an aeroplane when flying at a height of 5000m above the ground passes vertically above another aeroplane at an instant when the angle of elevation of two aeroplane from the same point on the ground are 60 ans 45 respectively. find the vertical distance between yhe two panes at that instsnt.
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The distance is 2116 .67 m
vertical distance between the two planes = 5000 - 5000√3 m = 2113 m
Step-by-step explanation:
an aeroplane flying at a height of 5000m above the ground
=> Tan 60° = Height of Plane / Horizontal Distance of Observing point from planes
=> √3 = 5000/Horizontal Distance of Observing point from planes
=> Horizontal Distance of Observing point from planes = 5000/√3
Tan45° = Height of Lower plane from Ground / Horizontal Distance of Observing point from planes
=> 1 = Height of Lower plane from Ground /(5000/√3)
Height of Lower plane from Ground = 5000/√3
vertical distance between the two planes = 5000 - 5000√3 m
= 2113 m
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