 Hello again! We are now going to start the process of learning how to solve triangles that are not right triangles. There are two primary tools to do this. One is called the law of sines, and the other is called the law of cosines. In this screencast, we're going to look at two examples that use the law of sines. So, here's a statement of the law of sines. Notice we're using our standard notation for a triangle in which the vertices of the triangle are capital letters A, B, and C. The corresponding angles then are alpha, beta, and gamma. And the side opposite angle alpha is side A. One opposite angle beta is B. And the one opposite angle gamma is C. And here's a statement of the law of sines. And again, this can seem like an awful lot to memorize, including kind of having to memorize how the diagram is set up, how the triangle is set up. And there's quite a bit of information there. But what I like to focus on in this is each individual ratio. For example, if you look at this ratio right here, you will notice that it involves the angle alpha and the side A. And if you look at that, then angle alpha is there. A is the length of the side opposite. And so, basically, what each of these ratios involve is the sign of an angle and the length of the side opposite that angle. And the other two, you would see the same thing. So again, if we look at this one, again, you can see the angle gamma and the opposite side is C. And of course, the second set of equations down here is just the reciprocal of those of the three that we have written there. So there's just equivalent, different way of writing the same thing. So I like to think about, in using the law of sines, is kind of a more general framework. And notice in this setup, there are four quantities involved. One is the angle theta, and it's opposite side of x, and the angle phi, and it's opposite side of y. So the idea here is that if you know three of these four quantities, you can use the law of sines to determine the fourth quantity. But again, regardless of how you write your ratios, notice that each one involves an angle and the length of the side opposite. Same thing with phi and y. And if we could set up a situation like that, that's when we can use the law of sines. So what I like to remember is always think of sine of an angle divided by the length of the opposite side. That ratio is the same for all three pairs, angle opposite side of the triangle. That way, I don't have to remember as much notation. I can remember the main concept. So what I would like you to do for this triangle that we're going to try to solve is pause the screencast and just see if you can at least set up one situation where you could use the law of sines. Notice again, there are three quantities given. We're given the measures of two angles and the length of one side. Okay. Now, one of the things we can do right away, too, is if we know two angles in the triangle, we can determine the third because we know the sum of the angles adds to 180 degrees. So again, we have to maybe introduce a name for that. We'll call that angle theta. And then we can say that theta plus 25 degrees plus 51.3 degrees is 180 degrees. And again, by solving that for theta, again, we get theta plus 76.3 degrees equals 180 degrees, or theta equals 103.7 degrees. So in reality, we have all three sides of, or I'm sorry, all three angles of the triangle plus one side. So we can set up ratios involving the pair of 25 degrees and 12 feet. So what we're going to use for a notation here, then, is we're going to call the length of this side x and the length of this side y. And with that, then, you can see we have ratios we can set up in particular. Since we're kind of trying to solve for x, it might be nice to put that in the numerator. x divided by the sine of 51.3 degrees. And again, you can see that's a ratio involving the sine of an angle and this length of the side opposite. The one that we know is 25 degrees and 12 feet. So we set that equal to 12 divided by the sine of 25 degrees. And now we've got an equation involving four quantities, three of which we can calculate. So we can solve for x. And basically what we do is multiply both sides of this equation by the denominator sine of 51.3 degrees. So we get a somewhat complicated computation, but the process is not too difficult. x is 12 times sine of 51.3 degrees divided by the sine of 25 degrees. And here's now where we do take out our calculator and do the computation. And again, remember, we're going to get a decimal approximation and I guess in this case, we're going to round off to the nearest hundredth of a foot. So I'll say 22.16 feet. That's what you should get for that computation. You might want to pause the screencast briefly and do that on your calculator to make sure that that is done correctly. Now we have two pairs. We have x in 51.3 degrees and 12 in 25 degrees. And what I like to do is use the original one, the one that we were given. And in fact, the way of determining y is almost the same as the other one. We can take y divided by the sine of theta and we've determined theta to be 103.7 degrees and that should be also equal to the ratio of 12 divided by the sine of 25 degrees. So you can see it's almost the same thing. We just have a sine of 103.7 degrees. We get y is equal to sine 12 times sine of 103.7 degrees divided by the sine of 25 degrees. And again, now's the time to take out the calculator and do that computation. If we do so, we should get y equals 27.59 feet. Okay, so here's the summary of kind of what we have done in this problem. And again, remember I've got the three unknowns labeled, but the whole starting point was this pair right here because we knew both the angle, the measure of the angle and the length of the side opposite that angle. That allowed us to use that ratio. We first determined the measure of the angle theta and then used the law of sines to determine length x and length y. Okay, here's another problem. Again, what you might try to do here is at least seeing what ratio we can use within this problem if we use the law of sines. So pause the screencast and try to set up the problem. You might have to introduce some notation to help with that. I will do that when you come back to it. And if you set it up, you know, likelihood, you might, you'll use slightly different notation than me. And again, it's not the notation that's important, but I will be consistent in my work here and continue to use whatever notation I introduce for you. Okay, we're back now. So what I'm going to do, again, the focus of my attention is right in these two quantities here. That's the key to the law of sines. We have an angle and a length opposite that. Notice in this case we actually have the length of two sides and only one angle. So what we're going to have to do is first solve from one of the angles. We have three unknown quantities here. So I'm going to call this side x the length of that side and then I'm just going to label these angles as alpha and beta. And you can see now, in one sense, we're going to have a hard time. There's one pair, namely the pair involving alpha and x, that we really can't use yet because we don't know either of those quantities. So our first step is going to be to determine the angle beta. But you can see that we do have one pair that we can use to set up our ratio. Since we're solving for an angle, I'm going to set up this ratio with sine of beta on the top of the numerator. And I'll get sine of beta over 2.5. And then I can use the quantities that I know. Sine of 60 degrees divided by 3.5. And again, now we have a ratio that we're an equation involving two ratios that we can solve for sine of beta. And again, to do that, I'm going to multiply both sides by 2.5 and we get sine of beta equals 2.5 sine of 60 divided by 3.5. And we can now use our calculator to at least first determine sine of beta. And one of the things we might consider here, we do know the exact value of sine of 60 degrees, but everything else that we're using is decimal approximations. So we might as well use our calculator to do all of the computations here. And if we do that, we come up with a result at sine of beta. Again, I'm going to write down seven digits. Don't round this off too much. Always use seven or eight digits, if not the full calculator display for that. And we get sine of beta equals 0.6185896. And here's where we have to be very careful using the inverse sine function. Remember that the inverse sine function is only going to return to us an angle between minus 90 and 90 degrees. In this case, since we're using a triangle, it's going to return something between 0 and 90 degrees. But it is possible for a triangle to have an angle greater than 90 degrees. There are two values that we would have to consider for beta. The easiest one is what one we get using the inverse sine function. And again, rounded to the nearest tenth of a degree, that comes out as 38.2 degrees. Now we have to use that as a reference angle equivalent to a reference arc and consider that the other possibility is 180 degrees minus 38.2 degrees. So we have another possible angle of 147.2 degrees. Sorry, that should be 141.2 degrees. So let's get that fixed up there and doing that subtraction. It should be 141.8 degrees. Okay, now, if both of these angles were possible, we would have two triangles that would solve this. We would have to give both of them. But please notice that 141.8 degrees plus 60 degrees, the other angle that we're given, is 201.8 degrees, which is greater than 180 degrees. So in this case, our second angle is not possible. We can't have two of the angles add up to more than 180 degrees. In fact, all three angles have to add up to 180 degrees. So in this case, we have only one solution for beta. And of course, the next thing we have to do is determine the angle alpha. So that's what we're going to try to do right now. There is, of course, what we're going to use is that the three angles have to add up to 180 degrees. And we'll be able to then solve for the side x. So basically what we have is alpha plus beta plus 60 degrees equals 180 degrees. So if you want alpha plus 38.2 degrees equals 120 degrees, or doing the necessary subtraction, we get alpha equals 81.8 degrees. And now, again, remember, we always have our given information of 3.5 and 60 degrees. And so I'm going to use that, as you can see in this equation here. And again, use the law of signs to set up the angle and its opposite side. So we get sine of alpha in the denominator and x in the numerator. And again, if we solve that equation for x using our value for alpha, it does come out that x is approximately 4.0 meters. And I would include the point zero because that indicates that we've rounded that value off to the nearest tenth. So we've done quite a bit of work here, but we finally have solved this triangle. We have all six sides determined. Three of them were given to us, and we determined the other three law of signs. There is a way we can go to check our work just to make sure we've done this correctly. And I would encourage you to kind of pause the screencast and do the three computations that I have indicated there. We could also do them as sine of 60 degrees divided by 3.5, sine of 38.2 degrees divided by 2.5 and so forth. I chose to do it this way. And if I do these computations, or you should be doing them, but what I get here, and again, perhaps including more decimal places than I need to, that rounds off to 4.0415. This one rounds off to 4.0426. And this one rounds off to 4.0413. And again, you can see they're not exactly equal, but that's because we're using many rounded off computations. But you can see that all three of them do round off to 4.04. And so we have strong evidence that we have solved this triangle correctly. And we could also now go back and probably use the same procedure to check our work with the first triangle that we solved here. Okay, that's it for now.