Answer
$c=7.6,A=52^\circ,B=32^\circ$
Work Step by Step
Step 1. Based on the given conditions, using the Law of Cosines, we have
$c^2=a^2+b^2-2ab\ cosC=(6)^2+(4)^2-2(6)(4)cos(96^\circ)\approx57.02$, thus $c\approx\sqrt {57.02}\approx7.6$
Step 2. To find the unknown angles, using the Law of Sines, we have
$\frac{SinB}{b}=\frac{SinC}{c}$, $SinB=\frac{4}{7.6}sin(96^\circ)\approx0.5268$, thus $B=sin^{-1}(0.5268)\approx32^\circ$
Step 3. We can find the angle as
$A=180^\circ-96^\circ-32^\circ=52^\circ$
Step 4. We solved the triangle with $c=7.6,A=52^\circ,B=32^\circ$
