Answer
The missing values are: $B=25{}^\circ,C=115{}^\circ \text{ and c}\approx 8.5$.
Work Step by Step
For any triangle,
The law of sines states that:
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\operatorname{sinC}}$
The law of cosines sates that:
$\begin{align}
& {{a}^{2}}={{b}^{2}}+{{c}^{2}}-2\cdot b\cdot c\cdot \cos A \\
& {{b}^{2}}={{a}^{2}}+{{c}^{2}}-2\cdot a\cdot c\cdot \cos B \\
& {{c}^{2}}={{a}^{2}}+{{b}^{2}}-2\cdot a\cdot b\cdot \cos C \\
\end{align}$
Using the law of sines we get,
$\begin{align}
& \frac{a}{\sin A}=\frac{b}{\sin B} \\
& \frac{6}{\sin 40{}^\circ }=\frac{4}{\sin B} \\
& 6\cdot \sin B=4\cdot \sin 40{}^\circ \\
& \sin B=\frac{4\cdot \sin 40{}^\circ }{6}
\end{align}$
We will simplify it further to get the measure of angle B.
$\begin{align}
& \sin B=0.42 \\
& B={{\sin }^{-1}}\left( 0.42 \right) \\
& B\approx 25{}^\circ \\
\end{align}$
Using the angle sum property we will obtain the measure of angle C.
$\begin{align}
& A+B+C=180{}^\circ \\
& 40{}^\circ +25{}^\circ +C=180{}^\circ \\
& C=180{}^\circ -65{}^\circ \\
& C=115{}^\circ
\end{align}$
Finally use the law if sines:
$\begin{align}
& \frac{a}{\sin A}=\frac{c}{\sin C} \\
& \frac{6}{\sin 40{}^\circ }=\frac{c}{\sin 115{}^\circ } \\
& 6\cdot \sin 115{}^\circ =c\cdot \sin 40{}^\circ \\
& c=\frac{6\cdot \sin 115{}^\circ }{\sin 40{}^\circ }
\end{align}$
This gives $c\approx 8.5$.
So, $B=25{}^\circ,C=115{}^\circ \text{ and c}\approx 8.5$.
Hence, $B=25{}^\circ,C=115{}^\circ \text{ and c}\approx 8.5$.
