Answer:
4.3° west of north
Step-by-step explanation:
The wind is blowing toward a direction 45° east of north, so will have an easterly component that is ...
(58 kph)sin(45°) = 41.0122 kph
We want the easterly component of the airplane's flight to be the opposite of this, so the angle (x) east of north is ...
(546 kph)sin(x) = -41.0122
sin(x) = -41.0122/546
x = arcsin(-41.0122/546) ≈ -4.31°
The pilot needs to fly 4.3° west of north to maintain a ground path that is straight north.