#### Answer

The angles of the triangle are as follows:
$A = 22.3^{\circ}, B = 108.2^{\circ},$ and $C = 49.5^{\circ}$
The lengths of the sides are as follows:
$a = 4, b = 10,$ and $c = 8$

#### Work Step by Step

Let $a = 4,$ let $b = 10,$ and let $c=8$
We can use the law of cosines to find $B$:
$b^2 = a^2+c^2-2ac~cos~B$
$2ac~cos~B = a^2+c^2-b^2$
$cos~B = \frac{a^2+c^2-b^2}{2ac}$
$B = arccos(\frac{a^2+c^2-b^2}{2ac})$
$B = arccos(\frac{4^2+8^2-10^2}{(2)(4)(8)})$
$B = arccos(-\frac{5}{16})$
$B = 108.2^{\circ}$
We can use the law of cosines to find $C$:
$c^2 = a^2+b^2-2ab~cos~C$
$2ab~cos~C = a^2+b^2-c^2$
$cos~C = \frac{a^2+b^2-c^2}{2ab}$
$C = arccos(\frac{a^2+b^2-c^2}{2ab})$
$C = arccos(\frac{4^2+10^2-8^2}{(2)(4)(10)})$
$C = arccos(\frac{13}{20})$
$C = 49.5^{\circ}$
We can find angle $A$:
$A+B+C = 180^{\circ}$
$A = 180^{\circ}-B-C$
$A = 180^{\circ}-108.2^{\circ}-49.5^{\circ}$
$A = 22.3^{\circ}$