## Precalculus (6th Edition) Blitzer

$9.9\ mi$ or $2.4\ mi$
Step 1. See figure. Based on the given conditions, we can identify that in triangle ABC, angle $B=90^\circ-62^\circ=28^\circ$, Step 2. Using the Law of Sines, we have $\frac{sinC}{7}=\frac{sin28^\circ}{5}$ Thus $sinC=\frac{7sin28^\circ}{5}\approx 0.6573$ and $C=sin^{-1}0.6573\approx41^\circ$ or $C=180^\circ-41^\circ=139^\circ$ Step 3. We can find the angle as $A=180^\circ-28^\circ-41^\circ=111^\circ$ or $A=180^\circ-28^\circ-139^\circ=13^\circ$ Step 4. For $A=111^\circ$, we have $\frac{sin111^\circ}{a}=\frac{sin28^\circ}{5}$, thus $a=\frac{5sin111^\circ}{sin28^\circ}\approx 9.9\ mi$ Step 5. For $A=13^\circ$, we have $\frac{sin13^\circ}{a}=\frac{sin28^\circ}{5}$ Thus $a=\frac{5sin13^\circ}{sin28^\circ}\approx 2.4\ mi$