Answer
a. $19.3\ mi$.
b. $S\ 58^\circ \ E$
Work Step by Step
a. Using the figure in the exercise, in triangle ABC, we have the angle
$B=90^\circ-40^\circ=50^\circ$
Using the Law of Cosines, we have
$b^2=25^2+13.5^2-2(25)(13.5)cos(50^\circ)\approx373$
which gives $b\approx19.3\ mi$.
b. Using the Law of Sines, we have
$\frac{sinA}{13,5}=\frac{sin50^\circ}{19.3}$
which gives
$sinA=\frac{13.5sin50^\circ}{19.3}\approx0.5355$
Thus, $A=asin(0.5355)\approx32^\circ$ and the bearing from A to C is
$S\ (90^\circ-32^\circ)\ E$ or $S\ 58^\circ \ E$
