Law of Cosines: Definition and Application

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  • 0:05 Triangles
  • 0:49 Law of Cosines
  • 2:50 Practice Problem #1
  • 4:26 Practice Problem #2
  • 5:51 Practice Problem #3
  • 7:37 Lesson Summary
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Lesson Transcript
Instructor: Jeff Calareso

Jeff teaches high school English, math and other subjects. He has a master's degree in writing and literature.

In this lesson, we'll learn how to solve problems involving three sides and one angle in a triangle. The Law of Cosines, a modification of the Pythagorean Theorem, will save the day.


Triangles - they can be troublesome. Love triangles? That's a phrase you never really use to describe a good situation. And the Bermuda triangle? The risk of disappearing into another dimension kind of spoils the otherwise amazing Bermuda vacation.

Fortunately, we have all kinds of great tools to help us with triangles in math problems. Let's say we know two angles and one side length of a triangle. How do we find the missing side? The Law of Sines! That's a/sin A = b/sin B = c/sin C.

The Law of Sines is great for problems involving two angles and two sides. But what about problems involving three sides and one angle? There's another law for that!

Law of Cosines

It's called the Law of Cosines. Before you think, 'Hey, what's with all these laws? Isn't that kind of restricting? I need to be free, man!' Hold on. This is one of those helpful laws that makes your life easier. It's like a law that requires chocolate chip cookies to be tasty. It's in your best interest, trust me.

Anyway, the law of cosines involves a triangle like this, where our sides are labeled a, b and c. And then we have one angle, like A here.

The law states that a2 = b2 + c2 - 2bc(cosA). So the square of this side equals the sum of the squares of the other two sides minus the cosine of the angle opposite times twice the product of the two other sides.

We can modify the formula to fit whatever angle we have. So it could also be b2 = a2 + c2 - 2ac(cosB) or c2 = a2 + b2 - 2ab(cosC). Just match the corresponding angle and side so they're on opposite sides of the equation.

Does that look a little familiar? Remember the Pythagorean Theorem? a2 + b2 = c2. Notice that the Law of Cosines is the same basic thing, just adding that -2ab(cosC). Why? Because the Pythagorean Theorem only works on right triangles. So this modification accounts for the obliqueness of the triangle. An oblique triangle is just any triangle that isn't a right triangle.

But how can we possibly remember this? Well, the Pythagorean Theorem involves a, b and c, right? The Law of Cosines is the same thing, but with a second a, b and c. So our 2ab(cosC), is our second a, b and c.

Practice Problem #1

Let's practice. Here's a triangle where we know angle A is 112 degrees. If we know side b is 4 and side c is 15, what is the approximate length of side a?

What's our Law of Cosines? a2 = b2 + c2 - 2bc(cosA). Let's just plug in what we know: a2 = 42 + 152 - 2(4)(15)(cos112). 42 is 16. 152 is 225. Add those to get 241.

The cosine of 112 is about -.37. 2 * 4 is 8. 8 * 15 is 120. 120 * -.37 is -44.4. But since we're supposed to subtract -44.4 from 241, we just add it. That gets us 285.4. So a2 = 285.4. Don't forget that it's a2, not a. That means we need to take the square root of 285.4, which is about 17.

Quick double-check: if side b is 4 and side c is 15, does it make sense that side a is 17? Well, a is clearly the widest angle, so a should be the longest side. Yeah, this seems to make sense. So we're good!

Practice Problem #2

Let's try another. In this triangle, we know that side a is 12 and side c is 5. We also know that angle B is 33. What is the approximate length of side b?

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