#### Answer

There are two possible triangles.
The angles of one triangle are as follows:
$A = 28^{\circ}10', B = 36^{\circ}10',$ and $C = 115^{\circ}40'$
The lengths of the sides are as follows:
$a = 21.2~yd, b = 26.5~yd,$ and $c = 40.5~yd$
The angles of the other possible triangle are as follows:
$A = 28^{\circ}10', B = 143^{\circ}50',$ and $C = 8^{\circ}$
The lengths of the sides are as follows:
$a = 21.2~yd, b = 26.5~yd,$ and $c = 6.3~yd$

#### Work Step by Step

Let $a = 21.2~yd$
Let $b = 26.5~yd$
Let angle $A = 28^{\circ}10'$
Let angle $B$ be the angle that subtends the side $b$. We can use the law of sines to find $B$:
$\frac{a}{sin~A} = \frac{b}{sin~B}$
$sin~B = \frac{b~sin~A}{a}$
$sin~B = \frac{(26.5)~sin~28^{\circ}10'}{21.2}$
$sin~B = 0.59$
$B = arcsin(0.59)$
$B = 36^{\circ}10'$
We can find angle C:
$A+B+C = 180^{\circ}$
$C = 180^{\circ}-A-B$
$C = 180^{\circ}-28^{\circ}10'-36^{\circ}10'$
$C = 115^{\circ}40'$
We can use the law of sines to find $c$:
$\frac{c}{sin~C} = \frac{a}{sin~A}$
$c = \frac{a~sin~C}{sin~A}$
$c = \frac{(21.2~yd)~sin~115^{\circ}40'}{sin~28^{\circ}10'}$
$c = 40.5~yd$
Note that there is another possible value for the angle $B$:
$B = 180^{\circ}- 36^{\circ}10' = 143^{\circ}50'$
We can find angle C:
$A+B+C = 180^{\circ}$
$C = 180^{\circ}-A-B$
$C = 180^{\circ}-28^{\circ}10'-143^{\circ}50'$
$C = 8^{\circ}$
We can use the law of sines to find $c$:
$\frac{c}{sin~C} = \frac{a}{sin~A}$
$c = \frac{a~sin~C}{sin~A}$
$c = \frac{(21.2~yd)~sin~8^{\circ}}{sin~28^{\circ}10'}$
$c = 6.3~yd$