## Precalculus (6th Edition) Blitzer

a. $19.3\ mi$. b. $S\ 58^\circ \ E$
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 $S\ 58^\circ \ E$