Answer
44.4$^o$ north of east.
Work Step by Step
Call east the positive x direction and north the positive y direction.
Let P denote the Plane, A the Air, and G the Ground. The pair “PG”, for example, represents the plane’s motion relative to the ground.
$$ \vec{v_{PG}} = \vec{v_{PA}} + \vec{v_{AG}} $$
See the diagram.
The vectors form a triangle. Apply the Law of Sines.
$$\frac{ v_{AG}}{sin \alpha} = \frac{ v_{PA}}{sin 128 ^{\circ} }$$
We find the small angle $\alpha$ to be 6.4$^o$. Therefore the heading of the plane should be 38 + 6.4 = 44.4$^o$ north of east.
