 In a previous video, we talked about how we can compute the area of a triangle if we have a side-angle-side situation. We also talked about angle-angle-side and we also talked about angle-side-angle. But there are some other triangle conditions we might want to consider. What about like angle-angle-angle? Well, angle-angle-angle doesn't uniquely determine a triangle. This will only determine a class of similar triangles. That is, the triangles are proportional to each other because their angles are the same, but the side links are not fixed. As such, the triangle could get arbitrarily large or arbitrarily small. So we have no way of describing the angle in general for angle-angle-angle situation. We need to know at least one of the sides. Another situation to bring up is the side-side angle situation, the so-called ambiguous case. The problem, of course, with the ambiguous case is that, as the name suggests, it's ambiguous. Given side-side angle, you might have no triangles. You could have one triangle. You have two triangles. So there's not going to be one unique formula for that. As you're investigating these triangles, you'd have to find some more information. Like, that is, pursue some of the other angles and see what you get there. The last possibility, which we will talk about in this video, is side-side-side. What if we know no information about the angles? Can we calculate the area of the triangle? And it seems to make sense because the area, one half base times height, that's something to do with the dimensions, the lengths of the triangle, not the angles. The two are related, of course, by trigonometry. But if we just have a purely side-side-side situation. Well, you basically do two options. You could use, like, the law of cosines. Law of cosines here. When you have side-side-side, you could then get an angle. And with an angle at hand, you basically have an angle, you have a side-angle-side, excuse me, situation, for which then you can find the area from that. So if you want to use a law of cosines to find an angle measurement, you can then do side-angle-side like we did earlier. That would be just great. But in this video, I want to offer an alternative method, something called Huron's formula, which, honestly, is the law of cosines in disguise. But because it's in disguise, you don't see any of the angles. It's actually a cute little formula to help you calculate the area of a triangle, if we just know the three sides. Now, Huron's formula is dependent on something called the semi-parameter, which is just half of the perimeter of the triangle. The perimeter of the triangle would be a plus b plus c, just the sum of the three sides. And then you take half of that, hence the semi-parameter. So it's half of the perimeter. Then with the semi-parameter defined, we'll call that s. a, b, c will be, again, the lengths of the three sides of the triangle, a, b, c. Then the area of the triangle will be given as the square root of s times s minus a times s minus b times s minus c. So you have the product of these four numbers, which you subtract the lengths from the semi-parameter. And that gives you the area of the triangle. It's a very easy formula to use. It's super easy to use, really. Let me show you an example here. Let's suppose we know the three sides of an oblique triangle. a is 12, b is 14, c is 8. So the first thing we're going to do is we're going to compute the perimeter of this. We're going to take 12 plus 14 plus 8. Of course, 12 and 8 make 20. And then you add 14 to that. You're going to end up with a perimeter of 34. The semi-parameter is half of that. So we have to take half of 34, which gives us here 17. All right. So this is going to come into play here. So next we have to do s minus a. So we're going to take the semi-parameter of minus 12, like so. This ends up with 5. We have to take s minus b. This is going to be the semi-parameter minus 14 there, which gives us 3. And then lastly, we have to do the semi-parameter minus c. So we're going to take 17 minus 8, which ends up with a 9. So then by Heron's formula, the area of the triangle is going to be the square root of s times s minus a times s minus b times s minus c. And we plug in all these values we just found out. So s, remember, was 17. s minus a was 5. s minus b was 3. s minus c was a 9. And so multiply those together. 17 times 5 is 85. 3 times 9 is 27. This is inside the square root. 85 times 27 is 2,295. And this would then be the exact area of our triangle. If we need an approximate value, not a big deal. We'll just, of course, throw it into our calculator right here. The approximation from your calculator would be 47.9. We'll round to the 10th place there. 47.9 meters squared is then the area of the triangle. Pretty easy to do. And this is a direct consequence of the trigonometry. But why? You know, why does Heron's formula work so effectively? Well, it turns out that using Heron's formula is basically using the law of cosines in disguise like I was suggesting to you on the previous slide. Well, as a reminder, the law of cosines told us that a squared equals b squared plus c squared minus 2bc times cosine of a. For which, if you were to solve for the angle, what you would have to do is you would want to solve for cosines. So move that to the right-hand side. Move the a to the, excuse me, move the cosine to the left from the right. Move the a squared to the right-hand side. You end up with 2bc cosine of a. This equals b squared plus c squared minus a squared. You can divide both sides by the 2bc. So they cancel over here. Do it on the right-hand side as well. And so we end up with this observation that cosine of a equals b squared plus c squared minus a squared over 2bc. All right? So we could solve for cosine using the law of cosines, right? Well, what does the semi-primiter have to do with anything? So s equals 1 half a plus b plus c. What's so special about that? Well, if the semi-primiter equals 1 half a plus b plus c, that means 2s, which is just the perimeter here, 2s is just equal to a plus b plus c, like so. And so let's look at some of these values. What if I take 2 times s minus a? Notice s minus a, this is one of the values that shows up in Heron's formula. Well, if you distribute, you'll get 2s minus 2a. So you end up with a plus b plus c minus 2a. So in the end, 2 times s minus a is equal to just b plus c minus a, all right? By a similar calculation, 2 times s minus b is going to equal a plus, excuse me, a plus c minus b. And similarly, 2 times s minus c is going to equal a plus b minus c. Again, similar calculations. And pay close attention. This a minus, or this 2 minus a, this s minus a, s minus b, and s minus c. This show up in the formula as well as s. So these numbers are going to come up in the appropriate place in the appropriate time. So the next thing I want to consider is actually consider the value 1 minus cosine of a. Well, by this formula right here, this is the same thing as 1 minus b squared plus c squared minus a squared over 2bc. Well, if you have a 1 minus, we want to kind of find a common denominator. So we can rewrite 1 as 2bc over 2bc. And then we can subtract from it the b squared plus c squared minus a squared. But we really would want to distribute that negative sign throughout all of this. So let me just do that right now. You'll end up with a negative b squared minus c squared plus a squared. It's a double negative there. So putting things in a slightly different order, you're going to get an a squared first. This will be all over 2bc still. You'll get a squared first. Then in this next group, you have a negative b squared. You're going to get a plus a 2bc. And then you get a minus c squared. Looking at those three terms, negative b squared plus 2bc minus c squared, I'm going to factor out a negative 1, in fact, and end up with here, b squared minus 2bc plus c squared like so. And why did I do that? Well, this right here is then a perfect square trinomial. This is the same thing as b minus c quantity squared right there. And so utilizing that factorization, we then get that this becomes a squared minus b minus c squared all over 2bc. Again, focusing on the numerator, you then have this a squared minus b minus c squared. You have a perfect square minus a perfect square. This is a difference of squares. And so this would factor using a difference of square formula as a minus b minus c, right? And then you're going to get a plus b minus c like so. Distributing the negative sign in the first case and dropping the redundant parentheses in the other case, we end up with a, you're going to get a negative b and a plus c. So I'm going to write the plus c first, a plus c minus b. And then you're going to get an a plus b minus c. And this all sits above 2bc. Well, if these terms look at all familiar to you, right? A plus c minus b, I feel like I've seen that somewhere before. Oh, a plus c minus b, that's just s minus b. And what about the other one? A plus b minus c, well, you see that one right here. That's just s minus c, all right? And so by making this observation here, we end up with that a plus c minus b. This is 2 times s minus b. And then a plus b minus c. That's going to be 2 times s minus c sitting above the 2bc, like so. The 2s are going to cancel out here. And so then if we simplify what we've discovered here, and not simplify, just kind of conclude. What do we observe? We start off with a 1 minus cosine a. Then we have this final form right here. What we've discovered is 1 minus cosine of a. This is equal to 2 times s minus b times s minus c, all over bc, like so. So that's a very important observation. So I want to make mention to you that if you made one slight modification here, if you switch this to a plus, and that switches to a plus, and then you follow all the consequences changes from that observation right there, then you will end up with the statement that 1 plus cosine of a is equal to 2s times s minus a over b minus c. And so I'd encourage you to pause the video and try that calculation out on your own if you want to try it. It's a good calculation there. So actually, s comes into play here. But we're not quite done. We're still not ready to use the area formula. Believe it or not, the next move we're going to use here is the half angle identity. What? How does half angle identity come into play here? Well, let's remember the half angle identities. We had that sine of a over 2 is equal to plus or minus the square root of 1 minus cosine of a over 2. Notice if you square both sides, you get sine squared of a over 2. This is going to equal 1 minus cosine of a over 2. Times both sides by 2, you get 1 minus cosine of a is equal to 2 sine squared of a over 2. Notice we have a 1 minus cosine. We have a 1 minus cosine. We can play around with that observation and conclude that 2 sine squared of a over 2 is equal to 2s minus b times s minus c over bc, for which the first thing to do is divide both sides by 2 to get rid of the 2s. And then take the square root of both sides. We then get the observation that sine of a over 2 is the same thing as the square root of s minus b times s minus c all over bc like so. So that's if we use the half angle identity for sine. How about the half angle identity for cosine? Well, remember that one says that cosine of a over 2, that's equal to plus or minus the square root of 1 plus cosine of a over 2. Doing the similar type of solving for the 1 plus cosine, you'll end up with 1 plus cosine of a. That one you'll see is equal to 2 cosine squared of a over 2. You then want to take this 1 plus cosine, equate it with this 1 plus cosine. Then you'll get that 2 cosine squared of a over 2 equals 2s s minus a over bc. If you divide by 2 and take the square root, you'll get a similar statement. You end up with cosine of a over 2 is equal to the square root of s times s minus a over bc. So these are some of the most intense trigonometric identities you've ever seen in your existence. Now we're in a situation to connect this with area. We're going to use the area formula we've introduced previously, area is equal to 1 half bc sine of a. We've introduced this previously. Now with this sine of a, I'm going to apply the double angle identity for sine of a, in which case this becomes 1 half bc times, we're going to get 2 sine of a over 2 times cosine of a over 2, for which we're then going to use these substitutions for sine of a over 2 and cosine of a over 2. You have 1 half times 2. They cancel out. So you're going to get bc times the square root of s minus b times s minus c over bc. Then you get the square root of s times s minus a over bc. So putting the square root together, we get bc times the square root of s s minus a s minus b s minus c s minus c is the last one there. And then you have a bc times a bc. You end up with a bc squared, for which then the square root basically can be simplified in the denominator. We just end up with a square root in the numerator and a bc in the bottom. But we have a bc on top, a bc on the bottom. They cancel out. And we now have Huron's formula, the square root of s times s minus a times s minus b times s minus c. So yes, starting with the law of cosines and playing around with some very, very challenging trigonometry identities, we're able to demonstrate that Huron's formula is, in fact, it does give us the area of the triangle derived from our original side angle sine formula. That used just some basic Sokotoa formulas to get this one and the law of cosines. So Huron's formula, it's difficult to prove, as we just saw, but easy to use, which these are some of our favorite formulas whatsoever. It's how easy is it for me to use that I love so much. And so the cool thing about mathematics is that we don't have to go through all these calculations each and every time we use Huron's formula. Because now it's been established, our doubt is now dead. The proof is established. We can now use Huron's formula with complete confidence. And we don't have to go through these lengthy calculations anymore. We can just use a slick, simple calculation known as Huron's formula.