An airplane, flying at a speed of 420 miles per hour, flies from point A in the direction 131 degrees for 90 min. and then flies in the direction 41 degrees for 30 min. in what direction does the plane need to fly in order to get back to point A?

Answer 1

the speed is irrelevant, since it does not change. So the relative times are the same as the relative distances.

90 min at 131° moves the plane 90(cos41°,-sin41°) = (67.92,-59.04)

30 min at 41° moves it an additional 30(cos41°,sin41°) = (22.64,19.68)

So, total displacement is (90.56,-39.36)

tanθ = 90.56/39.36 = 2.30
θ = 66.5°

the heading back to A is thus 360-66.5 = 293.5°