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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
    
Because if I do that, it's not testing the accuracy of the law. +1 because info and question useful. –  George Tomlinson Oct 15 '13 at 19:14
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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
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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
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@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

1 Answer 1

up vote 3 down vote accepted

OK, I did the Law of Cosines 3 times and came up with 60.647 , 20.404 and 98.949 respectively for angles A, B and C. Remember, the Law of Cosines does not have an ambiguous case, unlike the Law of Sines. I suspect (without further investigating) that his may be the culprit. My advice: Always use the Law of Cosines whenever you can. In this case, when all sides are known, clearly a case for Law of Cosines

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