Law of Sines - Online Tutor, Practice Problems & Exam Prep

To solve non-right triangles, the Law of Sines is essential. It states that the ratio of the sine of an angle to its opposite side is constant across all angles: $(\frac{\sin \left(A\right)}{a}=\frac{\sin \left(B\right)}{b}=\frac{\sin \left(C\right)}{c})$. This method allows for solving for missing sides and angles, particularly in SSA, SAA, and ASA configurations, which can yield one, two, or no solutions, highlighting the concept of ambiguous cases in triangle solving.

Intro to Law of Sines

If I gave you this right triangle on the left here and asked you to solve for the missing sides, you would be able to do it pretty quickly given everything you’ve learned in the course. Because you have a hypotenuse and an angle, you could use SOHCAHTOA to find that other side, which is equal to 3. Then you could use SOHCAHTOA again or the Pythagorean theorem to solve for the other side. Pretty straightforward. What if I gave you this triangle over here and asked you to do the same thing? The key difference here is that this angle is not 90 degrees, it's 70. So, this is not a right triangle but a non-right triangle, which means that this isn’t really a hypotenuse, and that means that you can’t use SOHCAHTOA or the Pythagorean theorem. It just doesn’t work. So, unlike right triangles, you cannot solve for missing sides in a non-right triangle by using SOHCAHTOA and the Pythagorean theorem. So, how do we actually solve for this? Well, I'm going to show you in this video how we use a different equation called the law of sines. And it sounds kind of scary at first, but I’m actually going to break it down for you, showing how it's really straightforward to use.

Let’s go ahead and get started. I’m just going to show you the law of sines. The equation here is the sine of a, big A over little a is equal to the sine of big B over little b, and that’s equal to the sine of big C over little c. A couple of notation points here: you’ll see that angles in these problems are always capital letters A, B, and C, and sides are always lowercase letters. The law of sines is really just three ratios, and it's comparing ratios of angles on the top to lengths on the bottom. It’s called the law of sines because all three terms involve sine, and there’s a pattern here: always an angle over a side, A over a, B over b, C over c, and so on.

To actually solve using the law of sines to find the missing variable, what happens here is you have three ratios. You just try to pick two out of the three in which you know or can figure out three out of four variables. What do I mean by this? In this problem, I know what big A is, big C, and little c. My job then is to choose two out of the three terms in which I know three out of four variables and I can solve that missing one. If I try to pick these two over here, for example, I only know one out of the four. That's definitely not going to work. I know nothing about big B or little b. If I pick these two over here, notice how it doesn’t actually include my target variable. Also, I only know two out of four variables, so I also can’t use this. So, basically, because I know nothing about B, I can just ignore it and choose the first and the third ratio. Over here, I have three out of four variables, and I can solve for this missing one.

I'm going to rewrite this. The sine of big A over a is equal to the sine of big C over c. Trying to solve for this little a over here, there are a couple of ways I can do this. I can cross multiply, but really what I like to do in these problems is when you're solving for a side, you could basically just flip this fraction. This becomes a over sine of A. Whatever you do to one side, you do to the other. So, this will be c over the sine of big C. And then you can move this sine of A up to the top and start plugging in numbers. So, c is equal to 6 and then sine of C, assuming angle C is 70 degrees. Then this is 6 over sine of 70 times the sine of A, which is 30 degrees. When you plug in all these numbers, you will get a is equal to 3.19.

So, a is 3.19, the side of the triangle. Notice how the answer makes perfect sense because in a right triangle with a hypotenuse of 6, when you just go straight down, that's only a length of 3. But in this triangle, you have to go a bit farther because of the diagonal. So, you get something that's a little bit longer than 3, it's 3.19. That's how you use the law of sines. One last point I want to make is that instead of seeing capital letters A, B, and C, you may see some Greek symbols like alpha, beta, and gamma, but it works the exact same way.

That's how you use the law of sines. Let me know if you have any questions, and let's get some practice.

Use the Law of Sines to find the length of side $a$ to two decimal places.



• A. 8.20
• B. 4.39
• C. 2.20
• D. 1.61

Use the Law of Sines to find the angle $B$ to the nearest tenth of a degree.



• A. 48.6°
• B. 77.2°
• C. 40.5°
• D. 35.3°

Solving SAA & ASA Triangles

So by now, we've seen how to use the law of sines to solve for missing sides in a non-right triangle. One of the most common types of problems you'll run into is where a problem will give you some angles and sides, and they'll ask you to classify and then solve a triangle by finding all of the other missing information, the other missing sides and angles. Now if that seems kind of intimidating, don't worry, because what I'm going to show you in this video is that, really, to solve these problems, all we need is the law of sines that we just learned, plus another familiar equation we've seen a bunch of times before. So I'm going to walk you through a step-by-step process of how to solve these two specific types of triangles. But first, I want to introduce you to the three types or categories of triangles for which you'll use the law of sines. Let's just jump right in. Alright? So you're going to see a bunch of acronyms in your books and classes like ASA, SAA, and SSA. It's kind of just a soup of letters of s's and a's. And, really, all these types or acronyms just have to deal with is the information that you're given at the start of a problem. The first one is called an angle-side-angle. This is where you were given two angles and the side between them. For example, you were given big A, big C, and little b. The next one is called the side-angle-angle. This is where you're given two angles and a side that's not between them but adjacent to them. So an example of this would be big A, big C, and little c. And then finally, the last one is called side-side-angle. This is where you're given, you guessed it, two sides and an angle. An example of this would be little b, little c, and big C. We're actually not really going to talk about this one at all in this video. We're going to talk about this one in a later video. All we're going to cover, really, are these two in this specific video. Alright? So without further ado, let's just jump right into our example here because the first thing it asks us to do is classify this triangle. So let's just jump right in. Alright? So I've got this information here. A = 30, C = 70 degrees, and then little c = 6. Alright? Regardless of if you can't sort of tell off the bat what type of triangle it is, the first thing you're always going to do is just try to draw or attempt to sketch the triangle because then you'll be able to see what it is from there. Alright? So I'm just going to go ahead and sketch this. Your sketch doesn't have to look exactly like mine, but what I always like to do is look at the angles first because it's easiest to visualize what's going on there. I know that one of the angles is going to be 30 degrees. So what I'm going to do is I'm going to draw a little sort of leg like this, and the 30-degree angle is going to look something like that. So this is going to be 30 degrees. Alright? Now the next one, the next angle that I know is going to be 70 degrees, which is going to look something like this but then a little bit sort of tilted in, not exactly a 90-degree angle. So it's going to look something like this. Alright? So this is going to be my 70 degrees. And I'm basically just going to keep on going with those lines until they sort of intersect like this. And that's going to be my other angle. So this is sort of a sketch of my triangle. This is going to be angle A, and that's angle C based on what I've given, which means that by default, this has to be B. Right? And remember, the rule for the sides, they always have to go opposite of their corresponding angles. So this is little c, which we know is 6. This is going to be little a, and this is going to be little b. So our job is to now solve for this triangle by solving for these missing variables, which we don't know over here, big B, little a, and little b. However, what we can do right now is we can actually classify this triangle because we've already drawn the sketch. If you notice here, what we've got is we've got an angle and angle and one of the sides that's not between the angles but adjacent. So that means that this is actually exactly what this type of triangle set. Right? So the example for this is ACC, and that's exactly what we were given inside this problem. But we just drew it out, and we kind of confirmed this. Right? So this is going to be an SAA triangle. Right? Side-angle-angle. Alright? So if we are given an angle-side-angle or SAA triangle, we're going to go ahead and stick to these steps. And I'm just going to jump right into the second one. So the easiest thing to solve for in these types of problems is actually this third angle over here. How do we solve for the third angle? Well, we're actually just going to go ahead and use a familiar equation that we've seen before, which is the fact that all of the angles in a triangle have to add up to 180 degrees. This is sometimes referred to as the angle sum formula. Basically, since we know these two, we can always solve for the third one, which is by subtracting this from 180. So here's what I'm going to do. I'm going to set that A + B + C is equal to 180 degrees. And so, therefore, I can find B just by subtracting the other two angles that I know from 180. So this is going to be 180 - 70 - 30, which is really just -100. So therefore, B is equal to this is going to be 80 degrees. A lot of times, you'll just be able to use really quick mental math to figure this out, but you could sort of solve this out if the numbers aren't so nice. We just figured out one of our missing variables is 80 degrees. Perfect. Great. So now that we have all of the angles solved, how do we solve for the two missing sides, a and b? Well, we're just going to go ahead and jump to our third step over here, which is we're going to use the law of sines. That's how we solve for missing sides in a non-right triangle. Alright? So we're just going to have to do this twice. And, really, in this third step, it actually doesn't matter which one you go ahead and solve for first. It doesn't matter. You could do a or b. Alright? So remember how the law of sines works? We're going to have to go ahead and set up our sine A over a = sine B over b = sine C over c. Remember, I'm looking for a in this specific example here. So I'm going to have to sort of pick two out of these three ratios in which I can figure out three out of four variables. Alright? If I try to look at B, what happens is I know the angle, but I don't know the side. I know the angle A, but I don't know the side. So I can't pick these two because that's only two out of four. So instead, I'm going to ignore this one, and I'm going to pick these two over here because I know everything about c. I know the angle and the side. Okay? So we've actually seen this exact sort of setup of variables before, but, really, what happens is

Classify the triangle, then solve: $A=60°,B=15°,c=6$.

• A. $SAA,a=6.69,b=22.4,C=105\degree$
• B. $ASA,a=6.69,b=22.4,C=105\degree$
• C. $ASA,a=5.38,b=1.61,C=105\degree$
• D. $SAA,a=5.38,b=1.61,C=105\degree$

An engineer wants to measure the distance to cross a river. If $B=30\degree$, $a=300$$ft$, $C=100\degree$ find the shortest distance (in $ft$) you’d have to travel to cross the river.

• A. $459.6ft$
• B. $195.8 ft$
• C. $152.3ft$
• D. $233.4ft$

Solving SSA Triangles ("Ambiguous" Case)

In the last few videos, we saw how to use the law of sines to solve these two types of triangles, ASA and SAA. Now we're going to take a look at how we can solve the last case that I mentioned, the SSA triangle, where you have two sides and an angle that isn't in between them. Now these types of problems are a little bit trickier because as you're solving them, you may get one of 3 possibilities. You may get no solution. You may get one solution. Or in some cases, you might actually get 2 solutions. This is why your books refer to this as the ambiguous case because you don't know what type of answer you'll get until you actually start working it out. This might seem really confusing at first, but I've solved a ton of these problems. And the way you solve them is pretty much the same every single time. So what I'm going to do is I'm going to break down this problem for you. We're going to work it out step by step, and I'll show you a good way to sort of get the right answer every single time. So let's just jump right in. And throughout the solution, we'll see the different possibilities that we might run into. Alright? So let's get started here. In this problem, we're going to solve for each angle in the triangle, not just the sides or not the sides. And so let's go ahead and get started. We've got a equals 6, b equals 8, and then big A is 41 degrees. So in other types of problems, the first thing you would do is try to sketch the triangle out just to sort of sketch it out and label the given information. But with these types of triangles, it's a little bit harder because you have 2 sides and an angle. So it's kind of hard to sketch out what the full triangle is going to look like because you only sort of know what one corner is going to look like. So I don't even bother trying to sketch it in the first part here. What I try to do first is is actually just use the law of sines to find a second angle. Alright? And that's going to be the first thing that you do in these problems. You use the law of sines to basically set up the sine of some angle is equal to a number. Alright? So if you look at this, what happens here is I have I have a and big A. So in other words, when I set up a law of sines, I have one ratio, and then I have b over here. So, basically, what this looks like here in step 1 is I'm just going to go ahead and set up a law of sines. I have the sine of big A over a is equal to the sine of big B over b. I don't even have to write out the c part because I know nothing about c or big C. Alright? So I've got sine of A over a equals sine of B over b. And what I can do here is I can look for this I can solve for this angle over here, the sine of B. Because that's the only variable in this case that I don't know. I have all of these other 3 over here. Okay? So, really, what happens is I'm just going to go ahead and start plugging in some numbers. I've got the sine of B is equal to when I cross multiply and bring this b up here. Basically, I'm going to start plugging in some numbers. It's going to be 8 times and then I have the sine of 41 degrees divided by 6. When you work this out, what you're going to get here is 0.875. Okay? So this is sort of, like, the first major milestone of this problem, which is that we have the sine of some angle is equal to some number here on the right side. Now, in some cases, this number will be less than 1. In some cases, it will be greater than 1. If you ever have a number that is greater than 1, then what's going to happen here is that you're basically done with the problem because you'll never be able to take the sine of some angle and have this number over here be greater than 1. The sine always goes from negative one to positive one. So if your number over here on the right side is ever bigger than 1, you're done with the problem. You don't even have to continue with the rest of the steps because there's no solution. However, what happens in our problem is that we did get a number that was less than 1. So we're going to continue on to step 2. Okay? Alright. So now that we're done with this first step over here, we're ready to move on with the second one. So the second step we're going to use is we're going to use the inverse sine to solve for that angle. We have the sine of B is equal to 0.875, so that means that B itself is really just going to be the inverse sine sine inverse of 0.875. If you work this out, what you're going to get is 61 degrees. Alright? However, what happens is when you take the inverse sine of some number, there's actually sort of an interesting thing that happens with the unit circle, which is that there are 2 possible angles that will actually get you this number over here when you take the sine. So when you solve for that angle B, there are actually always 2 possibilities that you have to consider. So the first one is when you just plug sine inverse straight into your calculator. You're going to get 61 degrees. And the second one, what I'm going to call here, is B2. B2 is going to be 180 degrees minus the angle that I get over here, so 61. And the reason for this is, if you don't believe me, you can basically go ahead and plug the sine of 61 and the sine of 119. And when you plug that into your calculator, you'll actually get 0.875 for both of those angles. And it's because of the way the way that sine works in the unit circle from 0 to a 180 degrees. So what happens here is that these two angles are both possibilities for what B ends up being. So that's what we do in step number 2. Alright? So we're going to solve for these two angles, right, angle 1 and angle 2. In this case, it's B1, B2. And, by the way, if these two things if if the number here on the right side is ever 1, that means that both of your angles are going to be 90 degrees, and, therefore, you're only just going to have one solution. But that's very rare. It doesn't happen very frequently. Okay? So that's basically done with step number 2. All we have to do now is we have to move on to step number 3. In step number 3, what we're going to do is we're going to look at this angle 2 a little bit more closely. Because what happens is, by now, we've already know we've already solved for 2 of the 3 angles in the triangle. We have A is equal to 41 and B is B2 is equal to 119. So what we're going to do here for this second angle is we're just going to add it to the given angle over here. So I have that A+2 is equal to this is going to be 41 +119. When I plug this in, what I'm going to get is 160 degrees. Alright? So, this is going to be 160 degrees over here. And what happens is I'm going to take a look at this 160 and I'm going to take that sum and compare it to a 180 degrees. Because what happens is if the sum of this ever ends up being more than or equal to 180, that means that the second triangle is not possible. If these two angles, when you add them up together, are already bigger than a 180 degrees, there's no way that 3rd angle could be something that's a positive number. So what happens is that angle is that second triangle isn't even possible, and you're only left with one solution. However, if the sum ends up being less than a 180 degrees, which is what happens in this triangle over here, then that means that the second angle is and second triangle is end up possible. It does end up being possible, and you actually end up with 2 solutions. So that's exactly what happened here. So here we have 2 solutions possible, and that's a huge sort of milestone in these types of problems. We know that there's 2 solutions. Therefore, there's actually 2 possible triangles that could fit the conditions that were given here. Again, that's why it's called the ambiguous cases because as you start working through this problem, you'll start to realize whether there's 0, 1, or 2 possible solutions here. Alright? So now that what we do is we just move on to step 4. Once you figure out how many solutions are possible in this triangle, now you're going to go ahead and solve for the remaining angles and sides by using law of sines and your angle sum formula. Okay? So that's really what is happening over here. So what I'm going to do is, in step 3, I'm still just going to add A to B1, which is just going to be, 41 plus 61 degrees. So 41 plus 61, and this is going to equal a 102 degrees. Okay? So what I'm going to do now is I'm going to solve for the remaining angles. So what happens here is I'm going to use A plus B1 plus C is a 180 degrees. Therefore, what happens is that C is the only remaining angle that I don't have. It's going to be 180 minus the other 2 that I just found, which is 102 degrees. So, what happens is this angle ends up being 78 degrees. That's one possibility for angle C. The other possibility is when I take C, is when I take these in my second triangle I add A plus B2 plus C is equal to 180 degrees. And what happens here is that the C ends up being 180 minus 160. So, that's the 2 angles summed up. That means that big C is equal to 20 degrees. So I want to sort of visualize what's happening here because now we've basically solved for all of the angles in this triangle. I'm going to call this 1 C1 and C2 because it's basically 2 possible triangles that we end up getting. Alright. So I want to actually draw out these triangles and visualize them now that we've figured out all the angles. For the triangle on the left, what this would look like is I'd have a triangle that has a 41-degree angle, something that looks about like this. And then we said that B1 was 61 degrees. In other words, that's going to be also an acute angle. It's going to look something like this. And so this is 61 degrees, and this ends up being 78 degrees. So all of these end up basically being acute angles. Right? Now, again, we we could figure out the sides. This would be A. This would be B1. This would be C1. And we can label out the rest of the sides and solve them if we wanted to, but I just want to sort of visualize the 2 triangles. Now what does the second triangle end up looking like? Well, it's the same angle A. So in other words, this is 41 degrees. But now B2, what we solved for, is it's an obtuse angle, a 119. So So this would actually look something like this. It would be look like a really, really, obtuse angle, maybe even steeper, something that looks like this. And then this angle C that you end up getting over here is going to be really small. That's 20 degrees. So this is a 119, and this ends up being 20. So it turns out that both of these triangles are actually solutions to these given initial conditions that we had here at the top of our problem. So this is would be angle A. This would be B2, and, therefore, this would be C2. Alright? And, again, you could figure out all the other sides. But that's really what's happening in this problem is we actually have 2 solutions, and these are the possibilities that you'll end up getting. Alright? So, hopefully, that made sense. This is a breakdown of how you solve these SSA triangles. Let's go ahead and get some practice.

Solving SSA Triangles ("Ambiguous" Case) Example 1

Everyone, in this example, we're going to solve a problem where we are given a triangle with a=1, b=4, and A=30° degrees. If it wasn't apparent at first, notice that two sides, a and b, and the angle A, which corresponds to one of those sides, are given. This is an indication of a Side-Side-Angle (SSA) triangle configuration.

To confirm this, you can quickly sketch a triangle where a would be opposite of angle A. This hint points to the triangle being a SSA setup. Remember, in these situations, this forms an ambiguous case because the exact shape of the triangle is not determinable from the given information alone.

The first step in solving these types of triangles is to use the Law of Sines without drawing the triangle, specifically to solve for another angle. The equation simplifies to sinAa=sinBb. Given that we only know A, a, b, and no information about c, the focus is solely on finding angle B.

Rearranging and substituting known values, the calculation becomes sinB=(4·sin(30°))1. Simplifying further, 4·sin(30°) equals 2, leading to sinB=2. However, as the sine of an angle cannot exceed 1, this results in no solution to the problem since 2 is impossible for the sine function.

The outcome of this scenario is that the triangle cannot be formed. This would mean that side a, despite being positioned at any angle, is too short to connect and complete a triangle with the given side length and angle. Thus, there is no solution, and the exploratory sketch only serves to visualize the impossibility of forming a triangle under these conditions.

Solving SSA Triangles ("Ambiguous" Case) Example 2

Hey, everyone. Let's take a look at this example. We have another triangle over here in which we're given 2 sides, \( b \) and \( c \). And then the other angle that we're given is one of the corresponding angles or one of the corresponding letters. This is, again, a side side angle triangle. Let's go ahead and stick to the first steps over here, which is we're going to set up a law of sines so that we can get an additional angle. Let's go ahead and get started. So this first step over here, we're going to set up a law of sines. Notice how, again, the only two letters involved here are \( b \) and \( c \). So when we actually set up our law of sines, we're going to completely disregard the \( a \) term because we don't know anything about \( a \) or little \( a \). So in other words, we just have \( \frac{\sin(b)}{b} = \frac{\sin(c)}{c} \). Alright? So again, what happens in these problems is you're going to have big \( b \) over here and little \( b \) and little \( c \), so you can go ahead and solve for \( c \). That's going to be that second angle that you find. Okay? So what we're going to do here, is once we cross multiply, move this to the other side. We're going to see is that \( \sin(c) \) is equal to, and I'm just going to start plugging in some numbers here. This is going to be \( 2 \times \sin(b) \), which, in other words, is \( \sin(29^\circ) \) divided by little \( b \), which is \( 4 \). Now if you go ahead and work this out, what you're going to get is you're going to get something like \( 0.242 \). And whenever you do these problems, again, I like to hold on to a few more decimal places, because otherwise, you get some rounding errors. But when you solve for this, we're going to see here is that we get \( 0.242 \). Remember in the first step, if you ever get something that's larger than 1, you're going to have no solution, but that's not what happened here. Right? So we actually are going to get something that's less than 1 or equal to 1. We move on to step 2, and we're perfectly fine to continue the problem. The second step here is we're going to use the inverse sine to solve for 2 possible angles here. So what's going on? Well, basically, what happens is if we solve for \( c \), remember, we can take the inverse sine. So we're just going to take the inverse sine of \( 0.242 \), what you're going to get here is you're going to get a number out of this. You're going to get the inverse sine is equal to \( 14^\circ \). Alright? But that's not the only angle that, when you plug it back into sine, you'll get this number for. So that's actually where we sort of split our problem into 2 parts. So, basically, what happens is that \( c_1 \) is equals to the inverse sine, which is \( 14^\circ \), but \( c_2 \) is just equal to \( 180^\circ - c_1 \). Right? So in other words, it's just minus the angle that we just found over here. So it's \( 180^\circ - 14 \), and then \( c_2 \), therefore, is equal to \( 166^\circ \). So we've got these 2 different angles over here that, when you plug them both back into the sine, you could double-check this, you'll actually get back to \( 0.242 \). Alright? So how do we figure out which one is possible? Well, actually, that leads to the third step. So after we're done with step 2, we're going to see that we have our 2 possible angles. And by the way, this didn't happen over here, which is that the sine of the angle is equal to 1. So again, this may happen, but it's pretty rare. We're just going to keep going on to step number 3. So, for step number 3, what we're going to do is for this angle here that we just calculated, the second one, we're going to add it to the given angle that we have, and there's a really good reason why. So here's what this means, add to the given angle. The only other given angle that we have is the fact that \( b \) is equal to \( 29^\circ \). So what does this mean? It means that if you add \( b+c_2 \), and this is just going to equal \( 29^\circ + 166^\circ \), so \( 29^\circ + 166^\circ \). If you work this out, what you're going to get is that this is equal to \( 195^\circ \). But how is it possible that two angles in a triangle add up to something that's greater than \( 180^\circ \)? That's impossible. So what happens here is that this first angle that you calculate, \( c_1 \), is always going to work. Right? There's no problem with that. Fourteen degrees is perfectly reasonable. But for this, \( c_2 \), where you get \( 166^\circ \), when you add it to the angle that you already know in the triangle, you're going to get something that's bigger than \( 180^\circ \). Therefore, this is impossible. So, basically, what happens here is that the second angle is completely impossible, and that means that we're only dealing with one solution. Alright? So what happens here is that we're so there's only one solution to this, to this, triangle over here. So I'm going to write here that there's no second solution. Alright? So this is basically part of the answer that there's only one triangle to focus on here, and that's going to be this one. So now we're going to move on to step number 4, because now that we figured out there's only one solution to this problem, we're going to solve the remaining angles and sides of all of the possible triangles that we have, and there's only one triangle here. Okay? So here's what we do. We're going to solve for the remaining angles. How do we do this? Well, in this case, I figured that \( c_1 \) is equal to \( 14^\circ \). What I can do is I could just add this to \( a + b + c_1 \) is equal to \( 180^\circ \). Right? So what we've got here is that I can solve for the missing angle, which, in this case, is just going to be that it's going to be \( a \) is equal to \( 180^\circ - (29^\circ + 14^\circ) \), and what you will get when you do this is you'll get \( 137^\circ \). Alright? So, again, basically, what I get here for this first triangle here is that \( a \) is equal to \( 137^\circ \), \( b \) is equal to \( 29^\circ \), and \( c \) is equal to \( 14^\circ \). All those are perfectly valid angles to have in a triangle. Okay? So this is basically what my third angle ends up being. So if I were to try to draw this out, I'm going to try to sketch this out. Here's basically what this triangle would look like. So I'm going to have one really, really tiny angle over here, which is going to be \( 14^\circ \). So I'm going to try to draw this like this, like this, and then I'm going to have one huge obtuse angle. So it's probably going to look something like this, like that. And then the other angle is also going to be relatively small as well. So this is going to be, like, \( 29^\circ \) over here. This is going to be my \( 14^\circ \), over here, and this is going to be my \( 137^\circ \) over here. So this is a, this is going to be \( b \), this is going to be \( c \), and therefore, what happens is that \( b \) is equal to \( 4 \). We already knew that. \( C \) is equal to \( 2 \). We already knew that as well. How do we solve for this last missing side over here for \( a \)? Well, we just use the law of sines for this. Alright? So I'm going to set up a law of sines, really, really straightforward. This is just going to be that the \( \frac{\sin(a)}{a} = \frac{\sin(b)}{b} \). I can use anything at this point because I already know everything else in the triangle. And what you're going to see here is if you solve for this really quickly and flip this, you're going to get that \( a \) is equal to \( \frac{\sin(137^\circ) \times 4}{\sin(29^\circ)} \). Alright? So if this if you wanted to fully solve out this triangle, you'd have to calculate this last remaining side over here, and what you're going to get is at \( 5.63 \). Alright? So this actually makes perfect sense that you get \( 5.63 \). Remember, all of these other sides are relatively small. This one's \( 2 \). This one's \( 4 \). This one's by far the largest side in this problem because it's opposite of the largest angle, and we get something that is bigger than everything else in the problem, which is \( 5.63 \). So that perfectly makes sense. All right. So that's how to solve these types of problems. It turns out that the second triangle wasn't even possible, and all we had was one solution. Alright.

Here’s what students ask on this topic:

What is the Law of Sines and how is it used to solve triangles?

The Law of Sines is a fundamental equation in trigonometry used to solve non-right triangles. It states that the ratio of the sine of an angle to its opposite side is constant for all three angles in a triangle. The formula is:

$\frac{\sin \left(A\right)}{a}=\frac{\sin \left(B\right)}{b}=\frac{\sin \left(C\right)}{c}$

To use the Law of Sines, you need to know at least one angle-side pair and one additional angle or side. You can then set up a proportion to solve for the unknown side or angle. This method is particularly useful for solving ASA, SAA, and SSA triangles.

How do you solve an SSA triangle using the Law of Sines?

Solving an SSA triangle with the Law of Sines can be tricky because it may yield one, two, or no solutions. Here’s a step-by-step process:

1. Set up the Law of Sines equation: $\frac{\sin \left(A\right)}{a}=\frac{\sin \left(B\right)}{b}$
2. Solve for $\sin \left(B\right)$: $\sin \left(B\right)=\frac{b\cdot \sin \left(A\right)}{a}$
3. Check if the value of $\sin \left(B\right)$ is between -1 and 1. If not, there is no solution.
4. Find angle B using the inverse sine function: $B=\sin ⁻¹\left(\sin \right(B\left)\right)$
5. Calculate the possible second angle: $B=180-B$
6. Check if the sum of the known angle and each possible B is less than 180°. If so, both solutions are valid.
7. Use the Law of Sines again to find the remaining sides and angles.

What are the different types of triangles where the Law of Sines can be applied?

The Law of Sines can be applied to three types of triangles:

1. ASA (Angle-Side-Angle): Two angles and the included side are known.
2. SAA (Side-Angle-Angle): Two angles and a non-included side are known.
3. SSA (Side-Side-Angle): Two sides and a non-included angle are known. This is also known as the ambiguous case because it can yield one, two, or no solutions.

In each case, the Law of Sines helps to find the missing sides and angles by setting up proportions based on the known values.

How do you solve for a missing side using the Law of Sines?

To solve for a missing side using the Law of Sines, follow these steps:

1. Identify the known angle-side pairs and the angle opposite the missing side.
2. Set up the Law of Sines equation: $\frac{\sin \left(A\right)}{a}=\frac{\sin \left(B\right)}{b}=\frac{\sin \left(C\right)}{c}$
3. Choose the ratio that includes the missing side and the known values.
4. Rearrange the equation to solve for the missing side: $a=\frac{\sin \left(A\right)\cdot c}{\sin }$
5. Plug in the known values and calculate the missing side.

This method allows you to find the length of a side when you know two angles and one side or two sides and one angle.

What is the ambiguous case in the Law of Sines?

The ambiguous case in the Law of Sines occurs when solving SSA (Side-Side-Angle) triangles. In this scenario, you have two sides and a non-included angle, which can lead to three possible outcomes:

1. No Solution: If the calculated sine value is greater than 1, no triangle can be formed.
2. One Solution: If the calculated angle and the given angle sum to less than 180°, only one triangle is possible.
3. Two Solutions: If the calculated angle and the given angle sum to less than 180°, and the supplementary angle also forms a valid triangle, two triangles are possible.

This ambiguity arises because the sine function can yield two different angles with the same sine value, leading to multiple possible triangle configurations.