 Welcome back. In this screencast, we're going to learn about the law of cosines. And this is another tool we can use to help us solve triangles. In fact, we will be able to use this law in places where we could not use the law of signs. So again, using our standard notation for triangles with angles alpha, beta, and gamma. And the corresponding sides A, B, and C. We get this statement here of the law of cosines in three different forms. And again, that can seem like an awful lot to remember. So I usually like to try to just remember one form of the law of cosines. And with kind of a different type of notation, that's not too hard to do. One thing should also be pointed out in some sense that might help is that the law of cosines in some sense is a generalization of the Pythagorean theorem. And in fact, if we focus on this one here, the first one, and say that the angle gamma is 90 degrees, we see that the cosine of gamma is zero. And so if we use that and write this form with cosine of gamma equal to zero, you can see we get C squared equals A squared plus B squared. But again, you have to remember that that equation here is only applicable when we do have a right triangle, in this case when the angle gamma is 90 degrees. Here's how I like to think about the law of cosines. And again, it's very similar to what we did with the law of sines. We got four quantities involved in this. Three of these are the lengths of the three sides. And within the law of cosines, there's one of the angles. We have the measure of that. And so that's kind of what I've set up in this diagram here. And what I like to focus on with this is, again, theta and z. z is the side opposite the angle theta. And so what we get here is that means theta is the angle between x and y. So if you notice, the z and the theta are on opposite sides of the equation. But on the right side, then, we use the lengths of the sides represented by x and y. And so what we see here is that we get z squared equals x squared plus y squared minus 2xy. And these x and y are the sides that form the angle theta. And again, the idea is if we know three of these four quantities, we can use the law of cosines to determine the value of the fourth side, of the fourth quantity. So here's a problem that we have, and it's stated there. You can read that. We've got the length of two sides of the triangle and the included angle. And so what I would like you to do here is pause the screencast and see if you can set up some notation to help in setting up the law of cosines and, in fact, try to solve for the time being. We'll focus only on solving for the length of this side here, the side opposite the 60-degree angle. So go ahead and try that. Okay, welcome back again. So what I've tried to do here then was give a letter to represent this and to kind of stay consistent with what I've stated there. I'll call it z. And you can see now we've got something representing all three sides. In fact, we know the values of the lengths of two sides, 3.5 meters and 2.5 meters. And z is the unknown quantity. But you can see there are three known quantities there. And so if we want, we can set up the law of cosines and write z squared equals 2.5 squared plus 3.5 squared. You can see what I'm using there are the sides of the angle 60 degrees, which I guess you could call theta here. And then again it's minus two times the lengths of the two sides. And I'm just going to have to write cosine of 60 degrees down here to have enough room to write that. But that now allows me to calculate z squared. And so if we do this calculation, and again we do have to take a little bit of time to do that, but we can just kind of go right from right to left or left to right on this and do the calculations. And you should come up with z squared equals 9.75. You might want to stop and try that on your calculator, but that would be the value that we would get with that. And so since z is the length of a side, it has to be the positive square root of 9.75. So right now we can get a decimal approximation for the length z. And in fact we will do that a little later on. It's always nice to have the more decimal approximations and something like a square root. But what we're going to be doing beyond this is finding the measures of the two remaining angles. In order to do that we're going to have to use the value of z. And so what I want to do in my future calculations is use this, use the full accuracy of that. So what I would suggest doing is doing that calculation on your calculator, whatever calculating device you have. Get a decimal value for square root of 9.75 and store it so that you have it to use. You can use the full accuracy of the calculator to do future calculations. Or however your calculator works, you want to have that full decimal approximation available to you in future computations. If I'm using a TI calculator, I would tend to store that value in the variable z, in the storage location z, just to keep the calculator stuff the same as the notation I'm using in the problem. So here is what we have. And again, it's kind of the state at what we're at now. And you can see here is a decimal approximation for z. And it's nice to be able to use that whole thing without having to key it in all the time. And we will get more accurate results for the other values if we can use that full value of that decimal approximation. And what we want to do now is think about finding the measures of the angles, alpha and beta. And here's a place where we do have a choice. We could use the law of signs, but remember there's sometimes one problem with the law of signs. The law of signs can sometimes give you two possible angles that you have to at least account for. Sometimes one of them is impossible. But if not, you could possibly get two solutions for alpha. And you have to be able to decide which one is the correct one to use. The law of cosines will not present that problem, primarily because the inverse cosine function will always return an angle between zero and 180 degrees. And that one angle will be alpha or beta, depending on which one we solve for. So what we're going to try to do here is solve for alpha. So what I look at here again is here are the two adjacent sides for alpha. So you can see I've got, in essence, if you want a side angle side. And of course, this then is the length of the opposite side. So when I write this out, the 2.5 squared goes on the left side of the equation. And on the right-hand side then, what we get is the two adjacent sides, 3.5 squared plus z squared. Here's an advantage of using the fully accuracy. We do know that z squared is 9.75. But then we would have 2 times 3.5 times z times cosine of alpha. And you can see the only unknown in this equation now is alpha. The first thing we're going to do is solve for cosine of alpha. And then, of course, use the inverse cosine function. So what I'm going to do to try to keep as many positive numbers in it, I'm going to take this term here and move it over to the left side of the equation. It's pretty easy to calculate that 2 times 3.5 is 7. So I get 7z. Now what I'm going to do then is move this one over to the right side of the equation. Oops, I'm sorry. We get 7z cosine alpha. And so now I'm going to have 3.5 squared plus z squared minus 2.5 squared. And that's minus 2.5 squared comes from moving this from the left side to the right side. And as you can see now, I can solve for cosine of alpha. The right side of the equation, remember, z squared is 9.75 what we would end up with is cosine of alpha is equal to this comes out to be 15.75 divided by 7 times z. And now you can see we can calculate cosine of alpha. We do know that z is equal to the square root of 9.75. And in fact, here you have to be a little careful on your calculator in doing this computation. But we should get something such as cosine of alpha. And again, you maybe don't have to write down all of these digits but you still want to use them in your calculation. Your calculator may give somewhat fewer digits, but that's okay. Basically, again, like on the TI calculator what I'm going to do is alpha is equal to the notation on the calculator is inverse cosine. And then what I'm going to do is use the answer register because that's going to have this value stored in it. And what we end up with is that alpha is approximately 43.90 degrees rounded off to the nearest 100th of a degree. And again, by including the 0.90 I'm indicating that this is accurate to the nearest 100th of a degree. So now we actually have the value for two angles and we can't use the fact that the three angles, the measures of the three angles have to add to 180 degrees. But sometimes I like to use that as a check of my work and I think that's a good idea that we should always strive to do. So what I would like to have you practice with here is use the law of cosines to try to determine the measure of the angle B or beta. And pause the screencast and work on that for a few minutes and see if you can find the measure of the angle B. This, what I have outlined on this screen here should give you the process that you should be able to use for that. So come back in a couple... Okay, so if I was solving for the angle B, this is how I would have it set up. And again, you can see in some sense the strategy is pretty much the same. These two sides here are the adjacent sides. So you can see they appear here, in here, and then of course in this side of the equation. The adjacent sides go with beta and the 3.5 is the opposite side. And so that's the initial setup for the law of cosines. And once I get that set up I can then algebraically solve for cosine of beta in much the same way I did for cosine of alpha and end up getting a value for cosine of beta and then doing cosine inverse of this number here. And we get beta is approximately 76.10 degrees. Nice thing about taking the extra time to do that is we now have a nice handy fact on our work. And in fact the check, again here is a summary of the results. I've rounded off Z to the nearest hundredth, but you can see if I take 43.9 degrees 43.90, let's say plus 76.10 degrees plus the given angle of 60 degrees that adds up to 180 degrees. And so I have a pretty good check on my work. It's probably not a perfect check but boy that should help convince me that the work I have done in solving this triangle is correct. Thanks for watching.