Answer
$b=4.3,A=48^\circ,C=100^\circ$
Work Step by Step
Step 1. Based on the given conditions, using the Law of Cosines, we have
$b^2=a^2+c^2-2ac\ cosB=(6)^2+(8)^2-2(6)(8)cos(32^\circ)\approx18.59$, thus $a\approx\sqrt {18.59}\approx4.3$
Step 2. To find the unknown angles, using the Law of Sines, we have
$\frac{SinB}{b}=\frac{SinA}{a}$, $SinA=\frac{6}{4.3}sin(32^\circ)\approx0.7394$, thus $A=sin^{-1}(0.7394)\approx48^\circ$
Step 3. We can find the angle as
$C=180^\circ-32^\circ-48^\circ=100^\circ$
Step 4. We solved the triangle with $b=4.3,A=48^\circ,C=100^\circ$
