Law of Cosines - Video Tutorials & Practice Problems

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1

concept

Intro to Law of Cosines

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If I gave you this triangle on the left and ask you to solve for side C, you'd be able to do it. Given everything we've learned about the law of science because we have a side and its opposing angle, all we need is another piece of information and we could set up a law of signs between two ratios in which we know three out of four variables. We've seen how to do this before. But sometimes in some triangles, you won't be given a side and its opposite angle here. We have A and B but not the angles here, we have angle C but not its corresponding side. Whenever this happens, you can't use the law of science. And I'll show you why in just a second, it might seem like we're gonna get stuck here. But what I'm gonna show you is that we'll need a different set of equations to solve these types of problems called the law of cosines. What I'm gonna show you is that it's actually a very straightforward set of equations. And most of the time when you use it, it's pretty much just plug and chug. So let's just jump right into this example and I'll show you how it works. All right. So we've seen how to solve these types of problems on the left where we set up a lot of signs, basically just gonna ignore side a because I don't know anything about it. And I've actually already got the answer sort of written out for you all. It is, is just C equals 4.9. So we can figure this out pretty quickly. Why doesn't it work for the triangle on the right side? Well, basically, let's just write out all of the law of signs, the whole entire thing. So I'm gonna write that sign of A over A is equal to sign of B over B. This is equal to sign of C over C. What do I know about this problem? I know the two sides A and B, but I don't know the angles here, I have the angle C but not the uh corresponding side. So in each one of the terms, I only know one out of the two variables. And what's worse is that because I only have one angle I can't solve for the other two angles because I have two unknowns there. So it seems like I'm kind of stuck here. I can't actually solve for side C. And that's because in these types of situations, you just can't use the law of science, you have too many unknowns. So what I'm gonna do here is, I'm just gonna go ahead and show you the law of cosines law of cosines really just is sort of a pattern and there's sort of three variations because the letters could be different and it depends on what you're trying to solve for, but the pattern is all the same. So for A and B, the equations are already written for C, I'm just gonna go ahead and give it to you C squared is gonna to be A squared plus B squared. In other words, it's just the other two letters um not or not C minus two times the two letters that you just wrote. So A and B times the cosine of the corresponding angle that's over here on the left side. So that's always gonna be sort of pattern for A and B. You can just double check it over here. All right. So really what this is the law of cosines is, it's just a set of equations which relates the squares, it relates the squares of the three triangle sides to a known angle. That's the most important thing here. You're gonna relate these things to an angle which you know, and in this case, we do because we have the angle big C, we know what that angle is. All right. So if I'm trying to solve or little C over here, I'm just gonna use this variation of the law of cosines, right. So we're gonna see that C squared is equal to, again, I'm just gonna write the whole thing out A squared plus B squared. So the other two variables minus two A B times the cosine of this corresponding angle over here pretty straightforward. The rest is basically just plug and chug because you have all of these numbers. So A is three. So this is just gonna be three squared B is two. So there's gonna be two squared minus two times three times two times the cosine of C which is equal to 60 degrees. Let's clean this up a little bit. So I'm just gonna get, get, this is uh uh nine plus four, which is 13 minus and then two times three is six times two is 12. So this is 12 times the cosine of 60 degrees. We can clean this up a little bit more because we know that cosine 60 is one half, this basically ends up being 13 minus six, which equals seven. So when you take the square roots, what you're gonna see here is that C is equal to the square root of seven. All right. So you obviously could have turned this into a decimal at any point. I just sort of kept it a little bit clean like a square roots, but that's basically it. So we just know that C is equal to the square root of seven. That's how you use the law of cosines. So I just want to point out a couple things here. You can also use a lot of cosines to solve for any missing angles as long as you have the other three sides. So you can totally do that. And the last thing I want to mention here is you may have noticed at some point when I was writing this, that this kind of looks like the Pythagorean theorem, we've got a squared plus B squared equals C squared. But then you have this other term over here. That's actually because the law of cosines or sorry, the Pythagorean theorem is really just a special case of the law of cosines. But that's just a fun fact. You don't really need to know that anyway. So thanks for watching and I'll see you in the next one.

2

Problem

Problem

A surveyor wishes to find the distance across a river while standing on a small island. If she measures distances of $a=30m$ to one shore, $c=60m$ to the opposite shore, and an angle of $B=100\degree$ between the two shores, find the distance between the two shores.

A

$69.4m$

B

$67.1m$

C

$90.6m$

D

$71.6m$

3

Problem

Problem

Use the Law of Cosines to find the angle $C$, rounded to the nearest tenth.

A

$109.5\degree$

B

$50.5\degree$

C

$111.9\degree$

D

$48.1\degree$

4

concept

Solving SAS & SSS Triangles

Video duration:

7m

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So in earlier videos, we learned the law of science and then we learn how to use the law of science to solve three different types of triangles. Well, similarly, now that we've learned the law of cosines, I'm gonna show you this video how we can use the law of cosines to solve the last two types of triangles. You'll run into the SAS and SSS triangles. I've come up with a list of steps to help you solve these types of problems. And basically what it is is we're just gonna use the law of cosines to solve for missing size and angles. Let's just go ahead and jump right into this problem and I'll show you how it works. So we have this triangle here in which we're given some information. Little A is four, little B, is three and big C is equal to 60 degrees. Now, if you haven't already drawn this triangle or if it's not already shown to you, the first thing you'll do is sketch the triangle, but I've already done that for you. And what we can see here is that we have two sides A and B but not their corresponding angles and then we one angle but not its corresponding side. So in other words, we never have a pair of a side and its corresponding angle. And if you look at this, what happens is we have a side and we have two sides and we have an angle that's between them. So this is an SAS type triangle. In fact, we've actually seen something like this in a previous video. So remember what happens in these problems is we actually can't use the law of signs because it just doesn't work. We're gonna have to use the law of code signs. So that actually brings us to our second step here, which is we're gonna use the law of cosines to solve for some different variables. If you're using an SAS triangle, if you have an SAS triangle, then you'll use this to find that third side. If you have an SSS triangle because you have all those sides, then you would just find any angle. All right. So we're gonna go ahead and work out two A over here because we're dealing with an SAS triangle. And I'm just gonna go ahead and find the third side, which is really just that little C variable. So if I'm solving for C and basically what I'm gonna do is go, go to my law of cosines over here and I'm gonna have to use this equation here. That's for C squared. And I'm just gonna write this out. Remember this is just A squared plus B squared minus two A B times the cosine of big C, we have every variable on the right side of this equation. So this is pretty simple, pretty straightforward. It's just plug and chug. So this is just gonna be C squared equals I've got A is equal to four. So this is four squared plus three squared plus two times four times three times the cosine of angle C which is 60 degrees. All right, go ahead and clean this up a little bit. If you work out these two numbers, this is 16 plus nine, which ends up being 25 and this is two times four times three, which ends up being 24. And then what happens is this is the cosine of 60 which we can again reduce because this is basically a half. So this works out to be 25 minus 12 and that just equals 13. So that's what C squared is equal to. It's equal to 13. And what that means here is that C itself C itself is equal to the square root of 13. Now, you can go ahead and round this or sort of approximate this. If you want this in your calculator, you get something like 3.6. But I was sort of just cautioned against doing this. It's nice to leave things in terms of square roots and you'll see why in just a second, basically you want to minimize the rounding errors. OK. So that's the first variable we're gonna solve for that C is equal to square roots of 13. So we're basically done with step two A. And now what happens is we're just gonna use the law of cosines again because no matter what type of triangle that you have by this point, whether it's an SAS or SSS, you have all three sides that are solved for and at least one angle. So what you're gonna do in this third step is you're gonna use a lot of cosines to find a second angle. So that's what we're gonna go ahead and do over here. It actually doesn't matter which angle that you solve for. It'll work out the same whether you're solving for A or B. I'm just gonna go ahead and pick, solve and, and solve for a. So if I want to find a, in this third step, then what I'm gonna do is I'm gonna have to go over here to my law of cosines and figure out which equation deals with the angle A and it's gonna be the first one. The first variation of this formula is gonna be uh that A squared. So we're gonna have the A squared. So let's do this over here is equal to A B squared plus C squared and this is minus two BC times the cosine of big A. Now, remember unlike step two A, where we actually solve for a side here. We're actually solving for an angle. And so what happens is everything to the left of this sort of variable, all of these variables over here, we actually know. So I can just go ahead and plug and chug this. My A squared is gonna be four squared equals, this is gonna be three squared plus C squared. All right. So what is C squared? You may be tempted if you've plugged this in as a, as if you've written this as a decimal to try and square this. But I'd like to warn you against that because again, you rounded this number. So if you square it, now you're going to introduce a bunch of rounding errors. We actually already know what, what C square is equal to because we solved for it. It's just 13. So I'm just gonna plug this over here. Uh That's why, again, sort of nice to leave things in terms of square roots. Um So this is a three squared plus 13 minus two times. This is three and then C by itself, not C squared, it's just equal to the square root of 13 times the cosine of a All right. So I'm just gonna go ahead and clean this up. I'm gonna subtract three squared from each side and 13. And then basically, we're gonna see here is, this is 16 minus nine minus 13. And then on the right side, I'm gonna get negative two times three. So it's gonna be ends up being negative six times the squared of 13 times the cosine of a all right. Now, what I'm gonna do is 16 minus nine minus 13. Ends up being negative 22. I'm sorry, this ends up being negative six. And then we're gonna basically divide by negative six radical 13. So that cancels out on the right side. So I'm gonna have to do this on the left and this is gonna equal the cosine of a. All right. So now, basically what I can do here is I can say that A is equal to the inverse cosine of. And what happens is this negative six will cancel out with the uh negative six, the bottom. And all you're left with is one over the square root of 13. All right. So when you plug this into your calculator one over the square root of 13 and take the inverse cosine which you're gonna get is 73.9 degrees. All right. So that is your angle or angle A that's gonna be this one over here 73.9. Notice how they actually kind of looks pretty consistent with the drawing here. Now, one thing you might be wondering is OK. Well, just like the sign inverse don't we have to then figure out the second angle because uh the inverse sign will give a 73.9 for this number. But there's also another angle that might get this, this number. And actually, no, because we're actually taking the inverse cosine here. So what I wanna sort of, what I have a note here is that in step three or actually two B as well, it's always better to use the law of cosines to solve four angles instead of the law of signs. And this is basically because the cosine inverse only yields one angle between zero and 100 and 80 degrees. You don't have to worry about there being a second angle, you have to go and test to make sure that it works. So actually, this is the only angle that works for this specific number. OK. So we don't have to do any extra work. All right. So now that we're done to step three, basically, I'm just gonna move on to step number four. And really this is sort of the, the most straightforward step because by this point, you have two of the angles. So you can always solve for the third one just by using the angle sum formula. So this is gonna be pretty straightforward. We've done this a bunch of times A plus B plus C is equal to 100 and 80 degrees. All the angles have to add up to 180. We're, we're now solving for the last angle, the one that you didn't solve for in step three. And in this case, that's gonna be B. So in other words, we're just gonna subtract both of these things from 1 81 80 minus 60 degrees minus 73.9 degrees. And what you should get over here is 43.1 degrees. All right. Um So that is 40 I'm sorry, this is not 43 4, 46.1 degrees. This is 46.1 degrees and that is your final angle that's missing. So 46.1. So with that, we've actually solved for all of the three remaining variables squared to 13, 73.9 degrees and 46.1. That's really all there is to it. So let me know if you have any questions and let's get some practice.