A pilot wants to fly from city A to city B, a distance of 394 km at an angle of 10.0 degrees west of north.

Question

The pilot heads directly toward city B, with an air speed of 203 km/h.
After flying for 1.0 hours, the pilot finds she is 17.5 km off course to the west of where she expected to be, assuming there was no wind.

In what direction should the pilot have aimed her plane to fly directly to city B without being blown off course? Answer in degrees west of north.

Answer 1

using similar triangles, you can see that after flying the full 394 km, the plane would be

394/203 * 17.5 = 34 km west of city B.

So, in the diagram, let
P = place where the plane would end up after 394km

Draw a line east from P to intersect AB at Q, and the line north from A at R.

PR = 394 cosP
AR = 394 sinP
QR = AR tan10°
PR = 34+QR, so
34 + 394 sinP tan10° = 394 cosP
angle P = 75.1°

We want angle A in triangle APQ, which is how far off-course the plane was heading.

AR = 394 sinP = 381
cos(angle PAR) = 381/394, so angle PAR = 14.75°

But <PAR = <PAQ+10°, so <PAQ = 4.75°

So, the plane should have headed 5.25° west of north, rather than 10°.