Answer
There are two possible triangles with the given parts.
One possible triangle has the following values:
$A = 55^{\circ}20', B = 94^{\circ}50'$, and $C = 29^{\circ}50'$
$a = 8.61~m, b = 10.44~m$, and $c = 5.21~m$
Another possible triangle has the following values:
$A = 124^{\circ}40', B = 25^{\circ}30'$, and $C = 29^{\circ}50'$
$a = 8.61~m, b = 4.51~m$, and $c = 5.21~m$
Work Step by Step
We can use the law of sines to find the angle $A$:
$\frac{c}{sin~C} = \frac{a}{sin~A}$
$sin~A = \frac{a~sin~C}{c}$
$sin~A = \frac{(8.61~m)~sin~(29^{\circ}50')}{5.21~m}$
$A = arcsin(0.8221)$
$A = 55^{\circ}20'$
We can find angle $B$:
$A+B+C = 180^{\circ}$
$B = 180^{\circ}-A-C$
$B = 180^{\circ}-55^{\circ}20'-29^{\circ}50'$
$B = 94^{\circ}50'$
We can find the length of side $b$:
$\frac{b}{sin~B} = \frac{c}{sin~C}$
$b = \frac{c~sin~B}{sin~C}$
$b = \frac{(5.21~m)~sin~(94^{\circ}50')}{sin~29^{\circ}50'}$
$b = 10.44~m$
Note that we can also find another angle for A.
$A = 180-55^{\circ}20' = 124^{\circ}40'$
We can find angle $B$:
$A+B+C = 180^{\circ}$
$B = 180^{\circ}-A-C$
$B = 180^{\circ}-124^{\circ}40'-29^{\circ}50'$
$B = 25^{\circ}30'$
We can find the length of side $b$:
$\frac{b}{sin~B} = \frac{c}{sin~C}$
$b = \frac{c~sin~B}{sin~C}$
$b = \frac{(5.21~m)~sin~(25^{\circ}30')}{sin~29^{\circ}50'}$
$b = 4.51~m$
