Question

I've been struggling a lot with interpreting what this problem wants me to find.

The plane departs an airport at 9:00 AM and flies 200 mph southwest. A second plane ldeparts the same airport at 9:30 AM and flies 320 mph northwest. Find the bearing of the second plane from the first one at 10:00 AM.

The reason: I've heard some people say a bearing has to be in one direction (north). I've also heard that a bearing has to be an acute angle and some insist there is an answer to this greater than 90º.

I drew out a diagram with a right triangle, legs of 200 and 160 (320 ÷ 2 for the half hour), separated my right triangle into two triangles with 45º angles, and went from there.

Where the hypotenuse and the 160 side meet, the angle is 51.34º. The 200 side makes a 38.66º angle (both came from arctan). I know for a fact there is an 83.66º angle that can be found.

Trouble is, I'm not sure if that's the actual bearing or not. Everything SEEMS correct there, but I've just heard so many other answers on this that I'm not sure who or what to believe!

Figured I'd see what everyone else thought. It's the "second from the first" that I'm struggling with semantically too. I can do the math just fine here.

Answer 1

6.34°

Explanation:

You want to know the bearing of a second plane from the first at 10 a.m. when the first leaves at 9 a.m. and flies 200 mph SW, while the second leaves at 9:30 a.m. and flies 320 mph NW.

Locations

The first plane flies for an hour at 200 mph, so is 200 miles from the airport on a bearing of 180° + 45° west of south, or 225°.

The second plane flies for half an hour at 320 mph, so is 160 miles from the airport on a bearing of 360° -45° west of north, or 315°.

Direction

The vector in the direction of the second plane from the first is found by subtracting the location of the first plane from the second:

vector of interest = (160∠315°) -(200∠225°)

A suitable calculator can tell you the difference is ...

vector of interest = 256∠6.34°

The second plane is on a bearing of 6.34° from the first.

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Additional comment

You are correct that an angle of 83.66° is involved. That would be the measure of the direction vector counterclockwise from the +x axis. You want the bearing angle, measured clockwise from the +y axis (North), so the angle you're looking for is 90° -83.66° = 6.34°.

If you work enough of these problems, you will find that it doesn't matter how you measure the angles, as long as you do it consistently. The second attachment shows the arithmetic using distance∠bearing. The effect is to swap the x, and y coordinates so that they are (N, E) coordinates, rather than the usual (E, N) coordinates. This doesn't have to be confusing. A diagram usually helps.

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