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### Course: Precalculus>Unit 2

Lesson 5: Law of cosines

# Solving for an angle with the law of cosines

Sal is given a triangle with all side lengths but no angle measure, and he finds one of the angle measures using the law of cosines. Created by Sal Khan.

## Want to join the conversation?

• Is the inverse of cosine (cos^-1) the same as arc cosine (arccos)?
(62 votes)
• Yes, arccos and the inverse of cosine (cos^-1) are the same thing.
(70 votes)
• In Sal's previous video, he said the formula was a^2=b^2+c^2-2ab cos Theta.
why is he now using c^2 instead of a^2?
(32 votes)
• The variables are reversible. You can put a^2, b^2, or c^2 on the left side of the equation, you just have to fix the other side of the equation to not contain the variable used on the left side.
(17 votes)
• At what is the actual value of cosine? How can we calculate the cosine of a number if we don't have a calculator?
(8 votes)
• You won't be asked to do that. The actual computation is far too difficult to do by hand in a reasonable about of time.

Instead, you will be expected to memorize the sines and cosines of some special angles. Other than that, either you will be allowed to use a calculator or you'll be given the values.

The actual computation for cosine (angles expressed with radians, not degrees):
cos x = ½ [ e^(-i*x) + e^(i*x)]
You will not be expected to do this until an advanced course in calculus.
(55 votes)
• At why didn't Sal just take the -6000 and add it to the other side, thus isolating theta ?
(5 votes)
• Well the term -6000 is together with the cosine of theta. For example if 10=5x you can't subtract 5 from each side to get x.
(36 votes)
• Around Mr. Khan moved the (cos) from one side of the problem to the other. When it moves wouldn't it have to be divided over? Why is he still multiplying cos-1 to the rest of the problem when he should be dividing it? I'm confused. Is there a secret rule I'm missing?
Is it that when (cos) is moved over it only becomes (cos-1) or something?
(5 votes)
• Your last sentence is correct. The cos⁻¹(x) is the inverse function to cosine(x). You could say it "undoes" the cosine function, so whereas cosine takes an angle and returns a ratio, cos⁻¹ takes a ratio and returns an angle.

You could regard what Sal did as taking cos⁻¹ of both sides, so we'd have
cos⁻¹(cos(θ)) = cos⁻¹((19/20)
So in the LHS we take the cosine of theta, and then take the inverse cosine, which is just theta, so we have
θ = cos⁻¹((19/20).

Also be aware that there are alternative names for the inverse trigonometric functions: cos⁻¹ is also called arcosine, sin⁻¹ is arcsine, and tan⁻¹ is arctangent.
(14 votes)
• At , do we always try to simplify the fraction?
(5 votes)
• A calculator will do that for you. If you put 15/24 into your calculator and press enter, you will get 5/8, which is the simplified form of 15/24.
(3 votes)
• Why is their no law or rule for tangent?
(3 votes)
• There is a Law of Tangents!

The Law of Tangents has been around since at least the 13th century, when Persian mathematician Nasir al-Din al-Tusi wrote about it in his book, Treatise on the Quadrilateral.

Law of Tangents:
(a-b)/(a+b)
= [tan ½(α - β)]/[tan ½(α + β)]

Hope this helps!
(6 votes)
• Is there a law of tangents?
(4 votes)
• A law of tangents does exist, but it is much less commonly known compared to law of cosine and law of sine. I was never taught the law of tangents because you can usually find all the information about the triangle (three sides and three angles) through law of sines or law of cosines. If you are interested in learning about it, a quick Google search should give you information about the law of tangents.
(4 votes)
• Why did Sal do 400-6100 when he could have done 6100-6000?
(4 votes)
• 6100 and 6000 are not like terms because of the variable with the 6000. So let x = cos(theta). You have 400=6100-6000x which is a two step equation. So to solve, you subtract and divide.
(4 votes)
• How do I know when to use the Law of Sines or the Law of Cosines?
(4 votes)
• You can usually figure it out based on what they give you.

If they give you two side measurements and 1 angle measurement and the angle measurement is NOT opposite of one of the given sides, then you have to use law of cosines to find the other side. If they give you 0 angles and 3 sides, then you have to use law of cosines to find one of the angles.

If they give you 1 angle and 2 sides and the given angle is opposite of one of the sides and the unknown angle is opposite of the other given side, then you can use law of sines.

It is kind of hard to explain without pictures, so check out these sites:
https://www.mathsisfun.com/algebra/trig-sine-law.html
https://www.mathsisfun.com/algebra/trig-cosine-law.html
(2 votes)

## Video transcript

Voiceover:Let's say you're studying some type of a little hill or rock formation right over here. And you're able to figure out the dimensions. You know that from this point to this point along the base, straight along level ground, is 60 meters. You know the steeper side, steeper I guess surface or edge of this cliff or whatever you wanna call it, is 20 meters. And then the longer side here, I guess the less steep side, is 50 meters long. So you're able to measure that. But now what you wanna do is use your knowledge of trigonometry, given this information, to figure out how steep is this side. What is the actual inclination relative to level ground? Or another way of thinking about it, what is this angle theta right over there? And I encourage you to pause the video and think about it on your own. Well it might be ringing a bell. Well you know three sides of a triangle and then we want to figure out an angle. And so the thing that jumps out in my head, well maybe the law of cosines could be useful. Let me just write out the law of cosines, before we try to apply it to this triangle right over here. So the law of cosines tells us that C-squared is equal to A-squared, plus B-squared, minus two A B, times the cosine of theta. And just to remind ourselves what the A, B's, and C's are, C is the side that's opposite the angle theta. So if I were to draw an arbitrary triangle right over here. And if this is our angle theta, then this determines that C is that side, and then A and B could be either of these two sides. So A could be that one and B could be that one. Or the other way around. As you can see, A and B essentially have the same role in this formula right over here. This could be B or this could be A. So what we wanna do is somehow relate this angle... We wanna figure out what theta is in our little hill example right over here. So if this is going to be theta, what is C going to be? Well C is going to be this 20 meter side. And then we could set either one of these to be A or B. We could say that this A is 50 meters and B is 60 meters. And now we could just apply the law of cosines. So the law of cosines tells us that 20-squared is equal to A-squared, so that's 50 squared, plus B-squared, plus 60 squared, minus two times A B. So minus two times 50, times 60, times 60, times the cosine of theta. This works out well for us because they've given us everything. There's really only one unknown. There's theta here. So let's see if we can solve for theta. So 20 squared, that is 400. 50 squared is 2,500. 60 squared is 3,600. And then 50 times... Let's see, two times 50 is 100, times 60, this is all equal to 6,000. So let's see, if we simplify this a little bit we're going to get 400 is equal to 2,500 plus 3,600. Let's see, that'd be 6,100. That's equal to 6,000... Let me do this in a new color. So when I add these two, I get 6,100. Did I do that right? Yeah. So it's 2,000 plus 3,000, plus 5,000. 500 plus 600 is 1,100. So I get 6,100 minus 6,000, times the cosine of theta. And let's see, now we can subtract 6,100 from both sides. So I'm just gonna subtract 6,100 from both sides so that I get closer to isolating the theta. So let's do that. So this is going to be negative 5,700. Is that right? 5,700 plus... Yes, that is right. Right, because if this was the other way around, if this was 6,100 minus 400 it would be positive 5,700. Alright. And then these two of course cancel out. And this is going to be equal to negative 6,000 times the cosine of theta. Now we can divide both sides by negative 6,000. And we get... I'm just gonna swap the sides. We get cosine of theta is equal to... Let's see we could divide the numerator and the denominator by essentially negative 100. So these are both going to become positive. So cosine of theta is equal to 57 over 60. And actually that can be simplified even more. Three goes into 57, is that 19 times? Yep, so this is actually... This could be simplified. This is equal to 19 over 20. We actually didn't have to do that simplification step because we're about to use our calculators, but that makes the math a little more tractable. Right, 3 goes into 57, yeah, 19 times. And so now we can take the inverse cosine of both sides. So we could get theta is equal to the inverse cosine, or the arc cosine, of 19 over 20. So let's get our calculator out and see if we get something that makes sense. So we wanna do the inverse cosine of 19 over 20. And we deserve a drum roll. We get 18.19 degrees. And I already verified that my calculator is in degree mode. So it gets 18.19 degrees. So if we wanted to round, this is approximately equal to 18.2 degrees, if we wanna round to the nearest tenth. So that essentially gives us a sense of how steep this slope actually is.