A plane is 48 miles west and 49 miles north of an airport. The pilot wants to fly directly to the airport. What bearing should the pilot take? Answer should be

Question

 I am just a little unsure about the bearing system. I have all of the other information, however, regarding the triangle that the problem makes on the cartesian plane

Answer 1

Bearings are measured as an acute angle from the north-south line.  If you put the airport at the origin (0,0) then the plane is at (-48,49).  To go directly to the airport the pilot must fly at an angle theta towards the origin, where theta is measured from the due south (negative y) direction.  Theta has the property that 

 

tan(theta) = 48/49  ==>  theta = arctan(48/49) ~ 44.4 degrees 

 

The bearing is written in 3 parts 

 

The first letter is N or S depending on whether the angle is measured from North or South

The second part is the measure of the angle

The third part is E or W depending on whether the angle is East or West of the north-south line

 

So in this case the bearing is S (44.4 degrees) E