Welcome back, everyone. So in this problem, we're gonna try to calculate the area of this triangle. Let's get started here. Remember that at the end of the day, an area is just 1/2 times a base times a height. So the area is equal to 1/2 times the base times the height, and the base of this triangle appears to be b=8. What about the height of this triangle? Remember that we're gonna try to draw from one of the corners down to where it intersects the base. The problem is if I try to do this here and try to figure out this height, there's a piece of this hypotenuse of this triangle that still keeps going up, and I'm losing all of this area if I try to plug it into 1/2 base times height. Alright? So it turns out that we can't really do this, so I kinda wanna explain what's going on in these triangles. But first, let's actually just take a look at the variables. We know that little b is 8, little a is 6, and the big C is equal to 120. We know that there are sort of three versions of the area formula of a triangle, and usually what happens in these problems is that as you look through your variables, 2 of them won't make sense. Let's take a look here. We have 1/2 b×c×sin(a). The problem is I don't know what a is in this problem or little c. So I have 2 unknowns. That equation is not gonna work. The second one is gonna be 1/2 of a×csin(b). Same problem. I don't know what c, little c is, or the big B. Both of these variables or angles are unknown. The only one that makes any sense at all is using this one over here because I actually know what all three of these values are, a, b, and C. But I want to show you why this works in this equation. So we're going to have one-half of a×b×sin(C). Remember, the area is always a base times the height. What's the base in this problem? It's really just that little b. So what I'm gonna do is I'm actually just gonna switch the variables around in this equation. This is gonna be one half of base of b, in which this little b here is the base. So then what's the height of this triangle? Well, it's really just gonna be this other piece over here, a×sin(C). Alright? But I wanna show you why that ends up being the height in this problem. Alright? So we already know that we can't sort of draw an a a little dash line straight up and have that as the base. But what we can do is we can go all the way to the corner over here of b and then basically just draw this line straight down even if it sticks out past the base of my triangle. Because what I can do is I could basically just pretend that the base of this triangle sort of keeps on moving on going on and sort of ongoing straight. So, really, this height over here is actually the height of my triangle. And notice that what we've done is we've made a smaller right triangle that's sort of as an extension of my obtuse triangle and the height of this right triangle is the same as the height of the obtuse triangle that I'm trying to find the area of. What is the height of this triangle related to the variables that I know? Well, I've got the hypotenuse, which is a. And just using SOHCAHTOA, this h would just be a×sin(whateverthisangleisoverhere). What is this angle? Well, it's basically just if we're assuming that this line here is gonna be supplemental, it's gonna be 180 degrees, then then this is really just gonna be 180 degrees minus a 120, which ends up just being 60 degrees. So this angle over here is 60 degrees. Therefore, the height would just be a×sin(60). In other words, it would really just be 6 times the sine of 60. Alright? Now it might seem like you have to do all this work to figure out this angle over here, but, actually, you don't. Because what I'm gonna show you here is that you could have done all this work to figure out this angle was 60, or you can actually just use 6 times the sine of the original angle C that you were given, which which is a 120. What I want you to do is remember an old identity that you may have learned, back in a previous chapter, which is the sine of any angle is equal to the sine of 180 minus that angle. So an example of this in this problem would just be so for example, the sine of 120 degrees is the same thing as the sine of 180 minus a 120, which is the sine of 60. So you don't have to actually do all of this work to figure out what the sine of this sort of smaller right right triangle is. Because what I'm gonna show you, whatever what I've shown you here, is that the sine of this 120, which was your original angle inside of your obtuse triangle, is actually the same as the sine of that supplementary angle. Alright? In fact, you can actually plug these things into your calculator, and what you'll see is that 6 times the sine of 60 ends up being 5.2, and 6 times 120 also ends up being 5.2. So what I'm trying to tell you in these problems here, what I'm really, what I'm trying to say is that you actually it might seem like you have to do all of this extra work to figure out the supplementary angle, but you actually don't. We could have just gone straight into using this equation, a×b×sin(C), because taking the sine of this angle with this hypotenuse is the same as taking the sign of this angle to figure out the height of this triangle. Okay? That's really what's going on here. So, really, I'm just gonna take the area over here, and this is gonna be 1/2. I know the base is 8. And what's the height? I just calculated the height of this triangle as 5.2. So this is going to be, sorry. This is going to be sin(C), which ends up being 5.2. And the area of this triangle ends up being 20.8 units. Alright. And that is the final answer. That is the area included inside of this obtuse triangle. Alright. Thanks for watching, and I'll see you in the next one.