 Welcome back to our lecture series, Math 1060, Trigonometry for Students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Misseldine. In lecture 29, we're going to continue our discussion of vector applications. That is, we're going to be looking at various force and velocity diagrams involving vectors and using our notions of trigonometry to help us answer questions related to these vector applications. In the previous video, we learned about something called static equilibrium. Static equilibrium is when you have an object that is not in motion, and that's because all the forces acting on that object add together to form the zero vector. We had an example of a little boy on a swing, whose sister was holding him up, that was the static equilibrium. We were able to solve that one by creating a right triangle and just using some basic Soca Toa trigonometry to help us out here. In this example, I want to consider a situation that similar, it's in static equilibrium, but not as simple as the previous example. This time, imagine that a traffic light that weighs 22 pounds is suspended by two wires shown in our diagram here. What you want to think of right here is our traffic light. We have red, yellow, green. I clearly didn't color code that very well, but it's being suspended by a wire of some kind. It's like you have this wire right here. You put a hook onto it. It's probably secured more than this, but the weight is going to sag the wire down like so. Your little dangly thing is sagging the wire down like so. Then we get a diagram that looks something like this, if we think of it just in terms of vectors. For which we'll take point A right here, that's the location of the traffic light. It has a weight of 22 pounds right here. The weight here is going to be 22 pounds. The direction is going to be downward because its weight is the force due to gravity. Let's consider these other points, B and C, as you see here on the diagram. This is our rope. This is where the ropes are connected, B and C right here. We could ask ourselves what is the tension along the wires, A, B, and along the wire, A, C, like so. When it comes to static equilibrium, since the light is not falling to the ground or going in another direction, the sum of all three of these arrows has to equal zero. Let's suppose that the angles formed with respect to the vertical are the following. When you take tension one and tension two, so tension one will be the tension from the light to the right, and then tension two will be the tension from the light to the left, like so. With respect to the vertical, these angles will form 60 degrees and 45 degree angles. To the right, you get the 60 degree angle. To the left, you get a 45 degree angle. Notice, of course, when you add those together, 60 degrees plus 45 degrees, you do not get 90 degrees. It's actually too big in that regard. So this isn't going to be a right triangle when we're done. But that doesn't mean that we can't handle this thing. It just means we have to change our perspective a little bit. What we're going to do is I want to redraw this diagram into a triangle to make life a little bit easier for us. Basically, what I want to do is I want to move this vector down below here. We're going to get a vector that looks like this. It's going to run parallel to it. This is T1. And then I'm going to move this vector over here. So you get something like this. Here's T2. Honestly, this is sometimes the most difficult part of these static equilibrium problems. Moving the vectors around to form a triangle or a parallelogram, whatever the shape turns out to be, making sure you move things in such a way, and you're going to keep things parallel. So the alternate interior angle theorem is going to come into play several times in this type of diagram. The point of this is we know that T1 plus T2 plus W is equal to the zero vector because we're in static equilibrium. They're going to all add up to B that in the end. So we're trying to form a triangle like we do right here. And so I have a secret picture of that down below. So when we move those things around, we're going to get the following. But how do the angles correlate in this situation? Well, we've got to take a look at that as well. So come back up, hide the secret picture. So in terms of angles, what's going to happen here? So think of this line right here. Let me draw it on the screen. My sketch was somewhat poor in that regard. But if we just basically move T2 down this direction, so it actually stays co-linear with its original location, we have this line right here. And therefore, these angles are going to be vertical angles. So they're going to be congruent to each other. So that gives us that this is a 45 degree angle. Also, if we take, say, this line associated to T2, we just translate it over here so that's parallel. What can we say about congruent angles in that situation as well? I'm going to insert a vertical line through this point right here just for the sake of it. So then we get by the alternative angle theorem again, this angle right here would correspond to this angle right here. So this is likewise 45 degrees. And to the right of that, this one would be 60 degrees like so. So then, with respect to these white lines that you see right here, right, with respect to those white lines, which are parallel, this red line transverses them as well, thus forming by corresponding angles, a 60 degree angle right there as well. So that's kind of messy. Let me come back here and summarize what we've discovered. So what we've discovered so far was that this angle right here is 45 degrees. This angle is also 45 degrees. This angle right here would be 60 degrees. And then the sum of the triangle, some of the angles of triangle need to add up to be 180 degrees. We've accounted for a 45 degree angle, a 60 degree angle. And so this last one would then have to be 75 degrees to add everything correctly. The weight, like we observed earlier, is going to be 22 pounds like so. And so with this information, we are going to try to solve for the tension of T1 and T2. So we need to find the tension of T1. We also want to find the tension of T2. How are we going to do that? This is not a right triangle anymore. What can we do to find the sides here? Well, the law of signs actually is an appropriate tool here because we know the length of this side. We know it's corresponding angle. This is an AOS, an angle opposite side. And so since we know this AOS, we can solve for the other AOSs. If we want to find T1, we'll then look at this AOS versus that AOS. And so the law of signs comes into play to help us find T1. So T1 over sine of 45 degrees, this will equal, of course, the weight divided by sine of 75 degrees, for which then we could try to compute these things, solving for T1. We end up with, well, clear the denominal time both sides by sine of 45 degrees. The weight of the light was 22 pounds. So we get that. We're going to get sine of 45 degrees right here. But sine of 45 degrees, we know what that is. That's root 2 over 2. So I'm just going to put that in there, root 2 over 2. And then we get sine of 75 degrees. Now sine of 75 degrees, we actually did do this once upon a time using the angle sum identities for sine here. We thought of sine of 75 as sine of 30 plus 45 degrees. And so remember previously, we got the square root of 6 plus the square root of 2 all over 4, like so. To make life a little bit easier for us, though, we're going to factor out from the denominator of the square root of 2 over 2. So we get 22 over root 2 over 2. And then if you factor out root 2 over 2 from the sine of 75 degrees, like so, that leaves behind the square root of 3 plus 1 over 2. And so then these square roots of 2 over 2 will cancel out. And then you're divided by fractions. So let's multiply by the reciprocal. And we end up with 22 times 2 over the square root of 3 plus 1. And while I don't think there should be mandated rationalizing denominators, this is a situation where rationalizing the denominator actually does simplify this expression here. So we're going to times the top and bottom by the square root of 3 minus 1. On the top, you're going to end up with 22 times 2 times the square root of 3 minus 1. In the denominator, you're going to end up with the square root of 3 times square root of 3, which is 3. And then you're going to get a minus 1, like so, which is a 2. This 2 will cancel with the factor of 2 in the denominator, thus giving us the exact value of 22 times the square root of 3 minus 1. And let's approximate this. And we would end up, we're on to the nearest pound. We end up with 16 pounds. And so that's the tension along the first wire T1. Well, to find the tension along the second wire T2, we're just going to do another law of sine situation. We're going to use this AOS this time. So we have that the tension along T2 over sine of 60 degrees. This is going to equal 22 over sine of 75 degrees. Clearing the denominators, or that's to say times both sides by sine of 60, we end up with 22 times sine of 60. Sine of 60 is root 3 over 2. And then the denominator, we have another root 2 over 2 times the square root of 3 plus 1 over 2, like so. Well, let's try to simplify this thing a little bit. In so doing, we can at least cancel out these divisors of 2 like so. And then multiplying by reciprocals again, we'll end up with a 22 times the square root of 3 divided by the square root of 2. And then we get 2 over the square root of 3 plus 1, like so. And we're going to do the same trick we did before. Let's times by the conjugates here, square root of 3 minus 1, square root of 3 minus 1, like so. So the square root of 3 plus 1 times square root of 3 minus 1, that gave us a 2 again. So that's going to cancel with those. So what do we have now? We have a 22 times the square root of 3 times the square root of 3 minus 1. This is above the square root of 2. I'm going to distribute the square root of 3 right here. And we're going to get 22 times 3 minus the square root of 3 all over the square root of 2. Yikes, where did the 2 go there? And then the last thing I want to do, again, trying to continue with this rationalizing here, I'm going to times the bottom by 2, square root of 2, the top by the square root of 2, like so. Distributing the square root of 2 on the top here, you're going to get 22 times 3 root 2 minus the square root of 6. In the bottom, you get a 2 now. 2 does go into 22, 11 times. So we end up with, at the exact value, 11 times 3 times the square root of 2 minus square root of 6, for which an approximation seems very appropriate. This is approximately 20 pounds of tension. So we were able to still handle this static equilibrium problem, but we didn't have a right triangle. We had an oblique triangle. And so when working with oblique triangles, the skills we developed in the previous chapter about the law of signs and law of cosines becomes very relevant. When you have a choice, I would suggest the law of signs. It's typically the easier approach. But sometimes the law of cosines might be appropriate, although we didn't see that in this exercise.