#### Answer

The angles of the triangle are as follows:
$A = 56.7^{\circ}, B = 55^{\circ},$ and $C = 68.3^{\circ}$
The lengths of the sides are as follows:
$a = 90, b = 88.2,$ and $c = 100$

#### Work Step by Step

Let $a=90$, and let $c = 100$.
We can use the law of cosines to find $b$:
$b^2 = a^2+c^2-2ac~cos~B$
$b = \sqrt{a^2+c^2-2ac~cos~B}$
$b = \sqrt{90^2+100^2-(2)(90)(100)~cos~55^{\circ}}$
$b = \sqrt{7775.62}$
$b = 88.2$
We can use the law of sines to find $A$:
$\frac{b}{sin~B} = \frac{a}{sin~A}$
$sin~A = \frac{a~sinB}{b}$
$A = arcsin(\frac{a~sinB}{b})$
$A = arcsin(\frac{90~sin(55^{\circ})}{88.2})$
$A = arcsin(0.83587)$
$A = 56.7^{\circ}$
We can find angle $C$:
$A+B+C = 180^{\circ}$
$C = 180^{\circ}-A-B$
$C = 180^{\circ}-56.7^{\circ}-55^{\circ}$
$C = 68.3^{\circ}$