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So let's see, where is CDA? CDA is this entire angle. It's this entire angle right over here. So I could put a bunch of arcs here if I want, just to show that it's different than all the other ones. So that's that angle right over there. So now we're really dealing with this larger right triangle. Let me highlight that in some.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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So I could put a bunch of arcs here if I want, just to show that it's different than all the other ones. So that's that angle right over there. So now we're really dealing with this larger right triangle. Let me highlight that in some. Let me highlight it in this pink color. So we're now dealing with this larger right triangle right over here. We care about the sine of this whole thing.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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Let me highlight that in some. Let me highlight it in this pink color. So we're now dealing with this larger right triangle right over here. We care about the sine of this whole thing. Remember, sine is opposite over hypotenuse. Sine is opposite over hypotenuse. So the opposite side is going to be side CA.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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We care about the sine of this whole thing. Remember, sine is opposite over hypotenuse. Sine is opposite over hypotenuse. So the opposite side is going to be side CA. So this is going to be equal to the length of CA over the hypotenuse, which is AD. So that is going to be over AD. Now once again, we don't see that as a choice here.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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So the opposite side is going to be side CA. So this is going to be equal to the length of CA over the hypotenuse, which is AD. So that is going to be over AD. Now once again, we don't see that as a choice here. But maybe we can express this ratio. Maybe this ratio is a trig function applied to one of the other angles. And they give us one of the angles.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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Now once again, we don't see that as a choice here. But maybe we can express this ratio. Maybe this ratio is a trig function applied to one of the other angles. And they give us one of the angles. They give us this angle right over here. I guess we could call this angle DAC. This is 30 degrees.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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And they give us one of the angles. They give us this angle right over here. I guess we could call this angle DAC. This is 30 degrees. So relative to this angle, what two sides are we taking the ratio of? We're taking now the ratio of, relative to this angle, the adjacent side over the hypotenuse. So this is the adjacent side over the hypotenuse.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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This is 30 degrees. So relative to this angle, what two sides are we taking the ratio of? We're taking now the ratio of, relative to this angle, the adjacent side over the hypotenuse. So this is the adjacent side over the hypotenuse. What deals with adjacent over hypotenuse? Well, cosine. So this is equal to the cosine of this angle.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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So this is the adjacent side over the hypotenuse. What deals with adjacent over hypotenuse? Well, cosine. So this is equal to the cosine of this angle. So this is equal to cosine of 30 degrees. Sine of CDA is equal to the cosine of this angle right over here. So this one is equal to this right over here.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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So this is equal to the cosine of this angle. So this is equal to cosine of 30 degrees. Sine of CDA is equal to the cosine of this angle right over here. So this one is equal to this right over here. So let me drag that in. So this one is equal to, so you can see that I just dragged it in, equal to that. And now we have one left.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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So this one is equal to this right over here. So let me drag that in. So this one is equal to, so you can see that I just dragged it in, equal to that. And now we have one left. We have one left. Home stretch. We should be getting excited.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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And now we have one left. We have one left. Home stretch. We should be getting excited. AE over EB. AE, let me use this color. Length of segment AE.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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We should be getting excited. AE over EB. AE, let me use this color. Length of segment AE. That's this length right over here. Let me make that stand out more. Let me do it in this red.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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Length of segment AE. That's this length right over here. Let me make that stand out more. Let me do it in this red. This color right over here. That's length of segment AE over length of segment EB. This is EB right over here.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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Let me do it in this red. This color right over here. That's length of segment AE over length of segment EB. This is EB right over here. This is EB. So now we are focused on this triangle, this right triangle right over here. Well, we know the measure of this angle over here.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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This is EB right over here. This is EB. So now we are focused on this triangle, this right triangle right over here. Well, we know the measure of this angle over here. We have double arcs. We have double arcs right over here. And they say this is 41 degrees.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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Well, we know the measure of this angle over here. We have double arcs. We have double arcs right over here. And they say this is 41 degrees. So we have double marks over here. And this is also going to be 41 degrees. So relative to this angle, what ratio is this?
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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And they say this is 41 degrees. So we have double marks over here. And this is also going to be 41 degrees. So relative to this angle, what ratio is this? This is the opposite over the hypotenuse. Opposite over the hypotenuse. This right over here is going to be sine of that angle, sine of 41 degrees.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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So relative to this angle, what ratio is this? This is the opposite over the hypotenuse. Opposite over the hypotenuse. This right over here is going to be sine of that angle, sine of 41 degrees. So it's equal to this first one right over there. So let's drag it. So this is going to be equal to sine of 41 degrees.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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This right over here is going to be sine of that angle, sine of 41 degrees. So it's equal to this first one right over there. So let's drag it. So this is going to be equal to sine of 41 degrees. So none of the ones actually ended up being equal to the tangent of 41 degrees. Now let's see if we actually got this right. I hope I did.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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And then we're asked, what is the sine of 32 degrees? And I encourage you to pause this video and try it on your own. And a hint is to look at this right triangle, one of the angles is already labeled 32 degrees, figure out what all of the angles are, and then use the fundamental definitions, your SOH CAH TOA definitions to see if you could figure out what sine of 32 degrees is. So I'm assuming you've given a go at it. Let's work it through now. So we know that the sum of the angles of a triangle add up to 180. Now in a right angle, one of the angles is 90 degrees.
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Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3
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So I'm assuming you've given a go at it. Let's work it through now. So we know that the sum of the angles of a triangle add up to 180. Now in a right angle, one of the angles is 90 degrees. So that means that the other two must add up to 90. These two add up to 90, plus another 90 is going to be 180 degrees. Or another way to think about it is that the other two non-right angles are going to be complementary.
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Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3
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Now in a right angle, one of the angles is 90 degrees. So that means that the other two must add up to 90. These two add up to 90, plus another 90 is going to be 180 degrees. Or another way to think about it is that the other two non-right angles are going to be complementary. So what plus 32 is equal to 90? Well, 90 minus 32 is 58. So this right over here is going to be 58 degrees.
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Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3
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Or another way to think about it is that the other two non-right angles are going to be complementary. So what plus 32 is equal to 90? Well, 90 minus 32 is 58. So this right over here is going to be 58 degrees. Well, why is that interesting? Well, we already know what the cosine of 58 degrees is equal to. But let's think about it in terms of ratios of the lengths of sides of this right triangle.
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Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3
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So this right over here is going to be 58 degrees. Well, why is that interesting? Well, we already know what the cosine of 58 degrees is equal to. But let's think about it in terms of ratios of the lengths of sides of this right triangle. Let's just write down SOH CAH TOA. So sine is opposite over hypotenuse. CAH, cosine is adjacent over hypotenuse.
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Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3
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But let's think about it in terms of ratios of the lengths of sides of this right triangle. Let's just write down SOH CAH TOA. So sine is opposite over hypotenuse. CAH, cosine is adjacent over hypotenuse. TOA, tangent is opposite over adjacent. So we could write down the cosine of 58 degrees, which we already know. If we think about it in terms of these fundamental ratios, cosine is adjacent over hypotenuse.
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Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3
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CAH, cosine is adjacent over hypotenuse. TOA, tangent is opposite over adjacent. So we could write down the cosine of 58 degrees, which we already know. If we think about it in terms of these fundamental ratios, cosine is adjacent over hypotenuse. This is the 58 degree angle. The side that is adjacent to it is, let me do it in this color, is side BC, right over here. It's one of the sides of the angle, the side of the angle that is not the hypotenuse.
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Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3
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If we think about it in terms of these fundamental ratios, cosine is adjacent over hypotenuse. This is the 58 degree angle. The side that is adjacent to it is, let me do it in this color, is side BC, right over here. It's one of the sides of the angle, the side of the angle that is not the hypotenuse. The other side, this over here, is the hypotenuse. So this is going to be the adjacent, the length of the adjacent side, BC, over the length of the hypotenuse. Over the length of the hypotenuse.
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Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3
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It's one of the sides of the angle, the side of the angle that is not the hypotenuse. The other side, this over here, is the hypotenuse. So this is going to be the adjacent, the length of the adjacent side, BC, over the length of the hypotenuse. Over the length of the hypotenuse. The length of the hypotenuse, well that is AB. Now let's think about what the sine of 32 degrees would be. So the sine of 32 degrees, well sine is opposite over hypotenuse.
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Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3
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Over the length of the hypotenuse. The length of the hypotenuse, well that is AB. Now let's think about what the sine of 32 degrees would be. So the sine of 32 degrees, well sine is opposite over hypotenuse. So now we're looking at this 32 degree angle. What side is opposite? Well it opens up onto BC.
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Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3
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So the sine of 32 degrees, well sine is opposite over hypotenuse. So now we're looking at this 32 degree angle. What side is opposite? Well it opens up onto BC. It opens up onto BC, just like that. And what's the hypotenuse? Well we've already, or the length of the hypotenuse, it's AB.
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Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3
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Well it opens up onto BC. It opens up onto BC, just like that. And what's the hypotenuse? Well we've already, or the length of the hypotenuse, it's AB. It's AB. Notice, the sine of 32 degrees is BC over AB. The cosine of 58 degrees is BC over AB.
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Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3
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Well we've already, or the length of the hypotenuse, it's AB. It's AB. Notice, the sine of 32 degrees is BC over AB. The cosine of 58 degrees is BC over AB. Or another way of thinking about it, the sine of this angle is the same thing as the cosine of this angle. So we could literally write the sine, I'm gonna do that in that pink color, the sine, the sine of 32 degrees is equal to the cosine, cosine of 58 degrees, which is roughly equal to 0.53. And this is a really, really useful property.
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Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3
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The cosine of 58 degrees is BC over AB. Or another way of thinking about it, the sine of this angle is the same thing as the cosine of this angle. So we could literally write the sine, I'm gonna do that in that pink color, the sine, the sine of 32 degrees is equal to the cosine, cosine of 58 degrees, which is roughly equal to 0.53. And this is a really, really useful property. The sine of an angle is equal to the cosine of its complement. So we could write this in general terms. We could write that the sine of some angle is equal to the cosine of its complement, is equal to the cosine of 90 minus theta.
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Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3
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And this is a really, really useful property. The sine of an angle is equal to the cosine of its complement. So we could write this in general terms. We could write that the sine of some angle is equal to the cosine of its complement, is equal to the cosine of 90 minus theta. Think about it. I could have done, I could change this entire problem. Instead of making this the sine of 32 degrees, I could make this the sine of 25 degrees.
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Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3
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We could write that the sine of some angle is equal to the cosine of its complement, is equal to the cosine of 90 minus theta. Think about it. I could have done, I could change this entire problem. Instead of making this the sine of 32 degrees, I could make this the sine of 25 degrees. And if someone gave you the cosine of what's 90 minus 25, if someone gave you the cosine of 65 degrees, then you could think about this as 25. The complement is going to be right over here. This would be 65 degrees.
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Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3
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And the simplest way of thinking about it is density is going to be some quantity per unit area. So for example, let's say that I have a football field right over here, and I have another identical football field right over here. Now they have the same area, but if I have, let's say, five people on this football field, actually six people on this football field, and I only have three people on this football field, the density of people per average unit area, or the density of people, I should say, per football field, is going to be higher in this left example. So it's always going to be quantity per area. Now with that out of the way, let's do a worked example that helps us understand this idea a little bit better. So here we're told the town of Tigersville has a population density of 13 cats per square kilometer. So they're giving us the density.
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Area density.mp3
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So it's always going to be quantity per area. Now with that out of the way, let's do a worked example that helps us understand this idea a little bit better. So here we're told the town of Tigersville has a population density of 13 cats per square kilometer. So they're giving us the density. I'll rewrite that, 13 cats, so the quantity is quantity in cats, per square kilometer. That's the density right over there. The town is shaped like a perfect isosceles trapezoid.
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Area density.mp3
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So they're giving us the density. I'll rewrite that, 13 cats, so the quantity is quantity in cats, per square kilometer. That's the density right over there. The town is shaped like a perfect isosceles trapezoid. So it looks something like this. It's a perfect isosceles trapezoid. It's gonna look something like that, with two parallel boundaries 12 kilometers apart.
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Area density.mp3
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The town is shaped like a perfect isosceles trapezoid. So it looks something like this. It's a perfect isosceles trapezoid. It's gonna look something like that, with two parallel boundaries 12 kilometers apart. So this distance right over here is 12 kilometers, one measuring eight kilometers. So this side over here is eight kilometers. The other is 16, that's the longer one over there.
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Area density.mp3
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It's gonna look something like that, with two parallel boundaries 12 kilometers apart. So this distance right over here is 12 kilometers, one measuring eight kilometers. So this side over here is eight kilometers. The other is 16, that's the longer one over there. How many cats are in Tigersville? So they give us the density here, and they give us, I think, enough information to figure out the area, and they want us to figure out how many cats we have, so what is the quantity? So pause this video and see if you can figure that out.
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Area density.mp3
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The other is 16, that's the longer one over there. How many cats are in Tigersville? So they give us the density here, and they give us, I think, enough information to figure out the area, and they want us to figure out how many cats we have, so what is the quantity? So pause this video and see if you can figure that out. Well, just as we said, the density is equal to quantity divided by area. If we multiply both sides of this equation by area, you get area times density is going to be equal to quantity. And we know the density, it's 13 cats per square kilometer, and we can figure out the area, and then just multiply the two.
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Area density.mp3
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So pause this video and see if you can figure that out. Well, just as we said, the density is equal to quantity divided by area. If we multiply both sides of this equation by area, you get area times density is going to be equal to quantity. And we know the density, it's 13 cats per square kilometer, and we can figure out the area, and then just multiply the two. So what's the area of this right over here? Well, the area of a trapezoid is going to be, let me write it here, area is going to be 12 kilometers, the height of the trapezoid, times the average of the two parallel sides, I guess you could say. So the average of those is going to be eight kilometers plus 16 kilometers over two.
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Area density.mp3
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And we know the density, it's 13 cats per square kilometer, and we can figure out the area, and then just multiply the two. So what's the area of this right over here? Well, the area of a trapezoid is going to be, let me write it here, area is going to be 12 kilometers, the height of the trapezoid, times the average of the two parallel sides, I guess you could say. So the average of those is going to be eight kilometers plus 16 kilometers over two. So this is going to be equal to 12 kilometers times eight plus 16 is 24, divided by two is 12, so times 12 kilometers. So this gives us 144 square kilometers. Now, we know we have 13 cats per square kilometer, so let me do this here in another color.
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Area density.mp3
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So the average of those is going to be eight kilometers plus 16 kilometers over two. So this is going to be equal to 12 kilometers times eight plus 16 is 24, divided by two is 12, so times 12 kilometers. So this gives us 144 square kilometers. Now, we know we have 13 cats per square kilometer, so let me do this here in another color. So if I multiply 13 cats per kilometer squared, and I multiply that times this business right over here, times 144 square kilometers, and you might also notice that the units cancel out the same way that variables might, so that cancels out with that. You're going to get 13 times 144, and then the units that you're left with is just cats. So 144 times 13, three times four is 12, three times four is 12, that gives us to 13, three times 100, 300 plus another 100 is 400.
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Area density.mp3
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Now, we know we have 13 cats per square kilometer, so let me do this here in another color. So if I multiply 13 cats per kilometer squared, and I multiply that times this business right over here, times 144 square kilometers, and you might also notice that the units cancel out the same way that variables might, so that cancels out with that. You're going to get 13 times 144, and then the units that you're left with is just cats. So 144 times 13, three times four is 12, three times four is 12, that gives us to 13, three times 100, 300 plus another 100 is 400. Now I'm just going to multiply 144 essentially by 10, which is just going to be 1440. And so if I add up all of that together, I'm going to jump down to here, I get 1872. So this is 1872 cats in total, and we are done.
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Area density.mp3
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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? 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.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3
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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. Well, if we don't remember, we can go back to SOHCAHTOA.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3
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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. 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.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3
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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. And actually, if we just wanna get the expression, explicitly solving for a, we can just multiply both sides of this equation times five.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3
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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. So now we can get our calculator out and figure out what this is to the nearest 10th.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3
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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. So this is, so a is approximately equal to 10.7.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3
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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. So we now know that this has length 10.7 approximately.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3
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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? Well, there's several ways to do it.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3
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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. We can come up with the third side.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3
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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. So this side wy is the hypotenuse.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3
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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. We could use adjacent and hypotenuse.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3
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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. So what trigonometric ratios deal with adjacent and hypotenuse?
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3
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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. And then we can try to solve for b.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3
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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. And then to solve for b, you can divide both sides by cosine of 65 degrees.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3
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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. We have to figure it out with our calculator, but this is just going to evaluate to some number.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3
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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. So let us now use our calculator to figure out the length of b.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3
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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. So b is approximately equal to, rounded 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.mp3
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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. 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.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3
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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. So I'll give you a few seconds to think about what the measure of angle w is.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3
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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. So all we need to do is, well, we could simplify the left-hand side right over here.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3
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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. And then we get angle w, if we subtract 155 from both sides, angle w is equal to 25 degrees.
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Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3
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We also know that FG is a perpendicular bisector of BC. So we've already shown that it's perpendicular, that this is a 90 degree angle, but it also bisects BC. So this length is equal to that length right over there. Then they tell us that arc AC, that's a curve on top, arc AC is part of circle B. So this is a circle centered at B. So if this is the center of the circle, this is part of that circle. It's really kind of the bottom left quarter of that circle.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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Then they tell us that arc AC, that's a curve on top, arc AC is part of circle B. So this is a circle centered at B. So if this is the center of the circle, this is part of that circle. It's really kind of the bottom left quarter of that circle. And then given that information, they want us to find what the measure of angle BED is. So what is BED? So it's BED.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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It's really kind of the bottom left quarter of that circle. And then given that information, they want us to find what the measure of angle BED is. So what is BED? So it's BED. So we need to figure out the measure of this angle right over here. And I encourage you to pause it and try it out. And you might imagine, well, you could pause it and try it without any hints.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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So it's BED. So we need to figure out the measure of this angle right over here. And I encourage you to pause it and try it out. And you might imagine, well, you could pause it and try it without any hints. And now I will give you a hint if you tried it the first time and you weren't able to do it and you should pause again after this hint, is try to draw some triangles that maybe split up this angle into a couple of different angles and it might be a little bit easier. You might be able to use some of what we know about triangles. Now, with that said, I will try to solve it.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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And you might imagine, well, you could pause it and try it without any hints. And now I will give you a hint if you tried it the first time and you weren't able to do it and you should pause again after this hint, is try to draw some triangles that maybe split up this angle into a couple of different angles and it might be a little bit easier. You might be able to use some of what we know about triangles. Now, with that said, I will try to solve it. And you should pause at any point where you think that you know exactly how to do this and try to do it yourself. So the trick here is to realize that this is a circle. And so any line that goes between B and any point on this arc is going to be equal to the radius of the circle.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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Now, with that said, I will try to solve it. And you should pause at any point where you think that you know exactly how to do this and try to do it yourself. So the trick here is to realize that this is a circle. And so any line that goes between B and any point on this arc is going to be equal to the radius of the circle. So AB is equal to the radius of the circle. BE is equal to the radius of the circle. And we can keep drawing other things that are equal to the radius of the circle.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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And so any line that goes between B and any point on this arc is going to be equal to the radius of the circle. So AB is equal to the radius of the circle. BE is equal to the radius of the circle. And we can keep drawing other things that are equal to the radius of the circle. BC is equal to the radius of the circle. So let's think about it a little bit. If we were to draw, and a lot of trickier geometry problems really all revolve around drawing the right lines or visualizing the right triangles.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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And we can keep drawing other things that are equal to the radius of the circle. BC is equal to the radius of the circle. So let's think about it a little bit. If we were to draw, and a lot of trickier geometry problems really all revolve around drawing the right lines or visualizing the right triangles. And I'll do one right here that might open up a lot for you in terms of thinking about how to do this problem. So let me draw segment EC. I'll draw that as straight as possible.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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If we were to draw, and a lot of trickier geometry problems really all revolve around drawing the right lines or visualizing the right triangles. And I'll do one right here that might open up a lot for you in terms of thinking about how to do this problem. So let me draw segment EC. I'll draw that as straight as possible. I can draw a better job of that. So segment EC. Now something becomes interesting.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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I'll draw that as straight as possible. I can draw a better job of that. So segment EC. Now something becomes interesting. Because what is the relationship between triangle EBG and triangle ECG? Well, they both definitely share this side right over here. They both share side EG.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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Now something becomes interesting. Because what is the relationship between triangle EBG and triangle ECG? Well, they both definitely share this side right over here. They both share side EG. And then BG is equal to GC. And they both have 90 degree angles. You have a 90 degree angle here.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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They both share side EG. And then BG is equal to GC. And they both have 90 degree angles. You have a 90 degree angle here. You have a 90 degree angle there. So you see by side, angle, side, side, angle, side, these two triangles are going to be congruent. So we know that triangle EBG is congruent to triangle ECG by side, angle, side, congruency.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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You have a 90 degree angle here. You have a 90 degree angle there. So you see by side, angle, side, side, angle, side, these two triangles are going to be congruent. So we know that triangle EBG is congruent to triangle ECG by side, angle, side, congruency. And that also tells us that all of the corresponding angles and sides are going to be the same. So that tells us right there that EC is equal to EB. So we know that EB is equal to EC.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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So we know that triangle EBG is congruent to triangle ECG by side, angle, side, congruency. And that also tells us that all of the corresponding angles and sides are going to be the same. So that tells us right there that EC is equal to EB. So we know that EB is equal to EC. And what also is equal to that length? Well, once again, this is the radius of the circle. BE is one radius of the circle going from the center to the arc.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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So we know that EB is equal to EC. And what also is equal to that length? Well, once again, this is the radius of the circle. BE is one radius of the circle going from the center to the arc. But so is BC. It is also a radius of the circle going from the center to the arc. So this is also equal to BC.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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BE is one radius of the circle going from the center to the arc. But so is BC. It is also a radius of the circle going from the center to the arc. So this is also equal to BC. So I could draw this out of the three things right here. I'm referring to the entire thing, not just one of the segments. All of BC.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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So this is also equal to BC. So I could draw this out of the three things right here. I'm referring to the entire thing, not just one of the segments. All of BC. So what kind of a triangle is this right over here? Triangle BEC is equilateral. And we know that because all three sides are equal.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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All of BC. So what kind of a triangle is this right over here? Triangle BEC is equilateral. And we know that because all three sides are equal. So that tells us that all of its angles are equal. So that tells us that the measure of angle BEC, we're not done yet, but it gets us close, is 60 degrees. So the measure of angle BEC right over there is 60 degrees.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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And we know that because all three sides are equal. So that tells us that all of its angles are equal. So that tells us that the measure of angle BEC, we're not done yet, but it gets us close, is 60 degrees. So the measure of angle BEC right over there is 60 degrees. So that gives us part of the problem. BEC is part of the angle BED. If we can just figure out the measure of angle CED now, if we can figure out this angle right over here, we just add that to 60 degrees and we're done.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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So the measure of angle BEC right over there is 60 degrees. So that gives us part of the problem. BEC is part of the angle BED. If we can just figure out the measure of angle CED now, if we can figure out this angle right over here, we just add that to 60 degrees and we're done. We've figured out the entire BED. Now let's think about how we can do this right over here. So there's a couple of interesting things that we already do know.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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If we can just figure out the measure of angle CED now, if we can figure out this angle right over here, we just add that to 60 degrees and we're done. We've figured out the entire BED. Now let's think about how we can do this right over here. So there's a couple of interesting things that we already do know. We know that this right over here is equal to the radius of the circle. And we also know that this length down here, this is a square. We know that this length down here is the same as this length up here.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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So there's a couple of interesting things that we already do know. We know that this right over here is equal to the radius of the circle. And we also know that this length down here, this is a square. We know that this length down here is the same as this length up here. That these are the exact same length. And this is equal to the radius of the circle. We already put these three slashes here.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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We know that this length down here is the same as this length up here. That these are the exact same length. And this is equal to the radius of the circle. We already put these three slashes here. BC is the same as that length. This is the same as that length. And so all four sides are going to be that same length because this is a square.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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We already put these three slashes here. BC is the same as that length. This is the same as that length. And so all four sides are going to be that same length because this is a square. So let me write this down. Because it's a square, I'll just write it this way, because it's a square, we know that CD is equal to BC, which is equal to, and we already established, which is equal to EC, which is equal to EB. But the important thing here is to realize that this and this are the same length.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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And so all four sides are going to be that same length because this is a square. So let me write this down. Because it's a square, I'll just write it this way, because it's a square, we know that CD is equal to BC, which is equal to, and we already established, which is equal to EC, which is equal to EB. But the important thing here is to realize that this and this are the same length. And the reason why that is interesting is it lets us know that this is an isosceles triangle. So whatever isosceles triangle, if you have your two legs of it, the two base angles are going to be congruent. So whatever angle this green angle is, this angle is going to be as well.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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But the important thing here is to realize that this and this are the same length. And the reason why that is interesting is it lets us know that this is an isosceles triangle. So whatever isosceles triangle, if you have your two legs of it, the two base angles are going to be congruent. So whatever angle this green angle is, this angle is going to be as well. So somehow we can figure out this angle right over here. We can subtract that from 180 and then divide by 2 to figure out these two because we know that they're the same. So how can we figure out this angle?
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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So whatever angle this green angle is, this angle is going to be as well. So somehow we can figure out this angle right over here. We can subtract that from 180 and then divide by 2 to figure out these two because we know that they're the same. So how can we figure out this angle? Well, we know all of the angles of this thing up. We can figure out the angles of all of this larger one up here. We know this is an equilateral triangle.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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So how can we figure out this angle? Well, we know all of the angles of this thing up. We can figure out the angles of all of this larger one up here. We know this is an equilateral triangle. So this over here has to be 60 degrees as well. That's 60 degrees, and that is also 60 degrees. In fact, I could write it over here, which is equal to the measure of angle BCE.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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We know this is an equilateral triangle. So this over here has to be 60 degrees as well. That's 60 degrees, and that is also 60 degrees. In fact, I could write it over here, which is equal to the measure of angle BCE. Measure of angle BCE. So if this is 60 degrees, we know we're dealing with a square, so this whole angle over here is a right angle. What is the measure of angle ECD?
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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In fact, I could write it over here, which is equal to the measure of angle BCE. Measure of angle BCE. So if this is 60 degrees, we know we're dealing with a square, so this whole angle over here is a right angle. What is the measure of angle ECD? What is this angle right over here? Let me do this in a new color. This right over here is going to have to be 30 degrees.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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What is the measure of angle ECD? What is this angle right over here? Let me do this in a new color. This right over here is going to have to be 30 degrees. So that is going to be 30 degrees. And now we're ready to solve for these two base angles. If we call these x, and we know they have to be the same, we have x plus x plus 30 degrees is going to be equal to 180 degrees.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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This right over here is going to have to be 30 degrees. So that is going to be 30 degrees. And now we're ready to solve for these two base angles. If we call these x, and we know they have to be the same, we have x plus x plus 30 degrees is going to be equal to 180 degrees. That's the sum of all of the interior angles of a triangle. So you get 2x plus 30 is equal to 180 degrees. Now you can subtract 30 from both sides, and we are left with 2x is equal to 150.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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If we call these x, and we know they have to be the same, we have x plus x plus 30 degrees is going to be equal to 180 degrees. That's the sum of all of the interior angles of a triangle. So you get 2x plus 30 is equal to 180 degrees. Now you can subtract 30 from both sides, and we are left with 2x is equal to 150. Divide both sides by 2, you get x is equal to 75. So we've figured out that x is equal to 75, and now we're at the home stretch. We need to figure out angle BED.
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Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3
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