Answer
$A\approx31^\circ, B\approx125^\circ, C\approx24^\circ$
Work Step by Step
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
$A\approx31^\circ, B\approx125^\circ, C\approx24^\circ$
