Answer
\[
8.12 \mathrm{m}=c, 43.63^{\circ}=A, 17.37^{\circ}=B
\]
Work Step by Step
\[
a=6.4 \mathrm{m}, b=2.8 \mathrm{m}, C=119^{\circ}
\]
We know the two sides and the angle between them. Thus we can use the Law of cosines to find the length of $c$.
\[
a^{2}+b^{2}-2 a b \cos C=c^{2}
\]
Substitute $a, b$ and $C$
\[
\begin{aligned}
&a^{2}+b^{2}-2 a b \cos C=c^{2} \\
&=6.4^{2}+2.8^{2}-2 \cdot 6.4 \cdot 2.8 \cos 119^{\circ} \\
&=40.96+7.84-35.84 \cdot(-0.48) \\
66=c^{2} & \\
&=48.8+17.2 \\
& \\
\begin{array}{l}
66=c^{2} \\
\sqrt{66}=c \\
8.12 \mathrm{m}=c
\end{array}
\end{aligned}
\]
Angle A is:
\[
\begin{array}{l}
\frac{\sin C}{c}=\frac{\sin A}{a} \\
\frac{\sin 119^{\circ}}{8.12} =\frac{\sin A}{6.4}\\
6.69 =\sin A\\
\frac{0.87}{8.12}=\frac{\sin A}{6.4} \\
\sin ^{-1} 0.69=\qquad A \\
0.69=\sin A \\
43.63^{\circ}=\qquad A
\end{array}
\]
The sum of all angles in a triangle is $180^{\circ}$
\[
180^{\circ}=A+B+C
\]
\[
\begin{array}{c}
180^{\circ}=119^{\circ} +43.63^{\circ}+B\\
B=180^{\circ}-43.63^{\circ}-119^{\circ} \\
\hline B=17.37^{\circ}
\end{array}
\]
