DIY Solar Panels

An airplane whose air speed is 620 km/h is supposed to fly in a straight path 35.0 degree north of east. But a steady 95 km/h wind is blowing from the north. In what direction should the plane head?
So if I draw two vectors:

vector 1

y= 620 sin 35 = 507.9
x= 620 cos 35 = 355.6

X Y
355.6 507.9
0 -95

355.6 412.9

Determine theta by using tan-1? Am I any closer to understanding it? I get 49 degrees and that isn’t correct.??
The answer is 42.2 North of East but i dont know how to get that

You’ve assumed that the plane will be traveling at a speed of 620 Km/h
in the direction 35NE.. This can’t be right because of the wind.
that speed is one of the unknowns.

however the airspeed in the unknown direction "theta" will be 620 Km/H.

airplane airspeed
y1 = 620 cos(theta)
x1 = 620 sin(theta)

Note that I used cosine for Y1
since due north —> i.e. 0 deg ===> Y1 = 620 cos(0) = 620.

wind
x2 = 0
y2 = -95. blowing toward the south.

resultant Vt = v1+v2 must be at 35NE

Xt = 620 sin(theta)
Yt = 620 cos(theta) – 95

tan35 = opposite/adjacent = Xt/Yt
Note that I used Xt/Yt since tan(0) is due north when Xt = 0.

Xt = Yt. tan35

Solve for theta.
620 sin(theta) = (620 cos(theta) – 95) . tan35