Law of Cosines (finding side x in a triangle)

• Feb 14th 2008, 08:26 PM
Melancholy
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:
$a^2 = b^2 + c^2 - 2bc (cos A)$
$b^2 = a^2 + c^2 - 2ac (cos B)$
$c^2 = a^2 + b^2 - 2ab (cos C)$

This is what I have so far:
$a^2 = 32^2 + 23^2 - 2(32)(23) (cos A)$
$a^2 = 1024 + 529 - 1472 (cos A)$
$a^2 = 1553 - 1472 (cos A)$
$a = [square root sign] 1553 - 1472 (cos A)$

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...
• Feb 14th 2008, 08:57 PM
Jhevon
Quote:

Originally Posted by Melancholy
Find x to the nearest tenth.
http://i208.photobucket.com/albums/b...esfindingx.jpg

Okay, so according to the Law of Cosines:
$a^2 = b^2 + c^2 - 2bc (cos A)$
$b^2 = a^2 + c^2 - 2ac (cos B)$
$c^2 = a^2 + b^2 - 2ab (cos C)$

This is what I have so far:
$a^2 = 32^2 + 23^2 - 2(32)(23) (cos A)$
$a^2 = 1024 + 529 - 1472 (cos A)$
$a^2 = 1553 - 1472 (cos A)$
$a = [square root sign] 1553 - 1472 (cos A)$

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...

first of all, type $$\sqrt{x}$$ to get $\sqrt{x}$

second of all, you know the value for A (it is $82^o$), so plug it in. you will then be able to solve for $a$