in this video you're going to learn how to work with the law of sines and the law of cosines so when do we use the law of sines in the law of cosines well it's when we don't have a right triangle we don't have that 90-degree angle so for the law of sines there's two different ways to write the formula you can either write at this top one or this second one here and let's talk about law of sines first each problem I'm going to show you I'm going to do two of sines and two of the law of cosines each one's a little bit different so you're gonna want to see each example when we're given a triangle let's just call it a generic triangle ABC notice that the capital letters are used to denote the angles and then the lowercase letters are used to denote the side lengths and notice that here's angle a across from angle a a side a across from angle B a side B across from angle C a side C and when we do the law of sines I like to use this top equation here you can think of this ratio the sine of angle a over its side opposite that ratio is equal to the sine of angle B over its side opposite and that's equal to the ratio of sine of angle C over its side opposite so you only need to use two of these three ratios to form a proportion and then you can solve for the missing quantity what I like to think of is I like to think of the angle and the sight across is like a pair so you've got a pair a pair and a pair like that so let me show you the first example like say here we want to find the measure of angle C well what I notice first off is that I have 70 degrees and I have the side across from that 70 degree angle I also have this side here 6 and I'm looking for angle C which is across from this side length of 6 so basically what we have is we have two pairs and so we only have one unknown quantity angle C so what I'm going to do is I'm going to say the sine of 70 degrees over its side opposite equals the sine of angle C over its side opposite which is 6 now all we have to do now is we need to get sine of angle C by itself so I'm going to multiply both sides by 6 that way the numerator denominator cancel and then what we're gonna do is solve for angle C is we're gonna have to take the sine inverse of this quantity let's do it one step at a time though so first things first we've got 6 times the sine of 70 degrees all divided by seven and that comes out to about 0.8 0-5 so 0.8 0-5 equals sign of C if we want to solve for angle C we're going to do the sine inverse of 0.8 0-5 and let's see what that comes out to fifty three point seven degrees so let's write that here are fifty three point seven degrees and you got it that's the measure of angle C now if you want to proceed from there then you can say well I know these two angles I can subtract from 180 to find this angle I could do law of sines again to find this missing side and then you've solved the triangle okay our second example using the law of sines we're trying to find this missing side right here X and again what we want to do is we want to think about the angle and the side across from that angle as like a pair so here I've got 50 degrees in ten here we have 35 degrees and X so we can use our law of sines to solve for this missing in quantity we've got 70 sine of 35 degrees over its side opposite equals sine of 50 degrees over its side opposite and now all we have to do is get this variable by itself now what we can do is we can cross multiply its x times the sine of 50 degrees equals 10 times the sine of 35 degrees now all we have to do instead of multiplying by sine of 50 is divide both sides by sine of 50 degrees that gives us X by itself and we can calculate that length so let's go ahead and do that we've got 10 sine of 35 divided by sine of 50 make sure your calculator is in degree mode since we're working with degrees and it comes out to approximately 7.5 and yeah the next two examples we're going to talk about the law of cosines we're gonna be working with these three formulas here at the bottom now if you know one of these formulas you'll automatically know the other twos so what I recommend is start off by memorizing this top one C squared equals a squared plus B squared just like Pythagorean theorem minus two times a times B times the cosine of angle C one thing you'll notice about these formulas is that see how this is side C and angle C see how they're across from each other or at opposite ends in the formula just like they're across from each other in the triangle if we use this middle form of see side B angle B they're across from each other in the formula they're across from each other in the triangle and we've got side a and angle a and notice then if we're working with this top one C C and C then we've got a and B a and B if we're working with me we've got a and C a and C if we're working with a we've got B and C B and C the other two that we're not using right so when do we use law of cosines that's the question a lot of students ask well you can use the law of cosines okay or you actually have to use law of cosines when they only give you three sides like side side side nothing else or they give you side angle side meaning the angles in between the two given sides if you try to use law of sines what will happen is you have too many unknowns you won't be able to solve because you've got like two variables or more so that's when you use law of cosines side side side or side angle side let me show you example number three here okay notice what we have we've got side angle side that case I was talking about right here we want to find the side across from angle C side C so we're going to use this top formula here and let's go ahead and write it out so we have C squared equals a squared plus B squared minus two times a times B times the cosine of angle C we don't know what C is so let's just leave that as a variable C okay a is the side across from angle a that's eight squared plus B that's the side across from angle B that's ten squared minus two times eight times ten times the cosine of angle C which is 42 degrees now notice how this is C squared here we just want to solve for side C not C squared so I'm gonna do is I'm gonna take the square root because the square and the square root those are inverses they cancel if we do that to the left side we also want to do it to the right side so let's go to our calculator let's put all that in and we're gonna take the square root so if you use parentheses you can do all this in one step eight squared plus 10 squared let's see minus two times eight times ten times the cosine of 42 again make sure your calculator is in degrees so that comes out to about six point seven and you've got that missing side okay example number four this is the where you have three sides side-side-side and you want to solve for a missing angle in this case we're trying to solve for angle B so what we're going to do since we're trying to solve for angle B let's use this middle equation where we have side B but not angle B and let's write that out so we've got B squared equals a squared plus C squared minus 2 AC cosine angle B okay B is a side across from angle B that's 15 squared a is a 1 across from angle a that's 11 squared C is 1 across from angle C that's 9 squared minus 2 times 11 times 9 times the cosine of angle D now somebody students make a little bit of arithmetic errors when they're trying to solve for angle B here so let me see if I can explain this let's go ahead and simplify a little bit here we've got 225 we have 121 plus 81 minus let's see 2 times 11 times 9 which is 198 cosine angle B now what students sometimes do is they try to combine these numbers here you can combine these but you really can't combine with this because this 198 is multiplied by cosine of via this is like all one group so what we're gonna do instead is we're gonna subtract 121 and we're going to subtract the 81 we're gonna get them to this left side of the equation so let's do that now so 225 minus 121 minus 81 so now we're at 23 equals negative 198 times the cosine of angle B we want to get cosine B by itself so let's go ahead and divide by negative 198 to both sides and let's see what that comes out to okay so we're looking at about negative point 1 1 6 equals cosine of angle B now remember whenever you're solving for the missing angle that's when you want to do the inverse sine or inverse cosine inverse tangent like that so the solver angle B we're going to take the cosine inverse of negative point 1 1 6 so let's go ahead and do that and what I tend to do is I tend to on my calculator use the previous answer key so I take that long decimal so you a little bit more accurate answer so it looks like angle B here is approximately ninety six point seven degrees and you got it if you want to see more examples what I didn't get into yet in this video he is when you're working with the law of science you have a potential sometimes to get one triangle two triangles or no triangles depending on the information that they give you so follow me over to that video right there where I talk about the ambiguous case using the law of sines and I'll see over in that video