SOLUTION: A plane is steering at S65°W at an air speed of 625 km/h. The wind is from the NW at 130 km/h. Find the ground speed and track of the plane. Include a vector diagram in your solut

    "S65°W" refers to a bearing or direction, meaning South 65 degrees West. 
    This notation indicates a direction starting from South and then rotating 65 degrees towards the West."


               S O L U T I O N


So, S65°W is the direction of  270° - 65° = 205° in the standard coordinate plane.

The direction of the wind "from NW" is 135° + 180° = 315° in the standard coordinate plane.


Thus, we are given two vectors

    the airspeed of the plane = 625*(cos(205°), sin(205°))  km/h;     (1)

    the wind speed vector     = 130*(cos(315°), sin(315°))  km/h.     (2)


Let the groundspeed of the plane be  (x,y).

The groundspeed is the sum of the vectors (1) and (2)


    x = 625*cos(205°) + 130*cos(315°) = 625*(-0.90630778703) + 130*(+0.70710678118) = -474.5184853 km/h;

    y = 625*sin(205°) + 130*sin(315°) = 625*(-0.42261826174) + 130*(-0.70710678118) = -356.0602951 km/h.


So, the groundspeed magnitude is   =  = 593.2509812 km/h.


The angle of the vector with the x-direction of the coordinate plane is


    a =  +  =  +  =  +  =  +  = 

      = 3.14159265 + 0.643732 = 3.78532465 radians = 216.883127 degrees.


ANSWER.  The groundspeed magnitude is about 593.251 km/h.

          The direction of the groundspeed is about 216.883 degrees counterclockwise from positive direction of x-axis,
          or 90 - 36.883 = 53.117 degrees from South to East  (S53.117°W).