# Talk:Law of Sines

## Contents

### Response to Checklist

Just as for Law of Cosines, for this page, I'm following the Checklist for writing helper pages. This is a page that is somewhere between a helper page and an image page I think. There are helper pages written to assist with this page, yet this page is there to ultimately provide some help for the Solving Triangles page.

I've given this a quick look over and written down my first remarks, but I'll be spending some more time on this tomorrow AnnaP 6/30

All done now! 7/1

### Message to Future

• This page links to the Solving Triangles page, which more thoroughly describes the applications of triangles and such.
• There may be some other topics linked to the law of sines (LOS) like Heron's formula. Topics like this, however, would make more sense here if there was more talk about perimeter and area. This page does already include some extension into the field of geometry with "A Geometric Extension".

### References and Footnotes

• Not much in this department. All of the images are credited to me in the references section, where the programs I used are also listed.

### Quality of prose and page structuring

• This page is very to the point and concise. What needs to be on the page is on the page, and it is set up to mirror the layout of the Law of Sines page.
• There are only two things that I can think of as possibly being superfluous on this page: one of the two derivations for the formula and the geometric extension. Both of the derivations are commonly used and very intuitive. The geometric extension is an interesting and useful link to other parts of geometry and is not seen in the everyday LOS lessons.
I think it's great you include both derivations and the geometric extension
• I think that this page follows a logical order when it comes to the content.
• I hate to say it, but you've done one of my pet peeves... In both of your derivations, you use the word "similarly" and follow it with an equation. To someone who's not use to reading math (eg, a high school student), that makes no sense. I suggest replacing the word with, "since we can go through the same process with the other variables..." or something like that. While "similarly" isn't nearly as bad as "it is easy to show..." and my personal favorite, "obviously" which seems to be exclusively used to justify things that are totally NOT obvious, it's more useful to write out more detail. Sure, what you've written is a totally acceptable proof for a class--it's just not accessible to lower level readers
You also should add a similar blurb at the end of your geometric extension.
Changed all three! Richard 7/1

Chris 7/1/11 While I generally agree with Anna that the page is to the point and concise, I have a suggestion about the first paragraph of the Using Area section at the top of the More Mathematical Description. Remove the first sentence "Let h(a) represent..." as you say it again in a better place three sentences after the Area formula. I also think that paragraph after the Area formula can be made more concise. My suggestion is as follows:

"A triangle can be oriented so that any of its sides is used as the base. In Figure (Image?) 1 below, a perpendicular line is dropped from vertex A to side a and labeled h(a). A similar process is done for Figure 2, this time using side b as the base."

I would then switch the order of your figures since base a is mentioned first. You will end up getting the sinB first, but I think that your text order matching your figure order is more important.

I shifted some things around that I think help to address this issue. Richard 7/1

### Integration of Images and Text

• The images on this page are placed intentionally so that they can be used as references for the proofs and examples.

Chris 7/12/11 As mentioned by Anna a while ago, the main image is of a triangle. The Law of Sines involves relationships within that triangle. What about adding the equation with the three ratios to the picture? Also, I would avoid the word "arbitrary" in the first sentence.

• So far, so good here. This links to all of the necessary triangle/trig pages.
Can you make the first time you use the word "Sine" link to Basic Trigonometric Functions? I know there is a way to make a mouse over a link in addition to being a mouse over... but I don't remember how to do that
It already is! There's a link in the intro sentence and a mouseover in the Basic Description. Do you think it would be more effective if I linked both? Switched them? Suggestions? Richard 7/1
I'd link both, though for the second one, doing it as a mouse over AND a link would be idea
Can't link and mouseover. I linked the one in the basic description and mouseover-ed the one in the intro. I think this is better. Richard 7/1
Figured it out (well Kate did) and fixed it! Richard 7/1

### Examples, Calculations, Applications, Proofs

• The example problem on this page really completes a simple and thorough explanation of the LOS.
Can you not call Example 1, "Example 1" since there is no "Example 2"?
Done! Richard 7/1

### Mathematical Accuracy and precision of language

• This page spells everything out in the proofs and the example with the mathematical formulas, and explains anything that may be confusing.
Can you make circumscribed into a helper page, even if you just leave it as a red link?
There we go!

### Layout

• The paragraphs are pretty short, the layout is pretty good and not unappealing.
• I adjusted the screen size to check for funky margins and this page looks pretty safe.

A couple of layout concerns(that I don't think pose too big an issue):

• One concern I may have with this page (just like for the law of cosines)is the main image, but I played around with it and there's little to do to make it both interesting and not weird. For now, I think this works for sure.
I agree that the image itself works, but your first sentence: "This image shows the law of sines, which relates the sine of a given angle to its opposite side length." isn't exactly right. The image doesn't show the law of sines... it shows a triangle. Can you instead say that it's the triangle used to show the law of sines (and the law of Cosines for that matter).
Chris 7/1/11 (I referred to this comment in the very top section of this page.) (moved below)
Chris 7/1/11 To me, this is clearly a helper page, and the opening sentence gets at that. I agree with Anna's point (in the layout section) that the image does not show the Law of Sines but instead shows a triangle. I would remove the top line entirely and begin with the Basic Description. I would then make the image smaller and align it with the paragraph that begins "Give a triangle with side lengths..."

Edited the opening sentence. Richard 7/1
• There is not much showing on the page with the mathematical description hidden. Though this does not seem like a problem to me, I still wasn't very sure...
• When you put fractions into equations, make your parentheses big. So instead of $(\frac{1}{2})$, you can have $\left( \frac{1}{2} \right)$. Just click on edit this page to see how to do that.
Done-zo! Richard 7/1

Chris 7/1/11 There is a warning on the editing page that the page is 40 kb long and that some browsers may have problems editing pages longer than 32 kb. You might consider splitting off the more recent comments from the older ones.

## Harrison's comments that he wants to make without trying to categorize them

Htasoff 21:29, 30 May 2011 (UTC)

### General

*I took the liberty of adding some more line spaces where your equations and/ or text got squished together. I figured that this is pretty non-invasive, so you wouldn't mind. It would also be a pain and an eye-sore if I just listed places in the article that could use an extra line break.

• Under your Proof section, your images get squished into your text. I recommend putting more room in between the images and the text, or putting the images on the right.
• When putting mathematics in parentheses, use $\left ( math stuff \right )$. this will show up as $\left ( math stuff \right )$. You might not be able to see the difference here, but it is far better when dealing with fractions, subscripts, superscripts, etc. See below"
$(\frac{1}{2})+(5^{81})-(x_4)$  goes to  $(\frac{1}{2})+(5^{81})-(x_4)$
$\left (\frac{1}{2} \right )+ \left (5^{81} \right )- \left (x_4 \right )$  goes to  $\left (\frac{1}{2} \right )+ \left (5^{81} \right )- \left (x_4 \right )$

• I highly recommend labeling both your Images and important equations, especially those you refer to in the text.
• I think we should discuss what to do with the Ambiguous Case. You reference "triangle congruence postulates" as the reason for the ambiguous case, however you actually prove it yourself. You may even be able to do away with the reference all together. I have the other cases (SSS, SAS, ASA, and AAS) on the Congruent triangles page. This is the SSA case, and the only one left is the AAA case. I'm not sure if the ambiguous case belongs on the congruent triangles page in stead or perhaps in addition to the law of sines page.

I would really love for others to weigh in on this and I still need to describe to you what links I can add for the law of sines. Richard 6/3

### Specific

*"that expresses more precisely this relationship" might be better as "that more precisely expresses this relationship." Not a biggy, just being nit-picky.

• "the ratio between a length of one side of a triangle and the sine of its opposite angle is equal for all three sides. " might be better as "the ratio between a length of one side of a triangle and the sine of the angle opposite it is equal for all three sides. "
• "To use the law of sines in triangulation, at least three parts of a triangle must be known: two angles and a side length. Whenever a side length and two angles are given, the law of sines can be helpful to complete the triangle."
• This is redundant. Try to compress it into a single sentence.

• "With two known side lengths and the measure of the angle between, there is no way to use the law of sines to solve the triangle because not one set of opposite angle measures and side lengths is provided. The law of sines is also not helpful when three angle measures or three side lengths are provided. Instead, the law of cosines is often used for triangulation."
• This is confusing. Try this instead: "When the lengths of two sides and the measure of the angle between are known, the law of sines cannot be used to solve the triangle. This is because no set of a side length and the measure of the angle opposite it is provided. The law of sines is also not helpful when three angle measures or three side lengths are provided. In these cases, the law of cosines is often used for triangulation."

*"Let $h$ represent height." should be"Let $h$ represent the height of the triangle."

• Define "vertex", perhaps with a balloon, when using it for the first time in "is the distance from a vertex to the opposite side, so that ..."
• Also, change "so that" to "such that". I had s talk with SM about this, and it reduces to: so that is used when describing processes, and such that is used when describing and clarifying nouns.

• "For example, when a is oriented as the base of the triangle, h runs perpendicular to a and is the distance from a to the vertex of A."
• Try: "For example, when $a$ is oriented as the base of the triangle, $h$ is perpendicular to it, and is the distance from $a$ to the vertex $A$."

• "More on Height" = the height of a moron. :-D

• Confusing notation: $Area_{base = b} = Area_{base = a}$
• I don't know what to suggest for this one. I'd usually use the pipes that you use in things like ${\frac{dx}{dt}}|_{t=12}$, but that's probably too advanced for the audience.

*"Substituting the formula for the area of a triangle,   $\frac{b h_{b}}{2} = \frac{a h_{a}}{2}$ "

• a. I highly recommend labeling important equations that you will be referring back to. By labeling the formula for the area of a triangle, not only will readers know exactly which formula you are referring to, but they will also be able to click on the little hyperlink and return to that formula if they forgot.
• b. Definitely define $h_{b}$ and $h_{a}$ before you use them. I had to give it some thought before I remembered that the former is the height when b is the base, and the latter, when a is the base. perhaps just use $height_{b}$ and $height_{a}$ to keep it consistent with your earlier equations. If you do this, still define them first. The same goes for all occurrences of $h_{b}$ and $h_{a}$.

• "Both $h_{b}$ and $h_{a}$ can be written in terms of side lengths $a, b, c$ and angles $A, B, C$. Therefore, we can substitute ${(c\sin A)}$ for $h_{b}$ and ${(c\sin B)}$ for $h_{a}$"
• After the first sentence you don't say why this is the case. State explicitly that $h_{b}={(c\sin A)}$ etc. and perhaps include pictures illustrating how this is true.

• "The same method follows for side c and angle C."
• After this statement you just put the final equation. Perhaps introduce it with "Thus: equation" or "In this way: equation" etc.
• I noticed that this happens twice, so the above statement applies to both instances.

• "Set both equations for h equal to each other to get ..."
• Instead of "set", say perhaps: "both equations now equal h, so they can, therefore, be set equal to each other to get ..."
• This makes it look less arbitrary

• "This triangle is known as an inscribed triangle."
• vague. Try: "When a triangle is drawn within a circle in this way, it is known as an inscribed triangle."

• "By the law of sines, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2r$. where r is the radius of the circumcircle."
• Why is this so? In all honesty, I myself can't see why this relationship is true.
• If you are going to prove this statement in the following Proof section, which it seems that you are, then state something to the effect of: "where r is the radius of the circumcircle. Which I will now prove."

• What is the little circle in the middle of your diagrams in the Geometric Extension section? If this is an "O" labeling the point that is the center of the circle, then make it bigger and say at the beginning that "O is the center of this circumcircle".
• In the bubble for circumcircle, it would be a lot more straight forward to simply say: circumcircle or better, circumcircle. The definition you you give, while more accurate, is also more nebulous.
• Define "oblique" the first time you use it.

• "According to properties of inscribed angles, two angles that cut the same arc are equivalent."
• Great clarification. Just put in the word "inscribed" so it reads: "two inscribed angles that cut the same arc are equivalent".

• "Multiplying both sides by $\overline{AD}$ and dividing by $\sin C$ gives us"
• I recommend simply saying: "Solving for $\overline{AD}$ gives us">

• Your Geometric Extension section runs into your Example Triangulation section. This part was confusing enough that I fixed it so that I could continue proof reading.
• For the solution to example 1 in Example Triangulation, I recommend solving for the unknown first and then plugging in the known values. What you have works, but it is not the way that unknowns are usually solved for.
• Your horizontal bar after $c = 20$ only underlines one column of the section.
• "If the above problem asked to find the radius of the circumcircle of $\vartriangle ABC$, the law of sines can help to find the diameter."
• "the law of sines can help to find the diameter." is superfluous and beside the point. Instead, just say: "... the law of sines will accomplish this task as well."
• Also, label the sides and angles of your triangle in this part with A, B, C and a, b, c, in addition to the given measurements. Otherwise, when you give the equations, no one knows which variable refers to which side.

• "Since the side length is shorter than the shortest possible distance between the base and the vertex of the triangle ..."
• Try: "Since the side length is shorter than the shortest possible distance between the base and the vertex opposite it ..."

• after your image for each of your scenarios in the Ambiguous Case section, explain what you are going to do before launching into math.
• "one in which the side of fixed length opposite the fixed angle and the side adjacent to the same angle form an acute angle, and one where those same two sides form an obtuse angle."
• I follow this, but it is pretty wordy. See if you can simplify the sentence.

• "Thus, there are no solutions when the side length $a$ is ..."
• For each of the times you say this, consider using the following: "when the length of side $a$ is ..."

• "In the picture below, no matter how the blue side swings, it's endpoint will touch the base of the triangle only twice."
• Try: "it's endpoint will touch the base of the triangle exactly twice."

• for supplementary try supplementary.

Thanks Harrison!

• It's a topic that really needs covering
• You say what the Law of Sines is used for, not just what it says
• You compare and contrast to the Law of Cosines and describe its relationship to SSA, building connections between this topic and other topics
• the sample triangulations are such a good idea! :) -Kate 18:54, 19 May 2011 (UTC)

Kind of similarly, if you use a variable, define it. In this page, a picture will often mostly take care of defining a variable, but there should be a bit of explanatory text that refers to the picture for each variable, anyway.

-Abram, 5/18

pictures are my next step -Richard 5/18

I did my best to fix all of the formula/math writing and I added several pictures. Richard 5/19

I think you need to figure out a way to format your equations and pictures so they don't appear in a random order. Last year, I used tables and they worked very well. They allow you to put pictures and text in ways the normal formatting won't allow you to do. See Compass & Straightedge Construction and the Impossible Constructions. I addition, for geometry type of writing, I find it very helpful to do bullet points. It allows the reader to follow your thought exactly and not confused and/or intimated by sentence preceding it or following it. It is like bite size food they can easily chew. - xd 01:53, 25 May 2011 (UTC)

Most pictures are already placed in columns where I thought they should go. Do you think I need to change where I'm putting my pictures? Richard 5/25

### Basic Description

Some suggestions:

• One thing I would consider is trying to find an image where you can see an application of finding the sine of an angle in the real world. Then you can use the introduction to explain the application and how you use the law of sines to solve it. Your main image could then be used in the basic description. If you need ideas we can talk on Monday, I should be around in the afternoon. If you like the page's organization the way it is, that's fine too. I just thought this might be a good way to hook people in.

I agree. I'm really not a fan of my main image as it is now.-Richard 5/23

• You might want to divide the basic description into three sections with small headings. The sections could be when the law of sines helps, when it doesn't tell you anything, and the ambiguous case. Your beginning is great, but the sections would probably be useful after you define triangulation. It would also allow you to elaborate on this section by possibly saying which exact cases fall under each category.

A couple people have given me comments about this and I'm working through a couple ideas. Abram suggested adding a helper page here for triangle congruence postulates? -Richard 5/23

Small question - how accessible do we want the first bit to be? What if someone comes to this page and doesn't know what sine is? I suppose we could just assume that users have enough trig knowledge to know what sine is, but we could probably also make a helper page for sine without too much trouble. -Kate, 5/17

Good point, Kate. Rather than defining "sine" in the text, though, maybe link to a Sine page, which you can create later. This page would just be a helper page, so it wouldn't take very long to create.
-Abram, 5/18
I call dibs on a Basic Trigonometric Functions helper page defining sine, cosine, & tangent. -Kate, 5/18

The basic description should say something meaningful even to people who don't know what the sine of an angle is. The way you talk about what the Law of Sines is used for in later paragraphs in that section is one good example of achieving this gold. I added a couple sentences in bold that I think also help achieve this goal. See what you think. -Abram, 5/18

thanks for the tips. I made "sine" a green mouse-over thing for the time being. I think I'm going to run into this same initial problem in the geometric examples part with all of the inscribed angles stuff -Richard 5/18

It seems to me like the last two paragraphs under basic description are trying to say the same thing, maybe you can combine them? -Kate 18:45, 19 May 2011 (UTC)

It seems like it might make organizing this page a lot easier if you were to have a separate page called "Triangle congruence postulates" or something. Imagine the content that page would have -- a statement of ASA and SAS and pictures of why they work, SSA and a picture of why it doesn't work, etc. --, and then simply refer to and link to that page as though it exists.

Your sort of begin to do this anyway, for instance in your mouse-over that refers to "SSA". How much easier would it be if you could do a similar thing throughout the page!

Later, you can create that page if you would like.

-Abram, 5/18

We want material to be accessible, but terminology should still be precise enough to give statements as much meaning as possible. A few examples of places to be more precise include:

• "The law of sines states that the relationship between a length of one side of a triangle and the sine of its opposite angle are equal for all three sides." What do you mean by "relationship"?
• "at least three parts of a triangle must be known" -- what do you mean by "part"?

-Abram, 5/18

Tried to address these so far. Let me know what you think. I'm not exactly sure if I'm a fan of hiding the formula in the basic description part though. -Richard 5/18

### Derivations

• In your "More on Height" section, you should explain how you're getting from one equation to the next. For example, you can start with
Since the area of the triangle is the same no matter how the triangle is oriented, the area of the triangle when we use $b$ as a base is the same as the area of the triangle when we use $a$ as a base.
$Area_{base = b} = Area_{base = a}$
Substituting the formula for the area of a triangle, we have
$\frac{b h_{b}}{2} = \frac{a h_{a}}{2}$
Then, since... (You get the picture). The same thing goes for the next section... You need to give brief explanations of how you're getting from one equation to the next.

Took a shot at this for all of my formula sections.If someone could take a look and let me know what you think, that'd be great. -Richard 5/23

I thought the proofs and the sample triangulation were much easier to read now. You should ask Prof Maurer to look at them, though, he's really good with proofs & writing them clearly. -Kate 14:49, 23 May 2011 (UTC)

• I would change the heading from "Using Definition of Sine" to "Using the Definition of Sine."

$(1/2)cb sinA = (1/2)casinB$

c sinA is the height and b is the base = c sinB is the height and a is the base

make the above explicit. Say that “c sinA is the height and b is the base” etc.

Also, try: (1/2) b (c sinA) = (1/2) a (c sinB)

Harrison 5/19/11

reversing the order of cb and ca, will make the equation read as (1/2) base x height.

For the second derivation, it would be helpful if you drew in or explained where h goes. -Kate 18:50, 19 May 2011 (UTC)

I agree with you guys in the height thing. Abram said it earlier too. But I'm not exactly sure how to go about doing it since there are three different ways I could make the height. Plus, for this second derivation, I use two different heights. Any suggestions?

I tried to address this with a new picture. let me know what you think.

Added the "More on Height" section and a picture to hopefully deal with this problem.

-Richard 5/19

It seems to me that, as the equations section is currently laid out, it would be really confusing to someone who didn't already understand basically why the law works. Can you set up your derivations more like actual "proofs"? That is, adding in some actual words and explanations of where the transitions in each step are coming from and why they're valid?

See Becky's comment above on same topic. My notes are up there. - Richard 5/23 Wow. Sorry. I thought I read through everything else before I posted this. -D

Also, on the topic of h, be careful about referring to it as a "distance," because you are often treating it as an actual side, rather than a measure; in such cases, it's not a "distance," even though its length is equal to the distance between the base and the upper vertex.

-Diana 15:15 5-20-11

But $h$ is a measure of length. The line in the picture shows which distance I'm referring to. All of the side lengths $a,b,c$ are also all measures. They are side lengths, not sides. - Richard 5/23

You're right. I think what bothered me was more the fact that you do use phrases like "side c," which complicates the issue of side length vs. side notation. But maybe I'm just being sophistic at this point. For some reason, it seemed unclear to me as I imagined reading it as someone who didn't already understand much geometry. -Diana 16:43 5-23

Why hiding the pictures? Without them, the first paragraph of the explanation does not make much sense. -xd 01:43, 25 May 2011 (UTC)

### Geometric Extension

I was just reading through the proof with the inscribed triangle, and it looks to me like the triangle that you've got in blue (the one whose side is a diameter) must also be a right triangle (or how else did you get sin D= AB/AD?) but that's not something you ever explicitly address. -Kate 22:36, 22 May 2011 (UTC)

• Your definitions of circumcircle and inscribed triangle are clear and the image is really helpful, but it's still a good idea to label them in the image. If you need help, I can show you how to do it in photoshop.
• I think your proof would be much clearer if there were pictures indicating each step. I can help you with this too on Monday.

Added the picture that we discussed in the geometric proof section, talked about the right angle. Let me know what you think. Richard 5/23

• I like the addition of the picture, but you need to explain the transition between the two pictures. After you have sinD=AB/AD, you should have a transition sentence. Something along the lines of "Next, we can add another triangle to this figure. The new triangle will be triangle ABC, and will share side AB with triangle ABD. Now, since both angles C and D cut the same arc..."
• Also, move the second picture up so that readers know they should be referring to it during this explanation.
• Before the next equation say something like "Substituting sinC for sinD gives us.."
• Another transition sentence to add: "Mutliplying both sides by sinC and dividing by AD gives us..."
• Great end to this section! Nice transitions.

Rebecca 01:38, 25 May 2011 (UTC)

changed the picture and added/edited those transitions. Let me know if it seems to work. Richard 5/26

### Example

• I really like that you give an example for people to solve. In the solution though, add short explanations between the equations again.

-Becky

made these changes and looked them over with Becky. Richard 5/23

• The columns are running together in one part... Can you space them out more?

changed the column spacing Richard 5/25 * I think your transitions are clear and articulate. Good job! I think the tone you use is atypical though. Like "Cross multiply to get", would normally be phrased "Cross multiplying gives us". I'm not sure why they do it this way, but it seems to be the norm. I would keep this in mind when you're adding them.

• I don't think the picture is in the best spot for this section.

Rebecca 01:39, 25 May 2011 (UTC)

which picture?, edited the transitions Richard 5/26

### The ambiguous case

I want to make sure that this part of the page is very clear. If people could read this over that'd be great! Richard 5/24

• I think you have an error in your first paragraph of this section. It's true that there is an unknown length and two unknown angles, but the swinging side cannot be connected to any point along the dashed side. It can be connected to one of two possible points corresponding to the angle, not to any point along the base. This might not have been what you mean, but i don't think the section or the picture is clear.
• You should make the fixed length swinging side long enough so that it touches the base in two places in the picture I think, since this is the most common case.
• I think you need to label the parts of this triangle in the picture as well. Its too hard to keep referring to things as the "third side" or the "base."
• The end of this section is very clear!

Rebecca 01:43, 25 May 2011 (UTC)

#### First Scenario: No solution

* First sentence is very confusing. You have too many fragmented thoughts.

• Second Scenario: One solution & Third Scenario: Two solutions sections are very clear!

Rebecca 01:44, 25 May 2011 (UTC)

edited first sentence Richard 5/25

xd 02:02, 25 May 2011 (UTC)

1. You need better transition between the previous section to this section. This should not be an independent section by itself.

I was actually thinking that this shouldn't be in the mathematical explanation section at all. If I move this and the example sections out, I think the remaining sections would leave just a mathematical explanation and the ambiguous case and the example would be more about computational aspects of the law of sines. Richard 5/25

2. I think the determinant of which kind of solution, i.e. 0, 1 or 2 is the swinging side with the fixed length instead of the height of tghe t

2. Determinant of solution -> there is no triangle to start with. So don't say "height of the triangle". Say distance between the vertex and the base line as shown in the picture.

• I added a few words to show that the swinging side compared to the height is the determinant
• but it ultimately is the height of the triangle. Would it be okay if I add a sentence or two to explain that?

Richard 5/25

## Second Proof

#### Version 2

According to the properties of inscribed angles, the central angle is twice the measure of an inscribed angle that subtends, or cuts, the same arc.

This means that $\angle{D} = \angle{2C}$.

Since the angles have equal measure, then so are their sines and cosines.

$\sin D = \sin 2C$
$\cos D = \cos 2C$

Using the Law of cosines to solve $\vartriangle BCD$, we can show that

$\frac{c}{\sin C} = 2r$

Plug in the appropriate values for $\vartriangle BCD$into the formula

$c^{2} = r^{2} + r^{2} - 2(r)(r) \cos D$

Substituting $\angle{D}$ for $\angle{2C}$ gives us

$c^{2} = 2 r^{2} - 2 r^{2} \cos 2C$

Using trigonometric identities, we know that

$1 - 2\sin^{2} C = \cos 2C$.

Substitution gives us

$c^{2} = 2 r^{2} - 2 r^{2} (1 - 2\sin^{2} C)$

Distribute to get

$c^{2} = 2r^{2} - 2r^{2} + 4r^{2} \sin^{2} C$

Simplify for

$c^{2} = 4r^{2} \sin^{2} C$

Dividing both sides by $\sin^{2} C$gives us

$\frac{c^{2}}{\sin^{2} C} = 4r^{2}$

Take the square root of both sides

$\sqrt{\frac{c^{2}}{\sin^{2} C}} = \sqrt{4r^{2}}$

Simplify for

$\frac{c}{\sin C} = 2r$

And by the law of sines,

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2r$

## The Ambiguous Case

Given two adjacent side lengths and an angle opposite one of them, there in no definite completion of a triangle. According to triangle congruence postulates, two triangles cannot be proved congruent given two adjacent side lengths and an angle opposite, or side-side-angle(SSA). With the SSA configuration, there is a fixed angle connecting the base of the triangle and one of the adjacent sides. The length of the base is unknown, denoted below by a dashed line. The length of the remaining side length is fixed, but neither adjacent angle is known. This means that this third side can be positioned in whatever way connects the upper point of the other side with the fixed length and any point along the indefinitely sized base.

Given any SSA configuration, there are three scenarios that can occur in triangle completion: no solution, one solution, or two solution. Which of the three scenarios occurs for any SSA configuration depends on the length of the swinging side compared to the height of the triangle. In any SSA configuration, the height can always be determined because the furthest vertex from the base is known. The height is the distance of the line segment that runs perpendicular to the base and goes through the uppermost point on the fixed side length that is adjacent to the angle.

We can use trigonometry to determine the value of height $h$.

$\sin A =\frac{opposite}{hypotenuse}$

Substituting the appropriate values,

$\sin A = \frac{h}{b}$

Multiply both sides by $b$ to get

$h = b \sin A$

Because the SSA configuration can cause these three different scenarios with different numbers of solutions, it is often referred to as the ambiguous case.

### First Scenario: No Solution

In the first scenario, the known side length that is opposite the fixed angle is shorter than the height, $h$. Because this leg of the triangle is shorter than the height, there is no solution. The shortest distance between a point and a given line is the the line segment that is perpendicular to the given line and goes through that point, which in this case is height $h$. Since the side length is shorter than the shortest possible distance between the base and the vertex of the triangle, the side opposite the fixed angle will never be able to reach the base of the triangle. Thus, there are no solutions when the side length $a$ is less than $h$ or $b \sin A$.

In the picture below, no matter how the orange side swings, it will never touch the base of the triangle. This triangle will never be complete.

$h = b \sin A$

Substituting in the appropriate measures,

$h = 10 \sin 30^\circ$

$h = 10 (\frac{1}{2})$
$h = 5$

Since the other given side length is $4$ and since $4<5$, there is no solution.

In summary, when $a < b \sin A$, there is no solution for a SSA configuration.

### Second Scenario: One Solution

In the second scenario, the given side length that is opposite the given angle is equal to the height, $h$. Because this leg of the triangle is equal to the height, there is only one solution. The height, as explained above, is the single shortest possible distance from the uppermost vertex to the base of the triangle. Since the leg of the triangle is the same length as the height, there is only one way to orient this leg to make the triangle complete: perpendicular to the base and through the vertex. Thus, there is only one solution when the length $a$ is equal to $h$ or $b \sin A$, and this automatically forms a right triangle.

In the picture below, no matter how the green side swings, it will only touch the base of the triangle once. This triangle will only be complete when the triangle becomes a right triangle.

$h = b \sin A$

Substituting in the appropriate measures,

$h = 10 \sin 30^\circ$

$h = 10 (\frac{1}{2})$
$h = 5$

Since the other given side length is $5$ and since $5=5$, there is only one solution which is a right triangle.

In summary, when $a = b \sin A$, there is just one solution for a SSA configuration.

### Third Scenario: Two Solutions

In the third scenario, the given side length that is opposite the given angle is greater than the height, $h$. Because this leg of the triangle is greater than the height, there are two solutions to complete the triangle. This side length will complete the triangle at exactly two points: one in which the side of fixed length opposite the fixed angle and the side adjacent to the same angle form an acute angle, and one where those same two sides form an obtuse angle. Thus, there are two solutions when the side length $a$ is greater than $h$ or $b \sin A$

In the picture below, no matter how the blue side swings, it's endpoint will touch the base of the triangle only twice. This SSA configuration will complete two separate triangles.

$h = b \sin A$

Substituting in the appropriate measures,

$h = 10 \sin 30^\circ$

$h = 10 (\frac{1}{2})$
$h = 5$

Since the other given side length is $6$ and since $6>5$, there is are two unique triangular completions.

In summary, when $a > b \sin A$, there are two solutions for a SSA configuration.

#### Determining Both Solutions

The ambiguous case often produce two possible completions of the triangle. In these two potential triangles, the corresponding angles between the swinging sides and the unknown sides are supplementary. To find both triangle, just solve for the first triangle normally, then find the supplement of the measure of the angle between the swinging side and the base and solve using that angle.