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Law of Sines

  1. Jun 25, 2008 #1
    Not exactly a homework question, even though it is related to my homework...

    So, the law is:

    (sin A/a) = (sin B/b) = (sin C/c)

    So, in certain problems we have to manipulate this law. For example our givens include angle C and side c.

    (sin A/a) = (sin C/c)

    a would have to equal [(c sin A)/sin C]

    Why can't a = [sin C/(c sin A)]

    If someone can explain this, I'd greatly appreciate it.

    Last edited: Jun 25, 2008
  2. jcsd
  3. Jun 25, 2008 #2


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    Huh? You just need to transpose 'a' on one side of the equation to get a = (c sin A) / sin C. As for why you can't get that expression is because that's not sin rule.
  4. Jun 26, 2008 #3


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    This is nothing more than axioms of Real Numbers and inverse operations, properties of equality. Everyone learns them in Introductory Algebra.
  5. Jun 26, 2008 #4


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    a and c are lengths. They might have units of, say, feet or meters. A and C are angles and while they might have units of degrees or radians. In any case "sin(A)" and "sin(C)" have no units and neither does sin(A)/sin(C). So in the formula a= (sin(A)/sin(C) c, a distance, c, with units, say, of meters, is multiplied by a number with no umits, leaving a to have meters as units as it should.

    If you wrote a= [sin(C)/c(sin(A))] instead, you now have c, with its units of meters, in the denominator- that would say that a have units of "1 over meters" which, what ever thaat might mean, cannot be a length.

    Certainly you should have learned basic algebra, solving equations, well before starting on trigonometry. If you have
    [tex]\frac{sin(A)}{a}= \frac{sin(C)}{c}[/tex]
    you might start by multiplying both sides by ac (sometimes called "cross-multiplying"):
    [tex]c sin(A)= a sin(C)[/tex]
    Now, since you want to solve for a, just divide both sides by sin(C):
    [tex]c [sin(A)/sin(C)]= a[/tex].

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