Understanding the Law of Sines: Explaining Side a Formula

In summary, you can't get a= (c sin A)/sin C because sin(A)/sin(C) has no units and neither does sin(A)/sin(C).
  • #1
thakid87
8
0
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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  • #2
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
This is nothing more than axioms of Real Numbers and inverse operations, properties of equality. Everyone learns them in Introductory Algebra.
 
  • #4
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.
 

Related to Understanding the Law of Sines: Explaining Side a Formula

1. What is the Law of Sines?

The Law of Sines is a mathematical relationship that helps to solve triangles that are not right triangles. It states that the ratio of the sine of an angle to the length of its opposite side is constant for all angles in a given triangle.

2. How do you use the Law of Sines to find a missing side?

To use the Law of Sines to find a missing side, you need to know at least two angles and one side of the triangle. Then, you can use the formula a/sinA = b/sinB = c/sinC, where a, b, and c represent the sides of the triangle, and A, B, and C represent the angles opposite those sides. Simply plug in the known values and solve for the missing side.

3. What is the significance of the "side a formula" in the Law of Sines?

The "side a formula" in the Law of Sines refers to the formula a/sinA = b/sinB = c/sinC, as mentioned in the previous answer. This formula is significant because it allows us to find missing sides and angles in non-right triangles, making it a powerful tool in solving various mathematical problems.

4. Can the Law of Sines be used for all types of triangles?

No, the Law of Sines can only be used for non-right triangles. This is because the sine function is only defined for acute and obtuse angles, not right angles. For right triangles, we use the Pythagorean theorem to find missing sides and angles.

5. How can understanding the Law of Sines be useful in real-world applications?

The Law of Sines has many real-world applications, such as in navigation, surveying, and engineering. For example, it can be used to find the height of a tall building without actually measuring it, or to determine the distance between two points on a map. It is also used in physics and astronomy to calculate the distance between celestial bodies.

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