Answer
$A\approx117^\circ, B\approx36^\circ, C\approx27^\circ$
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 $(6)^2=(4)^2+(3)^2-2(4)(3)cosA$, thus $cosA\approx-0.4583$ and $A=cos^{-1}(-0.4583)\approx117^\circ$
Step 2. Using the Law of Sines, we have
$\frac{SinB}{b}=\frac{SinA}{a}$, $SinB=\frac{4}{6}sin(117^\circ)\approx0.594$, thus $B=sin^{-1}(0.594)\approx36^\circ$
Step 3. We can find the angle as
$C=180^\circ-117^\circ-36^\circ=27^\circ$
Step 4. We solved the triangle with
$A\approx117^\circ, B\approx36^\circ, C\approx27^\circ$
