## Precalculus (6th Edition) Blitzer

$A\approx31^\circ, B\approx125^\circ, C\approx24^\circ$
Step 1. Based on the given conditions, using the Law of Cosines, we have $a^2=b^2+c^2-2bc\ cosA$, we have $(10)^2=(16)^2+(8)^2-2(16)(8)cosA$, thus $cosA\approx0.8594$ and $A=cos^{-1}(0.8594)\approx31^\circ$ Step 2. Using the Law of Sines, we have $\frac{SinC}{c}=\frac{SinA}{a}$, $SinB=\frac{8}{10}sin(31^\circ)\approx0.412$, thus $C=sin^{-1}(0.412)\approx24^\circ$ Step 3. We can find the angle as $B=180^\circ-31^\circ-24^\circ=125^\circ$ Step 4. We solved the triangle with $A\approx31^\circ, B\approx125^\circ, C\approx24^\circ$