Question 935239
<pre>
Let's find ALL the angles instead of just the largest one.

(BTW, the largest angle is the angle opposite the largest side)

Consider the sides to have the lengths 4,5, and 6.  
a=4, b=5, c=6

{{{drawing(400,265,-1,7,-1,4.3,
triangle(0,0,6,0,2.25,3.307189139,6,0),locate(0,0,B),
locate(6,0,A), locate(2.23,3.5,C),locate(2.6,0,c=6),locate(.65,2,a=4)
locate(4,2,b=5) )}}}

The law of cosines

{{{cos("ANGLE")}}}{{{""=""}}}{{{( 

(matrix(1,8,

SUM,OF,SQUARES,OF,SIDES,OF,THE,ANGLE))

-

(matrix(1,5,

SQUARE,OF,SIDE,OPPOSITE,ANGLE)) 

)

/

(matrix(1,7,
2,TIMES,PRODUCT,OF,SIDES,OF,ANGLE))}}}       )

To find angle A:

{{{cos(A)=(b^2+c^2-a^2)/(2bc)}}}

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

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

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

{{{cos(A)=3/4}}}

Use the calculator

{{{A=41.40962211°}}}

Now we can either switch over to the law of sines or 
continue with the law of cosines, which I think I'll
do:

{{{cos(B)=(a^2+c^2-b^2)/(2ac)}}}

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

{{{cos(B)=(16+36-25)/48}}}

{{{cos(B)=27/48}}}

{{{cos(A)=9/16}}}

Use the calculator

{{{A=55.77113367°}}}

Now we can just add angles A and B and subtract from
180° to find angle C.  But as a check we can also 
find it with the law of cosines then add them to see
if they have sum 180°.

{{{cos(C)=(a^2+b^2-c^2)/(2ab)}}}

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

{{{cos(C)=(16+25-36)/40}}}

{{{cos(C)=5/40}}}

{{{cos(C)=1/8}}}

Use the calculator

{{{C=82.81924422°}}}

Now we add them to see if we get 180°:

A=41.40962211°
B=55.77113367°
C=82.81924422°
--------------
 180.00000000°

Edwin</pre>