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## Geometry (all content)

# Proof of the law of sines

CCSS.Math:

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

## Want to join the conversation?

- does this law of sines work for all kinds of triangles?(112 votes)
- YES

if it had to be a right triangle, then you could apply the direct soh cah toa, you dont need any special laws...(9 votes)

- 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)
- 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)

- How could you put the law of sine into a word problem?(11 votes)
- 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)

- Anyone know if he did another video on him implementing the law of sines?(10 votes)
- Unfortunately no, Sal does have one on Cosine, and Law of Cosines. For more help on the Law of Sines, you could check out IXL. Hope this helps(5 votes)

- Does this work for an obtuse triangle as well?(4 votes)
- Yes, the law of sines works for any type of triangle!!(7 votes)

- 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)
- 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)

- 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)- 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)

- if in trig, side b =26sin47 divided by sin32 how does b=35.9(2 votes)
- 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)

- 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)
- 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)

- 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)
- 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.