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when a passenger jet begins its descent to the airport, it is 3900 feet from the ground. Its angle of descent is 6 degrees.

a) what is the plane's ground distance to the airport?

b) how far must the plane fly to reach the runway?

Guest Mar 7, 2018

Best Answer 

 #1
avatar+7155 
+2

Point  A  is the current location of the plane.

Point  B  is the point on the ground directly under the plane.

Point  C  is the location of the airport.

 

m∠BAC   =   90° - 6°   =   84°

 

a)

plane's ground distance to the airport =  BC

tan 84°  =   BC / AB

tan 84°  =   BC / 3900

3900 tan 84°  =  BC

BC  ≈  37106.021    feet

 

b)

distance between plane and runway   =   AC

cos 84°   =   AB / AC

cos 84°   =   3900 / AC

AC cos 84°  =  3900

AC  =  3900 / cos 84°

AC  ≈  37310.412    feet

hectictar  Mar 7, 2018
 #1
avatar+7155 
+2
Best Answer

Point  A  is the current location of the plane.

Point  B  is the point on the ground directly under the plane.

Point  C  is the location of the airport.

 

m∠BAC   =   90° - 6°   =   84°

 

a)

plane's ground distance to the airport =  BC

tan 84°  =   BC / AB

tan 84°  =   BC / 3900

3900 tan 84°  =  BC

BC  ≈  37106.021    feet

 

b)

distance between plane and runway   =   AC

cos 84°   =   AB / AC

cos 84°   =   3900 / AC

AC cos 84°  =  3900

AC  =  3900 / cos 84°

AC  ≈  37310.412    feet

hectictar  Mar 7, 2018

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