#### Answer

The angles are as follows:
$A = 61^{\circ}, B = 40.7^{\circ},$ and $C = 78.3^{\circ}$
The lengths of the sides are as follows:
$a = 5.4, b = 4,$ and $c = 6$

#### Work Step by Step

Let $b=4$, and let $c = 6$.
We can use the law of cosines to find $a$:
$a^2 = b^2+c^2-2bc~cos~A$
$a = \sqrt{b^2+c^2-2bc~cos~A}$
$a = \sqrt{4^2+6^2-(2)(4)(6)~cos~61^{\circ}}$
$a = \sqrt{28.729}$
$a = 5.4$
We can use the law of cosines to find $B$:
$b^2 = a^2+c^2-2ac~cos~B$
$2ac~cos~B = a^2+c^2-b^2$
$cos~B = \frac{a^2+c^2-b^2}{2ac}$
$B = arccos(\frac{a^2+c^2-b^2}{2ac})$
$B = arccos(\frac{(5.4)^2+6^2-4^2}{(2)(5.4)(6)})$
$B = arccos(0.7586)$
$B = 40.7^{\circ}$
We can find angle $C$:
$A+B+C = 180^{\circ}$
$C = 180^{\circ}-A-B$
$C = 180^{\circ}-61^{\circ}-40.7^{\circ}$
$C = 78.3^{\circ}$