#### Answer

The angles of the triangle are as follows:
$A = 33.6, B = 50.7^{\circ},$ and $C = 95.7^{\circ}$
The lengths of the sides are as follows:
$a = 5, b = 7,$ and $c = 9$

#### Work Step by Step

Let $a = 5,$ let $b = 7,$ and let $c=9$
We can use the law of cosines to find $B$:
$b^2 = a^2+c^2-2ac~cos~B$
$2ac~cos~B = a^2+c^2-b^2$
$cos~B = \frac{a^2+c^2-b^2}{2ac}$
$B = arccos(\frac{a^2+c^2-b^2}{2ac})$
$B = arccos(\frac{5^2+9^2-7^2}{(2)(5)(9)})$
$B = arccos(\frac{19}{30})$
$B = 50.7^{\circ}$
We can use the law of cosines to find $C$:
$c^2 = a^2+b^2-2ab~cos~C$
$2ab~cos~C = a^2+b^2-c^2$
$cos~C = \frac{a^2+b^2-c^2}{2ab}$
$C = arccos(\frac{a^2+b^2-c^2}{2ab})$
$C = arccos(\frac{5^2+7^2-9^2}{(2)(5)(7)})$
$C = arccos(-0.1)$
$C = 95.7^{\circ}$
We can find angle $A$:
$A+B+C = 180^{\circ}$
$A = 180^{\circ}-B-C$
$A = 180^{\circ}-50.7^{\circ}-95.7^{\circ}$
$A = 33.6^{\circ}$