

Explanation:

tan 60° = 5000/x

tan 45° = y/x

=> y= 5000/tan 60° = 5000/√3

∴Vertical distance between planes=5000− 5000/√3

=5000[ √3 - 1/ √3]

=2113.249

Answer 2

Answer:

$\boxed{\sf \: Vertical\:distance\:between\:two\:planes= \: \dfrac{5000 (3- \sqrt{3} )}{3} \: m \: }$

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 5000 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

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

$\sf \: \sqrt{3} = \dfrac{5000}{CD}$

$\sf \: CD = \dfrac{5000}{ \sqrt{3} }$

$\sf \: CD = \dfrac{5000}{ \sqrt{3} } \times \dfrac{ \sqrt{3} }{ \sqrt{3} }$

$\implies\sf \:\boxed{\sf \: \sf \: CD = \dfrac{5000 \sqrt{3} }{3} \: m \: } \: - - - (1)$

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

$\implies\sf \:\boxed{\sf \: \sf \: BD = \dfrac{5000 \sqrt{3} }{3} \: m \: } \:$

Now,

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

$\sf \: = \: AB$

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

$\sf \: = \: 5000 - \dfrac{5000 \sqrt{3} }{3}$

$\sf \: = \: \dfrac{15000 - 5000 \sqrt{3} }{3}$

$\sf \: = \: \dfrac{5000 (3- \sqrt{3} )}{3} \: m$

Hence,

$\implies\sf \: Vertical\:distance\:between\:two\:planes= \: \dfrac{5000 (3- \sqrt{3} )}{3} \: m$