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November 24, 2014

November 24, 2014

Posted by **bryan** on Thursday, November 29, 2012 at 12:24am.

_____mi/hr at ____° from due north

- trig -
**Reiny**, Thursday, November 29, 2012 at 8:40amI assume you are studying vectors.

Draw a line at 28° of length 200 (your resultant)

Draw a line at 38° of length 207, join its tail to the first line.

That smaller line will be the vector representing the wind

I see a triangle with sides 200 and 207, with a contained angle of 10°

Let the length of the smaller side be x

by cosine law:

x^2 = 200^2 + 207^2 - 2(200)(207)cos10°

etc.

x will be you speed of the wind, and from your diagram it should be straightforward to find the wind direction

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