PC 12.1 Law of Sines

You have a little Tesla going. Do you guys even know Tesla is? Tesla. 80s group, man. Big hair. All right, we're going to set up a situation. We're going to talk about the law of sines. Being section one, I'm going to bring out bean Lake from Marshall, Missouri. This is a real Lake. And it's not named after Mister Bean. It's just shaped like a bean. And that's kind of like Mister Bean. But anyways, we're going to set up a situation where bean and breast are out hiking. They get lost, and you find Mister Bean all the way over here. All right, so here's bean right here. Ta-da. Here's brust. And bean wants to swim across the Lake. Ah, well, mister brush, he's a trig whiz. So he measures the angle here, finds out to be 88°. He then paces 100 yards. So he's going to pace down here. And he measures the angle again, and gets 44°. So we get a 44, and we have an 88, can he actually figure out that distance across the Lake with that much information. That is the question and the answer is yes. We're going to use the law of sines. So what is the law of science? Well, let's observe a regular old triangle here. ABC. Draw a triangle ABC and drop an altitude. Remember the altitude forms a right angle. We're just going to do some observations, see if we can come up with a lot of sines. So if I look at and again, we're just observing things. Let's look at the sine of a remember sine works only for right triangles, and that's why I dropped that altitude down because now on the left hand side we have a right triangle. The sine is the opposite over the hypotenuse. So the sine of a is going to be H over C all right, let's look at the right. What's the sine of C, the sine of C is going to be H over a very similar, but there are a few differences. The a and the C are switched. Let's cross multiply. This is one of my favorite hobbies to do on a Friday night. We cross multiply. So we have H equals C times. I'm going to be careful how I write this because it can get confusing with all those letters. Remember, capital letters, the angle, lower cases, the side, on the right hand side, cross multiply H equals a times the sine of angle C now look, we have H is equal to this. And H is equal to that. Well, guess what? Substitution. We can put those things equal to each other. They're both equal to H, so they're equal to each other. So C is going to equal, I'm sorry, C sine of a is going to equal a times the sine of C all right, well we're almost done here with our observing. I'm going to divide both sides by AC, and the only reason why, to make it prettier, because we like pretty things. So I'm going to cross out those C's. They cancel cross out those a's and look what we have. The sine of a over a is going to be equal to the sine of C over C and when I started this problem, I just drew a random triangle. You know, I could have put C up here, and I could have put B down there. Well, what would that have given us? That would have given us the sign of B over B they're all equal to each other. Now, we know we can flip reciprocals. It's still equal, right? I mean, if we have two fifths, that's equal to four tenths, if I flip those and get 5 halves that's still equal to ten fourths. You can do that. So that's what brings us to this formula right here. It really doesn't matter which way you have it. You can do sine of a over a, or you can do a over the sine of a really doesn't matter, but this is our formula sine of a over a, sine of B over B, sine of C over C or a over sine of a B over sine of E C over sine of C that's a lot of words to get out there, but it's a pretty easy formula, and it works in the situation where you have angle angle side. If that information is given, or angle side angle, if you remember our triangles from geometry, angle angle side would be some angle, the next angle consecutively, the next side consecutively, if they give you that information, then you can use the loft signs. What about angle side angle? That's a different triangle. Angle, side, angle. The side has to be between the angles that they give you. The one thing that it also works for, but it's a pain in the angle side side is side side angle member. Side side angle is in geometry. That doesn't always work. That's the one situation you can not prove triangles congruent. Do you ever ask that question back in jail? Why doesn't side side angle work? You're going to learn today why side side angle didn't work. It still doesn't work. Never works. All right, so let's use the loft sides to solve the situation where we had when we have 88° here, and then he walked for a hundred yards, and then we had 44 degrees in this angle right here. All right, so using the law of sines, here, let's throw some letters on this bad boy. It doesn't matter which way the letters are. But I want to find this distance right here. Okay, now what do I need to know? Well, I need the sine of C over C all right, well that'll work for us. Sine of C over C is going to equal the sine of B over B well, that's not going to help me out because that's a distance. That's not the shortest path across the river. What I need is I need, let's see, I have a hundred here. So it would be nice to know that angle. Well, this angle is easy. We can add up those two. What do we get here? We get one 32. All right, so if they're at a 132, the sum of those two, 44 plus 88, then you subtract from one 80. Everybody know what I'm doing here. Some of the angles must equal to 180° there. Somebody teach you that. So what do we get here? What do you get 48? All right, so that's what this angle equals here. 48. That's angle a so we're going to get the sine of a over a let's substitute in. So the sine of C, we're going to get the sine of 44 over C or X here. You could use either one. Equals the sine of a sine of 48 over a, which is 100. All right, so let's cross multiply. I love cross multiplying. So sine of 48 times C so C times the sine of 48. Is going to equal 100 times the sine of 44, and then I divide both sides by the sine of 48. Both sides sine of 48, we get C is going to equal 100 times the sine of 44 divided by that quantity divided by the sine of 48. Let's figure that out. All right, but I got to calculate hundreds sine of 44, divided by the sine of 48. By the way, before you start this, make sure that your calculator is in the right mode. Because if it's not, you're going to get the wrong answers here. To do that, you hit the mode button and make sure you're in degrees. All right. So back to its hundred times the sine of 44 divided by the sine four 8. Yeah, you could do those each piece wise and get decimals. But we're going to get 93.4 7 5 5, pull that down here somewhere. All right, so if we pull that down, what are we going to call that? Well, I guess that's good enough to call it. Let's go to the nearest tenth on these. Sound like a good idea. So let's go so C is going to equal 93 .5. All right, so 93.5. What are you working in yards? Yards. All right, there's our answer there. That's how far across it is. That's the width of the Lake. Going straight across, pretty much straight across. And so that solves that problem. So being can make a decision, whether he wants to swim across that. All right, solving triangles. Now, when you solve triangles, it means finding every measure of every side and every angle. But the good news is they're going to give you they have to give you some of this information. So you really don't have to find 6 pieces of information. But at the end of it, you should know 6 pieces of information because they gave you three and you'll probably have to find three. Sometimes it's easy, like the first example. Let's look at that. Sine over side length equals sign oversight. That's easy. Cross multiply solve it. Sometimes it's not so easy. This side side angle situation. In fact, when site side angle happens, you have to draw a picture. And there might not always be a solution. In fact, there could be no solution. There could be exactly one solution, or there could be two solutions. What? That's right there why the geometry side side angle doesn't work improves. We need some conceptually elixir. Go ask mister breast if you have them right now for some conceptual elixir. I think he keeps running in the closet. Don't tell people. All right, so the case one, angle a is an obtuse angle. All right, the way I'm going to draw this, when they give you an angle here, let's put it always on the left. We're going to draw the triangle the same way. So if they tell you angle a is obtuse. All right, let's put angle a right here. And let's put B over here. So our triangle will probably look. You're going to have to excuse my great drawings here. But it's going to look something like that. So that angle there, a is obtuse. There are two possible solutions, all right? So let's draw the other one over here. Just so we're dealing with the same thing to start. All right, first situation, a is less than B all right, if this is angle B, this is the side opposite. So this is side B and then side a is up here somewhere. If a is less than B, that's going to swing down, it's never going to touch. All right? It's not going to touch the other side of the triangle. So guess what? This is a situation where there's no possible triangles. Say what? So if they give you side lengths, if they give you an obtuse angle, and they give you two side lengths, and then you draw it in such a way that a, the side opposite the obtuse angle is smaller than the other side, there's no possible triangles. You can not make that triangle, even if you wanted to. There's no way to do it. And you know what? Across from the largest angle should be the largest side. That's one of the reasons why. All right, so if B is larger than a, then you got problems. Let's look at the other situation. Suppose a is greater than B so it's the same situation where a is here and B is here. This would be side length B and then a is longer. Well, guess what? There's one possible triangle. You can make in that situation. And then that's all you have to worry about. You put the information and you'll solve it, you'll be good to go. That's the first case where a is obtuse. Well, that's easy. Let's go to the next case. If a is acute, there are four possible situations. Four, we're talking four. To consider them, we're going to look at the value of B sine a and I use B sine a loosely. That's only if it's a is the angle they give you. If they give you angle B, then these values will change, but we'll look at that as we do some examples. Okay, now remember, this is what happens when we have side side angle. So side side angle, here they're giving you just like we drew before. We have angle a so let's pretend like B's over here. They didn't give us B but we have an acute angle now. All right, so that would be side length B and then we'd have side a now we need to figure out some things. First of all, what is all right, this angle is given. So it doesn't change. So the length of a, that's going to matter. Is that long enough to come down and intersect this side here of AB, and besides C in the triangle, that would be side C right there. Is a long enough to come down and touch that picture this can revolve down. Well, we need to figure out how long it is from that vertex straight down, and that would make a right triangle. So if that makes the distance straight down, that is a right triangle, that's the altitude. I'm going to call that H all right, so what is the height equal to? Well, I know that just from the stuff we did before, we know that the sine of a is equal to the height over B, right? It's opposite over hypotenuse. So if I cross multiply the height is going to equal B times the sine of a ah, now I see where that comes from. Well, let's find the height of this triangle first. All right, so we don't have actual numbers now, but that's what we're going to do. If we find that the height is longer than, now let's do equal to first, because that's where we're at. If we find that the height is exactly equal to this other side, or the other way, this side opposite a is equal to the height. Then when it comes down, it'll intersect right there. It'll make one triangle. In fact, it'll make one right triangle. All right, so then we can just use sine cosine tangent, that kind of stuff. It would be pretty simple. Let's look at a different situation. Let's draw it again. Okay, so here we go. We have the same thing AB, a is given. It's acute. And we have the height here. All right, so we figure out remember this is actually the height. So we figure out the height. What if a is less than the height? Maybe it only goes to like here. It's going to swing down to there. Can it touch the other side? Remember that angle is fixed. No, it can't. It's not long enough. It's not the height is so long and then a is shorter. So there are no possible triangles. No possible triangles. You can't make a triangle. Again, when you figure out the height of this thing, suppose the height is 30, but this opposite side is only 28. It can not touch. It's not long enough. So that is a second situation that could happen. Let's see a third situation that could happen. All right, so suppose that a is greater than the height. The height. But it's also greater than this side here. All right, now look at this. If this sign is between a and where this intersects here, that point, that side is going to be shorter than side B if it's longer than it can not it can't intersect here. It'd be on the other side in that screw up our angle a so I guess it has to go this way then. And so there's only one triangle possible. So this is the situation is pretty easy. There's one triangle possible. Okay, so see what we're doing. We're examining the side lengths compared to the height of the triangle. The last one is right here, right? We redraw it again in this time, what do we have? A is, remember, that's that side here. A is, it's greater than the height, but it's less than that other side. All right, well, what does that mean? That means that it's going to come down like this, or it's going to come down like this. It's in between the length of B and whatever the height is. See how we have two different triangles that are possible. Depending on whether it goes, this way, all right, that's going to give you an obtuse triangle. Or if it goes this way. And that's going to be your acute triangle. This is why side side angle does not work if you're trying to prove two triangles congruent. Because you could have this angle and this side and this side here, and if that side length is between the height of the triangle and whatever the opposite side would be is here, then there are two distinct triangles that can be made. Therefore, you can not prove that two triangles are congruent. That takes you back to geometry, okay? We're not going to do any geometry proofs here. We got rid of that B, let's put it up here. But what we need to know is if that side length there, a is between side B and the height of the triangle, if it's between those two lengths, then we have two possible triangles that we can have. And that's the most work for you. Two possible triangles. All right, let's look at some examples here. All right, so determine the number of solutions and when possible solve the triangle. So first thing we need to do is determine the number of solutions. Let's draw this out. Always put your angle here on the right, and this is always what I do, but it seems to always work for me. Put a there. Here they're giving us side length B, so I'll put that on the right, just so we draw the same way every time. This is an acute angle. It's 63°. And then they tell us side length B and they tell us that a is 17. All right, first thing we need to do, we need to find the height of that triangle there. So we can see if 17 is long enough to intersect down here. All right, maybe the height of 17.5, and so this side length is in long enough, but we don't know until we actually find it. So what we're going to do, we're going to find the sine. The sine of a is going to equal. I'm going to call this the height for now. The height over 18, if I cross multiply, then I get the height equals. I'm going to sub that in right now. Height is going to equal 18 times the sine of 63. All right? So if you go through and you solve that all out, what are we going to get? All right, so trustee calculator here says 18 times the sine of 63 is 16.03. I'm going to call that 16.0. So we have the height is going to equal 16.0. All right, well, if you notice, if this is 16.0, all right, then this side length a here is a little bit longer than the height. But not longer than side length B here. All right, so that means that we're going to have two options. I'll draw it in green. Either this thing is going to come down. There, or it's going to come down in here somewhere. So let's look at the first situation. Let's look at that triangle. We need to draw this triangle. I always like to draw the triangle separately. But we have to answer the first question. Determine the number of solutions there are two possible triangles. So I'm going to write that two possible triangles. All right, so the first triangle is if B is acute. So let's write that out. Angle B is acute. Let's see what that looks like. So if angle B is a cue, we get a triangle that looks like this. All right, so angle B, we don't know the measure of, but we know that this measure is 63°. We also know that this side length is 18 and the side length here is 17. So now what I can do, let's put our letters on it. And label our angles, a, B, and C, so I know that the sine of a is over a is going to equal the sine of B over B so if I were to sub in we have the sine of 63, over 17, is going to equal the sine of B now I don't know what B is, but I know the opposite side length is 18. If I cross multiply and solve it, I get the sine of B is going to equal, well, let's do the whole thing. 17 times the sine of B is going to equal 18 times the sine of 63. I'm not going to show all these steps the next time I do this. This is a lot of work for me. All right, so we divide by 17. We're going to get the sine of B is going to equal ugliness. Let's figure out ugliness in the calculator. All right, so here's what ugliness is. We get the 18 times the sine of 63 divided by 17. We get .9434. All right, so the sine of B is going to equal .94 three four. Now, remember, we're not solving for a side length here. We're solving for the sine of B so how do you find out what B is? We need to find the inverse. We need to take inverse sine. We need to work backwards. All right, so let's write this out. The measure of angle B is going to equal whatever the inverse sine of this is. So use this in your calculator. Sine and the negative one of X if you forgot how to do that. Let's pull up the calculator. This is actually how I'm going to do it. This is your last answer, so I want second sign. It's above the sign button. So you want to find the inverse sine of this last answer. So I'm going to hit second, and then the answer button right there takes your last answer. You find the sign of it. 70.6. So the measure of angle B equals 70.6°. That is amazingly awesome in this situation. 70.6. Now we can start finding out some other stuff. So let's find out what angle C is. Well, how would we do that? We take 70.6, we add it. So 70.6 plus 63, and that's going to give us one 39 33.6. All right, if we subtract from one 80, we get let's see, one 80 minus that equals 46.4, I think. So what are we doing here? Remember all the angles of a triangle equal one 80, so we knew 63. We just figured out 70.6. We were at them all up. We subtract from one 80. We get 46.4. That is amazing. 46.4. We are almost done because the only thing we don't know. We know three angles. We know two sides. We've got to find this side length right here, which is C let's change colors so we know what we're doing. I know I'm kind of like running over into here. So the sine of 63 doesn't matter which angle you pick now, over 17, is going to equal the sine of 46.4. All over this opposite side C here. So we're going to call it C so cross multiply and divide. So it's going to be this times this. We're going to divide by the sine of 63. So C is going to equal 17 times the sine of 46.4 all over the sine of 63. So what is that going to equal C is going to equal roughly if you use your calculator and .8? That's what I get 13.8. That's for my calculator. So if I were going to write this all out, we got to write all of our answers down. So a equals. That was given to us. So a equals a equals 17, B equals 18, C equals 13.8. Angle a was given to us. That was 63°. Angle B, we figured out where was that somewhere down here? 70.6°. And then angle C, which we figured out last. And what do we get? It's 46. So this is a measure of angle C and make sure we label everything there. Is 46.4°. Hey, guess what the good news is. We're done with that triangle. Guess what the bad news is? We have another triangle to do and I ran out of room. I should have probably give you more room on the paper, huh? Well, blame mister Kelly. All right, so what's the other possibility? Well, this was all based on the premise. If I back it up here, that this side length here, this was an acute angle at B or the side length went this way, but it could have went this way. And if it did that, then we have an obtuse angle here, and we have a different angle measure right there. You're trying to look more like this. Before I erase all this stuff though, let's look. We solved the first one. We found this angle right here. This was angle B, we found angle B to be equal to what we got 70.6°. All right, well, I want to point something out to you. This triangle, I'm going to look at this triangle right here. This triangle is isosceles. Okay? So if this angle is 70.6, so is this angle? What we're interested in is that angle, which is on that side right there. So how are we going to find that? Well, they form a linear pair. This they're supplementary to each other. They equal one 80. How many ways can I say this? So this equals 70.6. That means that this angle on the inside is one O 9.4. This is 63, we can then find this angle to be 7.6°. All right? These are all in degrees. So this is 7.6, this angle is given to a 63°. I so now we have our new triangle. So now I'm going to erase everything now that we have the new triangle. Please make sure you understand why that is one 80 minus whatever that angle we found there before is. Okay, voila. Erased everything. Let's write down what we know so far. Here's our second triangle. What do we know? Well, we know the angle a is still 63°. We know the angle B is a 109.4°. We know angle C we figure that out is 7.6°. We also know a and B are still 17 and 18. So the good news is we're only missing C here. So I'm going to do the law of signs again. We're going to get the sine of a over a equals the sine of what do we want here at the top C over C all right, so the sine of 63 over 17 equals a sinus 7.6 over C, we cross multiply, we get 17 times the sine of 7.6 equals C times the sinus 63. We're going to divide both sides by the sine of 63, and we're going to get C is going to equal after they cancel 2.5. Let's see that calculator. Let's see how it all works out. K 17 sine of 7.6 divided by the sinus 6 three. We get 2.5 that's our last answer. So this is a pain in the butt. This is why it's side side angle. They should call it angle side side. I'm telling you, that's how much of a pain it is. But we get C equals 2.5. That was just that first triangle. This doesn't happen all the time, but you need to figure out when you're going to get two distinct triangles. I need to erase all this stuff. Pause the video if you needed that. All right, so now we're going to do the second example. Solve triangle you see if a equals all right, so now we have an obtuse. Obtuse is good news, because that means it's either zero or one. A is one 12 all right, so I'm going to put oh, it's C so I'm not going to be over there. So you got to be careful the way we draw this. Put a back here again. That's one 12. C is over here, because they give a side length C so side length C is 20. All right, a is 32. All right, so it's longer than this side 20 here. So we're good. That means we can make a triangle. If a was shorter than 20, then we'd be done. We just write no possible triangles. All right, so here's one here. What are we going to do first? Well, we know the sine of I'm going to start doing work over here. Sine of one 12 over 32 is going to equal the sine of C over 20. Sine a over a equals sine of C over C, cross multiply, divide by 32. We're going to get the sine of C is going to equal 20 times the sine of one 12 all over 32. If we work that out in the calculator, the sine of C is going to equal .5795 if we round. Okay, pause the video if you need that point. Well, let's just drag it here. Put it up there. So that equals .579, we'll say for a 9. I like to leave everything in the calculator. So as we're doing this, I need the inverse sign because that's what the sine of C is. If I want to find out what C is, I have to use inverse sine of my last answer, so I do second, and then that negative sign there brings up your last answer. And it's going to tell me 35.4. All right, so C is going to be approximately equal to 35.4. So 35.4, guess what? Now we can find this angle here. These all add up to 180°. So if you figure that out, you're going to get 32.6. And lastly, we'll do this in different color. We need to find what is this going to be. This is angle B so we need to find side length B here. So sign of B over B equals sine of a over a cross multiplied divided by sine of one 12, B is going to equal 18.6. Do we know everything now? So let's figure this out. What is a equal to 32? C is 20 B is what do we get here? We just figured that out 18.6. A is one 12. Angle B is 32.6, and angle C is 35.4 degrees. All right, two for you to try. Pause the video and do the last two. Ready, go. Okay, so this is the first situation for the first problem. We have two possible triangles because we look at H, which is 15 times the sine of 50. We figure it out there, blah, blah, blah. Here we go. Here's the answers we get here. That's the first possible triangle where B is acute. And if angle B's up two, we subtract from one 80 that other value of angle B to get 117.9. Then we can figure out a is 12.1. We use the law of signs. We get a equals 3.6. All right, there's the first example, two possible triangles. We have all the work right there. Sorry if I didn't leave enough room. All right, next example. Check this out. Side a is not long enough because side B is 50 inside a is only 33, so we have no possible triangles. That is it. Sorry, this video is kind of long. I promise the next one will be short. It's mister Kelly Bond hold remember. Nice to be important. One important to be nice.