## Precalculus (6th Edition) Blitzer

$N\ 12^\circ \ E$
Using the figure given in the exercise, based on the Law of Cosines, we have $7^2=5^2+6^2-2(5)(6)cos(A)$ which gives $cosA=0.2$ Thus angle $A=acos(0.2)\approx78^\circ$ and the bearing from A to C is $N\ (90^\circ-78^\circ)\ E$ or $N\ 12^\circ \ E$