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.
 A: According to the US Army definition of bearing, which this question seems to be using by the format, the layout is as follows:
$\hspace{3.5cm}$
Using the Law of Cosines, we get
$$
\begin{align}
\overline{AC}^2
&=\overline{AB}^2+\overline{BC}^2-2\overline{AB}\,\overline{BC}\cos(\angle ABC)\\
&=65^2+130^2-2\cdot65\cdot130\cos(86^\circ)
\end{align}
$$
A: You have $<ABC = 42+44 = 86^\circ$, and $AB = 65$, $BC = 130$. Use law of cosine to get the answer.
A: 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 $b^2 = a^2 + c^2 - 2acCosB$.
A: Start at $A$, go $42^o$ to B and then head $46^o$ to $C$. So the angle at $B$ is $(90-42)^o+46^o=94^o$,  $c=AB=65$ and $a=BC=130$.  Now use the law of cosines
$b^2=a^2+c^2-2ac\cos B$ to get $b=149.34$
