Question

An aeroplane when flying at a height 4000 meters from the ground passes vertically above the another aeroplane at an instant when the angle of elevation of the two planes from the same point on the ground are 60⁰and 45⁰respectively. 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,

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