Answer
$A\approx39^\circ, B\approx49^\circ, C\approx92^\circ$
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 $(10)^2=(12)^2+(16)^2-2(12)(16)cosA$, thus $cosA\approx0.7813$ and $A=cos^{-1}(0.7813)\approx39^\circ$
Step 2. Using the Law of Sines, we have
$\frac{SinB}{b}=\frac{SinA}{a}$, $SinB=\frac{12}{10}sin(39^\circ)\approx0.7552$, thus $B=sin^{-1}(0.7552)\approx49^\circ$
Step 3. We can find the angle as
$C=180^\circ-39^\circ-49^\circ=92^\circ$
Step 4. We solved the triangle with
$A\approx39^\circ, B\approx49^\circ, C\approx92^\circ,$
