# Law of Sines

1. Jun 25, 2008

### thakid87

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.

Last edited: Jun 25, 2008
2. Jun 25, 2008

### Defennder

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.

3. Jun 26, 2008

### symbolipoint

This is nothing more than axioms of Real Numbers and inverse operations, properties of equality. Everyone learns them in Introductory Algebra.

4. Jun 26, 2008

### HallsofIvy

Staff Emeritus
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
$$\frac{sin(A)}{a}= \frac{sin(C)}{c}$$
you might start by multiplying both sides by ac (sometimes called "cross-multiplying"):
$$c sin(A)= a sin(C)$$
Now, since you want to solve for a, just divide both sides by sin(C):
$$c [sin(A)/sin(C)]= a$$.