#### Answer

The angles of the triangle are as follows:
$A = 80^{\circ}40', B = 64^{\circ}50',$ and $C = 34^{\circ}30'$
The lengths of the sides are as follows:
$a = 156~cm, b = 143~cm,$ and $c = 89.6~cm$

#### Work Step by Step

We can use the law of cosines to find $a$:
$a^2 = b^2+c^2-2bc~cos~A$
$a = \sqrt{b^2+c^2-2bc~cos~A}$
$a = \sqrt{(143~cm)^2+(89.6~cm)^2-(2)(143~cm)(89.6~cm)~cos~80^{\circ}40'}$
$a = \sqrt{24321.4~cm^2}$
$a = 156~cm$
We can use the law of cosines to find $B$:
$b^2 = a^2+c^2-2ac~cos~B$
$2ac~cos~B = a^2+c^2-b^2$
$cos~B = \frac{a^2+c^2-b^2}{2ac}$
$B = arccos(\frac{a^2+c^2-b^2}{2ac})$
$B = arccos(\frac{(156)^2+(89.6)^2-(143)^2}{(2)(156)(89.6)})$
$B = arccos(0.426)$
$B = 64^{\circ}50'$
We can find angle $C$:
$A+B+C = 180^{\circ}$
$C = 180^{\circ}-A-B$
$C = 180^{\circ}-80^{\circ}40'-64^{\circ}50'$
$C = 34^{\circ}30'$