a plane can fly at 200km/h in still air and has to reach a destination 600km north-east. a 50km/h wind blows from the west. in which direction must the plane head?

The vector sum of the plane's velocity with respect to the air and the air velocity with respect to the ground must point northeast. What you want is the plane's velocity with respect to the air. The 600 km distance does not matter. Let Vy and Vx be the plane's air velocity components in the north and east directions.

For the resultant direction to be NE,
Vy = Vx - 50
200 cos A = 200 sin A - 50
where A is measured clockwise from north. That is the angle you want to compute.

cos A = sin A - 1/4
A = 55.2 degrees

Direction measured clockwise from North. Sail at angle T from North or Y axis.

Resultant must be 45 degrees (northeast)
East (x) velocity = 50 + 200 cos (90-T)
which is
50 + 200 sin T
North velocity = 200 cos T
Those east and north components are equal if the resultant is northeast.
50 + 200 sin T = 200 cos T
200 (cos T - sin T ) = 50
(cos T- sin T )= .25
I have to solve that by trial and error
T = 20 --> .597
T = 25 --> .484
so
T = 23 --> .529
T = 24 --> .507
north 24 degrees east
Sail at North 2 points east (a compass point is 11 1/4 degrees, 32 points of sail in a compass)

The wind is Blowing FROM the west

Damon is correct. I had the wind direction backwards

i don't understand why the 600km doesnt get incorperated into the calculations??

You need to keep to that flight direction no matter what the distance is. All they ask about is the direction that the plane flies. It is not unusual for homework and test problems to give you more information than you need to solve a problem. They are trying to prepare you for the real world.

thank you drwls :)