Physics
posted by Anonymous .
A pilot must fly his plane to a town which is 200 km from his starting point in a direction 30 degrees N of E. He must make the trip in 1/2 hr. An 80km/h wind blows in a direction 30 degrees E of S. Find the speed of the plane relative to the ground, the speed relative to air and the heading of the plane.
The answers are 400 km/h, 408 km/h and 41 degrees N of E. I don't know how to do this problem. I think I did my diagram wrong.

One method
Three velocity vectors: the wind, the plane, and the desired results. Since the plane must reach the town in 1/2 hour, the results must be 400km/hr at 30 degrees N of E. That defines the speed over the ground.
Draw the results vector.
Now draw the wind vector as given. Handily for this problem the wind and the results are at 90 degrees. Draw the planes vector from the end of the wind vector to the end of the results vector. The length is the plane's speed. The heading of the plane is the angle where the plane's vector crosses the xaxis. There are several ways to find this angle. Be careful. It is not equal to any of the angles in the triangle just solved, but you do need them to find it. 
The heading solved will be a heading relative to east (chosen since the original problem used that). You can adjust that to a heading with 0 degrees at true north if desired.
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