Understanding the Law of Sines: Explaining Side a Formula

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Homework Help Overview

The discussion revolves around the Law of Sines, specifically the manipulation of the formula relating angles and sides in a triangle. Participants are examining the expression for side 'a' in terms of angle 'C' and side 'c', questioning the validity of different algebraic manipulations of the formula.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the correct manipulation of the Law of Sines, particularly why certain algebraic forms are valid or invalid. There is a focus on the implications of units in the context of the formula and the properties of equality in algebra.

Discussion Status

The discussion is active, with participants providing insights into algebraic principles and questioning the original poster's understanding of the Law of Sines. Some participants offer clarifications on the correct form of the equation, while others emphasize the importance of units in the context of the problem.

Contextual Notes

There is an indication that the original poster is grappling with foundational algebra concepts as they relate to trigonometry, suggesting a potential gap in understanding that is being addressed through the discussion.

thakid87
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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.

Thanks.
 
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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.
 
This is nothing more than axioms of Real Numbers and inverse operations, properties of equality. Everyone learns them in Introductory Algebra.
 
thakid87 said:
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)]
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].

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

Thanks.
 

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