## Precalculus (6th Edition) Blitzer

$A\approx38^\circ, B\approx61^\circ, C\approx81^\circ$
Step 1. With the given numbers, we have $a=BC=4.3+3.0=7.3, b=AC=7.5+3.0=10.5,c=AB=7.5+4.3=11.8$ Step 2. Using the Law of Cosines, we have $c^2=a^2+b^2-2ab\ cosC$ or $11.8^2=7.3^2+10.5^2-2(7.3)(10.5)\ cosC$ which gives $cosC\approx0.1585$ and $C=acos(0.1585)\approx81^\circ$ Step 3. Using the Law of Sines, we have $\frac{sinA}{a}=\frac{sinC}{c}$ and $sinA=\frac{7.3sin(81^\circ)}{11.8}\approx0.6110$ thus $A=asin(0.6110)\approx38^\circ$ and $B\approx180^\circ-81^\circ-38^\circ=61^\circ$ Step 4. The solutions are $A\approx38^\circ, B\approx61^\circ, C\approx81^\circ$