If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Proof of the law of sines

Sal gives a simple proof of the Law of sines. Created by Sal Khan.

Want to join the conversation?

  • leaf green style avatar for user Jin Hee Kim
    does this law of sines work for all kinds of triangles?
    (112 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user monica.azalde
    When you are using law of sine for a triangle that is SSA, you can get the "ambiguous case" where there are 2 possibilities for the degrees, etc of the triangle. If I have already found the first triangle, then how do I find the second triangle?
    (36 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user mgraham
      First, it only shows up if the information you are given is 2 sides and 1 angle which is opposite one of the given sides.

      When you write and solve the law of sines, you end up with sinC=0.32 or something. You type sin^-1(0.32) in your calculator and you are given an acute angle. Actually there are two solutions to the equation sinC=0.32. One is acute (your calculator gave it to you) and the other solution is obtuse. To find the other solution, do 180deg - acuteangleanswer.

      The obtuse angle is the beginning of the second 'ambiguous' case, and you find the remaining unknown sides/angles using that obtuse angle.

      So say you are given sides a and b and angle A. You will get two answers for angle B (one acute and one obtuse). Each of these answers leads to two different angle C's and side c's.

      As a side note - to check what I said about acute/obtuse angles, type
      sin(30) and sin(150) into your calculator - you will notice they are equal.
      (31 votes)
  • piceratops sapling style avatar for user Thea Ruiz
    How could you put the law of sine into a word problem?
    (11 votes)
    Default Khan Academy avatar avatar for user
    • old spice man green style avatar for user sami00kan
      June wants to measure the distance of one side of a lake. The lake can be expressed as the triangle ABC. Angle a is opposite side BC, angle b is opposite side AC, and angle c is opposite side AB. She knows angle a= 54 degrees and angle b= 43 degrees. She also knows side AC= 106 feet. What is the measure of side BC?
      (22 votes)
  • piceratops ultimate style avatar for user Colin Warn
    Anyone know if he did another video on him implementing the law of sines?
    (10 votes)
    Default Khan Academy avatar avatar for user
  • aqualine ultimate style avatar for user Olivia Finch
    Does this work for an obtuse triangle as well?
    (4 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Fouad  Eltaha
    does anyone know where i can find videos for the double angle, half angle and product- sum formulas on this website or any other place?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • male robot hal style avatar for user Dana Cooke
      If you consider a and h as both being x in the addition rules for sine and cosine, you can easily figure out the double angle formulas.
      In other words:
      sin(2x) = sin(x+x) = sinxcosx + cosxsinx = 2sinxcosx
      and
      cos(2x) = cos(x+x) = cosxcosx - sinxsinx = (cosx)^2 - (sinx)^2
      (12 votes)
  • old spice man green style avatar for user Skywalker94
    What if you are trying to find the sine of the right angle in a triangle? That's opposite over hypotenuse, and the opposite side IS the hypotenuse. That should be one. Am I correct?

    What if you are trying to find the cosine of the right angle in a triangle? In that case, the hypotenuse is clear, but which of the other two is the adjacent? I don't know how to tell because in the case of a right angle in the triangle, the opposite side is the hypotenuse.

    Any help is much appreciated. :)
    (6 votes)
    Default Khan Academy avatar avatar for user
    • aqualine ultimate style avatar for user Nathan MacKay
      Think about the unit circle. Sine of theta is the y value of a point on the unit circle and cosine of theta is the x value of the point on the unit circle. 0 degrees is straight to the right and then as theta increases you move counter-clockwise. Thus when theta is 90 degrees you are at the point (0,1). Hence sin(90)=1 and cos(90)=0
      (4 votes)
  • blobby green style avatar for user cindy
    if in trig, side b =26sin47 divided by sin32 how does b=35.9
    (2 votes)
    Default Khan Academy avatar avatar for user
    • leafers seedling style avatar for user Shaina
      Cindy, 35.9 is a correct answer. Well, I tried solving this on my calculator, and I actually got 35.8831949. Its just that its rounded to the nearest tenths thats why instead of having it in 35.88, it becomes 35.9. You know there are some calculators that round off answers right away.
      (10 votes)
  • blobby green style avatar for user Devon Aphane
    How do I know that I'm supposed to write the law of sines as sinA/a=sinB/b or as a/sinA=b/sinB?
    (4 votes)
    Default Khan Academy avatar avatar for user
    • duskpin ultimate style avatar for user Polina Vitić
      It works either way! But I like to arrange it so that the unknown value is in the numerator of the fraction to the left of the equal sign.

      For example, if I don't know side b, I would write the equation like this:
      b / sinB = a / sinA

      Then it's really easy to rearrange the equation, plug in the values, and solve for b:
      b = (a ⋅ sinB) / sinA
      etc.

      If I didn't know ∠A, I would write (and rearrange) the equation like this:
      sinA / a = sinB / b
      sinA = (a ⋅ sinB) / b
      ∠A = sin⁻¹ [(a ⋅ sinB) / b]
      etc.

      Hope this helps!
      (6 votes)
  • aqualine ultimate style avatar for user Z
    Is there a reason why the law of sines works? I mean why the triangle has opposite sides and angles in equal ratios? I do get how he derived it but was wondering why the triangle has angles and their opposite sides in equal ratios to other angles and their opposite sides?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • female robot grace style avatar for user loumast17
      The derivation actually explains it.

      If you have any triangle and are comparing two of its angles and their corresponing sides, so like alpha and A or Beta and B, you can ue the third angle to drop a perpendicular and form two right triangles. Then both right triangles have the one side x. Then it's just a matter of using algebra. so sin(alpha) = x/B and sin(beta) = x/A. So in less math, splitting a triangle into two right triangles makes it so that perpendicular equals both A * sin(beta) and B * sin(alpha). Then you can further rearange this to get the law of sines as we know it.

      So to summarize, split a triangle in half with a line x and it makes two right triangles. As long as you know soh cah toa you can make a relationship relative to that new x line. And fromt here a relationship with each other.

      Let me know if that did not help.
      (4 votes)

Video transcript

I will now do a proof of the law of sines. So, let's see, let me draw an arbitrary triangle. That's one side right there. And then I've got another side here. I'll try to make it look a little strange so you realize it can apply to any triangle. And let's say we know the following information. We know this angle -- well, actually, I'm not going to say what we know or don't know, but the law of sines is just a relationship between different angles and different sides. Let's say that this angle right here is alpha. This side here is A. The length here is A. Let's say that this side here is beta, and that the length here is B. Beta is just B with a long end there. So let's see if we can find a relationship that connects A and B, and alpha and beta. So what can we do? And hopefully that relationship we find will be the law of sines. Otherwise, I would have to rename this video. So let me draw an altitude here. I think that's the proper term. If I just draw a line from this side coming straight down, and it's going to be perpendicular to this bottom side, which I haven't labeled, but I'll probably, if I have to label it, probably label it C, because that's A and B. And this is going to be a 90 degree angle. I don't know the length of that. I don't know anything about it. All I know is I went from this vertex and I dropped a line that's perpendicular to this other side. So what can we do with this line? Well let me just say that it has length x. The length of this line is x. Can we find a relationship between A, the length of this line x, and beta? Well, sure. Let's see. Let me find an appropriate color. OK. That's, I think, a good color. So what's the relationship? If we look at this angle right here, beta, x is opposite to it and A is the hypotenuse, if we look at this right triangle right here, right? So what deals with opposite and hypotenuse? Whenever we do trigonometry, we should always just right soh cah toa at the top of the page. Soh cah toa. So what deals with opposite of hypotenuse? Sine, right? Soh, and you should probably guess that, because I'm proving the law of sines. So the sine of beta is equal to the opposite over the hypotenuse. It's equal to this opposite, which is x, over the hypotenuse, which is A, in this case. And if we wanted to solve for x, and I'll just do that, because it'll be convenient later, we can multiply both sides of this equation by A and you get A sine of beta is equal to x. Fair enough. That got us someplace. Well, let's see if we can find a relationship between alpha, B, and x. Well, similarly, if we look at this right triangle, because this is also a right triangle, of course, x here, relative to alpha, is also the opposite side, and B now is the hypotenuse. So we can also write that sine of alpha -- let me do it in a different color -- is equal to opposite over hypotenuse. The opposite is x and the hypotenuse is B. And let's solve for x again, just to do it. Multiply both sides by B and you get B sine of alpha is equal to x. So now what do we have? We have two different ways that we solved for this thing that I dropped down from this side, this x, right? We have A sine of beta is equal to x. And then B sine of alpha is equal to x. Well, if they're both equal to x, then they're both equal to each other. So let me write that down. Let me write that down in a soothing color. So we know that A sine of beta is equal to x, which is also equal to B sine of beta -- sorry, B sine of alpha. If we divide both sides of this equation by A, what do we get? We get sine of beta, right, because the A on this side cancels out, is equal to B sine of alpha over A. And if we divide both sides of this equation by B, we get sine of beta over B is equal to sine of alpha over A. So this is the law of sines. The ratio between the sine of beta and its opposite side -- and it's the side that it corresponds to, this B -- is equal to the ratio of the sine of alpha and its opposite side. And a lot of times in the books, let's say, if this angle was theta, and this was C, then they would also write that's also equal to the sine of theta over C. And the proof of adding this here is identical. We've picked B arbitrarily, B as a side, we could have done the exact same thing with theta and C, but instead of dropping the altitude here, we would have had to drop one of the other altitudes. And I think you could figure out that part. But the important thing is we have this ratio. And of course, you could have written it -- since it's a ratio, you could flip both sides of the ratio -- you could write it B over the sine of B is equal to A over the sine of alpha. And this is useful, because if you know one side and its corresponding angle, the angle opposite it that kind of opens up into that side, and say you know the other side, then you could figure out the angle that opens up into it. If you know three of these things, you can figure out the fourth. And that's what's useful about the law of sines. So maybe now I will do a few law of sines word problems. I'll see you in the next video.