The nose of an ultralight plane is pointed south, and its airspeed indicator shows 28 m/s . the plane is in a 18 m/s wind blowing toward the southwest relative to the earth.

Here I think you have to find the velocity in x and y components where x is east and y is north
 So as air speed indicator shows the negative speed in y component and adding it in
  air speed while multiplying with the direction component we will get the velocity as velocity is a vector quantity so direction is also required
 v=-28 m/s y + 18 m/s (- x/sqrt(2) - y/sqrt(2))
 solving
  v= -12.7 m/s x-40.7 m/s y
 if magnitude of velocity or speed is required then
  speed= sqrt(12.7^2 + 40.7^2)
 speed= 42.63 m/s
 if angle is asked
  angle = arctan (40.7/12.7)
 angle = 72.67 degrees south of west

Ithink they do have the same properties just in less quantities i think