## Precalculus (6th Edition) Blitzer

two triangles: $C\approx83^\circ, B\approx48^\circ, b\approx93.5$ and $C\approx97^\circ, B\approx34^\circ, b\approx70.4$
Step 1. Draw a diagram as shown in the figure. Step 2. Using the Law of Sines, we have $\frac{sinC}{125}=\frac{sin49^\circ}{95}$, thus $sinC=\frac{125sin49^\circ}{95}\approx0.993$ Step 3. The angle C could be $C=sin^{-1}0.993\approx83^\circ$ or $C=180^\circ-83^\circ=97^\circ$ Step 4. For $C\approx83^\circ$, we have $B\approx180^\circ-83^\circ-49^\circ\approx48^\circ$ and $b=\frac{95sin48^\circ}{sin49^\circ}\approx93.5$ Step 5. For $C\approx97^\circ$, we have $B\approx180^\circ-97^\circ-49^\circ\approx34^\circ$ and $b=\frac{95sin34^\circ}{sin49^\circ}\approx70.4$ Step 6. We conclude that there are two triangles: $C\approx83^\circ, B\approx48^\circ, b\approx93.5$ and $C\approx97^\circ, B\approx34^\circ, b\approx70.4$