Answer
There are two possible triangles with the given parts.
$A = 46^{\circ}, B= 54^{\circ},$ and $C = 80^{\circ}$
$A = 26^{\circ}, B= 54^{\circ},$ and $C = 100^{\circ}$
Work Step by Step
We can use the law of sines to find the angle $B$:
$\frac{b}{sin~B} = \frac{c}{sin~C}$
$sin~C = \frac{c~sin~B}{b}$
$sin~C = \frac{(28)~sin~(54^{\circ})}{23}$
$C = arcsin(.9849)$
$C = 80^{\circ}$
We can find angle $A$:
$A+B+C = 180^{\circ}$
$A = 180^{\circ}-B-C$
$A = 180^{\circ}-54^{\circ}-80^{\circ}$
$A = 46^{\circ}$
Note that we can also find another angle for C.
$C = 180-80^{\circ} = 100^{\circ}$
We can find angle $A$:
$A+B+C = 180^{\circ}$
$A = 180^{\circ}-B-C$
$A = 180^{\circ}-54^{\circ}-100^{\circ}$
$A = 26^{\circ}$
