Answer
$A\approx19^\circ, B\approx100^\circ, C\approx61^\circ$
Work Step by Step
Step 1. Based on the given conditions, using the Law of Cosines, we have
$b^2=a^2+c^2-2ac\ cosB$, or $(9)^2=(3)^2+(8)^2-2(3)(8)cos(B)$, thus $cosB\approx-0.1667$ and $B=cos^{-1}(-0.1667)\approx100^\circ$
Step 2. Using the Law of Sines, we have
$\frac{SinA}{a}=\frac{SinB}{b}$, $SinA=\frac{3}{9}sin(100^\circ)\approx 0.3283$, thus $A=sin^{-1}(0.3283)\approx19^\circ$
Step 3. We can find the angle as
$C=180^\circ-100^\circ-19^\circ=61^\circ$
Step 4. We solved the triangle with
$A\approx19^\circ, B\approx100^\circ, C\approx61^\circ$
