 In this video, we're going to demonstrate how you can solve a oblique triangle under the AAS condition. So this is the angle angle side condition. What this means is we know two angles of the triangle and we know the side length that is not interior to them. So we have here our AAS wrapping around the triangle like this. You'll notice that the side S is not between the two angles. And so for this example, suppose the angles we know are is angle A is 30 degrees, angle B is 70 degrees, and then side length A is going to be 8 centimeters. So those are highlighted in our diagram in green. To solve this triangle, we need to find the side length C, the side length B, and the angle C right here. And so we're going to demonstrate how to do this. Now the first thing to do, and we're going to do this using the law of signs, the first thing to do is actually to find the missing angle C right here. Because we know that the measures of a triangle, so the measure of angle A plus the measure of angle B plus the measure of angle C for any triangle, whether it's right oblique, whatever, will always add up to be 180 degrees. This tells us that the measure of angle C will equal 180 degrees. Take away the measure of angle A, which is 30 degrees. Take away the measure of angle B, which is 70 degrees. Angle A and B together account for 100 degrees of the triangle. So the remaining side would have to be, excuse me, 80 degrees. We took away 100 degrees. And so we then can fill that in the picture here. Angle C is going to be 80 degrees. So if you know two angles, you basically know three angles. Because you can just subtract them from 180, the known angles to find the missing angles. So now how are we going to find angle B and angle C, or excuse me, side B and side C right here? This is where the law of signs comes into play. Because if we have an angle angle side condition, then that means we have an angle opposite side pair. That is we have an AOS. If we have an AOS, that means we can compute the value sine of A over little A. We can compute this because we know it. We have to compute sine of 30 degrees over 8. We can compute that ratio and that gives us that. Why is that useful? Well, this AOS, sine A over A, is going to equal the other three AOSs for this triangle. That is, by the law of signs, this will be the same thing as sine B over little B. This will be the same thing as sine C over little C. So if we just look at the first two for a moment, sine A over little A versus sine B over little B, we know angle A. We know it's side length A. We know angle B. We don't know little B. So we can solve for little B to help us out here. And that's exactly how we're going to solve these things right here. Now, I'm actually going to take the reciprocals of things to make it a little bit easier. We're going to take little A over sine A and we're going to take little B over sine B, like so. Little A we know is 8. We have sine of 30 degrees, like so. And then we have little B over sine of 70 degrees, like so. Notice to solve for B, we just have to times both sides by 70 degrees. We get B equals 8 times sine of 70 degrees over sine of 30. Sine of 30 degrees, we know from the unit circle diagram that's actually equal to 1 half. If you divide by 1 half, this is the same thing as times in by 2. So B is actually equal to 16 times sine of 70 degrees. This is the exact answer. Notice that sine of 7 degrees, it's not one we memorized. It's going to be some irrational number. Your calculator is going to be your best friend on this one. We can compute 16 times sine of 70 degrees on our calculator. Make sure it is in degree mode at this moment. In which case, if we round to the nearest tenth, this would equal 15.0 centimeters, like so. So that's going to be our approximation for that one. And that gave us side length B, so we get 15.0 for B right there. Well, how are we going to find C? We're going to do the same basic thing, but we're going to take the AOS A over sine A and we're going to compare that to C over sine C, like so. And so doing that one, we're going to have A over sine A and this is equal to C over sine C, like so. So little A, we know, is 8. And then that's going to be over sine of 30 again. And then little C, we don't know that one. And then we have to do that over sine of 80 degrees, like so. So again, times both sides by sine of 80 degrees. We get C equals 8 sine of 80 degrees over sine of 30. Which, remember, sine of 30 is again 1 half. If you divide by 1 half, that's times 1 by 2. So C is equal to 16 sine of 80 degrees. That gives us the exact value. If we approximate with that, with a calculator, 16 times sine of 80 degrees would give us, again, rounding to a tenth of a centimeter, we're going to get 15.8 centimeters. And that then gives us the missing side of our triangle that we should then label above. Just to finish the picture there, we're going to get 15.8 centimeters. And so now we've demonstrated how we can use the law of sines to solve the angle-angle side condition. Now, one comment I do want to point out here is that when we solve for B, we pretty much had to use the AOS of A, right? The angle opposite side, because that's the only angle opposite side we had. When we were solving for C, we had the AOS of A, but we also had the AOS of B. Now, I still decided to use the AOS A to help us find C. And the reason we had that is that whenever you have a choice between using information about a triangle, always use the information that was given. Notice how side length A was given. It was equal to 8 centimeters. We didn't have to worry about, is that right or wrong? That was just information given to us. We also didn't have to worry about rounding. Like 8, we can assume is exact, right? We didn't worry about the precision of the measurement. We'll just assume 8 is given to us. On the other hand, if we did something like B, B is approximately 15.0 centimeters. It's not exactly. We're just rounding for the sake of convenience. Now, the issue is if we use an answer which is only approximately right to help us find out other values, then the rounding error of B will spread and sort of infect C, which we eventually rounded it as well. But the thing is that if we use B to try to find C, then the rounding error of B can carry on into C and then make this one even worse. And so our approximations can get worse and worse over time if we use approximate answers. It's like taking the copy of a copy of a copy with an old copy machine or something like that, but the quality degrades over time. So it's best to use the original given information as much as possible. So for this exercise here of AAS, it's best to use the original AOS throughout the duration of the problem so you don't get this compounding error at the end.