Law of Cosines (finding side x in a triangle)

**Find x to the nearest tenth.**

http://i208.photobucket.com/albums/b...esfindingx.jpg

Okay, so according to the Law of Cosines:

$\displaystyle a^2 = b^2 + c^2 - 2bc (cos A)$

$\displaystyle b^2 = a^2 + c^2 - 2ac (cos B)$

$\displaystyle c^2 = a^2 + b^2 - 2ab (cos C)$

This is what I have so far:

$\displaystyle a^2 = 32^2 + 23^2 - 2(32)(23) (cos A)$

$\displaystyle a^2 = 1024 + 529 - 1472 (cos A)$

$\displaystyle a^2 = 1553 - 1472 (cos A)$

$\displaystyle a = [square root sign] 1553 - 1472 (cos A)$

First of all, is the math/formula I did/chose correct? Second, I'm having Algebra 1 issues: where the heck do I go from here? I tried putting this in my calculator: "[square root sign](cos)1553 - 1472" but it came out with the decimal 0.395517971. I have a TI-30XS calculator... and I really need help understand this... test tommorrow...