Physics
posted by Daniel on .
An airplane ﬂies at airspeed (relative to the air) of 430 km/h . The pilot wishes to ﬂy due North (relative to the ground) but there is a 43 km/h wind blowing Southwest (direction 225◦).
In what direction should the pilot head the plane (measured clockwise from North)?
Answer in units of degrees.

43/√2 = 30.4, so the wind's vector is (30.4,30.4)
If the plane flies on heading θ,
(30.4,30.4)+(430cos(θ45°),430sin(θ45°)) = (0,y)
so, we have 30.4 + 430cos(θ45) = 0
cos(θ45) = 30.4/430 = 4.05
so, the plane flies N41°E 
is 41 degrees east of north the exact degree number??? because a computer program wants the exact number and i cant round over or under 1%...