Find 3rd side, given two sides and bearings The bearing from A to B is N $42^\circ$ E. The bearing from B to C is S $44^\circ$ E. A small plane traveling $65$ miles per hour, takes $1$ hour to go from A to B and $2$ hours to go from B to C. Find the distance from A to C.

City A is furthest west.  City C is furthest east.  City B is furthest north.  So cities A,B,C make a triangle.  Let D denote that point on the base of the triangle that is directly south from B.  So the bearing from A to B is N 42∘ E, but by alternate interior angles, angle A,B,D is also 42∘.  The bearing from B to C is S 44∘ E.  So angle D,B,C is 44∘.  Thus angle A,B,C is 42∘ + 44∘ = 86∘.  So, despite "birdkiller's" comment, I think that 86∘ is correct.  The law of cosines is indeed the correct formula to apply.  Begin with $b^2 = a^2 + c^2 - 2acCosB$.
