Question 115249
 
The lengths of the sides of triangle ABC are AB=6, BC=4, and CA=5, then cos A=?
<pre><font size = 3><b>
{{{drawing(200,187.5,-1,6,-1,6, triangle(0,0,4,0,3.375,4.960783708),
locate(3.2,5.7,A), locate(-.4,.2,B), locate(4.2,.2,C),
locate(1.2,2.8,6), locate(2,0,4),locate(3.8,2.8,5)
 )}}}


The law of cosines for when you have all three sides and
no angle given is

                 (one ADJACENT side)² + (other ADJACENT side)² - (OPPOSITE)²
cosine(ANGLE) = -------------------------------------------------------------
                      2(one ADJACENT side)(other ADJACENT side)


The ANGLE we want to find the cosine of is A.
One ADJACENT side to angle A is AB which is 6 in length.
The other ADJACENT side to angle A is CA which is 5 in length.
The OPPOSITE side is the side across the triangle
from angle A, which is BC, 4 in length.

{{{cos(A)=(6^2+5^2-4^2)/(2*6*5)}}} 

{{{cos(A)=(36+25-16)/(60)}}}

{{{cos(A)=(45)/(60)}}}

{{{cos(A)=3/4)}}}
Edwin</pre>