Question

the angle of elevation of a jet plane from a point a on the ground is 60 degree after a flight of 15 second the angle of elevation changes to 30 degree if the jet plane is flying at a constant height of 1500 root 3 metre find the speed of the jet​

Answer 1

Answer:

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• Let A be the original position of the jet plane.
• And B be the position of the Jet plane after 15 seconds.

$=> \sf Speed = \dfrac{Distance \: \: \: \leftarrow(distance \: AB)}{Time \: \: \: (given)}$

$=>\sf Distance \: AB = Distance \: CD$

$=>\sf CD = PC - CD$

__________________

$=>\sf \tan( \theta) =\dfrac{Opposite \: side}{Adjacent \: Side}$

$\sf In \: \Delta ADP,$

$=>\sf \tan(60^{ \circ} ) =\dfrac{1500 \sqrt{3} }{PD}$

$=>\sf \sqrt{3} =\dfrac{1500 \sqrt{3} }{PD}$

$=>\sf PD ( \sqrt{3} ) =1500 \sqrt{3}$

$=>\sf PD=\dfrac{1500 \sqrt{3} }{ \sqrt{3} }$

$=>\sf PD=1500 \: m$

_________________...

$\sf In \: \Delta BCP,$

$=>\sf \tan(30^{ \circ} ) =\dfrac{1500 \sqrt{3} }{PC}$

$=>\sf \dfrac{1}{ \sqrt{3} } =\dfrac{1500 \sqrt{3} }{PC}$

$=>\sf PC = 1500 \sqrt{3} \left(\sqrt{3} \right)$

$=>\sf PC =4500 \: m$

____________________...

$=>\sf CD = 4500 - 1500$

$=>\sf CD = 3000 \: m$

$=>\sf AB = 3000 \: m$

• Therefore, Speed of jet plane is :]

$=>\sf Speed \: of \: jet \: plane = \dfrac{3000}{15}$

$=>\sf Speed \: of \: jet \: plane = 200 \:m/s$