Question 266816: When a 757 passenger jet begins its descent to the Ronald Regan International Airport in Washington, D.C, it is 3900 feet from the ground. Its angle of descent is 6 degrees.
A. What is the plane’s ground distance to the air port?
B. How far must the plane fly to reach the runway?
Found 2 solutions by stanbon, Alan3354:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! When a 757 passenger jet begins its descent to the Ronald Regan International Airport in Washington, D.C, it is 3900 feet from the ground. Its angle of descent is 6 degrees.
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Draw the picture: A right triangle with angle at the ground = 6 degrees
and side opposite = 3900 ft.
A. What is the plane’s ground distance to the air port?
tan(6 degrees) = 3900/x
x = 3900/tan(6 degrees) = 37,106 ft
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B. How far must the plane fly to reach the runway?
sin(6 degrees) = 3900/r
r = 3900/sin(6) = 37,310.4 ft
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Cheers,
Stan H.
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! When a 757 passenger jet begins its descent to the Ronald Regan International Airport in Washington, D.C, it is 3900 feet from the ground. Its angle of descent is 6 degrees.
A. What is the plane’s ground distance to the air port?
tan(6) = 3900/s
s = 3900/tan(6)
s = 37,106 feet
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B. How far must the plane fly to reach the runway?
sin(6) = 3900/hyp
hyp = 37,310 feet
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