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This side is that side right there. Since its two sides are equal, this is isosceles, so these two base angles must be the same. These two base angles must be the same. That and that must be the same, or if I were to draw it up here, that and that must be the exact same base angle. Now let me see. I already used theta. Maybe I'll use x for these angles.
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Proof Right triangles inscribed in circles High School Math Khan Academy.mp3
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That and that must be the same, or if I were to draw it up here, that and that must be the exact same base angle. Now let me see. I already used theta. Maybe I'll use x for these angles. This has to be x and that has to be x. What is x going to be equal to? x plus x plus 2 theta have to equal 180 degrees.
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Proof Right triangles inscribed in circles High School Math Khan Academy.mp3
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Maybe I'll use x for these angles. This has to be x and that has to be x. What is x going to be equal to? x plus x plus 2 theta have to equal 180 degrees. They're all in the same triangle. Let me write that down. We get x plus x plus 2 theta all have to be equal to 180 degrees.
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Proof Right triangles inscribed in circles High School Math Khan Academy.mp3
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x plus x plus 2 theta have to equal 180 degrees. They're all in the same triangle. Let me write that down. We get x plus x plus 2 theta all have to be equal to 180 degrees. We get 2x plus 2 theta is equal to 180 degrees. We get 2x is equal to 180 minus 2 theta. Divide both sides by 2, you get x is equal to 90 minus theta.
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Proof Right triangles inscribed in circles High School Math Khan Academy.mp3
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We get x plus x plus 2 theta all have to be equal to 180 degrees. We get 2x plus 2 theta is equal to 180 degrees. We get 2x is equal to 180 minus 2 theta. Divide both sides by 2, you get x is equal to 90 minus theta. x is equal to 90 minus theta. Let's see what else we could do with this. We could look at this triangle right here.
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Proof Right triangles inscribed in circles High School Math Khan Academy.mp3
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Divide both sides by 2, you get x is equal to 90 minus theta. x is equal to 90 minus theta. Let's see what else we could do with this. We could look at this triangle right here. This triangle, this side over here, also has this distance right here is also a radius of the circle. This distance over here, we've already labeled it, is a radius of a circle. Once again, this is also an isosceles triangle.
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Proof Right triangles inscribed in circles High School Math Khan Academy.mp3
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We could look at this triangle right here. This triangle, this side over here, also has this distance right here is also a radius of the circle. This distance over here, we've already labeled it, is a radius of a circle. Once again, this is also an isosceles triangle. These two sides are equal. These two base angles have to be equal. This is theta.
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Proof Right triangles inscribed in circles High School Math Khan Academy.mp3
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Once again, this is also an isosceles triangle. These two sides are equal. These two base angles have to be equal. This is theta. This is also going to be equal to theta. Actually, we use that information. We use that to actually show that first result about inscribed angles and the relation between them and central angles subtending the same arc.
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Proof Right triangles inscribed in circles High School Math Khan Academy.mp3
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This is theta. This is also going to be equal to theta. Actually, we use that information. We use that to actually show that first result about inscribed angles and the relation between them and central angles subtending the same arc. If this is theta, that's theta because this is an isosceles triangle. What is this whole angle over here? What is that whole angle over here?
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Proof Right triangles inscribed in circles High School Math Khan Academy.mp3
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We use that to actually show that first result about inscribed angles and the relation between them and central angles subtending the same arc. If this is theta, that's theta because this is an isosceles triangle. What is this whole angle over here? What is that whole angle over here? It's going to be theta plus 90 minus theta. That angle right there is going to be theta plus 90 minus theta. The thetas cancel out.
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Proof Right triangles inscribed in circles High School Math Khan Academy.mp3
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What is that whole angle over here? It's going to be theta plus 90 minus theta. That angle right there is going to be theta plus 90 minus theta. The thetas cancel out. No matter what, as long as one side of my triangle is a diameter and then the angle or the vertex of the angle opposite sits on the circumference, then this angle right here is going to be a right angle and this is going to be a right triangle. If I were to draw something random like this, if I were to just take a point right there like that and draw it just like that, this is a right angle. If I were to draw something like that and go out like that, this is a right angle.
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Proof Right triangles inscribed in circles High School Math Khan Academy.mp3
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What I want to do in this video is prove that the opposite angles of a parallelogram are congruent. So for example, we want to prove that CAB is congruent to BDC. So that that angle is equal to that angle, and that ABD, which is this angle, is congruent to DCA, which is this angle over here. And to do that, we just have to realize that we have some parallel lines and we have some transversals, and the parallel lines and the transversals actually switch roles. So let's just continue these so it looks a little bit more like transversals intersecting parallel lines. And really you could just pause it for yourself and try to prove it, because it really just comes out of alternate interior angles and corresponding angles of transversals intersecting parallel lines. So let's say that this angle right over here, let me do it in a new color since I've already used that yellow.
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Proof Opposite angles of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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And to do that, we just have to realize that we have some parallel lines and we have some transversals, and the parallel lines and the transversals actually switch roles. So let's just continue these so it looks a little bit more like transversals intersecting parallel lines. And really you could just pause it for yourself and try to prove it, because it really just comes out of alternate interior angles and corresponding angles of transversals intersecting parallel lines. So let's say that this angle right over here, let me do it in a new color since I've already used that yellow. So let's start right here with angle BDC. So angle BDC, and I'm just going to mark this up here, angle BDC right over here, it is an alternate interior angle with this angle right over here. And actually we could extend this point over here.
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Proof Opposite angles of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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So let's say that this angle right over here, let me do it in a new color since I've already used that yellow. So let's start right here with angle BDC. So angle BDC, and I'm just going to mark this up here, angle BDC right over here, it is an alternate interior angle with this angle right over here. And actually we could extend this point over here. I could call that point E if I want. So I could say angle CDB is congruent to angle EBD by alternate interior angles. This is a transversal, these two lines are parallel.
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Proof Opposite angles of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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And actually we could extend this point over here. I could call that point E if I want. So I could say angle CDB is congruent to angle EBD by alternate interior angles. This is a transversal, these two lines are parallel. AB or AE is parallel to CD. Fair enough. Now if we kind of change our thinking a little bit and instead we now view BD and AC as the parallel lines and now view AB as the transversal, then we see that angle EBD is going to be congruent to angle BAC because they are corresponding angles.
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Proof Opposite angles of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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This is a transversal, these two lines are parallel. AB or AE is parallel to CD. Fair enough. Now if we kind of change our thinking a little bit and instead we now view BD and AC as the parallel lines and now view AB as the transversal, then we see that angle EBD is going to be congruent to angle BAC because they are corresponding angles. So angle EBD is going to be congruent to angle BAC, or I could say CAB, they are corresponding angles. And so if this angle is congruent to that angle and that angle is congruent to that angle, then they are congruent to each other. So angle CDB is congruent to angle CAB.
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Proof Opposite angles of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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Now if we kind of change our thinking a little bit and instead we now view BD and AC as the parallel lines and now view AB as the transversal, then we see that angle EBD is going to be congruent to angle BAC because they are corresponding angles. So angle EBD is going to be congruent to angle BAC, or I could say CAB, they are corresponding angles. And so if this angle is congruent to that angle and that angle is congruent to that angle, then they are congruent to each other. So angle CDB is congruent to angle CAB. So we've proven this first part right over here. And then to prove that these two are congruent, we use the exact same logic. So first of all, we view this as a transversal, we view AC as a transversal of AB and CD.
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Proof Opposite angles of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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So angle CDB is congruent to angle CAB. So we've proven this first part right over here. And then to prove that these two are congruent, we use the exact same logic. So first of all, we view this as a transversal, we view AC as a transversal of AB and CD. Let me go here and let me create another point here. Let me call this point F right over here. So we know that angle ACD is going to be congruent to angle FAC because they are alternate interior angles.
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Proof Opposite angles of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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So first of all, we view this as a transversal, we view AC as a transversal of AB and CD. Let me go here and let me create another point here. Let me call this point F right over here. So we know that angle ACD is going to be congruent to angle FAC because they are alternate interior angles. And then we change our thinking a little bit and we view AC and BD as parallel lines and AB as a transversal. And then angle FAC is going to be congruent to angle ABD because they are corresponding angles. Angle F to angle ABD and they are corresponding angles.
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Proof Opposite angles of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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So we know that angle ACD is going to be congruent to angle FAC because they are alternate interior angles. And then we change our thinking a little bit and we view AC and BD as parallel lines and AB as a transversal. And then angle FAC is going to be congruent to angle ABD because they are corresponding angles. Angle F to angle ABD and they are corresponding angles. So the first time we view this as a transversal, AC is a transversal of AB and CD which are parallel lines. Now AB is a transversal and BD and AC are the parallel lines. And obviously if this is congruent to that and that is congruent to that, then these two have to be congruent to each other.
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Proof Opposite angles of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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And if we're talking about a right triangle, like the one that I've drawn here, one of them is going to be a right angle. And so we have two other angles to deal with. And what I want to explore in this video is the relationship between the sine of one of these angles and the cosine of the other, the cosine of one of these angles and the sine of the other. So to do that, let's just say that this angle, I guess we could call it angle A, let's say it's equal to theta. If this is equal to theta, if its measure is equal to theta degrees, say, what is the measure of angle B going to be? Well, the thing that will jump out at you, and we've looked at this in other problems, is the sum of the angles of a triangle are going to be 180 degrees. And this one right over here, it's a right triangle.
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Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3
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So to do that, let's just say that this angle, I guess we could call it angle A, let's say it's equal to theta. If this is equal to theta, if its measure is equal to theta degrees, say, what is the measure of angle B going to be? Well, the thing that will jump out at you, and we've looked at this in other problems, is the sum of the angles of a triangle are going to be 180 degrees. And this one right over here, it's a right triangle. So this right angle takes up 90 of those 180 degrees. So you have 90 degrees left. So these two are going to have to add up to 90 degrees.
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Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3
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And this one right over here, it's a right triangle. So this right angle takes up 90 of those 180 degrees. So you have 90 degrees left. So these two are going to have to add up to 90 degrees. This one and this one, angle A and angle B, are going to be complements of each other. They're going to be complementary. Or another way of thinking about it is B could be written as 90 minus theta.
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Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3
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So these two are going to have to add up to 90 degrees. This one and this one, angle A and angle B, are going to be complements of each other. They're going to be complementary. Or another way of thinking about it is B could be written as 90 minus theta. If you add theta to 90 minus theta, you're going to get, you're going to get, or theta to 90 degrees minus theta, you're going to get 90 degrees. Now why is this interesting? Well, let's think about, let's think about what the sine, let's think about what the sine of theta is equal to.
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Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3
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Or another way of thinking about it is B could be written as 90 minus theta. If you add theta to 90 minus theta, you're going to get, you're going to get, or theta to 90 degrees minus theta, you're going to get 90 degrees. Now why is this interesting? Well, let's think about, let's think about what the sine, let's think about what the sine of theta is equal to. Sine is opposite over hypotenuse. The opposite side is BC. So this is going to be the length of BC over the hypotenuse.
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Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3
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Well, let's think about, let's think about what the sine, let's think about what the sine of theta is equal to. Sine is opposite over hypotenuse. The opposite side is BC. So this is going to be the length of BC over the hypotenuse. The hypotenuse is side AB. So the length of BC over the length of AB. Over the length of AB.
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Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3
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So this is going to be the length of BC over the hypotenuse. The hypotenuse is side AB. So the length of BC over the length of AB. Over the length of AB. Now, what is that ratio if we were to look at this angle right over here? Well, for angle B, BC is the adjacent side, and AB is the hypotenuse. From angle B's perspective, this is the adjacent over the hypotenuse.
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Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3
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Over the length of AB. Now, what is that ratio if we were to look at this angle right over here? Well, for angle B, BC is the adjacent side, and AB is the hypotenuse. From angle B's perspective, this is the adjacent over the hypotenuse. Now what trig ratio is adjacent over hypotenuse? Well, that's cosine. So ka toa, let me write that down.
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Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3
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From angle B's perspective, this is the adjacent over the hypotenuse. Now what trig ratio is adjacent over hypotenuse? Well, that's cosine. So ka toa, let me write that down. Sine, doesn't hurt. Sine is opposite over hypotenuse. We see that right over there.
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Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3
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So ka toa, let me write that down. Sine, doesn't hurt. Sine is opposite over hypotenuse. We see that right over there. Cosine is adjacent over hypotenuse, ka, and toa. Tangent is opposite over adjacent. So from this angle's perspective, taking the length of BC, which is BC is its adjacent side, and the hypotenuse is still AB.
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Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3
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We see that right over there. Cosine is adjacent over hypotenuse, ka, and toa. Tangent is opposite over adjacent. So from this angle's perspective, taking the length of BC, which is BC is its adjacent side, and the hypotenuse is still AB. So from this angle's perspective, this is adjacent over hypotenuse. Or another way of thinking about it, it's the cosine of this angle. So that's going to be equal to the cosine of 90 degrees minus theta.
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Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3
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So from this angle's perspective, taking the length of BC, which is BC is its adjacent side, and the hypotenuse is still AB. So from this angle's perspective, this is adjacent over hypotenuse. Or another way of thinking about it, it's the cosine of this angle. So that's going to be equal to the cosine of 90 degrees minus theta. That's a pretty neat relationship. The sine of an angle is equal to the cosine of its complement. So one way to think about it, the sine of, we could just pick any arbitrary angle.
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Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3
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So that's going to be equal to the cosine of 90 degrees minus theta. That's a pretty neat relationship. The sine of an angle is equal to the cosine of its complement. So one way to think about it, the sine of, we could just pick any arbitrary angle. Let's say the sine of 60 degrees is going to be equal to the cosine of what? And I encourage you to pause the video and think about it. Well, it's going to be the cosine of 90 minus 60.
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Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3
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So one way to think about it, the sine of, we could just pick any arbitrary angle. Let's say the sine of 60 degrees is going to be equal to the cosine of what? And I encourage you to pause the video and think about it. Well, it's going to be the cosine of 90 minus 60. It's going to be the cosine of 30 degrees. 30 plus 60 is 90. And of course, you could go the other way around.
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Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3
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Well, it's going to be the cosine of 90 minus 60. It's going to be the cosine of 30 degrees. 30 plus 60 is 90. And of course, you could go the other way around. You could think about the cosine of theta is going to be equal to the adjacent side to theta, to angle A, I should say. So the adjacent side is right over here. That's AC.
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Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3
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And of course, you could go the other way around. You could think about the cosine of theta is going to be equal to the adjacent side to theta, to angle A, I should say. So the adjacent side is right over here. That's AC. So it's going to be AC over the hypotenuse, adjacent over hypotenuse. The hypotenuse is AB. But what is this ratio from angle B's point of view?
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Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3
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That's AC. So it's going to be AC over the hypotenuse, adjacent over hypotenuse. The hypotenuse is AB. But what is this ratio from angle B's point of view? Well, the sine of angle B is going to be its opposite side, AC over the hypotenuse, AB. So this right over here, from angle B's perspective, this is angle B's sine. So this is equal to the sine of 90 degrees minus theta.
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Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3
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Let's do some example problems using our newly acquired knowledge of isosceles and equilateral triangles. So over here I have a kind of a triangle within a triangle, and we need to figure out this orange angle right over here and this blue angle right over here. And we know that side AB is equal to, or segment AB is equal to segment BC, which is equal to segment CD, or we could also call that DC. So first of all we see that triangle ABC is isosceles, and because it's isosceles the two base angles are going to be congruent. This is one leg, this is the other leg right over there, so the two base angles are going to be congruent. So we know that this angle right over here is also 31 degrees. Well if we know two of the angles in a triangle we can always figure out the third angle.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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So first of all we see that triangle ABC is isosceles, and because it's isosceles the two base angles are going to be congruent. This is one leg, this is the other leg right over there, so the two base angles are going to be congruent. So we know that this angle right over here is also 31 degrees. Well if we know two of the angles in a triangle we can always figure out the third angle. They have to add up to 180 degrees. So if we call it, we could say 31 degrees plus 31 degrees plus the measure of angle ABC ABC is equal to 180 degrees. You can subtract 62, this right here is 62 degrees.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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Well if we know two of the angles in a triangle we can always figure out the third angle. They have to add up to 180 degrees. So if we call it, we could say 31 degrees plus 31 degrees plus the measure of angle ABC ABC is equal to 180 degrees. You can subtract 62, this right here is 62 degrees. You subtract 62 from both sides you get the measure of angle ABC is equal to, let's see, 180 minus 60 would be 120. You subtract another two you get 118 degrees. So this angle right over here is 118 degrees.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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You can subtract 62, this right here is 62 degrees. You subtract 62 from both sides you get the measure of angle ABC is equal to, let's see, 180 minus 60 would be 120. You subtract another two you get 118 degrees. So this angle right over here is 118 degrees. Well this angle right over here is supplementary to that 118 degrees. So that angle plus 118 is going to be equal to 180. We already know that that's 62 degrees.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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So this angle right over here is 118 degrees. Well this angle right over here is supplementary to that 118 degrees. So that angle plus 118 is going to be equal to 180. We already know that that's 62 degrees. 62 plus 118 is 180. So this right over here is 62 degrees. Now this angle is one of the base angles for triangle BCD.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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We already know that that's 62 degrees. 62 plus 118 is 180. So this right over here is 62 degrees. Now this angle is one of the base angles for triangle BCD. I didn't draw it that way but this side and this side are congruent. BC has the same length as CD. Those are the two legs of an isosceles triangle.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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Now this angle is one of the base angles for triangle BCD. I didn't draw it that way but this side and this side are congruent. BC has the same length as CD. Those are the two legs of an isosceles triangle. You can kind of imagine it was turned upside down. This is the vertex. This is one base angle.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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Those are the two legs of an isosceles triangle. You can kind of imagine it was turned upside down. This is the vertex. This is one base angle. This is the other base angle. Well the base angles are going to be congruent. They're going to be 62 degrees as well.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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This is one base angle. This is the other base angle. Well the base angles are going to be congruent. They're going to be 62 degrees as well. Then finally if you want to figure out this blue angle, the blue angle plus these two 62 degree angles are going to have to add up to 180 degrees. So you get 62 plus 62 plus the blue angle, which is the measure of angle BCD, is going to have to be equal to 180 degrees. These two characters, let's see, 62 plus 62 is 124.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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They're going to be 62 degrees as well. Then finally if you want to figure out this blue angle, the blue angle plus these two 62 degree angles are going to have to add up to 180 degrees. So you get 62 plus 62 plus the blue angle, which is the measure of angle BCD, is going to have to be equal to 180 degrees. These two characters, let's see, 62 plus 62 is 124. You subtract 124 from both sides. You get the measure of angle BCD is equal to, let's see, if you subtract 120 you get 60 and then you have to subtract another 4. So you get 56 degrees.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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These two characters, let's see, 62 plus 62 is 124. You subtract 124 from both sides. You get the measure of angle BCD is equal to, let's see, if you subtract 120 you get 60 and then you have to subtract another 4. So you get 56 degrees. So this is equal to 56 degrees. And we're done. Now we can do either of these.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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So you get 56 degrees. So this is equal to 56 degrees. And we're done. Now we can do either of these. Let's do this one right over here. So what is the measure of angle ABE? So they haven't even drawn segment BE here.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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Now we can do either of these. Let's do this one right over here. So what is the measure of angle ABE? So they haven't even drawn segment BE here. So let me draw that for us. So we have to figure out the measure of angle ABE. So we have a bunch of congruent segments here.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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So they haven't even drawn segment BE here. So let me draw that for us. So we have to figure out the measure of angle ABE. So we have a bunch of congruent segments here. In particular we see that triangle ABD, all of its sides are equal. So it's an equilateral triangle, which means all of the angles are equal. And if all of the angles are equal in a triangle, they all have to be 60 degrees.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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So we have a bunch of congruent segments here. In particular we see that triangle ABD, all of its sides are equal. So it's an equilateral triangle, which means all of the angles are equal. And if all of the angles are equal in a triangle, they all have to be 60 degrees. So all of these characters are going to be 60 degrees. Well, that's part of angle ABE, but we have to figure out this other part right over here. And to do that, we can see that we're actually dealing with an isosceles triangle kind of tipped over to the left.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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And if all of the angles are equal in a triangle, they all have to be 60 degrees. So all of these characters are going to be 60 degrees. Well, that's part of angle ABE, but we have to figure out this other part right over here. And to do that, we can see that we're actually dealing with an isosceles triangle kind of tipped over to the left. This is the vertex angle. This is one base angle. This is the other base angle.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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And to do that, we can see that we're actually dealing with an isosceles triangle kind of tipped over to the left. This is the vertex angle. This is one base angle. This is the other base angle. And the vertex angle right here is 90 degrees. And once again, we know it's isosceles because this side, segment BD is equal to segment DE. And once again, these two angles plus this angle right over here is going to have to add up to 180 degrees.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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This is the other base angle. And the vertex angle right here is 90 degrees. And once again, we know it's isosceles because this side, segment BD is equal to segment DE. And once again, these two angles plus this angle right over here is going to have to add up to 180 degrees. So if you call that an X, you get X plus X plus 90 is going to be 180 degrees. So you get 2X plus X plus 90 is going to be equal to 180 degrees. X plus X is the same thing as 2X plus 90 is equal to 180.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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And once again, these two angles plus this angle right over here is going to have to add up to 180 degrees. So if you call that an X, you get X plus X plus 90 is going to be 180 degrees. So you get 2X plus X plus 90 is going to be equal to 180 degrees. X plus X is the same thing as 2X plus 90 is equal to 180. And then we can subtract 90 from both sides. You get 2X is equal to 90. Or divide both sides by 2.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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X plus X is the same thing as 2X plus 90 is equal to 180. And then we can subtract 90 from both sides. You get 2X is equal to 90. Or divide both sides by 2. You get X is equal to 45 degrees. And then we're done because angle ABE is going to be equal to the 60 degrees plus the 45 degrees. So it's going to be this whole angle, which is what we care about.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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Or divide both sides by 2. You get X is equal to 45 degrees. And then we're done because angle ABE is going to be equal to the 60 degrees plus the 45 degrees. So it's going to be this whole angle, which is what we care about. Angle ABE is going to be 60 plus 45, which is 105 degrees. And now we have this last problem over here. This one looks a little bit simpler.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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So it's going to be this whole angle, which is what we care about. Angle ABE is going to be 60 plus 45, which is 105 degrees. And now we have this last problem over here. This one looks a little bit simpler. I have an isosceles triangle. This leg is equal to that leg. This is the vertex angle.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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This one looks a little bit simpler. I have an isosceles triangle. This leg is equal to that leg. This is the vertex angle. I have to figure out B. And the trick here is, wait, how do I figure out one side of a triangle if I only know one other side? Don't I need to know two other sides?
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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This is the vertex angle. I have to figure out B. And the trick here is, wait, how do I figure out one side of a triangle if I only know one other side? Don't I need to know two other sides? We'll do it the exact same way we just did that second part of that problem. If this is an isosceles triangle, which we know it is, then this angle is going to be equal to that angle there. And so if we call this X, then this is X as well.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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Don't I need to know two other sides? We'll do it the exact same way we just did that second part of that problem. If this is an isosceles triangle, which we know it is, then this angle is going to be equal to that angle there. And so if we call this X, then this is X as well. We get X plus X plus 36 degrees plus 36 is equal to 180. The two X's, when you add them up, you get 2X. And then I'll just skip steps here.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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And so if we call this X, then this is X as well. We get X plus X plus 36 degrees plus 36 is equal to 180. The two X's, when you add them up, you get 2X. And then I'll just skip steps here. 2X plus 36 is equal to 180. Subtract 36 from both sides. We get 2X.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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And then I'll just skip steps here. 2X plus 36 is equal to 180. Subtract 36 from both sides. We get 2X. That 2 looks a little bit funny. We get 2X is equal to 180 minus 30 is 150. And then you want to subtract another 6 from 150.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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We get 2X. That 2 looks a little bit funny. We get 2X is equal to 180 minus 30 is 150. And then you want to subtract another 6 from 150. It gets us to 144. Did I do that right? 180 minus 30 is 150.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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And then you want to subtract another 6 from 150. It gets us to 144. Did I do that right? 180 minus 30 is 150. Yep, 144. Divide both sides by 2. You get X is equal to 72 degrees.
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Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3
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We're asked to draw the line of reflection that reflects triangle ABC, so that's this blue triangle, onto triangle A prime, B prime, C prime, which is this red triangle right over here. And they give us a little line drawing tool in order to draw the line of reflection. So the way I'm gonna think about it is, well, when I just eyeball it, they look, it looks like I'm just flipped over some type of a horizontal line here. But let's see if we can actually construct a horizontal line where it does actually look like the line of reflection. So let's see, C and C prime, how far apart are they from each other? So if we go one, two, three, four, five, six down, so they are six apart. So let's see if we just put this three above C prime and three below C. Let's see if this horizontal line works as a line of reflection.
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Line of reflection example.mp3
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But let's see if we can actually construct a horizontal line where it does actually look like the line of reflection. So let's see, C and C prime, how far apart are they from each other? So if we go one, two, three, four, five, six down, so they are six apart. So let's see if we just put this three above C prime and three below C. Let's see if this horizontal line works as a line of reflection. So C, or C prime, is definitely the reflection of C across this line. C is exactly three units above it, and C prime is exactly three units below it. Let's see if it works for A and A prime.
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Line of reflection example.mp3
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So let's see if we just put this three above C prime and three below C. Let's see if this horizontal line works as a line of reflection. So C, or C prime, is definitely the reflection of C across this line. C is exactly three units above it, and C prime is exactly three units below it. Let's see if it works for A and A prime. A is one, two, three, four, five units above it. A prime is one, two, three, four, five units below it. So that's looking good.
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Line of reflection example.mp3
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Let's see if it works for A and A prime. A is one, two, three, four, five units above it. A prime is one, two, three, four, five units below it. So that's looking good. Now let's just check out B. So B, we can see it's at, the y-coordinate here is seven. This line right over here is y is equal to one.
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Line of reflection example.mp3
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So that's looking good. Now let's just check out B. So B, we can see it's at, the y-coordinate here is seven. This line right over here is y is equal to one. And so what we would have here is, let's see, this is, looks like it's six units above this line, and B prime is six units below the line. So this indeed works. We've just constructed the line of reflection that reflects the blue triangle, triangle ABC, onto triangle A prime, B prime, C prime.
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Line of reflection example.mp3
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So I'm just gonna think about how did each of these points have to be rotated to go from A to A prime or B to B prime or from C to C prime? So let's just start with A, so this is where A starts, remember we're rotating about the origin, that's why I'm drawing this line from the origin to A, and where does it get rotated to? Well, it gets rotated to right over here. So the rotation is going in the counterclockwise direction, so it's going to have a positive angle, so we can rule out these two right over here, and the key question is, is this 30 degrees or 60 degrees? And there's a bunch of ways that you could think about it. One, 60 degrees would be 2 3rds of a right angle, while 30 degrees would be 1 3rd of a right angle. A right angle would look something like this, so this looks much more like 2 3rds of a right angle, so I'll go with 60 degrees.
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Determining angle of rotation.mp3
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So the rotation is going in the counterclockwise direction, so it's going to have a positive angle, so we can rule out these two right over here, and the key question is, is this 30 degrees or 60 degrees? And there's a bunch of ways that you could think about it. One, 60 degrees would be 2 3rds of a right angle, while 30 degrees would be 1 3rd of a right angle. A right angle would look something like this, so this looks much more like 2 3rds of a right angle, so I'll go with 60 degrees. Another way to think about it is that 60 degrees is 1 3rd of 180 degrees, which this also looks like right over here. And if you do that with any of the points, you would see a similar thing. So just looking at A to A prime makes me feel good that this was a 60 degree rotation.
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Determining angle of rotation.mp3
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A right angle would look something like this, so this looks much more like 2 3rds of a right angle, so I'll go with 60 degrees. Another way to think about it is that 60 degrees is 1 3rd of 180 degrees, which this also looks like right over here. And if you do that with any of the points, you would see a similar thing. So just looking at A to A prime makes me feel good that this was a 60 degree rotation. Let's do another example. So we are told quadrilateral A prime, B prime, C prime, D prime in red here is the image of quadrilateral A, B, C, D in blue here under rotation about point Q. Determine the angle of rotation.
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Determining angle of rotation.mp3
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So just looking at A to A prime makes me feel good that this was a 60 degree rotation. Let's do another example. So we are told quadrilateral A prime, B prime, C prime, D prime in red here is the image of quadrilateral A, B, C, D in blue here under rotation about point Q. Determine the angle of rotation. So once again, pause this video and see if you can figure it out. Well, I'm gonna tackle this the same way. I don't have a coordinate plane here, but it's the same notion.
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Determining angle of rotation.mp3
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Determine the angle of rotation. So once again, pause this video and see if you can figure it out. Well, I'm gonna tackle this the same way. I don't have a coordinate plane here, but it's the same notion. I can take some initial point and then look at its image and think about, well, how much did I have to rotate it? I could do B to B prime, although this might be a little bit too close. So I'm going from B to, let me do a new color here just because this color is too close to, I'll use black.
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Determining angle of rotation.mp3
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I don't have a coordinate plane here, but it's the same notion. I can take some initial point and then look at its image and think about, well, how much did I have to rotate it? I could do B to B prime, although this might be a little bit too close. So I'm going from B to, let me do a new color here just because this color is too close to, I'll use black. So we're going from B to B prime right over here. We are going clockwise, so it's going to be a negative rotation, so we can rule that and that out. And it looks like a right angle.
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Determining angle of rotation.mp3
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So I'm going from B to, let me do a new color here just because this color is too close to, I'll use black. So we're going from B to B prime right over here. We are going clockwise, so it's going to be a negative rotation, so we can rule that and that out. And it looks like a right angle. This looks like a right angle, so I feel good about picking negative 90 degrees. We could try another point and feel good that that also meets that negative 90 degrees. Let's say D to D prime.
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Determining angle of rotation.mp3
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And it looks like a right angle. This looks like a right angle, so I feel good about picking negative 90 degrees. We could try another point and feel good that that also meets that negative 90 degrees. Let's say D to D prime. So this is where D is initially. This is where D is, and this is where D prime is. And once again, we are moving clockwise, so it's a negative rotation, and this looks like a right angle, definitely more like a right angle than a 60-degree angle, and so this would be negative 90 degrees.
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Determining angle of rotation.mp3
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We have x plus one over nine minus x is equal to 2 3rds. Pause this video and see if you can try this before we work through it together. All right, now let's work through this together. Now, the first thing that we might want to do, there's several ways that you could approach this, but the one thing I like to do is get rid of this x here in the denominator, and the easiest way I can think of doing that is by multiplying both sides of this equation by nine minus x. Now, when you do that, it's important that you then put the qualifier that the x cannot be equal to the value that would have made this denominator zero, because clearly if somehow you do all this algebraic manipulation and you got x is equal to nine, that still wouldn't be a valid solution, because if you were to substitute nine back into the original equation, you would be dividing by zero in the denominator. So let's just put that right over here. X cannot be equal to nine.
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Rational equations intro Algebra 2 Khan Academy.mp3
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Now, the first thing that we might want to do, there's several ways that you could approach this, but the one thing I like to do is get rid of this x here in the denominator, and the easiest way I can think of doing that is by multiplying both sides of this equation by nine minus x. Now, when you do that, it's important that you then put the qualifier that the x cannot be equal to the value that would have made this denominator zero, because clearly if somehow you do all this algebraic manipulation and you got x is equal to nine, that still wouldn't be a valid solution, because if you were to substitute nine back into the original equation, you would be dividing by zero in the denominator. So let's just put that right over here. X cannot be equal to nine. And so then we can safely move ahead with our algebraic manipulations. So on the left-hand side, as long as x does not equal nine, if we multiply and divide by nine minus x, they cancel out, and we'll just be left with an x plus one. And on the right-hand side, if you multiply 2 3rds times nine minus x, we get 2 3rds times nine is six, and then 2 3rds times negative x is negative 2 3rds x.
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Rational equations intro Algebra 2 Khan Academy.mp3
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X cannot be equal to nine. And so then we can safely move ahead with our algebraic manipulations. So on the left-hand side, as long as x does not equal nine, if we multiply and divide by nine minus x, they cancel out, and we'll just be left with an x plus one. And on the right-hand side, if you multiply 2 3rds times nine minus x, we get 2 3rds times nine is six, and then 2 3rds times negative x is negative 2 3rds x. And once again, let's remind ourselves that x cannot be equal to nine. And then we can get all of our x's on the same side. So let's put that on the left.
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Rational equations intro Algebra 2 Khan Academy.mp3
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And on the right-hand side, if you multiply 2 3rds times nine minus x, we get 2 3rds times nine is six, and then 2 3rds times negative x is negative 2 3rds x. And once again, let's remind ourselves that x cannot be equal to nine. And then we can get all of our x's on the same side. So let's put that on the left. So let's add 2 3rds x to both sides. So plus 2 3rds, 2 3rds x, plus 2 3rds x. And then what do we have?
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Rational equations intro Algebra 2 Khan Academy.mp3
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So let's put that on the left. So let's add 2 3rds x to both sides. So plus 2 3rds, 2 3rds x, plus 2 3rds x. And then what do we have? Well, on the left-hand side, we have one x, which is the same thing as 3 3rds x plus 2 3rds x is going to give us 5 3rds x plus one is equal to six, and then these characters cancel out. And then we can just subtract one from both sides, and we get 5 3rds x, 5 3rds x is equal to five. And then last but not least, we can multiply both sides of this equation times the reciprocal of 5 3rds, which is, of course, 3 5ths, and I'm doing that, so I just have an x isolated on the left-hand side.
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Rational equations intro Algebra 2 Khan Academy.mp3
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And then what do we have? Well, on the left-hand side, we have one x, which is the same thing as 3 3rds x plus 2 3rds x is going to give us 5 3rds x plus one is equal to six, and then these characters cancel out. And then we can just subtract one from both sides, and we get 5 3rds x, 5 3rds x is equal to five. And then last but not least, we can multiply both sides of this equation times the reciprocal of 5 3rds, which is, of course, 3 5ths, and I'm doing that, so I just have an x isolated on the left-hand side. So times 3 5ths, and we are left with 3 5ths times 5 3rds is, of course, equal to one, so we're left with x is equal to five times 3 5ths is three. And so we're feeling pretty good about x equals three, but we have to make sure that that's consistent with our original expression. Well, if we look up here, if you substitute back x equals three, you don't get a zero in the denominator.
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Rational equations intro Algebra 2 Khan Academy.mp3
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So we have a circle right over here, and the first question we'll ask ourselves is what are the coordinates of the center of that circle? Well, we can eyeball that. We can see, look, it looks like the center, it looks like the circle is centered on that point right over there, and the coordinates of that point, the x-coordinate is negative four, and the y-coordinate is negative seven. So the center of that circle would be the point negative four comma negative seven. Now let's say on top of that, someone were to tell us, someone were to tell us that this point, negative five comma negative nine, is also on the circle. So negative five comma negative nine is on the circle. So based on this information, the coordinate of the center and a point that sits on the circle, can we figure out the radius?
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Features of a circle from its graph Mathematics II High School Math Khan Academy.mp3
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So the center of that circle would be the point negative four comma negative seven. Now let's say on top of that, someone were to tell us, someone were to tell us that this point, negative five comma negative nine, is also on the circle. So negative five comma negative nine is on the circle. So based on this information, the coordinate of the center and a point that sits on the circle, can we figure out the radius? Well, the radius is just the distance between the center of the circle and any point on the circle. In fact, one of the most typical definitions of a circle is all of the points that are the same distance or that are the radius away from another point, from the, and that other point would be the center of the circle. So how do we find out the distance between these two points?
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Features of a circle from its graph Mathematics II High School Math Khan Academy.mp3
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So based on this information, the coordinate of the center and a point that sits on the circle, can we figure out the radius? Well, the radius is just the distance between the center of the circle and any point on the circle. In fact, one of the most typical definitions of a circle is all of the points that are the same distance or that are the radius away from another point, from the, and that other point would be the center of the circle. So how do we find out the distance between these two points? Between these two points, so the length of that orange line? Well, we can use the distance formula, which is essentially the Pythagorean theorem. The distance squared, so if this is, if the length of that is the distance, so we could say the distance squared is going to be equal to, is going to be equal to our change in x squared, so that right there is our change in x, I don't have to write really small, but that's our change in x, our change in x squared plus our change in y squared, our change in y squared, change in y squared.
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Features of a circle from its graph Mathematics II High School Math Khan Academy.mp3
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So how do we find out the distance between these two points? Between these two points, so the length of that orange line? Well, we can use the distance formula, which is essentially the Pythagorean theorem. The distance squared, so if this is, if the length of that is the distance, so we could say the distance squared is going to be equal to, is going to be equal to our change in x squared, so that right there is our change in x, I don't have to write really small, but that's our change in x, our change in x squared plus our change in y squared, our change in y squared, change in y squared. Now what is our change in x? Our change in x, and you can even eyeball it here, it looks like it's one, but let's verify it. We could view this point as the, it doesn't matter which one you use, the start or the end, as long as you're consistent.
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Features of a circle from its graph Mathematics II High School Math Khan Academy.mp3
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The distance squared, so if this is, if the length of that is the distance, so we could say the distance squared is going to be equal to, is going to be equal to our change in x squared, so that right there is our change in x, I don't have to write really small, but that's our change in x, our change in x squared plus our change in y squared, our change in y squared, change in y squared. Now what is our change in x? Our change in x, and you can even eyeball it here, it looks like it's one, but let's verify it. We could view this point as the, it doesn't matter which one you use, the start or the end, as long as you're consistent. So let's see, if we view this as the end, we'd say negative five, it'd be negative five, minus negative four, minus negative four, and so this would be equal to negative one. So when you go from the center to this outer point, negative five comma nine, negative five comma negative nine, you go one back in the x direction. Now the actual, this distance would just be the absolute value of that, but it doesn't matter that this is a negative because we're about to square it and so that negative sign will go away.
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Features of a circle from its graph Mathematics II High School Math Khan Academy.mp3
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We could view this point as the, it doesn't matter which one you use, the start or the end, as long as you're consistent. So let's see, if we view this as the end, we'd say negative five, it'd be negative five, minus negative four, minus negative four, and so this would be equal to negative one. So when you go from the center to this outer point, negative five comma nine, negative five comma negative nine, you go one back in the x direction. Now the actual, this distance would just be the absolute value of that, but it doesn't matter that this is a negative because we're about to square it and so that negative sign will go away. Now what is our change in y? Our change in y? Well if this is the finishing y, negative nine, minus negative seven, minus our initial y, is equal to negative two.
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Features of a circle from its graph Mathematics II High School Math Khan Academy.mp3
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Now the actual, this distance would just be the absolute value of that, but it doesn't matter that this is a negative because we're about to square it and so that negative sign will go away. Now what is our change in y? Our change in y? Well if this is the finishing y, negative nine, minus negative seven, minus our initial y, is equal to negative two. And notice, just to go from that point, that y to that y, we go to negative two. So actually we could call the length of that side as the absolute value of our change in y. And we could view this as the absolute value of our change in x.
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Features of a circle from its graph Mathematics II High School Math Khan Academy.mp3
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Well if this is the finishing y, negative nine, minus negative seven, minus our initial y, is equal to negative two. And notice, just to go from that point, that y to that y, we go to negative two. So actually we could call the length of that side as the absolute value of our change in y. And we could view this as the absolute value of our change in x. And it doesn't really matter because once we square them, the negatives go away. So our distance squared, our distance squared, I really could call this the radius squared, is going to be equal to our change in x squared. Well it's negative one squared, which is just going to be one, plus our change in y squared.
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Features of a circle from its graph Mathematics II High School Math Khan Academy.mp3
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And we could view this as the absolute value of our change in x. And it doesn't really matter because once we square them, the negatives go away. So our distance squared, our distance squared, I really could call this the radius squared, is going to be equal to our change in x squared. Well it's negative one squared, which is just going to be one, plus our change in y squared. Negative two squared is just positive four. One plus four. And so you have your distance squared is equal to five, or that the distance is equal to the square root of five.
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Features of a circle from its graph Mathematics II High School Math Khan Academy.mp3
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Well it's negative one squared, which is just going to be one, plus our change in y squared. Negative two squared is just positive four. One plus four. And so you have your distance squared is equal to five, or that the distance is equal to the square root of five. And I could have just called this variable the radius. So we could say the radius is equal to the square root of five. And we're done.
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Features of a circle from its graph Mathematics II High School Math Khan Academy.mp3
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And what I'm curious about is all of the points on my screen right over here that are exactly 2 centimeters away from A. So 2 centimeters on my screen is about that far. So clearly, if I started A and I go 2 centimeters in that direction, this point right over there is 2 centimeters from A. If I call that point point B, then I could say line segment AB is 2 centimeters. The length is 2 centimeters. Remember, this would refer to the actual line segment. I could say this looks nice.
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Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3
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If I call that point point B, then I could say line segment AB is 2 centimeters. The length is 2 centimeters. Remember, this would refer to the actual line segment. I could say this looks nice. But if I talk about its length, I would get rid of that line on top. And I would just say AB is equal to 2. If I wanted to put units, I could say 2 centimeters.
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Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3
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I could say this looks nice. But if I talk about its length, I would get rid of that line on top. And I would just say AB is equal to 2. If I wanted to put units, I could say 2 centimeters. But I'm not curious just about B. I want to think about all the points, the set of all of the points that are exactly 2 centimeters away from A. So I could go 2 centimeters in the other direction, maybe get to point C right over here. So AC is also going to be equal to 2 centimeters.
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Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3
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If I wanted to put units, I could say 2 centimeters. But I'm not curious just about B. I want to think about all the points, the set of all of the points that are exactly 2 centimeters away from A. So I could go 2 centimeters in the other direction, maybe get to point C right over here. So AC is also going to be equal to 2 centimeters. But I could go 2 centimeters in any direction. And so if I find the set of all of the points that are exactly 2 centimeters away from A, I will get a very familiar-looking shape like this. And I'm trying to draw it freehand.
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Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3
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So AC is also going to be equal to 2 centimeters. But I could go 2 centimeters in any direction. And so if I find the set of all of the points that are exactly 2 centimeters away from A, I will get a very familiar-looking shape like this. And I'm trying to draw it freehand. So I would get a shape that looks like this. And actually, let me draw it in. I don't want to make you think it's only the points where that there's white.
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Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3
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