#### Answer

There are two possible triangles with the given parts.
One triangle has the following values:
$A = 38^{\circ}40', B = 49^{\circ}20',$ and $C = 92^{\circ}$
$a = 9.72~m, b = 11.8~m$, and $c = 15.5~m$
Another triangle has the following values:
$A = 38^{\circ}40', B = 130^{\circ}40',$ and $C = 10^{\circ}40'$
$a = 9.72~m, b = 11.8~m$, and $c = 2.88~m$

#### Work Step by Step

We can use the law of sines to find the angle $B$:
$\frac{a}{sin~A} = \frac{b}{sin~B}$
$sin~B = \frac{b~sin~A}{a}$
$sin~B = \frac{(11.8~m)~sin~(38^{\circ}40')}{9.72~m}$
$B = arcsin(0.7585)$
$B = 49^{\circ}20'$
We can find angle $C$:
$A+B+C = 180^{\circ}$
$C = 180^{\circ}-A-B$
$C = 180^{\circ}-38^{\circ}40'-49^{\circ}20'$
$C = 92^{\circ}$
We can find the length of side $c$:
$\frac{a}{sin~A} = \frac{c}{sin~C}$
$c = \frac{a~sin~C}{sin~A}$
$c = \frac{(9.72~m)~sin~(92^{\circ})}{sin~38^{\circ}40'}$
$c = 15.5~m$
Note that we can also find another angle for B.
$B = 180-49^{\circ}20' = 130^{\circ}40'$
We can find angle $C$:
$A+B+C = 180^{\circ}$
$C = 180^{\circ}-A-B$
$C = 180^{\circ}-38^{\circ}40'-130^{\circ}40'$
$C = 10^{\circ}40'$
We can find the length of side $c$:
$\frac{a}{sin~A} = \frac{c}{sin~C}$
$c = \frac{a~sin~C}{sin~A}$
$c = \frac{(9.72~m)~sin~(10^{\circ}40')}{sin~38^{\circ}40'}$
$c = 2.88~m$