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?
+9479 
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
+9479
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
