Answer
a. $23.1\ mi$.
b. $S\ 68^\circ \ E$
Work Step by Step
a. Using the figure as shown, in triangle ABC, we have the angle
$B=90^\circ-45^\circ=45^\circ$
Using the Law of Cosines, we have
$b^2=30^2+12^2-2(30)(12)cos(45^\circ)\approx534.9$
which gives $b\approx23.1\ mi$.
b. Using the Law of Sines, we have
$\frac{sinA}{12}=\frac{sin45^\circ}{23.1}$
which gives
$sinA=\frac{12sin45^\circ}{23.1}\approx0.3673$
Thus, $A=asin(0.3673)\approx22^\circ$ and the bearing from A to C is
$S\ (90^\circ-22^\circ)\ E$ or $S\ 68^\circ \ E$
