Answer
$b\approx4.7, B\approx75^\circ, C\approx55^\circ$
Work Step by Step
Step 1. Based on the given conditions, using the Law of Cosines, we have
$b^2=a^2+c^2-2ac\ \cos B$, or $b^2=(6)^2+(5)^2-2(6)(5)\cos(50^\circ)\approx 22.43$, thus $b=\sqrt {22.43}\approx 4.7$
Step 2. Using the Law of Sines, we have
$\frac{\sin B}{b}=\frac{\sin C}{c}$, $\sin C=\frac{5}{4.7}\sin(50^\circ)\approx 0.8149$, thus $C=\sin^{-1}(0.8149)\approx55^\circ$
Step 3. We can find the angle as
$A=180^\circ-50^\circ-55^\circ=75^\circ$
Step 4. We solved the triangle with
$b\approx4.7, A\approx75^\circ, C\approx55^\circ$
