What did I do wrong when calculating the angles of this triangle, it doesn't equal to 180

Here is the original question: http://i46.tinypic.com/117fxo5.png

So for a I got 3. I got this by doing:

a^2 = b^2 + c^2 - 2bcCosA
a^2 = 113 - 104
a^2 = 9
a = 3

So a = 3, b = 7, c = 8, and angle A = 21.7867893.

Now to solve for angle B:

b^2 = a^2 + C^2 - 2acCosB
7^2 = 3^2 + 8^2 - 2(3)(8)cosB
49 = 9 + 64 - 48cosB
49 = 73 - 48cosB
0.5 = cosB
cos^(-1)(1/2) = B
B = 60

Now to solve for angle C

c^2 = a^2 + b^2 - 2abCosC
c^2 = 3^2 + 7^2 - 2(3)(7)cosC
64 = 9 + 49 - 42cosC
64 = 58 - 42cosC
6 = -42cosC
1/7 = cosC
C = 81.7867893

Add up all the angles:

21.7867893 + 60 + 81.7867893 = 163.5735786

It doesn't equal to 180, what did I do wrong?

hi! you seriously almost got it correct
the only thing you got wrong was the last two steps of angle C. the rest was all correct.
the second to last step is
-1/7=Cos C
to find the value of C do
inverse cosine of -1/7 (don't forget the negative sign)
cos^(-1)(-1/7) = C
the value of C is 98.2132107 instead of 81.7867893 ANSWER
if you add that to A and B it does equal 180
:)
(btw, what grade math is this. tenth grade?)

now that you have finished this problem, can you please answer my quick survey? thanks.
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You lost your minus sign in the last calculation. It should be -1/7 = cos C.