## Precalculus (6th Edition) Blitzer

$A=31^\circ, B=122^\circ, C=27^\circ$
Step 1. Using the given figure in the exercise and the Law of Cosines, we have $b^2=a^2+c^2-2ac\ cosB$ Thus $cosB=\frac{17^2+15^2-28^2}{2(17)(15)}\approx-0.5294$ and $B=acos(-0.5294)\approx122^\circ$ Step 2. Using the Law of Sines, we have $\frac{sinA}{a}=\frac{sinB}{b}$ and $sinA=\frac{17sin122^\circ}{28}\approx0.5149$ and $A=asin(0.5149)\approx31^\circ$ Step 3. Angle $C=180^\circ-B-A\approx27^\circ$ Step 4. The solutions are $A=31^\circ, B=122^\circ, C=27^\circ$