Question 1179989
The resultant vector will have direction due north and magnitude equal to the 
speed: 400 mi/2.5 h = 160 miles per hr
The air speed and heading of the plane must be able to counteract the speed 
and direction of the wind. The direction of the wind is 58 deg E of N, which 
is 32 deg N of E.  The plane must head west of north to counteract this. Adding
the vectors head to tail, we get the resultant vector. The coordinates of the
end point of the resultant vector are given by (x,y) = (-11cos(32),160-11sin(32))
Thus the air speed is given by sqrt(x^2+y^2) = 154.45
The direction is given by atan(x/y) = -3.46 deg, or 3.46 deg W of N
Ans: speed = 154.45 mph, direction 3.46 deg W of N
See attached diagram.
*[illustration vectors.png]