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This doesn't have a 35, it has a 90. But triangle two here has a 35 degree angle, has a 90 degree angle, and has a 55 degree angle. And if you did the math, knowing that 35 plus 90 plus this have to add up to 180 degrees, you would see that this too has a measure of 55 degrees. And so given that all of our angle measures are the same between triangle PNM and triangle number two right over here, we know that these two are similar triangles. And so the ratios between corresponding sides are going to be the same. We could either take the ratio across triangles, or we could say the ratio within when we just look at one triangle. And so if you look at PN over MN, let me try to color code it.
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Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3
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And so given that all of our angle measures are the same between triangle PNM and triangle number two right over here, we know that these two are similar triangles. And so the ratios between corresponding sides are going to be the same. We could either take the ratio across triangles, or we could say the ratio within when we just look at one triangle. And so if you look at PN over MN, let me try to color code it. So PN right over here, that corresponds to the side that's opposite the 35 degree angle. So that would correspond to this side right over here on triangle two. And then MN, that's this that I'm coloring in this bluish color.
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Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3
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And so if you look at PN over MN, let me try to color code it. So PN right over here, that corresponds to the side that's opposite the 35 degree angle. So that would correspond to this side right over here on triangle two. And then MN, that's this that I'm coloring in this bluish color. Not so well. Probably spent more time coloring. That's opposite the 55 degree angle.
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Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3
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And then MN, that's this that I'm coloring in this bluish color. Not so well. Probably spent more time coloring. That's opposite the 55 degree angle. And so opposite the 55 degree angle would be right over there. Now, since these triangles are similar, the ratio of the length of the red side over the length of the blue side is going to be the same in either triangle. So PN, let me write it this way.
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Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3
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That's opposite the 55 degree angle. And so opposite the 55 degree angle would be right over there. Now, since these triangles are similar, the ratio of the length of the red side over the length of the blue side is going to be the same in either triangle. So PN, let me write it this way. The length of segment PN over the length of segment MN is going to be equivalent to 5.7 over 8.2. Because this ratio is going to be the same for the corresponding sides, regardless of which triangle you look at. So if you take the side that's opposite 35 degrees, that's 5.7 over 8.2.
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Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3
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So PN, let me write it this way. The length of segment PN over the length of segment MN is going to be equivalent to 5.7 over 8.2. Because this ratio is going to be the same for the corresponding sides, regardless of which triangle you look at. So if you take the side that's opposite 35 degrees, that's 5.7 over 8.2. Now, to be very clear, it doesn't mean that somehow the length of this side is 5.7 or that the length of this side is 8.2. We would only be able to make that conclusion if they were congruent. But with similarity, we know that the ratios, if we look at the ratio of the red side to the blue side on each of those triangles, that would be the same.
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Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3
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So if you take the side that's opposite 35 degrees, that's 5.7 over 8.2. Now, to be very clear, it doesn't mean that somehow the length of this side is 5.7 or that the length of this side is 8.2. We would only be able to make that conclusion if they were congruent. But with similarity, we know that the ratios, if we look at the ratio of the red side to the blue side on each of those triangles, that would be the same. And so this gives us that ratio. And let's see, 5.7 over 8.2. Which of these choices get close to that?
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Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3
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But with similarity, we know that the ratios, if we look at the ratio of the red side to the blue side on each of those triangles, that would be the same. And so this gives us that ratio. And let's see, 5.7 over 8.2. Which of these choices get close to that? Well, we could say that this is roughly, if I am approximating it, let's see, it's going to be larger than 0.57 because 8.2 is less than 10. And so we are going to rule this choice out. And 5.7 is less than 8.2, so it can't be over one.
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Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3
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Which of these choices get close to that? Well, we could say that this is roughly, if I am approximating it, let's see, it's going to be larger than 0.57 because 8.2 is less than 10. And so we are going to rule this choice out. And 5.7 is less than 8.2, so it can't be over one. And so we have to think between these two choices. Well, the simplest thing I can do is actually just try to start dividing it by hand. So 8.2 goes into 5.7 the same number of times as 82 goes into 57.
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Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3
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And 5.7 is less than 8.2, so it can't be over one. And so we have to think between these two choices. Well, the simplest thing I can do is actually just try to start dividing it by hand. So 8.2 goes into 5.7 the same number of times as 82 goes into 57. And I'll add some decimals here. So it doesn't go into 57, but how many times does 82 go into 570? I would assume it's about six times, maybe seven times, looks like.
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Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3
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So 8.2 goes into 5.7 the same number of times as 82 goes into 57. And I'll add some decimals here. So it doesn't go into 57, but how many times does 82 go into 570? I would assume it's about six times, maybe seven times, looks like. So seven times two is 14, and then seven times eight is 56, I guess it's 57. So it's actually a little less than 0.7. This got, made me go a little bit too high.
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Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3
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We're told draw the image of triangle ABC under a dilation whose center is P and scale factor is 1 4th. And what we see here is the widget on Khan Academy where we can do that. So we have this figure, this triangle, ABC, A, B, C, right over here. And what we wanna do is dilate it. So that means scaling it up or down. And the center of that dilation is this point P. So one way to think about it is, let's think about the distance between point P and each of these points, and we wanna scale it by 1 4th. So the distance is gonna be 1 4th of what it was before.
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Dilating shapes shrinking Performing transformations High school geometry Khan Academy.mp3
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And what we wanna do is dilate it. So that means scaling it up or down. And the center of that dilation is this point P. So one way to think about it is, let's think about the distance between point P and each of these points, and we wanna scale it by 1 4th. So the distance is gonna be 1 4th of what it was before. So for example, this point right over here, if we just even look diagonally from P to A, we can see that we are crossing one square, two squares, three squares, four squares. So if we have a scale factor of 1 4th, instead of crossing four squares diagonally, we would only cross one square diagonally. So I'll put the corresponding point to A right over there.
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Dilating shapes shrinking Performing transformations High school geometry Khan Academy.mp3
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So the distance is gonna be 1 4th of what it was before. So for example, this point right over here, if we just even look diagonally from P to A, we can see that we are crossing one square, two squares, three squares, four squares. So if we have a scale factor of 1 4th, instead of crossing four squares diagonally, we would only cross one square diagonally. So I'll put the corresponding point to A right over there. Now what about for point C? It's not quite as obvious, but one way we could think about it is, we could think about how far are we going horizontally from P to C, and then how far do we go vertically? So horizontally, we're going one, two, three, four, five, six, seven, eight of these units, and then vertically, we're going one, two, three, four.
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Dilating shapes shrinking Performing transformations High school geometry Khan Academy.mp3
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So I'll put the corresponding point to A right over there. Now what about for point C? It's not quite as obvious, but one way we could think about it is, we could think about how far are we going horizontally from P to C, and then how far do we go vertically? So horizontally, we're going one, two, three, four, five, six, seven, eight of these units, and then vertically, we're going one, two, three, four. So we're going to the left eight and up four. Now if we have a scale factor of 1 4th, we just multiply each of those by 1 4th. So instead of going to the left eight, we would go to the left two.
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Dilating shapes shrinking Performing transformations High school geometry Khan Academy.mp3
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So horizontally, we're going one, two, three, four, five, six, seven, eight of these units, and then vertically, we're going one, two, three, four. So we're going to the left eight and up four. Now if we have a scale factor of 1 4th, we just multiply each of those by 1 4th. So instead of going to the left eight, we would go to the left two. Eight times 1 4th is two. Instead of going up four, we would go up one. So this would be the corresponding point to point C. And then we'll do the same thing for point B.
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Dilating shapes shrinking Performing transformations High school geometry Khan Academy.mp3
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So instead of going to the left eight, we would go to the left two. Eight times 1 4th is two. Instead of going up four, we would go up one. So this would be the corresponding point to point C. And then we'll do the same thing for point B. When we go from P to B, we're going one, two, three, four, five, six, seven, eight up, and we're going four to the left. So if we have a scale factor of 1 4th, instead of going eight up, we'll go two up, and instead of going four to the left, we will go one to the left. So there you have it.
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Dilating shapes shrinking Performing transformations High school geometry Khan Academy.mp3
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We have a bunch of triangles here and some lengths of sides and a couple of right angles. Maybe we can establish similarity between some of the triangles. There's actually three different triangles that I can see here. This triangle, this triangle, and this larger triangle. If we can establish some similarity here, maybe we can use ratios between sides somehow to figure out what BC is. When you look at it, you have a right angle right over here. In triangle BDC, you have one right angle.
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Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3
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This triangle, this triangle, and this larger triangle. If we can establish some similarity here, maybe we can use ratios between sides somehow to figure out what BC is. When you look at it, you have a right angle right over here. In triangle BDC, you have one right angle. In triangle ABC, you have another right angle. If we can show that they have another angle or another corresponding set of angles that are congruent to each other, then we can show that they're similar. Actually, both of those triangles, both BDC and ABC, both share this angle right over here.
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Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3
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In triangle BDC, you have one right angle. In triangle ABC, you have another right angle. If we can show that they have another angle or another corresponding set of angles that are congruent to each other, then we can show that they're similar. Actually, both of those triangles, both BDC and ABC, both share this angle right over here. If they share that angle, then they definitely share two angles. They both share that angle right over there. Let me do that in a different color just to make it different than those right angles.
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Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3
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Actually, both of those triangles, both BDC and ABC, both share this angle right over here. If they share that angle, then they definitely share two angles. They both share that angle right over there. Let me do that in a different color just to make it different than those right angles. They both share that angle there. We know that two triangles that have at least two of their angles, that have at least two congruent angles, they are going to be similar triangles. We know that triangle, I'll write this, triangle ABC, we went from the unlabeled angle to the yellow right angle to the orange angle.
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Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3
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Let me do that in a different color just to make it different than those right angles. They both share that angle there. We know that two triangles that have at least two of their angles, that have at least two congruent angles, they are going to be similar triangles. We know that triangle, I'll write this, triangle ABC, we went from the unlabeled angle to the yellow right angle to the orange angle. Let me write it this way. We went from the unlabeled angle right over here to the orange angle, or sorry, to the yellow angle, I'm having trouble with colors, to the orange angle, ABC. We want to do this very carefully here because the same points or the same vertices might not play the same role in both triangles.
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Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3
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We know that triangle, I'll write this, triangle ABC, we went from the unlabeled angle to the yellow right angle to the orange angle. Let me write it this way. We went from the unlabeled angle right over here to the orange angle, or sorry, to the yellow angle, I'm having trouble with colors, to the orange angle, ABC. We want to do this very carefully here because the same points or the same vertices might not play the same role in both triangles. We want to make sure we're getting the similarity right. White vertex to the 90 degree angle vertex to the orange vertex, that is going to be similar to triangle. Which is the one that is neither right angle, so we're looking at the smaller triangle right over here, which is the one that is neither a right angle or the orange angle?
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Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3
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We want to do this very carefully here because the same points or the same vertices might not play the same role in both triangles. We want to make sure we're getting the similarity right. White vertex to the 90 degree angle vertex to the orange vertex, that is going to be similar to triangle. Which is the one that is neither right angle, so we're looking at the smaller triangle right over here, which is the one that is neither a right angle or the orange angle? It's going to be vertex B. Vertex B had the right angle when you think about the larger triangle, but we haven't thought about just that little angle right over there. We're going to start at vertex B, then we're going to go to the right angle. The right angle is vertex D, and then we go to vertex C, which is in orange.
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Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3
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Which is the one that is neither right angle, so we're looking at the smaller triangle right over here, which is the one that is neither a right angle or the orange angle? It's going to be vertex B. Vertex B had the right angle when you think about the larger triangle, but we haven't thought about just that little angle right over there. We're going to start at vertex B, then we're going to go to the right angle. The right angle is vertex D, and then we go to vertex C, which is in orange. We have shown that they are similar. Now that we know that they are similar, we can attempt to take ratios between the sides. Let's think about it.
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Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3
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The right angle is vertex D, and then we go to vertex C, which is in orange. We have shown that they are similar. Now that we know that they are similar, we can attempt to take ratios between the sides. Let's think about it. We know what the length of AC is. AC is going to be equal to 8, 6 plus 2. We know that AC, what's the corresponding side on this triangle right over here?
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Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3
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Let's think about it. We know what the length of AC is. AC is going to be equal to 8, 6 plus 2. We know that AC, what's the corresponding side on this triangle right over here? You can literally look at the letters. A and C is going to correspond to BC. The first and the third, first and the third.
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Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3
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We know that AC, what's the corresponding side on this triangle right over here? You can literally look at the letters. A and C is going to correspond to BC. The first and the third, first and the third. AC is going to correspond to BC. This is interesting because we're already involving BC. What is going to correspond to, and then if we look at BC on the larger triangle, so if we look at BC on the larger triangle, BC is going to correspond to what on the smaller triangle?
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Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3
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The first and the third, first and the third. AC is going to correspond to BC. This is interesting because we're already involving BC. What is going to correspond to, and then if we look at BC on the larger triangle, so if we look at BC on the larger triangle, BC is going to correspond to what on the smaller triangle? It's going to correspond to DC. It's good because we know what AC is, and we know what DC is, and so we can solve for BC. I want to take one more step to show you what we just did here because BC is playing two different roles.
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Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3
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What is going to correspond to, and then if we look at BC on the larger triangle, so if we look at BC on the larger triangle, BC is going to correspond to what on the smaller triangle? It's going to correspond to DC. It's good because we know what AC is, and we know what DC is, and so we can solve for BC. I want to take one more step to show you what we just did here because BC is playing two different roles. On this first statement right over here, we're thinking of BC. BC on our smaller triangle corresponds to AC on our larger triangle. Then in the second statement, BC on our larger triangle corresponds to DC on our smaller triangle.
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Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3
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I want to take one more step to show you what we just did here because BC is playing two different roles. On this first statement right over here, we're thinking of BC. BC on our smaller triangle corresponds to AC on our larger triangle. Then in the second statement, BC on our larger triangle corresponds to DC on our smaller triangle. In both of these cases, these are our larger triangles, and then this is from the smaller triangle right over here, corresponding sides. This is a cool problem because BC plays two different roles in both triangles. Now we have enough information to solve for BC.
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Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3
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Then in the second statement, BC on our larger triangle corresponds to DC on our smaller triangle. In both of these cases, these are our larger triangles, and then this is from the smaller triangle right over here, corresponding sides. This is a cool problem because BC plays two different roles in both triangles. Now we have enough information to solve for BC. We know that AC is equal to 9. We know that AC is equal to 8. 6 plus 2 is 8.
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Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3
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Now we have enough information to solve for BC. We know that AC is equal to 9. We know that AC is equal to 8. 6 plus 2 is 8. We know that DC is equal to 2. That's given. Now we can cross multiply.
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Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3
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6 plus 2 is 8. We know that DC is equal to 2. That's given. Now we can cross multiply. 8 times 2 is 16, is equal to BC times BC, is equal to BC squared. BC is going to be equal to the principal root of 16, which is 4. BC is equal to 4.
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Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3
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Now we can cross multiply. 8 times 2 is 16, is equal to BC times BC, is equal to BC squared. BC is going to be equal to the principal root of 16, which is 4. BC is equal to 4. BC is equal to 4, and we're done. The hardest part about this problem is just realizing that BC plays two different roles and just keeping your head straight on those two different roles. Just to make it clear, let me actually draw these two triangles separately.
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Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3
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BC is equal to 4. BC is equal to 4, and we're done. The hardest part about this problem is just realizing that BC plays two different roles and just keeping your head straight on those two different roles. Just to make it clear, let me actually draw these two triangles separately. If I drew ABC separately, it would look like this. It would look like this. This is my triangle ABC.
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Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3
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Just to make it clear, let me actually draw these two triangles separately. If I drew ABC separately, it would look like this. It would look like this. This is my triangle ABC. Then this is a right angle. This is our orange angle. We know that the length of this side right over here is 8, and we know that the length of this side when we figure it out through this problem is 4.
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Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3
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This is my triangle ABC. Then this is a right angle. This is our orange angle. We know that the length of this side right over here is 8, and we know that the length of this side when we figure it out through this problem is 4. Then if we wanted to draw BDC, we would draw it like this. BDC looks like this. This is BDC.
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Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3
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We know that the length of this side right over here is 8, and we know that the length of this side when we figure it out through this problem is 4. Then if we wanted to draw BDC, we would draw it like this. BDC looks like this. This is BDC. That's a little bit easier to visualize because this is our right angle. This is our orange angle. This is 4, and this right over here is 2.
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Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3
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This is BDC. That's a little bit easier to visualize because this is our right angle. This is our orange angle. This is 4, and this right over here is 2. I did it this way to show you that you kind of have to flip this triangle over and rotate it just to have kind of a similar orientation. Then it might make it look a little bit clearer. If you found this part confusing, I encourage you to try to flip and rotate BDC in such a way that it seems to look a lot like ABC.
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Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3
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Give the lengths to the nearest tenth. So when they say solve the right triangle, we can assume that they're saying, hey, figure out the lengths of all the sides, so whatever a is equal to, whatever b is equal to, and also what are all the angles of the right triangle? They've given two of them, we might have to figure out this third right over here. So there's multiple ways to tackle this, but we'll just go and we'll just try to tackle side xw first, try to figure out what a is, and I'll give you a hint. You can use a calculator, and using a calculator, you can use your trigonometric functions that we've looked at a good bit now. So I'll give you a few seconds to think about how to figure out what a is. Well, what do we know?
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3
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So there's multiple ways to tackle this, but we'll just go and we'll just try to tackle side xw first, try to figure out what a is, and I'll give you a hint. You can use a calculator, and using a calculator, you can use your trigonometric functions that we've looked at a good bit now. So I'll give you a few seconds to think about how to figure out what a is. Well, what do we know? We know this angle y right over here, we know the side adjacent to angle y, and length a, this is the side, that's the length of the side that is opposite, that is opposite to angle y. So what trigonometric ratio, what trigonometric ratio deals with the opposite and the adjacent? So if we're looking at angle y, relative to angle y, this is the opposite, and this right over here is the adjacent.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3
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Well, what do we know? We know this angle y right over here, we know the side adjacent to angle y, and length a, this is the side, that's the length of the side that is opposite, that is opposite to angle y. So what trigonometric ratio, what trigonometric ratio deals with the opposite and the adjacent? So if we're looking at angle y, relative to angle y, this is the opposite, and this right over here is the adjacent. Well, if we don't remember, we can go back to SOHCAHTOA. SOHCAHTOA. Sine deals with opposite and hypotenuse, cosine deals with adjacent and hypotenuse, tangent deals with opposite over adjacent, opposite over adjacent.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3
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So if we're looking at angle y, relative to angle y, this is the opposite, and this right over here is the adjacent. Well, if we don't remember, we can go back to SOHCAHTOA. SOHCAHTOA. Sine deals with opposite and hypotenuse, cosine deals with adjacent and hypotenuse, tangent deals with opposite over adjacent, opposite over adjacent. So we can say that the tangent, the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a, over the length of the adjacent side, which they gave us in the diagram, which has length, which has length five. And you might say, well, how do I figure out a? Well, we can use our calculator to evaluate what the tangent of 65 degrees are, and then we can solve for a.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3
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Sine deals with opposite and hypotenuse, cosine deals with adjacent and hypotenuse, tangent deals with opposite over adjacent, opposite over adjacent. So we can say that the tangent, the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a, over the length of the adjacent side, which they gave us in the diagram, which has length, which has length five. And you might say, well, how do I figure out a? Well, we can use our calculator to evaluate what the tangent of 65 degrees are, and then we can solve for a. And actually, if we just wanna get the expression, explicitly solving for a, we can just multiply both sides of this equation times five. So let's do that. Five times, times five, these cancel out, and we are left with, if we flip the equal around, we're left with a is equal to five times the tangent of 65, of 65 degrees.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3
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Well, we can use our calculator to evaluate what the tangent of 65 degrees are, and then we can solve for a. And actually, if we just wanna get the expression, explicitly solving for a, we can just multiply both sides of this equation times five. So let's do that. Five times, times five, these cancel out, and we are left with, if we flip the equal around, we're left with a is equal to five times the tangent of 65, of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest 10th. My handy TI-85 out, and I have five times the, not the, the tangent, no, I didn't need to press that second right over there, just the regular tangent, of 65 degrees. And I am, I get, if I round to the nearest 10th like they asked me to, I get 10.7.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3
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Five times, times five, these cancel out, and we are left with, if we flip the equal around, we're left with a is equal to five times the tangent of 65, of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest 10th. My handy TI-85 out, and I have five times the, not the, the tangent, no, I didn't need to press that second right over there, just the regular tangent, of 65 degrees. And I am, I get, if I round to the nearest 10th like they asked me to, I get 10.7. So this is, so a is approximately equal to 10.7. I say approximately, because I rounded it, I rounded it down. I didn't, this is not the exact number, but a is equal to 10.7.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3
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And I am, I get, if I round to the nearest 10th like they asked me to, I get 10.7. So this is, so a is approximately equal to 10.7. I say approximately, because I rounded it, I rounded it down. I didn't, this is not the exact number, but a is equal to 10.7. So we now know that this has length 10.7 approximately. There's several ways that we can try to tackle b, and I'll let you pick the way you want to, but then I'll just do it the way I would like to. So my next question to you is, what is the length of the side yw, or what is the value of b?
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3
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I didn't, this is not the exact number, but a is equal to 10.7. So we now know that this has length 10.7 approximately. There's several ways that we can try to tackle b, and I'll let you pick the way you want to, but then I'll just do it the way I would like to. So my next question to you is, what is the length of the side yw, or what is the value of b? Well, there's several ways to do it. This is a hypotenuse, so we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse, or we could just use the Pythagorean theorem. We know two sides of a right triangle.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3
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So my next question to you is, what is the length of the side yw, or what is the value of b? Well, there's several ways to do it. This is a hypotenuse, so we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse, or we could just use the Pythagorean theorem. We know two sides of a right triangle. We can come up with the third side. I will go with using trigonometric ratios, since that's what we've been working on a good bit. So this length b, that's the length of the hypotenuse.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3
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We know two sides of a right triangle. We can come up with the third side. I will go with using trigonometric ratios, since that's what we've been working on a good bit. So this length b, that's the length of the hypotenuse. So this side wy is the hypotenuse. And so what trigonometric ratios, and we can decide what we want to use. We could use opposite and hypotenuse.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3
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So this length b, that's the length of the hypotenuse. So this side wy is the hypotenuse. And so what trigonometric ratios, and we can decide what we want to use. We could use opposite and hypotenuse. We could use adjacent and hypotenuse. Since we know that xy is exactly five, we don't have to deal with this approximation. Let's use that side.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3
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We could use opposite and hypotenuse. We could use adjacent and hypotenuse. Since we know that xy is exactly five, we don't have to deal with this approximation. Let's use that side. So what trigonometric ratios deal with adjacent and hypotenuse? Well, we see from Sohcahtoa, cosine deals with adjacent over hypotenuse. So we could say that the cosine of 65 degrees, cosine of 65 degrees, is equal to the length of the adjacent side, which is five, over the length of the hypotenuse, which has length b.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3
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Let's use that side. So what trigonometric ratios deal with adjacent and hypotenuse? Well, we see from Sohcahtoa, cosine deals with adjacent over hypotenuse. So we could say that the cosine of 65 degrees, cosine of 65 degrees, is equal to the length of the adjacent side, which is five, over the length of the hypotenuse, which has length b. And then we can try to solve for b. You multiply both sides times b. You're left with b times cosine of 65 degrees is equal to five.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3
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So we could say that the cosine of 65 degrees, cosine of 65 degrees, is equal to the length of the adjacent side, which is five, over the length of the hypotenuse, which has length b. And then we can try to solve for b. You multiply both sides times b. You're left with b times cosine of 65 degrees is equal to five. And then to solve for b, you can divide both sides by cosine of 65 degrees. This is just a number here. So we're just dividing.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3
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You're left with b times cosine of 65 degrees is equal to five. And then to solve for b, you can divide both sides by cosine of 65 degrees. This is just a number here. So we're just dividing. We have to figure it out with our calculator, but this is just going to evaluate to some number. So we can divide both sides by that, by cosine of 65 degrees, cosine of 65 degrees. And we're left with b is equal to five over the cosine of 65 degrees.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3
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So we're just dividing. We have to figure it out with our calculator, but this is just going to evaluate to some number. So we can divide both sides by that, by cosine of 65 degrees, cosine of 65 degrees. And we're left with b is equal to five over the cosine of 65 degrees. So let us now use our calculator to figure out the length of b. Length of b is five divided by cosine of 65 degrees. And I get, if I round to the nearest tenth, 11.8.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3
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And we're left with b is equal to five over the cosine of 65 degrees. So let us now use our calculator to figure out the length of b. Length of b is five divided by cosine of 65 degrees. And I get, if I round to the nearest tenth, 11.8. So b is approximately equal to, rounded to the nearest tenth, 11.8. So b is equal to 11.8. And then we're almost done solving this right triangle.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3
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And I get, if I round to the nearest tenth, 11.8. So b is approximately equal to, rounded to the nearest tenth, 11.8. So b is equal to 11.8. And then we're almost done solving this right triangle. And you could have figured this out using the Pythagorean theorem as well, saying that five squared plus 10.7 squared should be equal to b squared. And hopefully you would get the exact same answer. And the last thing we have to figure out is the measure of angle w. Angle w right over here.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3
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And then we're almost done solving this right triangle. And you could have figured this out using the Pythagorean theorem as well, saying that five squared plus 10.7 squared should be equal to b squared. And hopefully you would get the exact same answer. And the last thing we have to figure out is the measure of angle w. Angle w right over here. So I'll give you a few seconds to think about what the measure of angle w is. Well, here we just have to remember that the sum of the angles of a triangle add up to 180 degrees. So angle w plus 65 degrees, that's this angle right up here, plus the right angle, this is a right triangle, they're going to add up to 180 degrees.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3
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And the last thing we have to figure out is the measure of angle w. Angle w right over here. So I'll give you a few seconds to think about what the measure of angle w is. Well, here we just have to remember that the sum of the angles of a triangle add up to 180 degrees. So angle w plus 65 degrees, that's this angle right up here, plus the right angle, this is a right triangle, they're going to add up to 180 degrees. So all we need to do is, well, we could simplify the left-hand side right over here. 65 plus 90 is 155. So angle w plus 155 degrees is equal to 180 degrees.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3
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What we've got over here is a triangle where all three sides have the same length, or all three sides are congruent to each other. And a triangle like this we call equilateral. This is an equilateral triangle. Now what I want to do is prove that if all three sides are the same, then we know that all three angles are going to have the same measure. So let's think how we can do this. Well, first of all, we could just look at, we know that AB is equal to AC. So let's just pretend that we don't even know that this also happens to be equal to BC.
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Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3
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Now what I want to do is prove that if all three sides are the same, then we know that all three angles are going to have the same measure. So let's think how we can do this. Well, first of all, we could just look at, we know that AB is equal to AC. So let's just pretend that we don't even know that this also happens to be equal to BC. And we know for isosceles triangles, if two legs have the same length, then the base angles have the same length. So let's write this down. We know that angle ABC is going to be congruent to angle ACB.
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Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3
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So let's just pretend that we don't even know that this also happens to be equal to BC. And we know for isosceles triangles, if two legs have the same length, then the base angles have the same length. So let's write this down. We know that angle ABC is going to be congruent to angle ACB. So let me write this down. We know angle ABC is congruent to angle ACB because, so maybe this is my statement right over here, statement, and then we have reason. And the reason here, and I'll write it in just kind of shorthand, is that they're base angles of, I guess you could say, an isosceles because we know that this side is equal to that side.
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Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3
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We know that angle ABC is going to be congruent to angle ACB. So let me write this down. We know angle ABC is congruent to angle ACB because, so maybe this is my statement right over here, statement, and then we have reason. And the reason here, and I'll write it in just kind of shorthand, is that they're base angles of, I guess you could say, an isosceles because we know that this side is equal to that side. Now obviously this is an equilateral. All of the sides are equal, but the fact that these two legs are equal show that the base angles are equal. So we say two legs equal imply base angles are going to be equal.
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Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3
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And the reason here, and I'll write it in just kind of shorthand, is that they're base angles of, I guess you could say, an isosceles because we know that this side is equal to that side. Now obviously this is an equilateral. All of the sides are equal, but the fact that these two legs are equal show that the base angles are equal. So we say two legs equal imply base angles are going to be equal. And that just comes from what we actually did in the last video with isosceles triangles. But we could also view this triangle the other way. We could also say that maybe this angle over here is the vertex angle, maybe these two are the base angles.
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Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3
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So we say two legs equal imply base angles are going to be equal. And that just comes from what we actually did in the last video with isosceles triangles. But we could also view this triangle the other way. We could also say that maybe this angle over here is the vertex angle, maybe these two are the base angles. Because then you have a situation where this side and this side are congruent to each other, and then that angle and that angle are going to be base angles. So you could say angle CAB is going to be congruent to angle ABC really for the same reason. So now looking at different legs here and different base angles, this would now be the base in this example.
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Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3
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We could also say that maybe this angle over here is the vertex angle, maybe these two are the base angles. Because then you have a situation where this side and this side are congruent to each other, and then that angle and that angle are going to be base angles. So you could say angle CAB is going to be congruent to angle ABC really for the same reason. So now looking at different legs here and different base angles, this would now be the base in this example. You could imagine turning an isosceles triangle on its side, but it's the exact same logic. So let's just review what I talked about. These two sides are equal, which imply these two base angles are equal.
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Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3
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So now looking at different legs here and different base angles, this would now be the base in this example. You could imagine turning an isosceles triangle on its side, but it's the exact same logic. So let's just review what I talked about. These two sides are equal, which imply these two base angles are equal. These two sides being equal imply these two base angles are equal. Well, if ABC is congruent to ACB and it's congruent to CAB, all of these angles are congruent to each other. So then we get angle ABC is congruent to angle ACB, which is congruent to angle CAB.
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Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3
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These two sides are equal, which imply these two base angles are equal. These two sides being equal imply these two base angles are equal. Well, if ABC is congruent to ACB and it's congruent to CAB, all of these angles are congruent to each other. So then we get angle ABC is congruent to angle ACB, which is congruent to angle CAB. And that pretty much gives us all of the angles. So if you have an equilateral triangle, it's actually an equiangular triangle as well. All of the angles are going to be the same.
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Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3
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So then we get angle ABC is congruent to angle ACB, which is congruent to angle CAB. And that pretty much gives us all of the angles. So if you have an equilateral triangle, it's actually an equiangular triangle as well. All of the angles are going to be the same. And you actually know what that measure is if you have three things that are the same. So let's call that x, x, x, and they add up to 180. You get x plus x plus x is equal to 180, or 3x is equal to 180.
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Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3
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All of the angles are going to be the same. And you actually know what that measure is if you have three things that are the same. So let's call that x, x, x, and they add up to 180. You get x plus x plus x is equal to 180, or 3x is equal to 180. Divide both sides by 3. You get x is equal to 60 degrees. So in an equilateral triangle, not only are they all the same angles, but they're all equal to exactly 60 degree angles.
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Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3
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You get x plus x plus x is equal to 180, or 3x is equal to 180. Divide both sides by 3. You get x is equal to 60 degrees. So in an equilateral triangle, not only are they all the same angles, but they're all equal to exactly 60 degree angles. Now let's think about it the other way around. Let's say I have a triangle. Let's say we've got ourselves a triangle where all of the angles are the same.
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Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3
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So in an equilateral triangle, not only are they all the same angles, but they're all equal to exactly 60 degree angles. Now let's think about it the other way around. Let's say I have a triangle. Let's say we've got ourselves a triangle where all of the angles are the same. So let's say that's point x, point y, and point z. And we know that all the angles are the same. So we know that this angle is congruent to this angle is congruent to that angle.
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Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3
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Let's say we've got ourselves a triangle where all of the angles are the same. So let's say that's point x, point y, and point z. And we know that all the angles are the same. So we know that this angle is congruent to this angle is congruent to that angle. So what we showed in the last video on isosceles triangles is that if two base angles are the same, then the corresponding legs are also going to be the same. So we know, for example, that yx is congruent to yz. We know yx is congruent to yz.
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Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3
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So we know that this angle is congruent to this angle is congruent to that angle. So what we showed in the last video on isosceles triangles is that if two base angles are the same, then the corresponding legs are also going to be the same. So we know, for example, that yx is congruent to yz. We know yx is congruent to yz. And we know that because the base angles are congruent. We also know that yz is congruent to xz by the same argument. But here we're dealing with different base angles.
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Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3
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We know yx is congruent to yz. And we know that because the base angles are congruent. We also know that yz is congruent to xz by the same argument. But here we're dealing with different base angles. So now once again you can view this as almost an isosceles triangle turned on its side. This is the vertex angle right over here. These are the two base angles.
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Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3
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But here we're dealing with different base angles. So now once again you can view this as almost an isosceles triangle turned on its side. This is the vertex angle right over here. These are the two base angles. This would be the base now. And we know that because these two base angles are congruent by the same logic. In this first case, the base angles were this angle and that angle.
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Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3
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These are the two base angles. This would be the base now. And we know that because these two base angles are congruent by the same logic. In this first case, the base angles were this angle and that angle. In the second case, the base angles are that angle and that angle. And actually let me write it down. The base angles in this first case, let me do that same magenta, are y, x, z.
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Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3
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In this first case, the base angles were this angle and that angle. In the second case, the base angles are that angle and that angle. And actually let me write it down. The base angles in this first case, let me do that same magenta, are y, x, z. So this angle y, x, z is congruent to angle y, z, x. That was in the first case. These were the base angles.
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Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3
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The base angles in this first case, let me do that same magenta, are y, x, z. So this angle y, x, z is congruent to angle y, z, x. That was in the first case. These were the base angles. So based on the proof we saw in the last video, that implies these sides are congruent. Here we have these two base angles. So here we're saying angle, let me do that in green, angle x, y, z is congruent to angle y, x, z.
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Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3
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These were the base angles. So based on the proof we saw in the last video, that implies these sides are congruent. Here we have these two base angles. So here we're saying angle, let me do that in green, angle x, y, z is congruent to angle y, x, z. And so that implies that these two guys right over here are congruent. Well there we've proved it. We said that this side, y, x, is congruent to y, z.
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Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3
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So here we're saying angle, let me do that in green, angle x, y, z is congruent to angle y, x, z. And so that implies that these two guys right over here are congruent. Well there we've proved it. We said that this side, y, x, is congruent to y, z. And we've shown that y, z is congruent to x, z. So all of the sides are congruent to each other. So once again, if you have all the angles equal, and they're going to have to be 60 degrees, then you know that all of the sides are going to be equal as well.
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Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3
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Find the coordinates of point B on segment, line segment AC, such that the ratio of AB to AC is three to five. So pause this video and see if you can figure that out. All right, now let's work through this together, and to help us visualize, let's plot these points. So first, let us plot point A, which is at negative one comma four. So negative one comma one, two, three, four. So that right over there is point A. And then let's think about point C, which is at four comma negative six.
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Dividing line segments according to ratio.mp3
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So first, let us plot point A, which is at negative one comma four. So negative one comma one, two, three, four. So that right over there is point A. And then let's think about point C, which is at four comma negative six. So one, two, three, four, comma negative six. Negative one, negative two, negative three, negative four, negative five, negative six. Just like that.
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Dividing line segments according to ratio.mp3
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And then let's think about point C, which is at four comma negative six. So one, two, three, four, comma negative six. Negative one, negative two, negative three, negative four, negative five, negative six. Just like that. And so the segment AC, I get my ruler tool out here, segment AC is going to look like that. And the ratio between the distance of A to B and A to C is three to five. Or another way to think about it is, B is going to be 3 5ths along the way from A to C. Now, the way that I think about it is, in order to be 3 5ths along the way from A to C, you have to be 3 5ths along the way in the X direction and 3 5ths along the way in the Y direction.
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Dividing line segments according to ratio.mp3
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Just like that. And so the segment AC, I get my ruler tool out here, segment AC is going to look like that. And the ratio between the distance of A to B and A to C is three to five. Or another way to think about it is, B is going to be 3 5ths along the way from A to C. Now, the way that I think about it is, in order to be 3 5ths along the way from A to C, you have to be 3 5ths along the way in the X direction and 3 5ths along the way in the Y direction. So let's think about the X direction first. We are going from X equals negative one to X equals four to go from this point to that point. Our change in X is one, two, three, four, five.
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Dividing line segments according to ratio.mp3
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Or another way to think about it is, B is going to be 3 5ths along the way from A to C. Now, the way that I think about it is, in order to be 3 5ths along the way from A to C, you have to be 3 5ths along the way in the X direction and 3 5ths along the way in the Y direction. So let's think about the X direction first. We are going from X equals negative one to X equals four to go from this point to that point. Our change in X is one, two, three, four, five. And so if we wanna go 3 5ths of that, we went a total of five, 3 5ths of that is going just three. So that is going to be B's X coordinate. And then we can look on the Y coordinate side.
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Dividing line segments according to ratio.mp3
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Our change in X is one, two, three, four, five. And so if we wanna go 3 5ths of that, we went a total of five, 3 5ths of that is going just three. So that is going to be B's X coordinate. And then we can look on the Y coordinate side. To go from A to C, we are going from four to negative six. So we're going down by one, two, three, four, five, six, seven, eight, nine, 10. And so 3 5ths of 10 would be six.
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Dividing line segments according to ratio.mp3
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And then we can look on the Y coordinate side. To go from A to C, we are going from four to negative six. So we're going down by one, two, three, four, five, six, seven, eight, nine, 10. And so 3 5ths of 10 would be six. So B's coordinate is going to be one, two, three, four, five, six down. So just like that, we were able to figure out the X and the Y coordinates for point B, which would be right over here. And you could look at this directly and say, look, B is going to have the coordinates.
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Dividing line segments according to ratio.mp3
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And so 3 5ths of 10 would be six. So B's coordinate is going to be one, two, three, four, five, six down. So just like that, we were able to figure out the X and the Y coordinates for point B, which would be right over here. And you could look at this directly and say, look, B is going to have the coordinates. This looks like this is two comma negative two, which we were able to do with the graph paper. So another way you could think about it even algebraically is the coordinates of B, we could think about it as starting with the coordinates of A. So negative one comma four.
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Dividing line segments according to ratio.mp3
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And you could look at this directly and say, look, B is going to have the coordinates. This looks like this is two comma negative two, which we were able to do with the graph paper. So another way you could think about it even algebraically is the coordinates of B, we could think about it as starting with the coordinates of A. So negative one comma four. But we're gonna move 3 5ths along the way in each of these dimensions towards C. So it's going to be plus 3 5ths times how far we've gone in the X direction. So in the X direction to go from A to C, we're going from negative one to four. And so that distance is four minus negative one.
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Dividing line segments according to ratio.mp3
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So negative one comma four. But we're gonna move 3 5ths along the way in each of these dimensions towards C. So it's going to be plus 3 5ths times how far we've gone in the X direction. So in the X direction to go from A to C, we're going from negative one to four. And so that distance is four minus negative one. And this of course is going to be equal to five. And then on the Y dimension, this is going to be our A's Y coordinate plus 3 5ths times the distance that we travel in the Y direction. And here we're going from four to negative six.
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Dividing line segments according to ratio.mp3
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And so that distance is four minus negative one. And this of course is going to be equal to five. And then on the Y dimension, this is going to be our A's Y coordinate plus 3 5ths times the distance that we travel in the Y direction. And here we're going from four to negative six. So we say negative six minus four, that is negative 10. And so the coordinates of B are going to be negative one plus 3 5ths times five is going to be plus three. And then four plus 3 5ths times negative 10.
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Dividing line segments according to ratio.mp3
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And here we're going from four to negative six. So we say negative six minus four, that is negative 10. And so the coordinates of B are going to be negative one plus 3 5ths times five is going to be plus three. And then four plus 3 5ths times negative 10. Well, 3 5ths times negative 10 is negative six. And so that gets us to comma negative two. And we are done, which is exactly what we got right over there.
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Dividing line segments according to ratio.mp3
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And they say that line B contains the point 6, negative 7. And they tell us lines A and B are perpendicular. So that means that slope of B must be negative inverse of slope of A. So what we'll do is we'll figure out the slope of A, then take the negative inverse of it, then we'll know the slope of B, then we can use this point right here to fill in the gaps and figure out B's y-intercept. So what's the slope of A? This is already in slope-intercept form. The slope of A is right there.
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Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3
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So what we'll do is we'll figure out the slope of A, then take the negative inverse of it, then we'll know the slope of B, then we can use this point right here to fill in the gaps and figure out B's y-intercept. So what's the slope of A? This is already in slope-intercept form. The slope of A is right there. It's the 2, mx plus b. So the slope here is equal to 2. So the slope of A is 2.
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Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3
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The slope of A is right there. It's the 2, mx plus b. So the slope here is equal to 2. So the slope of A is 2. What is the slope of B? So what is B's slope going to have to be? Well, it's perpendicular to A, so it's going to be the negative inverse of this.
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Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3
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So the slope of A is 2. What is the slope of B? So what is B's slope going to have to be? Well, it's perpendicular to A, so it's going to be the negative inverse of this. The inverse of 2 is 1 half. The negative inverse of that is negative 1 half. So B's slope is negative 1 half.
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Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3
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Well, it's perpendicular to A, so it's going to be the negative inverse of this. The inverse of 2 is 1 half. The negative inverse of that is negative 1 half. So B's slope is negative 1 half. So we know that B's equation has to be y is equal to its slope m times x plus some y-intercept. We still don't know what the y-intercept of B is, but we can use this information to figure it out. We know that y is equal to negative 7 when x is equal to 6, negative 1 half times 6 plus b. I just know that this is on the point.
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Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3
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