Let the position of the airport be A, the point 160 miles north be B, and the point 85 miles east of that (i.e. the location of the plane) be C. Then ABC forms a right-angled triangle.
<span>tan BAC = 85/160 </span>
<span>BAC = arctan (85/160) </span>
<span>= arctan 0.531 (approx.) </span>
<span>= 27.98° (approx.) </span>
<span>So C is N 27.98° E from A. </span>
<span>The plane should fly S 27.98° W to get to the airport.
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