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

Articles, Blog 73 Comments

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

can figure out everything else about this triangle just

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

everything else about it. Or if you actually had

two sides and an angle, you also would be able to figure out everything else about the triangle.

Narek KazarianPost authorGreat memory refresh for Law of Sines. Thank you Sal

EduardoKicksPost authorcan u figure everything about the triangle if u have only sides

Tasneem AlkamPost authorThanks a lot you explained it really well

tarfa spacePost author1/4 or 1?

MelissaPost authorilysm

Andrea RevelesPost authorMay God bless your soul. I needed so much help. Thank you so much 🙂

Lorenzo PanePost authorThank 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 VindrotPost authorI am too stupid for this! how do I use sin on a calculator properly?

bloodykills 750Post authorOh god my brain

Muhammad NafisPost authorhow could people dislike khan academy

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

onedgivesmelifePost authorTopped the class bc of this THANK YOUUU

Grim ReaperPost authorcan someone tell me when you assume two solutions?

3Post authorBest video, thank you!!

Dars CollPost authorThis is the hardest

Ebruhh KPost authorIm failing math. ☹️️ this shit is top hard

Steven CPost authorWhen I still use Kahn Academy in college.

David MateusPost authorchicken massala

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

cottoncandy295Post authorI don't get the reciprocal part… Is there another way?

Reese Elijah MoraldePost authorWhy do the angles in the triangle have to add up to 180??

Reese Elijah MoraldePost authorwhen 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 MotilaPost authorWhen doing the equation why didnt u just use times then divide and get missing value its legit much easier

Michael TimmonsPost authori honestly hate this stuff

Living TribunalPost authorhow did it became 1/4?

Mr. SmithPost authorHe sounds like a youtuber ImMarksmen

Gumball The MonkeyPost authorStill don't get it.

Nathan OkapiPost authorlmao 5:39 he said law of cosines

good video though, helped me out.

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

superredpanda12Post authorCan you figure out everything if you only have two sides with no angles?

KINGOFGUITAR2101Post authorIf he had kept the one half over 2 it would be a different answer no?

Joshua ThiessenPost authorsorry what???

Griffin LeggieriPost authorWhy don't you just cross multiply?

minxxdia 》Post authorThanks! This video made my day 😉

Mohammed AlnawfalPost authorhe makes simple steps so complex

always ferdaPost authorI 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 ferdaPost authorHow did he get root 2/2 ?

YakultPost authorWtheck did it turned into 1/4

Hella BummyPost authorwouldn'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 UsernamePost authorSpaced out during discussion. This is actually surprisingly simple.

BroThasJeffreyPost authorThank 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

whatsupbudbudPost authorWould 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 manPost authorOnce you get the flu math becomes impossible

Angela beverlyPost authorThank you so so much this is great

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

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

BAYPALSPost authorjust 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 APost authorU did it way to complicated and just did unnecessary extra work.

TIM1VERPost author(3:10) sinus's are annoying (look at what he is writing.)

KHfan890Post authorhow did you get 1 over 4?

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

M00NL0RD36Post authorThe 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)

tony sherifPost authorthats so helpful

ericPost authorI 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 PowersPost authorthis went over my head. I guess I'm just bad at shapes.

Isaiah KingPost authorYea 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

calob vilmotPost authorBless

StopTrynaKillMePost authorwho else has a test tomorrow?

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

Elliott WoodPost authorhe mentions proving the law of sines, where is that?

BJ JeffriesPost authorIs 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.

Caitlin MariePost authorI love your handwriting

Rubiks CubingPost authorDid anyone notice the mistake he said at 5:38 he said law of cosines

gemini wonderPost authorI can't believe that there is almost 1million views of Law of Sines

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

zyrsPost authorUSING FRACTIONS ISN'T EASIER THAN MIXED NUMBERS STOP THAT

Samuel FeasterPost author5:37 “law of cosines”

B WPost authorAnyone know how he got the TI calculator onto his desktop?

Bros Before doggosPost authorBut I got 3.40 on b instead of his answer what did I do wrong

Aditya BagaskaraPost authorI've just learned this now

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

Adam SkeadPost authorOh 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 NightsPost authorall 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)