Airplane landing on a runway

1. The problem statement, all variables and given/known data

3. The attempt at a solution
I tried to use vector addition. I got the correct groundspeed (65.9). However I could not get the angle. The answer key says the angle is 87.7 degrees but I can't see how I can get that with my diagram. I think I must be interpreting something wrong.

http://i.imgur.com/LNscV.jpg
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 Recognitions: Homework Help Basically you need to sketch out the scalene triangle and work out the interior angles. Some of them you can get from what you are given, like the angle between the runway and the wind, but I suspect the one you want will need the sine rule or the cosine rule. Ten you have to convert the interior angle to a bearing. If you get stuck - try doing a careful scale drawing with a ruler and protractor :) Note: for a bearing of 87.7 degrees, the plane must be approaching the runway from the opposite side to that shown in the diagram (or fly in backwards!)

 Quote by Simon Bridge Basically you need to sketch out the scalene triangle and work out the interior angles. Some of them you can get from what you are given, but I suspect the one you want will need the sine rule or the cosine rule. If you get stuck - try doing a careful scale drawing with a ruler and protractor :)
I got all 3 of the angles. Are those the ones you are referring to?

Recognitions:
Homework Help

Airplane landing on a runway

All three internal angles - inside the corners of the triangle?
<looks again> Oh I see ... so you are stuck on how to turn the angle into a bearing?

You have the plane angled 7.715 degrees north of the runway, and the runway is bearing 260 degrees ... so the plane is bearing ...? Don't worry about what the model answer says: what makes sense?
 I believe the official answer is wrong by 180 degrees. Looks like they did.. 260 + 7.7 + 180 = 87.7 I believe the correct answer should be 260 + 7.7 = 267.7
 Recognitions: Homework Help It would be good for OP's confidence to have that realisation instead of just being told though ;) Part of the power of science is being able to overturn an authority.

 Quote by Simon Bridge It would be good for OP's confidence to have that realisation instead of just being told though ;) Part of the power of science is being able to overturn an authority.
Hahaha indeed, but thanks for the help everyone! I agree with why it is 267.7 but just in case I have posted the solution written by our professor:

In his diagram he shows that u vector points towards 267.7 however his answer ends up to be 87.7.
 Recognitions: Homework Help The x axis may well "point towards 80deg", but, in that case, the plane is travelling in the -x direction! The -x axis is pointing at 260deg.

 Quote by Simon Bridge The x axis may well "point towards 80deg", but, in that case, the plane is travelling in the -x direction! The -x axis is pointing at 260deg.
Ah okay, so he just mixed up the direction of u. Thanks a lot!
 I think it is the way the aviation do with direction. Non-direction beacon (ndb) is used for aircraft to navigate TO the airport. If you're heading to the airport, you heading according to NDB is heading to zero degree. Here in the example, heading to runway 26(heading 260) equal to zero degree heading which means you are aligned with the runway. Please correct me if i'm wrong.
 Recognitions: Homework Help Yeah - you could answer the question ... the aircraft is heading 7.7 degrees starboard of the runway direction on approach from the east end. Sometimes you have a beacon, sometimes you use compass bearings to line you up. This question defined a 0deg direction ... nothing to do with ndb, prof just mixed up the directions. If the approach was from the west, the the heading would be 7.7 degrees port of the runway for a bearing of 82.3 degrees from north.

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