The course taken by the pilot is okay. The wind is blowing from the about northwest (from 300 degrees, which is based from the north). The "horizontal" component of the wind is going eastward. To counteract that, the plane's velocity should have a horizontal component that is going westward. Hence the 357-degree direction, which is about north-northwest or almost northward but slightly westward too, is correct.

Draw the figure on paper.

a) First, the velocity vectors in relation to line AB.

In the northwest quadrant, or in the quadrant from 270deg to 360deg, say, AB is the north axis.

Draw the wind vector coming from the 300deg, or 30deg above the 270-deg axis or west axis, pointing to A.

Draw the plane vector going 357deg, or 3deg west of AB, originating at A.

Here, the angle between the wind and plane vectors is 90 -30 -3 = 57deg.

b) Then, the closed triangle of the 3 distances: AB, wind's distance after time t hours, and plane's distance after time t hours.

It is an obtuse triangle, with these:

---vertical side = AB = 280 km.

---bottom side = wt ----where w is wind's speed in kph, and t is time in hours.

---righthand side = 320t ----the plane's distance after time t.

---angle between wt and 320t = 57 deg.

---angle between wt and AB = 120 deg.

---angle between AB and 320t = 3 deg.

So it is a triangle with known 3 angles and a side.

Hence, the Law of Sines will find the two unknown sides wt and 320t.

280/sin(57deg) = 320t/sin(120deg)

Cross multiply,

280sin(120deg) = 320t*sin(57deg)

t = [280sin(120deg)] / [320sin(57deg)]

t = 0.90354 hr

t = 0.90354 *60 = 54.2 minutes -----the length of flight, answer.

280/sin(57deg) = wt/sin(3deg)

Cross multiply,

280sin(3deg) = wt*sin(57deg)

w = [280sin(3deg)] / [0.90354sin(57deg)]

w = 19.34 km/hr ---------------------------wind's speed, answer.