Answer
\[
\text { Triangle } 1: B=70.83^{\circ}, C=58.17^{\circ}, c=7.08 \mathrm{ft}
\]
\[
\text { Triangle } 2: B=109.12^{\circ}, C=19.88^{\circ}, c=2.83 \mathrm{ft}
\]
Work Step by Step
Given
\[
A=51^{\circ}, a=6.5 \mathrm{ft}, b=7.9 \mathrm{ft}
\]
We know two sides and one angle; thus we will use the law of sines to find the reference angle.
\[
\begin{aligned}
&\frac{b}{\sin B}=\frac{a}{\sin A} \\
\frac{6.5}{\sin 51^{\circ}} &=\frac{7.9}{\sin B} \\
&=\frac{0.78 \cdot 7.9}{6.5} \\
&\frac{\sin 51^{\circ}}{6.5} \cdot 7.9 =\sin B \\
&=\frac{6.14}{6.5} \\
& \sin B=0.94
\end{aligned}
\]
Becatise sin $B$ is positive in Quadrant I and Quadrant II, we have two angles. Thus the two angles are
\[
\begin{array}{ll}
B_{1}=\sin ^{-1} 0.94 & B_{2}=180^{\circ}-70.83^{\circ} \\
109.12=70.83^{\circ} & =B_{1}
\end{array}
\]
We know two angles. Thus let's find the remaining angle
\[
\begin{array}{c}
180^{\circ}=A+B+C \\
51^{\circ}+70.83^{\circ}+C=180^{\circ} \\
180^{\circ}-121.83^{\circ}=C \\
{\left[C=58.17^{\circ}\right.}
\end{array}
\]
We know three angles and two sides. Thus we use law of sines to find the remaining side $c$
\[
\begin{aligned}
& \frac{c}{\sin C} =\frac{a}{\sin A} \\
& \frac{c}{\sin 58.17^{\circ}}=\frac{6.5}{\sin 51^{\circ}} \\
=& \frac{6.5 \cdot 0.85}{0.78} \\
=& \frac{5.5}{\sin 51^{\circ}} \cdot \sin 58.17^{\circ} \\
&=0.78 \\
\hline & \\
&=7.08 \mathrm{ft}
\end{aligned}
\]
Triangle 2
Find the missing parts while considering $B$ = $109.12^{\circ}$
We know two angles.Thus let's find the remaining angle
\[
\begin{array}{c}
180^{\circ}=A+B+C \\
180^{\circ}=109.12^{\circ}+51^{\circ}+C \\
180^{\circ}-160.12^{\circ}=C \\
{\left[C=19.88^{\circ}\right.}
\end{array}
\]
We know three angles and two sides. Thus we use law of sines to find the remaining side:
\[
\begin{aligned}
&\frac{c}{\sin C}=\frac{a}{\sin A} \\
&\frac{c}{\sin 19.88^{\circ}}=\frac{6.5}{\sin 51^{\circ}} \\
&=\frac{6.5 \cdot 0.34}{0.78} \\
&=\frac{6.5}{\sin 51^{\circ}} \cdot \sin 19.88^{\circ} \\
& \frac{2.21}{0.78} \\
&=2.83 \mathrm{ft}
\end{aligned}
\]
