Answer :
Assuming east is positive x, north is positive y, break down the airplane's velocity into a vector in the x-y plane.
The plane is traveling 340 mph at a bearing of 210 degrees or (340cos(210), 340 sin(210)).
The wind is blowing from the west (to the east) at 50 mph or (50,0).
The actual ground speed would be the vector sum of those two (340 cos(210)+50, 340 sin(210)) or (-294.4+50,-170) or (-244.4,-170).
The magnitude of that velocity vector is
For direction
Since this angle is measured from the (-y) axis, we add 180 to get the actual heading.
The plane is traveling at 297.8 mph at a heading of 214.8 degrees.
The plane is traveling 340 mph at a bearing of 210 degrees or (340cos(210), 340 sin(210)).
The wind is blowing from the west (to the east) at 50 mph or (50,0).
The actual ground speed would be the vector sum of those two (340 cos(210)+50, 340 sin(210)) or (-294.4+50,-170) or (-244.4,-170).
The magnitude of that velocity vector is
For direction
Since this angle is measured from the (-y) axis, we add 180 to get the actual heading.
The plane is traveling at 297.8 mph at a heading of 214.8 degrees.