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

#1**+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**+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