#### Answer

$N\ 12^\circ \ E$

#### Work Step by Step

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 $N\ 12^\circ \ E$