#### Answer

The angles of the triangle are as follows:
$A = 4.8^{\circ}, B = 168.2^{\circ},$ and $C = 7.0^{\circ}$
The lengths of the sides are as follows:
$a = 15.1~cm, b = 34.1~cm,$ and $c = 19.2~cm$

#### Work Step by Step

We can use the law of cosines to find $b$:
$b^2 = a^2+c^2-2ac~cos~B$
$b = \sqrt{a^2+c^2-2ac~cos~B}$
$b = \sqrt{(15.1~cm)^2+(19.2~cm)^2-(2)(15.1~cm)(19.2~cm)~cos~168.2^{\circ}}$
$b = \sqrt{1164.236~cm^2}$
$b = 34.1~cm$
We can use the law of cosines to find $C$:
$c^2 = a^2+b^2-2ab~cos~C$
$2ab~cos~C = a^2+b^2-c^2$
$cos~C = \frac{a^2+b^2-c^2}{2ab}$
$C = arccos(\frac{a^2+b^2-c^2}{2ab})$
$C = arccos(\frac{15.1^2+34.1^2-19.2^2}{(2)(15.1)(34.1)})$
$C = arccos(0.99258)$
$C = 7.0^{\circ}$
We can find angle $A$:
$A+B+C = 180^{\circ}$
$A = 180^{\circ}-B-C$
$A = 180^{\circ}-168.2^{\circ}-7.0^{\circ}$
$A = 4.8^{\circ}$