Answer
$A\approx21^\circ, B\approx32^\circ, C\approx127^\circ$
Work Step by Step
Step 1. Based on the given conditions, using the Law of Cosines, we have
$c^2=a^2+b^2-2ab\ cosC$, or $(9)^2=(4)^2+(6)^2-2(4)(6)cos(C)$, thus $cosC\approx-0.6042$ and $C=cos^{-1}(-0.6042)\approx127^\circ$
Step 2. Using the Law of Sines, we have
$\frac{SinA}{a}=\frac{SinC}{c}$, $SinA=\frac{4}{9}sin(127^\circ)\approx 0.3549$, thus $A=sin^{-1}(0.3549)\approx21^\circ$
Step 3. We can find the angle as
$B=180^\circ-127^\circ-21^\circ=32^\circ$
Step 4. We solved the triangle with
$A\approx21^\circ, B\approx32^\circ, C\approx127^\circ$
