Common Core Geometry Unit 9 Lesson 6 Tangents, Secants, and Their Angles
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Learning the Common Core Geometry Unit 9 Lesson 6 Tangents, Secants, and Their Angles by eMathInsructions
Hello and welcome to another common core geometry video by E math instruction. My name is Kirk weiler and today we'll be doing unit number 9 lesson number 6 on tangent secants and their angles. All right, so so far in unit 9 we've seen chords, diameters, radii, in the last lesson we saw a tangent. Today, we're going to introduce a new idea, and that's the idea of a secant. We'll introduce that right away. And then we're going to look at how tangents and secants interact with the arcs of circles to create particular angles. So let's jump into it right now. All right? First things first, let's talk about a secant. A secant is a line that when extended in both directions intersects the circle exactly twice, exactly twice. Now that may sound a lot like a chord, right? Because a chord intersects a circle exactly twice. The big difference between a chord and a secant is that a chord is specifically a line segment. It has a beginning and an end, and one could say that a chord will always be part of a secant. All right? But let's take a look at an image.
All right. In this image, we've got circle O, and we've got our friend the tangent, right? And the tangent intersects the circle only once at point F there and then goes merrily on its way. So BC is a tangent. On the other hand, BA or AB is a secant because it's hidden, the circle at these two points D and E now, segment DE, or ED, either way, that is a chord. Okay? But the actual a, B, that is a secant. All right? And what we're going to be really concentrating a lot on today is that angle out here. All right? So let's jump into that. And see where we're going with this. All right, exercise number one. This is just kind of a little bit of a skill building exercise. Then we'll kind of get into the angles and things like that. Exercise one. Given the diagram below with secants AB and AC and tangent DE answer the following questions. What two arcs are intercepted by AB and AC when they form angle a so this is going to be important because the angles that are formed by secants and tangents are all about the arcs that they intercept or intersect depending on which word your teacher is using. And what I'd like to do is I'd like to kind of get into a different color, make my line nice and thick so that you can see it. And let's talk about a together, okay? What two arcs are intercepted by AB? And AC when they form angle a well, this arc is intercepted. And as well.
This arc. Is intercepted. If you will, the arcs that are kind of trapped between those two secants. So it's easy enough, those would be arc FG. And arc BK. Or KB. Either way. All right, FG and BK are intercepted by those two secants. Let's take a look at letter B it says what arc is intercepted by secant AC. Secant AC and tangent DE when they form angle a KD. Now this is important, okay? Before I just said, hey, you got these two secants, what arcs do they intercept? Now I'm specifically saying, we've got secant AC, and we've got tangent DE, and now I'm thinking about angle a, K, D, a KD, what arc is intercepted by those two. Well, let me go different color. It's going to be. The src. It's going to be arc FK. All right? Again, it's the arc that gets sandwiched between that secant and this tangent specifically to make this angle. All right. So let's go back here. That would be arc KF or fk. Either way. Arc F all right, finally, what arc do these same segments intercept when they form angle a KE. So that's this angle. A K E why don't you take a shot at that one. Pause the video now if you need to. All right. Well, let's do it. One last color. We'll go with green. It's going to overlap one of the arcs I drew before. But the arc that gets trapped sort of between AC and DE in order to make this angle. Would be this arc. Would be arc, wow. It looks like it would be arc KF again. But this now is a major arc, not a minor arc.
So I have to name it with at least three letters. All right? We're going to begin with F, we're going to end with K or begin with K and with F but we have to choose one, at least one of those points in between. And there's a couple. There's point G, there's point H, we could do them all. In fact, to illustrate that, let's do them all. All right. And in fact, we could have arc K, H, G, F in large letters. Like that. Or you could just have three of them. You've got to have the K and the F, beginning and ending, or F and K, beginning and ending. It doesn't matter. But you could have something like, let's try to make this smaller again. You could easily have something like K, G, F, or K, H, F those would be okay as well. But you've got to have three points labeled because it's a major arc instead of being a minor arc. All right. Now, there's nothing yet about that that helps us with angles. But we're going to need to be able to visualize and identify those arcs in order to apply a theorem that's coming up right now. So let's take a look. All right. Angles made by a cord and a tangent, a cord. And a tangent. So the measure of an angle formed by a cord and a tangent will be equal to one half the intercepted arc. Then I have this sound familiar. What does that sound familiar to? Well, it sounds a lot like inscribed angles, right? And angle has the vertex on the circle, and whose two rays intersect the circle, its measure will be always one half.
The measure of its intercepted arc. All right? The same thing is going to be true for an angle that is formed by a chord and a tangent. These are a little bit more difficult to visualize, though. All right. So let's take a look at a picture. Here we've got this tangent, and we've got this chord and certainly if that chord had been extended so that it was a secant we could say exactly the same thing. But if I look at the measure of angle a, C, E, the sub two, ACE, it's going to be one half the measure of arc a D.C., and likewise, if I look at this acute angle down here, the measure of angle a, C, B, is going to be one half of arc AC, all right? And that's all this theorem says. It's again, it's all about the arcs that are being intercepted by this tangent. And this secant or chord. It really doesn't matter which because I could easily extend this further and make it into a secant. Anyway, by the way, we're going to present multiple multiple theorems in this lesson today. Almost all of them are going to have one half in them. Actually, I think they all have one half in them. But we won't prove them until the next lesson, in fact, the next lesson we're just going to devote to proving these theorems. Today, we're just going to state them and use them, and in the next lesson we'll prove them.
It's just too much to do in all in one lesson. All right, so let's get a little bit of a workout on that one theorem. Take a look, exercise number two. It says encircle O, the measure of arc AEB is 252°. Chord AB and tangent D.C. are drawn. Determine the measure of arc AB. Well, we know that the measure of arc AEB is 252°. So why don't you go ahead and find the measure of arc AB? All right, well, it's pretty simple, right? Any time you have a major and a minor arc that complete a full circle, obviously, they have to add up to 360°. So the measure of arc AB must be 360° minus 252°. Which I believe is 108°. Three. Yeah, a 108°. Simple enough. Now it asks us for the measure of angle B, a, C and for that specifically, specifically, the measure of angle BAC will always be one half of the intercepted arc AB. All right? So that's simple enough. We can say that measure of angle BAC let me get rid of this. Is one half the measure of arc AB so that's one half of a 108°. And that's 54°. And that's it. Right? So it's an easy enough theorem to apply. Let me step out of the way now, so you have a moment to take a look and then we will move on to another problem. All right. Let's do it. Now, let's talk about angles made by secants and tangents. The measure of an angle formed by two secants, two tangents, or a secant and a tangent intersecting outside a circle. Can be found by one half the difference between the measures of the two intercepted arcs.
So there's that one half again. And any time you have a secant and a tangent, we'll look at many, many pictures in a moment. Any time you have a secant and a tangent that intersect outside of a circle, as opposed to let's say a chord and a tangent intersecting right on the circle, that's what we were just looking at. Then you can always find the measure of that exterior angle, an angle outside of the circle. By finding one half the difference in the intercepted arcs. So let's take a look at what this really means. Let's take a look at two secants. All right? Here we go. The two secant situation, right? I've got this secant and that secant. It's intersecting arc AB and CD. And the measure of angle P will be one half the difference in those two intercepted arcs. And then we take one half of it. You could also think about it as CD minus a, B divided by two altogether. It all depends on how you need to use the theorem. Let's take a look though, and when we have a secant and a tangent. All right? In that case, the picture looks more like this. The two arcs that are intercepted are arc AB here. And arc AC down here. And again, the measure of that exterior angle P is going to be one half arc AC minus arc AB. All right.
And finally, if we have two tangents, then the picture kind of looks like that. And it almost looks like a hat sitting on top of someone's face that's very, very circular. And if I've got those two tangents, and it intersects minor arc AB and major arc ACB, right? Then I'll just do that major arc minus the minor arc, take one half of the difference. All right. Again, there is no reason for you to say, well, that makes a lot of sense. That theorem, that is. But we'll prove them in the next lesson. Now what we want to do is just do a bunch of different problems involving those relationships. Okay? Let's take a look at exercise number three. Two secants intersect outside of a circle to form an angle that measures 38°. The smaller of the two arcs intercepted measures 65°. What is the measure of the larger arc in receptive? All right, so let's draw a picture. All right? We've got a nice picture. All right, let's put an exterior point. Oftentimes that's called point P but you can quite frankly label things almost any way you want in this picture because we're not given any specific information about this vertex or that vertex. All we know is we've got two secants drawn. Okay. We know that the angle formed this thing, the measure of angle P is 38°. Okay. And the smaller of the two arcs, maybe I'll label these actually with some letters. Maybe called this a and B may be called this C and D, right? We know the smaller of the two arcs, a, B, has a measurement of 65°. And what we want to know is the measure of the larger of the two arcs. We want to know that. All right.
So what that theorem said was that the measure of angle P is always going to be the measure of arc CD minus the measure of arc AB, all divided by two. All divided by two. So take a look, let's substitute in everything we know. We know that the measure of angle P or that exterior angle is 38°. We don't know the measure of arc CD, so I'm just going to leave that. We do know the measure of arc AB is 65°. So we just have that equation. And again, what's beautiful, of course, about this equation is that the only thing we don't know, the measure of arc CD is what we're looking for. So now we can solve for it by multiplying both sides by two fairly easy, right? Two times 38 is 76°. Equals the measure of arc CD -65°. And now, of course, we can just add 65 to both sides. All right, and what we find is we find the measure of arc CD, 76 plus 65, if I do my math right, is 141°. So that's it. Now granted, this particular problem is not a straightforward application of the theorem. And what I mean by that is a straightforward application of theorem would say, well, I'm going to give you this R, I'm going to give you this arc and I want you to calculate that angle. In this case, we're given the angle. We're given one of the arcs and we have to calculate the other arc. And again, it is simple enough.
The key here is being able to draw good picture. Writing down the theorem knowing the theorem well enough that you can kind of write it down. Substituting in all the things that you already know and then solving for the thing that you don't know. All right? Let me step out of the way for a moment and give you a minute to write down anything you need to. All right, let's keep moving on. Let's see what we have next. Exercise number four. In the diagram of circle C shown, tangent AD is drawn and secant AB is also drawn such through that it passes through C, all right? And that's pretty important. Anytime we've got secants or chords or anything like that that pass through the center of the circle, right? Then it's something that we should kind of pay attention to. It's probably gonna matter. If the measure of arc ED is 42°, then find the measure of angle a so in this case, we don't have secants, but we have a tangent, and we've got a secant. So let's make sure we can write down the theorem using this. The measure of angle a right is going to be once again highlight maybe the two arcs. The two arcs that are intercepted by those things are that one. Let me change color. So we can tell the difference. And that one. Okay? Those two we have. So what do we got? Let me write down the theorem. The measure of angle a is going to be the measure of arc BD minus the measure of R ED, all divided by two. Okay?
Now this time at least we actually are trying to find the measure of angle a so it's a straightforward kind of just put the numbers in, do the calculation, get the answer, and we're done with it. But one thing that's a little bit problematic is we know that the measure of ED is 42°, but we don't know what the measure of arc BD is yet. We don't know what the measure of the red arc is. So how can we figure that out? Ah, well, that's where the fact that the secant passes through the center of the circle is so important. Because that means that the measure of arc EB, right? Is a 180° because it's a half circle. And therefore we can get the measure of arc BD, right? We can figure out the measure of arc BD. Is simply a 180° minus the 42°. Which again, if I do the subtraction correct in my head, it's a 138°. Fair enough. And now we have everything. Now I can say that the measure of angle a is going to be BD, which is a 138°. Minus ED, which is 42°, all divided by two. Again, making sure I get my subtraction right. I get 96° divided by two. And that means the measure of angle a must be 48°. Again, assuming I've done all my subtraction right. That's my Achilles heel. I can miss subtraction problems more than anything else. Anyway, take a moment right down anything you need to, and then we'll keep moving on. All right, let's do it. Last problem.
We had to have the two tangent scenario. That's what we've got going on here. So let's read how it works. In the following circle, tangents are drawn from point E to the circle, such that the ratio of the measure of arc ACB to the measure of arc AB is 5 to one. Determine the measure of angle E all right, well again, simple enough to really think about the theorem, right? Again, we've got these two arcs that are intercepted. One is arc AB. And the other one is arc a CB wow, there we go. Whoops, didn't quite do it. Good enough, right? And then what we've got is we've got the theorem that just says, all right, the measure of angle E is going to be the measure of arc ACB. Minus the measure of arc AB, all divided by two. That's simple enough. Except we don't either. We don't know ACB or AB. We don't know the measure of either arc. But what we do know is that when we add them together, we get 360°, because they are a major in a minor arc that put together, create the entire circle. But we still need both of their measures and the key here. Here will be that 5 to one ratio. When we see something like that, we should immediately jump into this idea. The measure of arc ACB, we're going to let B 5 X and the measure of arc AB we're just going to be X let the X and the reason we're going to do that is that no matter what X is, if X is two, if X is ten, if X is 50, it doesn't matter, we're still going to get that 5 to one ratio because 5 X to one X is 5 to one. But now we need to solve for X and we can do that by simply saying when I take that larger arc and I add the smaller arc to it, I must get 360°.
See if they didn't form the full circle, then that 5 to one ratio would be irrelevant. I couldn't do much with it. But now I can take those two, put them together, get 6 X equals 360. Divide both sides by 6. X then is 60. Okay. That now allows me to say that the measure of arc ACB. Which is 5 times X is 5 times 60. Or 300°. And the measure of arc AB is just that single X, so just 60°, and now I can do my theorem, right? I wonder if I can grab the whole theorem. Oh, I could have, but and it got away from me. No, now I only have part of it. All right. That's all right. Let's just do it right here. The measure of angle E is going to be ACB, which is 300. Minus AB, 60, all divided by two. That's going to give me 240°, divided by two. And that leaves me with a nice obtuse angle of a 120°. And that's it. All right, go ahead and pause the video now and copy down anything you need to and then we'll wrap up. All right, let's do it. So what we saw today, we were a couple of new things. One, we just saw the idea of a secant. A line which when extended forever in two directions, intersects the circle twice. What we then went on to see was how the angles exterior angles to a circle between a secant, a tangent, a secant, and a secant or a tangent and a tangent relate to the arcs that are intercepted, and primarily we keep getting that one half thrown in there.
All of those theorems, though, basically boil down to the same thing. Whether we're talking about two secants, two tangents or a secant and tangent. Two arcs are intercepted. And when we find their difference and then take half of it, we get the measure of that exterior angle. We also saw, and this isn't minor, we saw that the angle that was created by a chord in a tangent at the point of tangency is one half the intercepted arc, just like inscribed angles. And we'll need that in the next lesson, because the next lesson we're going to spend the entirety of the lesson simply proving the theorems that we worked with in today's lesson. Wow, that sounds like fun. But it's not going to be T table proofs. It's actually going to be some algebraic proofs that hopefully you'll find a little bit different and therefore a little bit interesting. For now, I just want to thank you for joining me for another common core geometry lesson by E math instruction. My name is Kirk weiler, and until next time, keep thinking and keep solving problems.