Finding Angle B in Triangle ABC with Side Lengths a, b, c

[disregardthat]

Homework Statement

In triangle ABC where you only know the sides: a, b and c I must find angle B.
a=8
b=6
c=12

Homework Equations

Law of cosines: c^2 = a^2 + b^2 -2ab*cos(C)
When angle C is at the opposite of side c, (same for a and b)

The Attempt at a Solution

12^2 = 6^2 + 8^2 -2*6*8*cos(C)

144 = 36+64-96cos(C)
44=-96cos(C)
-(44/96) = cos(C)
- (11/24) = cos(C)

(I know the answer is supposed to be 117.4)
And cos(117.4)=-(11/24)

The problem is:
How do I find the angle? The explanation says: "Use calculator (degree mode)"
How do I do that?

I have an TI-84 Plus

 

[Dick]

Look for 'arccos' or 'cos^(-1)'.

 

[turdferguson]

Are you sure about that? Check your formula again. The largest angle must be opposite the largest side, and side b is not the largest side

 

[disregardthat]

Look for 'arccos' or 'cos^(-1)'.

Thanks, I found it, and I filled in the number, and I got the answer.

 

[disregardthat]

Look for 'arccos' or 'cos^(-1)'.

Thanks, I found it, and I filled in the number, and I got the answer.