Why does there seem to be so much error in the laws of sines and cosines?

I've been computing the angles of a triangle with sides a = 17, b = 6 and c = 15 using the law of cosines to find the first angle and then the law of sines to find the other 2. I follow the convention of naming the angles opposite these sides A, B and C respectively. Here are my results:

$C = \arccos( \frac {6^2+17^2-15^2}{2(6)(17)}) = 60.647$ degrees to 3 d.p.

$B = \arcsin( \frac {6 \sin C}{15}) = 20.405$ degrees to 3 d.p.

$A = \arcsin( \frac {17 \sin B}{6}) = 81.051$ degrees to 3 d.p.

Clearly, adding these should give $180$ degrees, but it gives 162 degrees to 3 s.f. Assuming I haven't made any mistakes, the error seems quite high and I'm just wondering if anyone knows why this is? It seems high enough to challenge the validity of the laws.

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Maybe it's because of the ambiguous case, that arises when using the Law of sines. Why don't you do the Law of Cosines two times and subtract from 180° to find the third angle? No need for Law of Sines here. – imranfat Oct 15 '13 at 19:07
The laws are acurate and so is your calculator for the trig terms, but the ambiguous case is the issue – imranfat Oct 15 '13 at 19:16
Note that the error is about $18$ degrees, which is just the difference between $81$ degrees and $180-81=99$ degrees. Those angles have the same sine. You are picking the wrong one. – Ross Millikan Oct 15 '13 at 19:19
Hey Ross, congrats BTW on your repuation points. In essence, isn't picking the "wrong" one inherently related to the ambiguous case due to sinx = sin(180-x) ? – imranfat Oct 15 '13 at 19:21
@imranfat: The arcsin function is defined (so as to be single valued) to return values between $-90$ and $+90$ degrees. The error was made in going from $\sin A= ()$ to $A=\arcsin ()$. Those are not equivalent. It is the same as going from $x^2=2$ to $x=\sqrt 2$ and missing the $\pm$ sign. The calculator is useful and returned the correct answer to the question it was asked. – Ross Millikan Oct 15 '13 at 19:27