#### Answer

The angles of the triangle are as follows:
$A = 91^{\circ}50', B = 45^{\circ}40',$ and $C = 42^{\circ}30'$
The lengths of the sides are as follows:
$a = 27.6~cm, b = 19.8~cm,$ and $c = 18.7~cm$

#### Work Step by Step

$a = 27.6~cm$
$b = 19.8~cm$
The angle $C = 42^{\circ}30'$
We can use the law of cosines to find $c$:
$c = \sqrt{a^2+b^2-2ab~cos~C}$
$c = \sqrt{(27.6~cm)^2+(19.8~cm)^2-(2)(27.6~cm)(19.8~cm)~cos~42^{\circ}30'}$
$c = \sqrt{347.985~cm^2}$
$c = 18.7~cm$
We can use the law of sines to find angle $B$:
$\frac{c}{sin~C} = \frac{b}{sin~B}$
$sin~B = \frac{b~sin~C}{c}$
$B = arcsin(\frac{b~sin~C}{c})$
$B = arcsin(\frac{(19.8)~sin~42^{\circ}30'}{18.7})$
$B = arcsin(0.715)$
$B = 45^{\circ}40'$
We can find angle $A$:
$A+B+C=180^{\circ}$
$A = 180^{\circ}-B-C$
$A = 180^{\circ}-45^{\circ}40'-42^{\circ}30'$
$A = 91^{\circ}50'$