Answer
There are two possible triangles for these values.
One triangle is as follows:
$A = 51.20^{\circ}~$, $~B = 69.09^{\circ}~$, $~C=59.71^{\circ}$
$a = 7208~cm~$, $~b = 8640~cm~$, $c = 7986~cm$
The other triangle is as follows:
$A = 51.20^{\circ}~$, $~B = 8.51^{\circ}~$, $~C=120.29^{\circ}$
$a = 7208~cm~$, $~b = 1369~cm~$, $c = 7986~cm$
Work Step by Step
We can use the law of sines to find the angle $C$:
$\frac{a}{sin~A} = \frac{c}{sin~C}$
$sin~C = \frac{c~sin~A}{a}$
$sin~C = \frac{(7986~cm)~sin~(51.20^{\circ})}{7208~cm}$
$sin~C = 0.8635$
$C = arcsin(0.8635)$
$C = 59.71^{\circ}$
We can find angle $B$:
$A+B+C = 180^{\circ}$
$B = 180^{\circ}-A-C$
$B = 180^{\circ}-51.20^{\circ}-59.71^{\circ}$
$B = 69.09^{\circ}$
We can find the length of side $b$:
$\frac{b}{sin~B} = \frac{a}{sin~A}$
$b = \frac{a~sin~B}{sin~A}$
$b = \frac{(7208~cm)~sin~(69.09^{\circ})}{sin~(51.20^{\circ})}$
$b = 8640~cm$
Note that we can also find another angle for C.
$C = 180-59.71^{\circ} = 120.29^{\circ}$
We can find angle $B$:
$A+B+C = 180^{\circ}$
$B = 180^{\circ}-A-C$
$B = 180^{\circ}-51.20^{\circ}-120.29^{\circ}$
$B = 8.51^{\circ}$
We can find the length of side $b$:
$\frac{b}{sin~B} = \frac{a}{sin~A}$
$b = \frac{a~sin~B}{sin~A}$
$b = \frac{(7208~cm)~sin~(8.51^{\circ})}{sin~(51.20^{\circ})}$
$b = 1369~cm$
