Why does:
$$\frac{4.5}{\sin40^\circ} \not= \frac{3+3}{\sin(180^\circ - 58^\circ)}$$
Am I using the rule wrong? Any incorrect assumptions?
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You may be dealing with the ambiguous case of the law of sines. See here:http://en.wikipedia.org/wiki/Law_of_sines#The_ambiguous_case – noobProgrammer Apr 05 '13 at 17:34
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1@noobProgrammer this isn't the ambiguous case, since all 3 angles are known. – icurays1 Apr 05 '13 at 17:37
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1I would say that you could try drawing this type of triangle, taking 40 degrees, 6 m and 4.5 m, but you would get around 58.99 degrees for the acute angle at P instead of 58 degrees. So, I think that a lot of the values in that drawing have been rounded off. – Jerry Apr 05 '13 at 17:37
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Your application of the sine rule is correct. You'll find that your two numbers are in fact equal to two decimal. The problem comes from the lengths (or angles) being rounded. For example, the correct length of the side given length $6\,\text{m}$ should be $5.94\,\text{m}$ to two decimal places.
Fly by Night
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@Wk_of_Angmar No worries, glad I could help. What were you trying to work out? – Fly by Night Apr 05 '13 at 17:53
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The resultant moment about P: http://math.stackexchange.com/questions/352390/calculating-the-resultant-moment-about-a-point-p – Wk_of_Angmar Apr 05 '13 at 18:02