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.