 In this video, we're going to solve a problem where a plane is flying with an airspeed of 170 miles per hour with a heading of 52.5 degrees. It's true course on the other hand will have a heading of 64.1 degrees. If the wind currents are our constant 40 miles per hour, what are the possibilities for the ground speed of the plane? Now we introduced the ideas of airspeed, heading, ground speed, all of this type of stuff in a previous video. If you're not familiar with this terminology, take a look at that video first, but we're going to proceed forward, assuming you have that knowledge here. So when it comes to these headings, as we decided upon earlier, these headings are always with reference to a clockwise rotation from north. So if the plane has an airspeed of 170 and a heading of 52.5, that means the airplanes, its intended path, right, will be this vector you see right here. It's measurement from north is 52.5 degrees. The length there would be 170 degrees. All right. What do we know about the true course? We know its direction. We don't know the ground speed. That is the magnitude of the true course there. So we have this vector right here going along, right? The direction is 64.1 degrees, but we don't know the ground speed. And so we want to consider what that's going to be. All right. So we also know something about the wind, right? The wind is going at 40 miles per hour, but we actually don't know what the direction the wind is blowing in, which causes somewhat of a difficulty here. But if the wind is blowing at 40 miles per hour with this other information, what could what could be the possible ground speed and say, what are the possibilities plural? Is that a possibility? Well, notice what we have right here. OK, because we know the heading and the true course, these directions, we can calculate the difference between those, right? 64.1 minus 52.5 degrees. That is equal to 11.6 degrees. Let's call that angle theta, which we represented here on the diagram. We have that right there. Notice we have a side. We have a side and we have an angle. So we are currently in the side, side angle situation, the so-called ambiguous case with the ambiguous case could have no possible triangles, one possible triangle or two possible triangles. So there in fact, there is some ambiguity on what the possible ground speeds could be. We don't have enough information. Maybe, maybe there is only one possibility we'll see here. So we have to investigate this a little bit more now in order to investigate this further, we're going to want to use the law of signs. That's that's the best approach when considering the ambiguous case because we know this angle theta, we know the wind speed and that gives us an angle opposite side pair in AOS. So if we want to think of X over here, we could try to compare it with this angle beta right there, right, which we don't know that one is at all. But we do know this right here and there is an alpha. So that's really what we need to try first. We need to do this AOS, which is given and we need to solve for alpha in this situation to see what's going on here. So we're going to take sine of alpha over 170 and set that equal to sine of theta over 40, like so to solve for alpha. We're going to times both sides by 170 and take arc sine. So alpha is equal to arc sine of 170 times sine of 11.6 degrees over 40, for which when you put this in the calculator, you're going to end up with sine inverse of this quantity. This will give your calculator, will give you 58.7 degrees. Now, since your calculator did not give an error, that means there is a solution here, but there could also be a second solution, right? Remember arc sine can't tell the difference between angle and its supplement. It can't tell the difference between acute and obtuse angles. So we have to take 58.7. We also have to take 180 degrees minus 58.7 degrees, which is 121.3 degrees. That's a second possibility that we have to consider. Now, the first situation, right? Let's investigate that one. So the first situation is when alpha is 58.7 degrees, like so. So since we already know theta, we can then solve for beta in this situation. So beta would equal 180, take away theta, which was 11.6, take away alpha, which is 58.7. This would mean that beta is 109.7 degrees, like so. So that's acceptable. Nothing wrong with this so far. And so applying the law of sines, we get that X over sine of beta is equal to 40 over sine of theta, 11.6, right? Solve for X, we end up with X is equal to 40 times sine of 109.7 degrees over sine of 11.6 degrees. Use your calculator to approximate this thing and you would end up, make sure you're in degree mode, 187 miles per hour. All right. So that is a possible ground speed for the airplane. But that's one of two possibilities. Maybe we have to investigate the obtuse case. Maybe 121.3 degrees works. It actually seems quite feasible. So if alpha was equal to 121.3 degrees, we have to still compute beta. Beta would be 180 degrees minus theta, which is 11.6 minus alpha, which in this case is equal to 121.3. In that situation, you actually do get a positive value for beta. You get 47.1 degrees. So it turns out there are two possibilities. Once we get beta, we're still going to do the law of sines. So X over sine beta is equal to 40 over sine theta, like so. Solving for X, you still get X equals 40 times sine of beta. Beta, of course, is different this time. Beta is going to be a lot smaller, so you get 47.1 degrees. And then this is going to sit over sine of 11.6 as well. Use your calculator, again, make sure you're in degree mode. And this time you're going to get 146 miles per hour, like so. So the ground speed is much smaller that time, 187 versus 146. Why the big difference there? Well, remember that the airspeed of the plane was 170. So one of them, you're getting a bonus. One of them, you're getting a penalty. And let's think about the possibilities there for a moment. So the two choices for alpha were 58.7 degrees, as the picture seems to suggest, right? It's an acute angle. The other possibility would be something where A or alpha is obtuse. And so you actually get something like the following, like so. In which case, if alpha is much bigger, then that kind of forces this vector to be a lot smaller, because instead of the original one where the wind is kind of pointing off to this direction and you can get much more distance, right? So in one direction, the wind is kind of pointing in the direction, thus giving a bonus. And then the other one, it's pushing against, which is why the speed is a little bit retarded there. And so that brings us to the end of lecture 29, which we've been talking about some of the applications of vectors with respect to direction and static equilibrium and some other things like that, right? It turns out that the trigonometry we've learned can be very, very useful, whether we're using right triangle trigonometry or this oblique triangle trigonometry, law of sines, law of cosines. We can do it, but it can get really tricky as you see in examples like this in the next lecture. We'll actually introduce a simplification of this process known as an algebraic representation of the vectors. It still uses trigonometry, but it dramatically simplifies it that we don't have to be working with this oblique triangle trigonometry anymore. And I think you're going to appreciate it. Thanks for watching this video, though. If you learned anything, please give it a like and subscribe to the channel to see more videos like this in the future. If you don't mind and post any comments, excuse me, post any questions in the comments below. If you have any, I'll be glad to answer them. Bye, everyone.