Question

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

Answer 1

$\huge \bf{ \red{hey}}$

$\huge{ \mathfrak{ \pink{here \: is \: answer}}}$

$\bf{question - }$

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

Now

$\bf{solution - }$

$\sf{given - }$

angle of elevations are

60° and 45°

and

height is 4000m

$\bf{ \green{step \: by \: step \: explanation}}$

picture is in attachment

so

the vertical distance

sin 60°= AB/AC

sin60°=4000/AC

$\frac{ \sqrt{3} }{2} = \frac{4000}{ac}$

$\sqrt{3}ac = 8000$

$ac = \frac{8000}{ \sqrt{3} }$

$ac = \frac{8000 \times \sqrt{3} }{ \sqrt{3} \times \sqrt{3} }$

$\sf{ac = \frac{8000 \sqrt{3} }{ 3}}$

and

now

sin45°=AB/AD

sin45°=4000/AD

$\frac{1}{ \sqrt{2} } = \frac{4000}{ad}$

$\sf{ad = 4000 \sqrt{2}}$

Answer 2

Answer:

$\boxed{\sf \: Vertical\:distance\:between\:two\:planes= \:1690.67 \: m \: }$

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,

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

$\sf \: = \: AB$

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

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

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

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

$\sf \: = \: \dfrac{4000(3 - 1.732)}{3}$

$\sf \: = \: \dfrac{4000 \times 1.268}{3}$

$\sf \: = \: \dfrac{5072}{3}$

$\sf \: = \: 1690.67 \: m$

Hence,

$\implies\sf \: Vertical\:distance\:between\:two\:planes= \: 1690.67 \: m$