 In this video, I want to demonstrate how we can use the law of sines to solve oblique triangles that satisfy the ASA condition. That is, we know two angles, and so for this one, let's say we know angles B and C, and we know the side length that's between the two given angles. So for this example, we know angle B, let's say it's 34 degrees. We know angle C, let's suppose it's 82 degrees. And then, let's suppose that the side length A, which is between them, measures at 5.6 centimeters. We need to find out what is the missing angle A, and what are the two missing sides, B and C, right here. All right, so how do we do this? We've learned previously how to solve the AS condition. So if you have angle A, angle B, and side length A, then we can solve this using AS. Okay, it turns out ASA is basically the same thing, because our first goal is to find angle A right here. Because you'll notice we know angle B, we know angle C, and the measure of any two, excuse me, the measure of any three angles of a triangle. So you take the measure of angle A, the measure of angle B, plus the measure of angle C, this always adds up to 180 degrees. So if you know two of the angles of a triangle, you can find the measure of the last one. The measure of angle A is just going to be 180 degrees, take away 34 degrees, because that's B, take away 82 degrees, because that's C. And so if we take 34 and 82 away from 180, that's going to leave behind 64 degrees, just a simple subtraction right there. All right, and so we can then see that the missing side is 64 degrees. Notice now, once we compute angle A, we now actually have an AOS that is an angle opposite side pair. If you have an AOS, you can use the law of sines, which remember the law of sines tells us that sine of A over sine, excuse me, sine of A over A is equal to sine of B over B, and then sine, this is also equal to sine C over little C right there. So these are the three possible AOSs, angles in their opposite sides there, this ratio, sine of the angle divided by its opposite side doesn't depend on the angle. So if ever you have an angle opposite side, so you know like say capital A and little A, well we also know the remaining sides. So you have like little, you have capital B here, oh little B, we don't know it, we can solve it using the law of sines, and you can do the same thing for little C right here. Although I'm going to prefer to take the reciprocals of these things. Let's put these fractions upside down. Take little A over sine of A, and then take little B over sine of B, like so. So little A was given as 5.6, sine of A, well that's 64 degrees, you have then little B over sine of B, sine of B is 34 degrees right there. So if you solve for B, you just have to times both sides by sine of 34 degrees. So little B is equal to 5.6 sine of 34 degrees over sine of 64 degrees, and that would be the exact answer. We're going to use a calculator to approximate this. So putting your calculator 5.6 times by sine of 34 degrees, probably hit equals at that moment, then divide by sine of 64 degrees. Do make sure your calculator is in degree mode so that you get this calculation correct. The approximation we're going to get from a calculator, we're going to round to one decimal place. This is going to be 3.5 centimeters, and so I'm going to add that label to my triangle right here. B turned out to be 3.5. Alright, we're going to replicate this process to find C here, in which case we get A over sine of A is equal to little C over sine of C. Using the data we have, little A is 5.6. Let me draw that 6 again. We get sine of 64 degrees. Little C is unknown, but capital C is 84 degrees, so we get sine of 84 down below. This tells us that little C is equal to, if we cross multiply, then divide by the sine of 64 there. We're going to get C is equal to 5.6 sine of 84 degrees over sine of 64 degrees. That's the exact answer. Put this now into your calculator to get an approximation. Again, take 5.6 times it by sine of 84 degrees, hit equals, and then divide by sine of 64. And your approximation rounding to the nearest tenth of a centimeter, you're going to get 6.2 centimeters right, like so. So C turns out to be 6.2. And therefore, we've now solved the angle side angle triangle condition using the law of sines. Now one thing I want to point out here is that when we solve for B, the only angle opposite side we had was A over A, like we see here on our picture. So we needed to use that one to solve for B. But when we solve for C, we had the AOS of A, but we also had the AOS of B. Why didn't we use B? Well, the reason we didn't use B is that while we had an exact measurement for angle B, we don't have an exact measurement for angle, for side length B. I mean, we do the exact values right here, but that's very cumbersome to use in our formula. We could do that to get an exact value for C. Absolutely, we could do that. If we use the approximation of 3.5, that's the thing, it's an approximation. If we use the approximation to help us compute the other value, C, for which we approximated later on, turns out the rounding error can compound later on the problem. Because B was imprecise, C will be less precise than it was before. And so the issue is when you're trying to solve for missing side of the triangle, make sure you always use the exact value. So try to use the information you're given as much as possible. Now, in this problem, we did have to use the angle measure A, which we were not given. But notice angle A is not approximated here. We can assume that the measurements of angle B and C are precise. We don't worry about the measurements there, we'll just take it as an assumption and go from that. How do we compute angle A? Well, we subtracted angles B and C from 180 degrees. So the subtraction, there's no rounding involved here. So even though angle A was not given to us, we can assume that the measurement of angle A is precise because we just got from subtraction. The sign lengths B and C though are approximations. So it's best not to use approximations to find new approximations if it can be avoided. So we should focus on the AOS of A throughout the duration of this problem to avoid these compounding rounding errors.