Law of Sines - Video Tutorials & Practice Problems

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1

concept

Intro to Law of Sines

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4m

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If I gave you this right triangle on the left here and asked you to solve for the missing sides, you'd 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 Sokoto it to find that other side in which that side is just equal to three, then you could use Sokoto 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 isn't 90 degrees, it's 70. So this is not a right triangle. It's a non right triangle, which means that this isn't really a hypotenuse. And that means that you can't use Sokoto 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 SOGA to in the Pythagorean theorem. So how do we actually solve for this? Well, what I'm gonna show you in this video is we use a different equation called the Law of Science and it sounds kind of scary at first, but I'm actually gonna break it down for you and show you that it's really straightforward to use. Let's go ahead and get started. All right. So I'm actually gonna, I'm just gonna show you the law of signs. It's an equation here. It's the sign of a big A over little A is equal to the sign of Big B over little B that's equal to the sign of Big C over Little C. So just a couple of notation systems here. So what you're gonna see here is that angles in these problems are always capital letters A B and C and sides are always lowercase letters. So really what we see here is that the law of science is really just three ratios and it's comparing ratios of angles on the top two lengths. So it's called the law of signs because all the three terms involve sign and there's a pattern here. It's always an angle over a side A over A B over BC, over C and so on. All right. So then how do I actually solve and use this law of science to solve for that missing variable? Basically, what happens here is you have three ratios, you're just gonna try to pick two out of the three in which you know, it would know or can figure out three out of four variables. What do I mean by this? Basically what happens here is in this problem? I know what big A is big C and little C, my job is, I have to pick two out of the three sort of terms in which I know three out of four variables and I can solve that missing one. So if I try to pick these two over here, for example, what happens is I only know one out of the four. So that's definitely not gonna work. I know nothing about big B or little B if I try to pick these two over here, notice how it doesn't actually include my target variable, but 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'm just gonna ignore it and I'm just gonna pick the first and the third ratio, those are gonna be my two out of three. So notice how here I have three out of four variables and I can solve this missing one. All right. So I'm just gonna rewrite this. The sign of big A over A is equal to the sign of big C over C trying to solve for this little A over here. There's a couple of ways I can do this, I can cross multiply. But really what I like to do these problems is when you're solving for a side, you could basically just flip this fraction. This becomes a over sign of a, whatever you do to one side to do. To the other. So this is gonna be C over the sign of big C. And then, now what you can do is you can move this sign of A, up to the top and you can just start plugging in numbers. So a over here, the reason you have to do this is because otherwise you're solving for a denominator. So if you plug in these numbers, which you're gonna get AC over sine C, in other words, C is equal to the six and then sign of C, we'll see, we'll see here that the sign of angle C is 70 degrees. So in other words, six over angle s of 70 times the sign of A. So this is gonna be the sign of and this is gonna be 30 degrees over here. All right, when you actually plug in all of these numbers, which you're gonna see here is that you're gonna get A is equal to 3.19. All right. So we just get 3.19. And that is the side that is this side over here. 3.19. All right. That's how you use the law of signs. Notice how the answer makes perfect sense because here in the right triangle with a hypotenuse of six, if you just go straight down, that's only a length of three. But in this triangle, you kind of have to go a little bit farther because this is sort of like this weird diagonal thing, you get something that's a little bit longer than 33.19. All right. So it totally makes sense. One last point I want to make here, by the way 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. All right. That's how you use the law of signs. Let me know if you have any questions and let's get some practice.

2

Problem

Problem

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

3

Problem

Problem

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°

4

concept

Solving SAA & ASA Triangles

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7m

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So by now, we've seen how to use the law of science 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 gonna show you in this video that really to solve these problems, all we need is the law of science that we just learned plus another familiar equation. We've seen a bunch of times before. So I'm gonna 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 Science. Let's just jump right in. All right. So you're gonna see a bunch of acronyms in your books and classes like a SAS A a and SS A. It's kind of just a soup of letters of Ss 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 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 a 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 a side side angle. This is where given you guessed it two sides in an angle. An example of this would be little B, little C and big C. We're actually not really gonna talk about this one any in at all in this video. We're going to talk about this one in a later video. All we're gonna cover really is this these two in this specific video? All right. So without further ado, let's just jump right into our example here because the first thing we ask us to do is classify this triangle. So let's just jump right in. All right. So I've got this information here. A equals 30 C equals 70 degrees and then little C equals six. All right, regardless of if, if you can't sort of tell uh off the bat what type of triangle it is the first thing you're always gonna 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. All right. So, uh I'm just gonna 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 gonna be 30 degrees. So what I'm gonna do is I'm gonna draw a little sort of leg like this and the 30 degree angle is gonna look something like that. So this is gonna be 30 degrees. All right. Now, the next one, the next angle that I know is gonna be 70 degrees, which is gonna look something like this, but then a little bit sort of tilted in not exactly a 90 degree angle. So it's gonna look something like this. All right. So this is gonna be 70 degrees and I basically just sort of gonna keep on, keep on going with those lines until they sort of intersect like this. And that's gonna be my other angle. So this is sort of a sketch of my triangle. This is gonna 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 six. This is gonna be little A and this is gonna 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, but 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, an 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 uh right. So the example for this is, is AC C 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 gonna be an sa a triangle, right side angle angle, all right. So if we are given a angle, side angle or sa A triangle, we're gonna go ahead and stick to these steps and I'm just gonna 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 gonna 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 100 and 80 degrees. This is sometimes referred to as the angle some 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 gonna do, I'm gonna do in step two, I'm just gonna set that A plus B plus C is equal to 100 and 80 degrees. And so therefore, I can find B just by subtracting the other two angles that I know from 180. So it's gonna 180 minus 70 minus 30 which is really just minus 100. So therefore, B is equal to, this is gonna be 80 degrees. All right. A lot of times we 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. So we just figured out one of our missing variables is 80 degrees. All right. Perfect. Great. So now that we have all of the angles sort of solved, how do we solve for the two missing sides? A and B? Well, we're just gonna go ahead and jump to our third step over here, which is, we're gonna use the law of sign, that's how we software missing sides in a non right triangle. All right. So we're just gonna 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. All right. So remember how the law of signs works, we're gonna have to go ahead and set up our sin A equals A equals SB over B equals sc over C. Remember I'm looking for a, in this specific uh in, in this uh example over here. So I'm gonna have to sort of pick two out of these three ratios in which I can figure out three out of four variables. All right, 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 gonna ignore this one and I'm gonna pick these two over here because I know everything about C I know the angle and the side. OK. So we've actually seen this exact sort of setup of variables before. But really what happens is I'm just gonna flip this upside down. So flip this becomes a over sine A equals C over sine C. And I'm just gonna go ahead and plug in some numbers, you can cross multiply A is just gonna equal but little C is equal to six over the sine of big C which is 70 times the sign of angle A which remember is 30 degrees. We've actually seen this exact number before and this works out to 3.19. So this is 3.19. That's what this other angle ends up being. And that is we're one step closer. All right. So now we have to use the law of signs again, uh in which we're going to solve for that other missing side. OK. So now we're just gonna do the exact same thing again. We're just gonna set up for uh solving for B. So I'm just gonna sort of go a little bit faster here. If we're solving for B, we don't care about A and I'm just gonna do sign of B over B equals sign of C over C. You also used A, in this case, that's totally fine. I'm just gonna use these two ratios. All right. So I'm just gonna go ahead and flip this, this equals B equals. And then when I flip and then move this to the other side, this is just gonna become six over the sine of 70 right? We just saw how to do that six over S 70 times the s of B and the sign of angle B is gonna be the sign of 80. When you work the sandwich, you're gonna, for little B is, you're gonna get 6.29. All right. So that's your final number, right? So we just figured out the angle and the two missing sides, it was these numbers over here. Therefore, we've actually completely solved this triangle. So there's a step by step process, you draw, you use the angle some formula and then you use the law of signs to figure out the missing sides. That's it for this one, folks. Thanks very much for watching and let's get some practice.

5

Problem

Problem

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

$SAA,a=5.38,b=1.61,C=105\degree$

6

Problem

Problem

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

$233.4ft$

7

concept

Solving SSA Triangles ("Ambiguous" Case)

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9m

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In the last few videos, we saw how to use the law of science to solve these two types of triangles A S A and SA A. Now we're gonna take a look at how we can solve. The last case that I mentioned the SS A 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 three possibilities, you may get no solution, you may get one solution or in some cases you might actually get two 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 gonna do is I'm gonna break down this problem for you. We're gonna 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. All right. 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 six B equals eight 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 two sides and an angle. So it's kind of hard to sketch out what the full triangle is gonna look like because you only sort of know what one corner is gonna look like. So I don't even bother trying to sketch it in the first part here. What I try to do first is I actually just use the law of signs to find a second angle. All right. And that's gonna be the first thing that you do in these problems to use the law of signs to basically set up the sign of some angle is equal to a number. All right. So if you look at this hap, what happens here is that I have a and big A, some of the words when I set up a law of signs, I have one ratio and then I have B over here. So basically what this looks like here in step one is I'm just gonna go ahead and set up a law of signs. I have that. The sign of big A over A is equal to the sign 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. All right. So get sine of A over A equals sine of B over B. And what I can do here is I can look for this, uh I can solve for this uh angle over here. The S of B because that's the only variable in this case that I don't know I have all of these other three over here. OK. So really what happens is I'm just gonna go ahead and start plugging in some numbers. I've got the sign of B is equal to when I cross multiply and bring this B up here. Basically, I'm gonna start plugging in some numbers. It's gonna be eight times and then I have the sign of 41 degrees divided by six. When you work the sandwich you're gonna get here is 0.8 75. OK. So this is sort of like the first major milestone of this problem, which is that we have the sign of some angle is equal to some number here on the right side. Now, in some cases, this number will be less than one. In some cases, it will be greater than one. If you ever have a number that is greater than one. And what's gonna happen here is that you're basically done with the problem because you'll never be able to take the sign of some angle and have this number over here be greater than one sign always goes from negative one to positive one. So if your number over here on the right side is ever bigger than one, 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 one. So we're gonna continue on to step two. OK. All right. So now that we're done with this first step over here, we're ready to move on with the second one. So in the second step we're gonna use is we're gonna use the inverse sign to solve for that angle. We have the sign of B is equal to 0.875. So that means that B itself is really just gonna be the inverse sign an inverse of 0.8 75. If you work the sound, what you're gonna get is 61 degrees. All right. However, what happens is when you take the inverse sign of some number, there's actually sort of an interesting thing that happens with the unit circle, which is that there's two possible angles that will actually get you this number over here when you take the sign. So when you solve for that angle B, there's actually always two possibilities that you have to consider. So the first one is when you just plug sign inverse straight into your calculator, you're gonna get 61 degrees. And the second one, what I'm gonna call here is B one and two B two is gonna be 100 and 80 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 sign of 61 and a sign of 119. And when you plug that into your calculator, you'll actually get 0.8 75 for both of those angles. And it's because of the way the way that sign works in the unit circle from 0 to 100 and 80 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 two. All right. So we're gonna solve for these two angles, right angle one and angle two. In this case, it's B one B two. And by the way, if these two things, uh if, if the number here on the right side is ever one, that means that both of your angles are gonna be 90 degrees and therefore, you're only just gonna have one solution, but that's very rare. It doesn't happen very frequently. OK. That's basically done with step number two. All we have to do now is we have to do, move on to step number three and step number three. What we're gonna do is we're gonna look at this angle two, a little bit more closely because what happens is by now we've already known we've already solved or two of the three angles in the triangle, we have A is equal to 41 and B is B two is equal to 119. So we're gonna do here for this second angle is we're just gonna add it to the given angle over here. So I have that A plus B two is equal to, this is gonna be 41 plus 119. When I plug this in, what I'm gonna get is 100 and 60 degrees. All right. So, um this is gonna be 100 and 60 degrees over here. And what happens is I'm gonna take a look at this 160 I'm gonna take that sum and compare it to 100 and 80 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 100 and 80 degrees. There's no way that third 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 100 and 80 degrees, which is what happens in this triangle over here, then that means that the second angle is, and second triangle is end up poss does end up being possible and you actually end up with two solutions. So that's exactly what happened here. So here we have two solutions possible. And that's a huge sort of milestone in this type, these types of problems we know that there's two solutions. Therefore, there's actually two 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 01 or two possible solutions here. All right. So now that what we do is we just move on to step four. Once you figure out how many solutions are possible in this triangle, now you're gonna go ahead and solve for the remaining angles and sides by using some uh they using love signs and your angle, some formula. OK. So that's really what is happening over here. So what I'm gonna do is in step three, I'm still just gonna add A to B one which is just gonna be a 41 plus 61 degrees. So 41 plus 61 and this is gonna equal 100 and two degrees. OK? So what I'm gonna do now is I'm gonna solve for the remaining angles. So what happens here is I'm gonna use A plus B one plus C is 100 and 80 degrees. Therefore, what happens is that C is the only remaining angle that I don't have is gonna be 100 and 80 minus. The other two that I just found, which is 100 and two 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 uh is when I take these, the, in my second triangle, I add A plus B two plus C is equal to 100 and 80 degrees. And what happens here is that the C ends up being 100 and 80 minus 100 and 60. So that's the two sum of the two angles summed up. That means that big C is equal to 20 degrees. So I wanna sort of visualize what's happening here because now we've basically solved for all of the angles in this triangle. I'm gonna call this one C one and C two because it's basically two po triangles that we end up getting. All right. So I want to actually draw out these triangles and visualize them now that we 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 we said that B one was 60 degrees. In other words, that's gonna be also a uh an acute angle. It's gonna 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 B one, this would be C one and we could label out the rest of the sides. And so them if we wanted to, but I just want to sort of visualize the two triangles. Now, what is the second triangle end up looking like? Well, it's the same angle. A. So in other words, this is 41 degrees. But now B two, what we solved for is it's an obtuse angle. 100 and 19. So this would actually look something like this. It would be look like a really, really uh obtuse angle, maybe even steeper something that looks like this. And then this angle c that you end up getting over here is gonna be really small, that's 20 degrees. So this is 100 and 19 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 would be angle A, this would be B two and therefore, this would be C two, all right. And again, you could figure out all the other sides. But that's really what's happening in this problem is we actually have two solutions. And these are the possibilities that you end up getting. All right. So hopefully, that makes sense. This is a breakdown of how you solve these ss a triangles. Let's go ahead and get some practice.

8

example

Solving SSA Triangles ("Ambiguous" Case) Example 1

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4m

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All right, everyone. So in this example, we're gonna do this problem here where we're giving a triangle, A equals one B equals four and big A equals 30 degrees. Now, right off the bat, if you didn't notice what you see here is that you have two sides that are given A and B and then the angle which is big A is the corresponding or counterparts to one of those sides that we were given. This is a dead giveaway that this is an SS A triangle. If you're ever unsure, just go ahead and just really quickly draw a quick little sketch of a triangle because this could be side A, for example, this would be side B and A, the angle is gonna be opposite of its side. So basically, this is side, side angle like this. All right. That's a good way to little uh to check. Now, remember in these types of problems, this is the ambiguous case, we actually don't really know what the triangle is gonna look like, but that's just a little sketch just to kind of like confirm that this is exact. This is in fact what you're dealing with So let's go ahead and get into the steps here, have actually how to solve these types of triangles. Remember that the first thing you wanna do is without actually drawing the triangle, you set up a law of signs to solve for a second angle. All right. So let's go ahead and do this here. If I solve for the A second angle, uh this is gonna go ahead, I'm gonna go ahead and list this as my first step here. This is gonna be the, the sign of A over A equals the sign of B over B. I don't have to write the C part because we can actually clearly tell here that all we have is lowercase letters for A and B and then uppercase A, we have no information about C. So we're not even gonna bother there. All right. So we're gonna do here is we have 3 to 4 variables. I know these three and I wanna go ahead and solve for another angle, which in this case is gonna be this angle over here. This is, this is gonna be a capital B. All right. So I'm actually gonna go, what I'm gonna do is I'm gonna sort of flip this and I'm gonna move B to the other side and I'm gonna start plugging in some numbers. So in other words, the sign of B is the true. If you switch these equations, you have B that goes to the opposite side and you're gonna multiply. So in other words, this is gonna be little B which is four times the sign of angle A which is 30 degrees that's 30 divided by and then uh A which is just equal to one. All right. So actually, fortunately the math works out to be really nice here because four times the sign of 30 ends up just being two. OK. So in other words, we have the sign of some angle, which is the value two. So let's take a look at this step over here for step one. Remember what happens is the entire point is you're gonna try to set up the sign of an angle is equal to some number. And then what happens is we look at this number and if it's greater than one, what happens is that we're done with the problem because there is no solution. So this is basically impossible. You can't have the sign of some angle have a value that is greater than one because remember Sino goes from negative one to positive one. So this is impossible. All right. So this is basically the, we're actually done with the problem. We don't even have to do anything else here because the answer to this problem is that there is no solution. So what I wanna do in this, in this uh sort of uh last part over here is I actually wanna kind of wanna visualize what's going on here what would this triangle even look like? All right. So again, we're told that A is equal to 30 degrees. I always like to start with the angles because they're a little bit easier to visualize, right? So this would be something like this, right? So I've got an angle that's uh looks something like this. This is gonna be my 30 degree angle. And in this case, what happens is since this is angle A, this is gonna be my B which is gonna be four. And that means that the side that is opposite to the angle A has to be little A but we know little A this A over here is equal to one. Basically what's happening in this problem is that this A over here, this little A is too short to actually connect all the way down and form a triangle no matter where you position this A, no matter what angle it sort of forms because don't know what angle that is, it doesn't matter because anywhere you draw this A, it won't be long enough to actually complete a triangle. That's actually why you get no solution here. So this is what the sort of triangle or lack thereof would actually look like. OK. So I kind of wanted to visualize this, but that's the answer to this problem. There is no solution and we don't have to go any further.

9

example

Solving SSA Triangles ("Ambiguous" Case) Example 2

Video duration:

7m

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Everyone. Let's take a look at this example. We have another triangle over here in which we're given two 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 gonna set up a law of signs so that we can get an additional angle. Let's go ahead and get started. So this first step over here, we're gonna set up a law of signs. Notice how again the only two letters involved in here are B and C. So when we actually set up our law of signs, we're actually gonna completely disregard the A term because we don't know anything about A or little A. So in other words, we just have sign of B over little B is equals to the sign of C over little C. All right. So again, what happens in these problems is you're gonna have big B over here and little B and little C. So you can go ahead and solve for C, that's gonna be that second angle that you find. OK. So what we're gonna do here is once we cross multiply moves to the other side, we're gonna see is that sine of C is equal to. And I'm just gonna start plugging in some numbers here. This is gonna be two times the sine of B which the words is the sign of 29 degrees divided by little B which is four. Now, if you go ahead and work this out, what you're gonna get is you're gonna get something like 0.2 42. And whatever 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 gonna see here is that we get 0.2 42. Remember for the first step, if you ever get something that's larger than one, you're gonna have no solution, but that's not what happened here, right? So we actually are gonna get something that's less than one or equal to one. We move on to step two and we're perfectly fine to continue the problem. The second step here is we're gonna use the inverse sign to solve for two possible angles here. So what's going on? Well, basically what happens is if we solve for C remember we can take the inverse sign. So we're just gonna take the inverse sign of two of 0.2 42. And what you're gonna get here is you're gonna get a number out of this, you're gonna get the, the inverse sign is equal to 14 degrees. All right. But that's not the only angle that when you plug it back into S you'll get this number for. So that's actually where we sort of split our problem into two parts. So basically what happens is that C one is equal to the inverse sign which is 14 degrees, but C two is just equal to 100 and 80 minus C one, right? So in other words, it's just minus the angle that we just found over here. So it's 100 and 80 minus 14 and then C two therefore is equal to 100 and 66 degrees. So we've got these two different angles over here that when you plug them both back into the sign, you could double check this, you'll actually get back to 0.2 42. All right. So how do we figure out which one is possible? Well, actually, that leads to this the third step. So after we're done with step two, we're gonna see that we have our two possible angles and by the way, this didn't happen over here, which is that the sign of the angle is equal to one. So again, this may happen, but it's pretty rare. We're just gonna keep going on to step number three. So for step number three, what we're gonna do is for this angle here that we just calculated. The second one, we're gonna 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. So what does this mean? It means that if you add B plus C two and this is just gonna equal 29 plus 100 and 66 degrees. So 29 degrees plus 100 and 66 degrees, if you work the sandwich, you're gonna get is that this is equal to 100 and 95. But how is it possible that two angles in a triangle add up to something that's greater than 100 and 95 degrees? That's impossible. So what happens here is that this first angle that you calculate? C one is always gonna work right? There's no problem with that 14 degrees is perfectly reasonable. But for this C two where you get 100 and 66 degrees, when you add it to the angle that you already know in the triangle, you're gonna get something that's bigger than 100 and 80. 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. All right. So what happens here is that we're So there's only one solution to this. Um So this uh triangle over here. So I'm gonna write here that there's no second solution. All right. So this is basically part of the answer that there's only one triangle to focus on here and that's gonna be this one. So now we're gonna move on to step number four, because now that we figured out there's only one solution to this problem. We're gonna solve the remaining angles and sides of all of the possible triangles that we have and there's only one triangle here. OK. So here's what we do we're gonna solve for the remaining angles. How do we do this? Well, in this case, I figured that C one is equal to 14 degrees. So what I can do is I could just add this to A plus B plus C one is equal to 100 and 80 degrees, right? So what we've got here is that I can solve for the missing angle uh which in this case is just going to be uh this is just gonna be that it's gonna be the A is equal to 100 and 80 minus the two angles that I know which is gonna be um 14 degrees and 29. So this is gonna be one minus 29 minus 14 from 100 and 80. And what you, you will get when you do this is you'll get 100 and 37 degrees all right. So again, basically what I get here for the first triangle here is that uh A is equal to 100 and 37 degrees. B is equal to 29 and C is equal to 14 degrees. All those are perfectly valid angles to have in a triangle. OK. So this is basically what my third angle ends up being. So if I were to try to draw this out, I'm gonna try to sketch this out. Here's basically what this triangle would look like. So I'm gonna have one really, really tiny angle over here, which is gonna be 14 degrees. So I'm gonna try to draw this like this like this and then I'm gonna have one huge obtuse angle. So it's probably gonna look something like this like that's, and then the other angle is also gonna be relatively small as well. So this is gonna be like 29 degrees over here. Uh This is gonna be my 14 degrees uh over here and this is gonna be my 137 degrees over here. So this is A, this is gonna be B, this is gonna be C and therefore, what happens is that B is equal to four. We already knew that uh C is equal to two. 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 signs for this. All right. So we're gonna set up a lot of signs really, really straightforward. This is just gonna be that the sign of A over A is equal to the sign of B over B I can use anything at this point because I already know everything else in the triangle. And what you're gonna see here is if you solve for this really quickly and uh flip this, you're gonna get that A is equal to uh the sign of big A which is gonna be 100 and 37 degrees times. And this is gonna be four divided by the sign of 29. All right. So 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 gonna get is at 5.63. All right. So this actually makes perfect sense that you get 5.63. Remember all of these other sides are relatively small. This one's two, this one is four. This one is 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. All right.