Law of Cosines - Online Tutor, Practice Problems & Exam Prep

The law of cosines is essential for solving triangles when the law of sines is not applicable, particularly in SAS and SSS scenarios. The equation $C_2=A_2+B_2-2ABcos\left(C\right)$ allows for straightforward calculations of unknown sides and angles. This method minimizes rounding errors by keeping values in radical form, ensuring accuracy in geometric problem-solving.

Intro to Law of Cosines

If I give 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 sines. 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 sines between 2 ratios in which we know 3 out of 4 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 sines, 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. Alright?

So we've seen how to solve these types of problems on the left where we set up a law of sines. 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 = 4.9. So we could 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 sines, the whole entire thing. So I'm gonna write that sinaa = sinbb = sincc. 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 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 2 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 sines. 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.

The law of cosines really just is sort of a pattern, and there's sort of 3 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 equation is already written. For c, I'm just gonna go ahead and give it to you. c2 = a2 + b2 - 2abcosc . In other words, it's just the other two letters, that are not c, minus 2 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 the sort of pattern for a and b. You can just double-check it over here. Alright?

So, really, what this is, the law of cosines, is it's just a set of equations which 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. Alright? So if I'm trying to solve for little c over here, I'm just gonna use this variation of the law of cosines. Alright. So we're gonna see that c2 = a2 + b2 - 2abcosc again, I'm just gonna write the whole thing out. A squared plus b squared, so the other two variables, minus 2 ab 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 3, so this is just gonna be 3 squared. B is 2, so this is gonna be 2 squared minus 2 times 3 times 2 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 this is, 9+4, which is 13, minus, and then 2 times 3 is 6, times 2 is 12. So this is 12 times the cosine of 60 degrees. We can clean this up a bit more because we know that cosine 60 is 1/2. This basically ends up being 13 minus 6, which equals 7. So when you take the square root, what you're going to see here is that c is equal to the square root of 7. Alright. So you obviously could have turned this into a decimal at any point. I just sort of kept it a bit clean like a square root, but that's basically it. So we just know that C is equal to the square root of 7. That's how you use the law of cosines. So I just want to point out a couple of things here. You can also use the law 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.

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. $71.6m$

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. $48.1\degree$

Solving SAS & SSS Triangles

So in earlier videos, we learned the law of sines, and then we learned how to use the law of sines to solve 3 different types of triangles. Well, similarly, now that we've learned the law of cosines, I'm going to show you in this video how we can use the law of cos 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 sides 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 4, little b is 3, 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 2 sides, a and b, but not their corresponding angles, and then we have one angle, bu...

Let me know if you have any questions, and let's get some practice.

Solve the triangle: $b=5,c=3,A=100\degree$.

• A. $a=6.05,B=50.4\degree,C=29.6\degree$
• B. $a=6.26,B=28.2\degree,C=51.8\degree$
• C. $a=6.26,B=51.8\degree,C=28.2\degree$
• D. $a=6.05,B=29.6\degree,C=50.4\degree$

Solve the triangle: $a=5$, $b=15$, $c=13$.

• A. $A=18.9\degree,B=103.8\degree,C=57.3\degree$
• B. $A=13.8\degree,B=108.9\degree,C=57.3\degree$
• C. $A=13.8\degree,B=103.8\degree,C=62.4\degree$
• D. $A=18.9\degree,B=98.7\degree,C=62.4\degree$

Here’s what students ask on this topic:

What is the Law of Cosines and when is it used?

The Law of Cosines is a formula used to solve triangles when the Law of Sines is not applicable, particularly in SAS (Side-Angle-Side) and SSS (Side-Side-Side) scenarios. The formula is given by:

$c^2=a^2+b^2-2ab\cos \left(C\right)$

It relates the lengths of the sides of a triangle to the cosine of one of its angles. This method is particularly useful for finding unknown sides or angles in non-right triangles.

How do you use the Law of Cosines to find an unknown side?

To find an unknown side using the Law of Cosines, you need to know the lengths of the other two sides and the included angle. The formula is:

$c^2=a^2+b^2-2ab\cos \left(C\right)$

Plug in the known values for $a$, $b$, and $C$, then solve for $c$. For example, if $a$ = 3, $b$ = 4, and $C$ = 60°, the calculation would be:

$c^2=32^{}+42^{}-234\cos (60°)$

Solving this gives $c=\sqrt{7}$.

How do you use the Law of Cosines to find an unknown angle?

To find an unknown angle using the Law of Cosines, you need to know the lengths of all three sides. The formula is:

$\cos \left(C\right)=\frac{a^2+b^2-c^2}{2ab}$

Rearrange the formula to solve for the angle $C$. For example, if $a$ = 3, $b$ = 4, and $c$ = 5, the calculation would be:

$\cos \left(C\right)=\frac{32^{}+42^{}-52^{}}{234}$

Solving this gives $C=cos-1\left(\frac{2}{24}\right)$, which simplifies to $C=60°$.

What are the steps to solve a triangle using the Law of Cosines?

To solve a triangle using the Law of Cosines, follow these steps:

1. Identify the type of triangle (SAS or SSS).
2. For SAS, use the Law of Cosines to find the unknown side:

$c^2=a^2+b^2-2ab\cos \left(C\right)$

3. For SSS, use the Law of Cosines to find an unknown angle:

$\cos \left(C\right)=\frac{a^2+b^2-c^2}{2ab}$

4. Use the Law of Cosines again to find another angle if needed.
5. Use the angle sum property of triangles to find the remaining angle:

$A+B+C=180°$

How is the Law of Cosines related to the Pythagorean Theorem?

The Law of Cosines generalizes the Pythagorean Theorem. For a right triangle, where the angle $C$ is 90°, the cosine term becomes zero:

$\cos (90°)=0$

Thus, the Law of Cosines simplifies to:

$c^2=a^2+b^2$

This is the Pythagorean Theorem. Therefore, the Pythagorean Theorem is a special case of the Law of Cosines, applicable only to right triangles.