# Law of sines | Trig identities and examples | Trigonometry | Khan Academy

Voiceover:We’ve got a triangle here where we know two of the angles
and one of the sides. And what I claim, is that I
with this information. You give me two angles and a side, and I can figure out what the other two sides are going to be. And I can, of course,
figure out the third angle. So, let’s try to figure that out. And the way that we’re going to do it, we’re going to use something
called the Law of Sines. In a future video, I will
prove the Law of Sines. But here, I am just going to show you how we can actually apply it. And it’s a fairly straightforward idea. The Law of Sines just tells us that the ratio between
the sine of an angle, and the side opposite to it, is going to be constant for any
of the angles in a triangle. So for example, for this
triangle right over here. This is a 30 degree angle, This is a 45 degree angle. They have to add up to 180. So this right over here
has to be a, let’s see, it’s going to be 180 minus 45 minus 30. That’s 180 minus 75, so
this is going to equal 105 degree angle, right over here. And so applying the Law of Sines, actually let me label the different sides. Let’s call this side right over here, side A or has length A. And let’s call this side,
right over here, has length B. So the Law of Sines tells us that the ratio between
the sine of an angle, and that the opposite side is going to be constant through this triangle. So it tells us that sine of
this angle, sine of 30 degrees over the length of the side opposite, is going to be equal to
sine of a 105 degrees, over the length of the
side opposite to it. Which is going to be equal
to sine of 45 degrees. equal to the length of the side opposite. So sine of 45 degrees over B. And so if we wanted to figure out A, we could solve this
equation right over here. And if we wanted to solve for B, we could just set this equal
to that right over there. So let’s solve each of these. So what is the sine of 30 degrees? Well, you might just remember
it from your unit circles or from even 30, 60, 90
triangles and that’s 1/2. And if you don’t remember it, you can use a calculator to verify that. I have already verified that this is in degree mode, so it’s 0.5. So this is going to be
equal to 1/2 over two. So another way of thinking about it, that’s going to be equal to 1/4, this piece is equal to 1/4 is equal to sine of a 105 degrees over A. Let me write this, this is equal to sine of 105 degrees over A. And actually, we could also say, since we could actually
do both at the same time, that this is equal to that. That 1/4 is equal to sine
of 45 degrees over B. Actually, sine of 45 degrees
is another one of those that is easy to jump out of unit circles. You might remember it’s
square of two over two. Let’s just write, that’s
square root of two over two. And you can use a calculator, but you’ll get some decimal
value right over there. But either case, in
either of these equations, let’s solve for A then let’s solve for B. So one thing we could do is we could take the reciprocal of both
sides of this equation. The reciprocal of 1/4 is four. And the reciprocal of this right-hand side is A over the sine of 105 degrees. And then to solve for A,
we could just multiply both sides times the
sine of a 105 degrees. So we get four times the sine
of 105 degrees is equal to A. Let’s get our calculator out, so four times the sine of 105 gives us, it’s approximately equal to, let’s just round to the
nearest 100th, 3.86. So A is approximately equal to 3.86. Which looks about right if this is two, and I have made my angles appropriately, that looks like about 3.86. Let’s figure out what B is. We could once again take the reciprocal of both sides of this and we get four is equal to B over
square root of two over two, we could multiply both sides times square root of two over two. And we would get B is equal to four times the square
root of two over two. Come to think of it, B is four times the sine of 45 degrees. Let’s figure out what that is. If we wanted actual numerical value, we could just write this
as two square roots of two. But let’s actually
figure out what that is. Two square roots of two is equal to 2.83. So B is approximately equal to 2.83. So [I’m] be clear, this
four divided by two is two square roots of two, which is 2.8. Which is approximately equal to 2.83 if we round to the nearest 100th, 2.83, which also seems pretty reasonable here. So the key of the Law of Cosines is if you have two angles and a side, you’re able to figure out
two sides and an angle, you also would be able to figure out everything else about the triangle.

## 73 thoughts on “Law of sines | Trig identities and examples | Trigonometry | Khan Academy”

• ### Narek Kazarian Post author

Great memory refresh for Law of Sines. Thank you Sal

• ### EduardoKicks Post author

can u figure everything about the triangle if u have only sides

• ### Tasneem Alkam Post author

Thanks a lot you explained it really well

1/4 or 1?

ilysm

• ### Andrea Reveles Post author

May God bless your soul. I needed so much help. Thank you so much 🙂

• ### Lorenzo Pane Post author

Thank you so much! One thing is at the end, Sal says that the key to the law of "cosines" is if you have…… Did he mean sines?

• ### Anton Vindrot Post author

I am too stupid for this! how do I use sin on a calculator properly?

• ### bloodykills 750 Post author

Oh god my brain

• ### Muhammad Nafis Post author

how could people dislike khan academy

• ### Kakapo Papal Post author

khan academy you can solve with less rules and easier ways and this is very hard to learn from you

• ### onedgivesmelife Post author

Topped the class bc of this THANK YOUUU

• ### Grim Reaper Post author

can someone tell me when you assume two solutions?

• ### 3 Post author

Best video, thank you!!

• ### Dars Coll Post author

This is the hardest

• ### Ebruhh K Post author

Im failing math. ☹️️ this shit is top hard

• ### Steven C Post author

When I still use Kahn Academy in college.

• ### David Mateus Post author

chicken massala

• ### Masta Harashibu Post author

What do you do when none of the angles are well-known unit circle values? Like sin74 and sin69?

• ### cottoncandy295 Post author

I don't get the reciprocal part… Is there another way?

• ### Reese Elijah Moralde Post author

Why do the angles in the triangle have to add up to 180??

• ### Reese Elijah Moralde Post author

when equating sin(30)/2 and sin(105)/a, i noticed that you evaluated sin(30) and left sin(105) alone. My question is, what if i also evaluated sin(105) and proceed to solving for a, do i still get the same value for a?

• ### Atila Motila Post author

When doing the equation why didnt u just use times then divide and get missing value its legit much easier

• ### Michael Timmons Post author

i honestly hate this stuff

• ### Living Tribunal Post author

how did it became 1/4?

• ### Mr. Smith Post author

He sounds like a youtuber ImMarksmen

• ### Gumball The Monkey Post author

Still don't get it.

• ### Nathan Okapi Post author

lmao 5:39 he said law of cosines
good video though, helped me out.

• ### Danny Boy Jango Post author

My trig professor is the wort kind of teacher. Teacher's should not only be qualified to teach, but should also love teaching.

• ### superredpanda12 Post author

Can you figure out everything if you only have two sides with no angles?

• ### KINGOFGUITAR2101 Post author

If he had kept the one half over 2 it would be a different answer no?

• ### Joshua Thiessen Post author

sorry what???

• ### Griffin Leggieri Post author

Why don't you just cross multiply?

• ### minxxdia 》 Post author

Thanks! This video made my day 😉

• ### Mohammed Alnawfal Post author

he makes simple steps so complex

• ### always ferda Post author

I don't understand why we have to learn this to graduate none of it has real life applications even as an engineer how often are you gonna need to figure out what the angle of a triangle is you'll just measure it out lol great video tho

• ### always ferda Post author

How did he get root 2/2 ?

• ### Yakult Post author

Wtheck did it turned into 1/4

• ### Hella Bummy Post author

wouldn't you have to multiply the 2 square root of 2 by 4 at the end though? what happens to that 4?

• ### Not a Cool Username Post author

Spaced out during discussion. This is actually surprisingly simple.

• ### BroThasJeffrey Post author

Thank you so much man I had a test for trigonometry and this was the only thing I dind't get at first. I got 90% on it

• ### whatsupbudbud Post author

Would be nice if you'd explain the last part after multiplying the whole equation by "square root of 2 divided by 2". We ended up with b=4 times square root of 2 divided by 2. I got the correct end result when I: 1) Got the square root of 2. 2) Divided that result by 2. 3) And multiplied that by 4. But the author of the video just divided 4 by 2. Why?

• ### Moth man Post author

Once you get the flu math becomes impossible

• ### Angela beverly Post author

Thank you so so much this is great

• ### 정철 Post author

Well, it's very easy to understand. Korean high school students learn this in their sophomore.

• ### Lil Reddy Post author

Can you cross multiple to get sin30(A)=Sin105(2) But then what would you do from there

• ### BAYPALS Post author

just wondering what grade you guys are learning this? how old are you ? just curious as iam behind and learning this right now. I’m 16 years old. is that a normal age to learn this ?

• ### Aryane A Post author

U did it way to complicated and just did unnecessary extra work.

• ### TIM1VER Post author

(3:10) sinus's are annoying (look at what he is writing.)

• ### KHfan890 Post author

how did you get 1 over 4?

• ### Akash shinde Post author

Make the diagrams more accurate and clear to visible, also have to work on font size.

• ### M00NL0RD36 Post author

The flaw is it teaches me with 30 degrees. That is easy because sin30 is known to be 0.5, and this doesn't teach us to do things that are, not 30, 45, 60, etc? (I feel like such a nerd writing this)

• ### eric Post author

I like kahn but be wary he kind of explained this one poorly.

It is much easier to cross multiply than multiplying the reciprocal than doing what he did.

Also, he did not mention that when you want to use the law of sines, you use the ratio of the sine which has two definite values from the get go to solve for the one that does not have anything specified.

• ### John Powers Post author

this went over my head. I guess I'm just bad at shapes.

• ### Isaiah King Post author

Yea but shouldn’t he have just crossed multiplied and got 2sin105 then divide that by 30 on the calculator??? Then inverses it to get 3.68

Bless

• ### StopTrynaKillMe Post author

who else has a test tomorrow?

• ### IamYams Post author

I text book says the equation is a/sinA=b/sinB=c/sinC

• ### Elliott Wood Post author

he mentions proving the law of sines, where is that?

• ### BJ Jeffries Post author

Is there any reason why you didn’t just “cross multiply” once you set your ratios equal to each other? I can follow your method but my students get lost every time. Just wondering if I’m missing the reason behind that approach.

• ### Rubiks Cubing Post author

Did anyone notice the mistake he said at 5:38 he said law of cosines

• ### gemini wonder Post author

I can't believe that there is almost 1million views of Law of Sines

• ### Hodgepodge Post author

Does it matter which order you set it up? Also my teacher suggested you cross multiple and divide.

• ### zyrs Post author

USING FRACTIONS ISN'T EASIER THAN MIXED NUMBERS STOP THAT

• ### Samuel Feaster Post author

5:37 “law of cosines”

• ### B W Post author

Anyone know how he got the TI calculator onto his desktop?

• ### Bros Before doggos Post author

But I got 3.40 on b instead of his answer what did I do wrong

I've just learned this now

• ### Jaidev Ramcharan Post author

I'm here because I have my math exam tomorrow. 😭

Oh I see your using one side to solve the other, in homework for math I assumed I'd have to solve one side before I can solve the one I wanted to solve first

• ### Green Nights Post author

all over the net, what I see makes me want to say, quit beatin round the bush !!!! and just DO IT !!!
and show your work. uhm like this:

function ssa_obtuse_in(s,ss,a) {
/// input the obtuse angle
return ssa_acute(s,ss,PI – a);
}
function ssa_obtuse(s,ss,a) {
/// return the obtuse angle
return PI – ssa_acute(s,ss,a);
}
/// Side Side Angle /// in/out obtuse difference –
function ssa_acute(s,ss,a) {
return Math.asin(ss*Math.sin(a)/s);
} // angle opposite side2
/// for all others, angles can be acute or obtuse
function sas_cos(s,a,ss) {
with (Math) {
return sqrt((pow(s,2)+pow(ss,2))-(2*s*ss*cos(a)));
} // Side Angle Side – solve length of missing side
}

function aas_sin(a,b,s) {
return Math.sin(b) * (s / Math.sin(a));
} // AAS solve opposing side of angle2 (b)

function asa_sin(a,s,b) {
return Math.sin(b) * (s / Math.sin(PI-(a+b)));
} // ASA solve opposing side of angle2 (b)

function sss_cos(a,b,c) {
with (Math) {
return acos((pow(a,2)+pow(c,2)-pow(b,2))/(2*a*c));
}
} // SSS solve angle opposite side2 (b)