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An expedition was sent to find how high the water had risen. The people measured the edge of the pyramid that's above the water and found it was 72 meters long. So this distance right over here is 72 meters. They knew that the entire length of the edge is 180 meters when it's not flooded, so this entire length is 180 meters. They also knew that the vertical height of the pyramid is 139 meters, so this is 139 meters. What is the level of the water above the ground? So the ground is right over here at the base of the pyramid, and so they want the level of the water above the ground.
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How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3
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They knew that the entire length of the edge is 180 meters when it's not flooded, so this entire length is 180 meters. They also knew that the vertical height of the pyramid is 139 meters, so this is 139 meters. What is the level of the water above the ground? So the ground is right over here at the base of the pyramid, and so they want the level of the water above the ground. So that's this height right over here, so let's just call that h. We want to figure out what h is. Round your answer, if necessary, to two decimal places. So what do we know and what do we not know?
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How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3
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So the ground is right over here at the base of the pyramid, and so they want the level of the water above the ground. So that's this height right over here, so let's just call that h. We want to figure out what h is. Round your answer, if necessary, to two decimal places. So what do we know and what do we not know? So they've labeled this little angle here theta, and this, of course, is a right angle. So this angle here at the base of the pyramid, this is going to be the complement of theta. It's going to be 90 degrees minus theta.
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How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3
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So what do we know and what do we not know? So they've labeled this little angle here theta, and this, of course, is a right angle. So this angle here at the base of the pyramid, this is going to be the complement of theta. It's going to be 90 degrees minus theta. And using that information, we can also figure out that this angle up here is also going to be theta. If that looks a little bit strange to you, let me just draw it out here and make it a little bit clearer. If we have a right triangle where this angle right over here is 90 minus theta, and we wanted to figure out what this is up here, let's say this is x, well, we could say x plus 90 minus theta plus 90 degrees is going to be equal to, well, the sum of the angles of a triangle are going to be 180 degrees.
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How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3
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It's going to be 90 degrees minus theta. And using that information, we can also figure out that this angle up here is also going to be theta. If that looks a little bit strange to you, let me just draw it out here and make it a little bit clearer. If we have a right triangle where this angle right over here is 90 minus theta, and we wanted to figure out what this is up here, let's say this is x, well, we could say x plus 90 minus theta plus 90 degrees is going to be equal to, well, the sum of the angles of a triangle are going to be 180 degrees. Well, if we subtract 180 from both sides, so that's 180 from the left, 180 from the right, we get x minus theta is equal to 0, or if you add theta to both sides, x is equal to theta. So this thing up here is going to be theta as well. So this is also going to be theta.
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How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3
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If we have a right triangle where this angle right over here is 90 minus theta, and we wanted to figure out what this is up here, let's say this is x, well, we could say x plus 90 minus theta plus 90 degrees is going to be equal to, well, the sum of the angles of a triangle are going to be 180 degrees. Well, if we subtract 180 from both sides, so that's 180 from the left, 180 from the right, we get x minus theta is equal to 0, or if you add theta to both sides, x is equal to theta. So this thing up here is going to be theta as well. So this is also going to be theta. And what else do we know? Well, we know that this is 72. We know that the whole thing is 180.
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How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3
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So this is also going to be theta. And what else do we know? Well, we know that this is 72. We know that the whole thing is 180. So this is 72 and the whole thing is 180. The part of this edge that's below the water, this distance right over here, let me draw it without cluttering the picture too much, doing the black color, this distance right over here is going to be 108. 108 plus 72 is 180.
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How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3
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We know that the whole thing is 180. So this is 72 and the whole thing is 180. The part of this edge that's below the water, this distance right over here, let me draw it without cluttering the picture too much, doing the black color, this distance right over here is going to be 108. 108 plus 72 is 180. So what does this do for us? We need to figure out this height. We know that this right over here is a right triangle.
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How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3
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108 plus 72 is 180. So what does this do for us? We need to figure out this height. We know that this right over here is a right triangle. This right over here is a right triangle. I could color this in just to make it a little bit clearer. This thing in yellow right over here is a right triangle.
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How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3
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We know that this right over here is a right triangle. This right over here is a right triangle. I could color this in just to make it a little bit clearer. This thing in yellow right over here is a right triangle. If we look at that right triangle and if we wanted to solve for h and solve for h using a trig ratio based on this angle theta right over here, we know that relative to this angle theta, this side of length h is an adjacent side and this length of 108 right over here along the edge, that's the hypotenuse of this yellow triangle that I just highlighted in. So which trig ratio involves an adjacent side and a hypotenuse? We just write SOH CAH TOA.
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How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3
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This thing in yellow right over here is a right triangle. If we look at that right triangle and if we wanted to solve for h and solve for h using a trig ratio based on this angle theta right over here, we know that relative to this angle theta, this side of length h is an adjacent side and this length of 108 right over here along the edge, that's the hypotenuse of this yellow triangle that I just highlighted in. So which trig ratio involves an adjacent side and a hypotenuse? We just write SOH CAH TOA. Sine is opposite over hypotenuse. That would be this distance over the hypotenuse. Cosine is adjacent over hypotenuse.
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How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3
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We just write SOH CAH TOA. Sine is opposite over hypotenuse. That would be this distance over the hypotenuse. Cosine is adjacent over hypotenuse. We get the cosine of theta is going to be equal to the height that we care about, that's the adjacent side of this right triangle, over the length of the hypotenuse, over 108. That doesn't help us yet because we don't know what the cosine of theta is. But there's a clue here.
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How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3
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Cosine is adjacent over hypotenuse. We get the cosine of theta is going to be equal to the height that we care about, that's the adjacent side of this right triangle, over the length of the hypotenuse, over 108. That doesn't help us yet because we don't know what the cosine of theta is. But there's a clue here. Theta is also sitting up here, so maybe if we can figure out what cosine of theta is based up here, then we can solve for h. If we look at this theta, what is the cosine of theta? Now we're looking at a different right triangle. We're looking at this entire right triangle now.
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How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3
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But there's a clue here. Theta is also sitting up here, so maybe if we can figure out what cosine of theta is based up here, then we can solve for h. If we look at this theta, what is the cosine of theta? Now we're looking at a different right triangle. We're looking at this entire right triangle now. Based on that entire right triangle, what is cosine of theta? Well, cosine of theta, once again, is equal to adjacent over hypotenuse. The adjacent length is this length right over here.
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How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3
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We're looking at this entire right triangle now. Based on that entire right triangle, what is cosine of theta? Well, cosine of theta, once again, is equal to adjacent over hypotenuse. The adjacent length is this length right over here. We already know that's 139 meters, so it's going to be equal to 139 meters. And what's the length of the hypotenuse? Well, the hypotenuse is this length right over here.
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How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3
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The adjacent length is this length right over here. We already know that's 139 meters, so it's going to be equal to 139 meters. And what's the length of the hypotenuse? Well, the hypotenuse is this length right over here. It's 72 plus 108, or we already have it labeled here. It's 180. We can assume that this pyramid is an isosceles triangle.
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How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3
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Well, the hypotenuse is this length right over here. It's 72 plus 108, or we already have it labeled here. It's 180. We can assume that this pyramid is an isosceles triangle. So 180 on that side, 180 on that side. So the cosine is adjacent, 139, over the hypotenuse, which is 180. These thetas are the same theta.
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How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3
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We can assume that this pyramid is an isosceles triangle. So 180 on that side, 180 on that side. So the cosine is adjacent, 139, over the hypotenuse, which is 180. These thetas are the same theta. We just showed that. Now we have cosine of theta is h over 108. Cosine of theta is 139 over 180.
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How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3
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These thetas are the same theta. We just showed that. Now we have cosine of theta is h over 108. Cosine of theta is 139 over 180. Or we could say that h over 108, which is cosine of theta, also is equal to 139 over 180. Both of these things are equal to cosine of theta. Now to solve for h, we just multiply both sides by 108.
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How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3
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Cosine of theta is 139 over 180. Or we could say that h over 108, which is cosine of theta, also is equal to 139 over 180. Both of these things are equal to cosine of theta. Now to solve for h, we just multiply both sides by 108. So h is equal to 139 times 108 over 180. So let's get our calculator out and calculate that. So that is going to be 139 times 108 divided by 180 gets us to 83.4 meters.
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How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3
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construct a regular hexagon inscribed inside the circle. So what I'm going to do, first I'm going to draw the diameter of the circle. Actually, I'm going to go beyond the diameter of the circle. I'm just going to draw a line that goes through the center of the circle and just keeps on going. I'm going to make it flat so it goes directly through. So this one right over here goes directly through the center. Now what I'm going to do is I'm going to construct a circle that's the exact same dimensions of this circle that's already been drawn.
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Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3
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I'm just going to draw a line that goes through the center of the circle and just keeps on going. I'm going to make it flat so it goes directly through. So this one right over here goes directly through the center. Now what I'm going to do is I'm going to construct a circle that's the exact same dimensions of this circle that's already been drawn. So let me put this one right over here. Let me make it the same dimensions. Now what I'm going to do is I'm going to move it over so that this new circle intersects the center of the old circle.
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Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3
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Now what I'm going to do is I'm going to construct a circle that's the exact same dimensions of this circle that's already been drawn. So let me put this one right over here. Let me make it the same dimensions. Now what I'm going to do is I'm going to move it over so that this new circle intersects the center of the old circle. These circles are the same size. Notice this center intersects the old circle and the new circle itself intersects the center of the old circle. Now the reason why this is interesting, we already know that this distance, the distance between these two centers, this is equal to a radius.
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Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3
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Now what I'm going to do is I'm going to move it over so that this new circle intersects the center of the old circle. These circles are the same size. Notice this center intersects the old circle and the new circle itself intersects the center of the old circle. Now the reason why this is interesting, we already know that this distance, the distance between these two centers, this is equal to a radius. We also know that this distance right over here is equal to a radius. It's equal to the radius of our new circle right over here. We also know that this distance right over here is equal to a radius of our old circle and that they both have the same radius.
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Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3
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Now the reason why this is interesting, we already know that this distance, the distance between these two centers, this is equal to a radius. We also know that this distance right over here is equal to a radius. It's equal to the radius of our new circle right over here. We also know that this distance right over here is equal to a radius of our old circle and that they both have the same radius. So this is a radius between these two points. Between these two points is a radius. And then between these two points is a radius.
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Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3
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We also know that this distance right over here is equal to a radius of our old circle and that they both have the same radius. So this is a radius between these two points. Between these two points is a radius. And then between these two points is a radius. So now I have constructed an equilateral triangle. And essentially I have to just do this six times and then I'm going to have a hexagon inscribed inside the circle. Let me do it again.
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Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3
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And then between these two points is a radius. So now I have constructed an equilateral triangle. And essentially I have to just do this six times and then I'm going to have a hexagon inscribed inside the circle. Let me do it again. So go from here to here. This is a radius of my new circle, which is the same as the radius of the old circle. And I could go from here to here.
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Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3
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Let me do it again. So go from here to here. This is a radius of my new circle, which is the same as the radius of the old circle. And I could go from here to here. That's the radius of my old circle. So I have another equilateral triangle. Radius, radius, radius.
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Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3
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And I could go from here to here. That's the radius of my old circle. So I have another equilateral triangle. Radius, radius, radius. Another equilateral triangle. I just have to do this four more times. So let me go to my original, let's see, let me make sure I can, well it's actually going to be hard for me to, let me just add another circle here to do it on the other side.
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Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3
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Radius, radius, radius. Another equilateral triangle. I just have to do this four more times. So let me go to my original, let's see, let me make sure I can, well it's actually going to be hard for me to, let me just add another circle here to do it on the other side. So if I put the center of it right, I want to move it a little bit right over, I want to make it the same size. So that's close enough. And let's see, that looks pretty close.
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Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3
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So let me go to my original, let's see, let me make sure I can, well it's actually going to be hard for me to, let me just add another circle here to do it on the other side. So if I put the center of it right, I want to move it a little bit right over, I want to make it the same size. So that's close enough. And let's see, that looks pretty close. That's the same size. Now let me move it over over here. Now I want this center to be on the circle, right like that.
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Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3
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And let's see, that looks pretty close. That's the same size. Now let me move it over over here. Now I want this center to be on the circle, right like that. And now I'm ready to draw some more equilateral triangles. So add a, and really I don't have to even draw the inside of it. Now I see my six vertices for my hexagon.
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Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3
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Now I want this center to be on the circle, right like that. And now I'm ready to draw some more equilateral triangles. So add a, and really I don't have to even draw the inside of it. Now I see my six vertices for my hexagon. Here, here, here, here, here, here. And I think you're satisfied now that you could break this up into six equilateral triangles. So let's do that.
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Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3
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Now I see my six vertices for my hexagon. Here, here, here, here, here, here. And I think you're satisfied now that you could break this up into six equilateral triangles. So let's do that. So this would be the base of one of those equilateral triangles. And actually let me move these, let me move this one out of the way. I could move this one right over here because I really just care about the hexagon itself.
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Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3
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So let's do that. So this would be the base of one of those equilateral triangles. And actually let me move these, let me move this one out of the way. I could move this one right over here because I really just care about the hexagon itself. And I could move this right over here. But we know that these are all the lengths of the radius anyway. Actually I'm not even having to change the length there.
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Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3
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I could move this one right over here because I really just care about the hexagon itself. And I could move this right over here. But we know that these are all the lengths of the radius anyway. Actually I'm not even having to change the length there. Then I have to just connect one more right down here. So let me add another straight edge, connect those two points, and I would have done it. I would have constructed my, I've constructed my regular hexagon inscribed in the circle.
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Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3
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So that's labeled right over there. AB is equal to 1. And then they tell us that BE and BD trisect angle ABC. So BE and BD trisect angle ABC. So trisect means dividing it into three equal angles. So that means that this angle is equal to this angle is equal to that angle. And what they want us to figure out is what is the perimeter of triangle BED?
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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So BE and BD trisect angle ABC. So trisect means dividing it into three equal angles. So that means that this angle is equal to this angle is equal to that angle. And what they want us to figure out is what is the perimeter of triangle BED? Triangle BED. So it's kind of this middle triangle in the rectangle right over here. So at first, it seems like a pretty hard problem.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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And what they want us to figure out is what is the perimeter of triangle BED? Triangle BED. So it's kind of this middle triangle in the rectangle right over here. So at first, it seems like a pretty hard problem. Because you're like, well, what is the width of this rectangle? How can I even start on this? They've only given us one side here.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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So at first, it seems like a pretty hard problem. Because you're like, well, what is the width of this rectangle? How can I even start on this? They've only given us one side here. But they've actually given us a lot of information, given that we do know that this is a rectangle. We have four sides. And then we have four angles.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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They've only given us one side here. But they've actually given us a lot of information, given that we do know that this is a rectangle. We have four sides. And then we have four angles. And the sides are all parallel to each other. And that the angles are all 90 degrees, which is more than enough information to know that this is definitely a rectangle. And so one thing we do know is that opposite sides of a rectangle are the same length.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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And then we have four angles. And the sides are all parallel to each other. And that the angles are all 90 degrees, which is more than enough information to know that this is definitely a rectangle. And so one thing we do know is that opposite sides of a rectangle are the same length. So if this side is 1, then this side right over there is also 1. The other thing we know is that this angle is trisected. Now, we know what the measure of this angle is.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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And so one thing we do know is that opposite sides of a rectangle are the same length. So if this side is 1, then this side right over there is also 1. The other thing we know is that this angle is trisected. Now, we know what the measure of this angle is. It was a right angle. It was a 90 degree angle. So if it's divided into three equal parts, that tells us that this angle right over here is 30 degrees.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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Now, we know what the measure of this angle is. It was a right angle. It was a 90 degree angle. So if it's divided into three equal parts, that tells us that this angle right over here is 30 degrees. This angle right over here is 30 degrees. And then this angle right over here is 30 degrees. And then we see that we're dealing with a couple of 30-60-90 triangles.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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So if it's divided into three equal parts, that tells us that this angle right over here is 30 degrees. This angle right over here is 30 degrees. And then this angle right over here is 30 degrees. And then we see that we're dealing with a couple of 30-60-90 triangles. This one is 30, 90. So this other side right over here needs to be 60 degrees. So that side right over there needs to be 60 degrees.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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And then we see that we're dealing with a couple of 30-60-90 triangles. This one is 30, 90. So this other side right over here needs to be 60 degrees. So that side right over there needs to be 60 degrees. This triangle right over here, you have 30, you have 90. So this one has to be 60 degrees. They have to add up to 180.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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So that side right over there needs to be 60 degrees. This triangle right over here, you have 30, you have 90. So this one has to be 60 degrees. They have to add up to 180. 30-60-90 triangle. And you can also figure out the measures of this triangle, although it's not going to be a right triangle. But knowing what we know about 30-60-90 triangles, if we just have one side of them, we can actually figure out the other sides.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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They have to add up to 180. 30-60-90 triangle. And you can also figure out the measures of this triangle, although it's not going to be a right triangle. But knowing what we know about 30-60-90 triangles, if we just have one side of them, we can actually figure out the other sides. So for example, here we have the shortest side. We have the side opposite the 30 degree side. Now, if the 30 degree side is 1, then the 60 degree side is going to be square root of 3 times that.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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But knowing what we know about 30-60-90 triangles, if we just have one side of them, we can actually figure out the other sides. So for example, here we have the shortest side. We have the side opposite the 30 degree side. Now, if the 30 degree side is 1, then the 60 degree side is going to be square root of 3 times that. So this length right over here is going to be square root of 3. And that's pretty useful, because we now just figured out the length of the entire base of this rectangle right over there. And we just used our knowledge of 30-60-90 triangles.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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Now, if the 30 degree side is 1, then the 60 degree side is going to be square root of 3 times that. So this length right over here is going to be square root of 3. And that's pretty useful, because we now just figured out the length of the entire base of this rectangle right over there. And we just used our knowledge of 30-60-90 triangles. If you see that was a little bit mysterious how I came up with that, I encourage you to watch that video. We know that 30-60-90 triangles, their sides are in the ratio of 1 to square root of 3 to 2. So if this is 1, this is the 30 degree side, this is going to be square root of 3 times that.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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And we just used our knowledge of 30-60-90 triangles. If you see that was a little bit mysterious how I came up with that, I encourage you to watch that video. We know that 30-60-90 triangles, their sides are in the ratio of 1 to square root of 3 to 2. So if this is 1, this is the 30 degree side, this is going to be square root of 3 times that. And then the hypotenuse right over here is going to be 2 times that. So this length right over here is going to be 2 times the side right over here. So 2 times 1 is just 2.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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So if this is 1, this is the 30 degree side, this is going to be square root of 3 times that. And then the hypotenuse right over here is going to be 2 times that. So this length right over here is going to be 2 times the side right over here. So 2 times 1 is just 2. So that's pretty interesting. Let's see if we can do something similar with this side right over here. Here the 1 is not the side opposite the 30 degree side.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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So 2 times 1 is just 2. So that's pretty interesting. Let's see if we can do something similar with this side right over here. Here the 1 is not the side opposite the 30 degree side. Here the 1 is the side opposite the 60 degree side. This is the 1 opposite the 60 degree side. So once again, if we multiply this side times square root of 3, we should get this side right over here.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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Here the 1 is not the side opposite the 30 degree side. Here the 1 is the side opposite the 60 degree side. This is the 1 opposite the 60 degree side. So once again, if we multiply this side times square root of 3, we should get this side right over here. Remember, this 1, so right over here, this 1, this is the 60 degree side. So this has to be 1 square root of 3's of this side. Let me write this down.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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So once again, if we multiply this side times square root of 3, we should get this side right over here. Remember, this 1, so right over here, this 1, this is the 60 degree side. So this has to be 1 square root of 3's of this side. Let me write this down. 1 over the square root of 3. And the whole reason I was able to get this is, well, whatever this side, if I multiply it by the square root of 3, I should get this side right over here. I should get the 60 degree side, the side opposite the 60 degree angle.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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Let me write this down. 1 over the square root of 3. And the whole reason I was able to get this is, well, whatever this side, if I multiply it by the square root of 3, I should get this side right over here. I should get the 60 degree side, the side opposite the 60 degree angle. Or if I take the 60 degree side, if I divide it by the square root of 3, I should get the short side, the 30 degree side. So if I start with the 60 degree side and divide by the square root of 3, I get that right over there. And then the hypotenuse is always going to be twice the length of the side opposite the 30 degree angle.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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I should get the 60 degree side, the side opposite the 60 degree angle. Or if I take the 60 degree side, if I divide it by the square root of 3, I should get the short side, the 30 degree side. So if I start with the 60 degree side and divide by the square root of 3, I get that right over there. And then the hypotenuse is always going to be twice the length of the side opposite the 30 degree angle. So this is the side opposite the 30 degree angle. The hypotenuse is always twice that. So this is the side opposite the 30 degree angle.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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And then the hypotenuse is always going to be twice the length of the side opposite the 30 degree angle. So this is the side opposite the 30 degree angle. The hypotenuse is always twice that. So this is the side opposite the 30 degree angle. The hypotenuse is going to be twice that. It is going to be 2. over the square root of 3. So we're doing pretty good.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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So this is the side opposite the 30 degree angle. The hypotenuse is going to be twice that. It is going to be 2. over the square root of 3. So we're doing pretty good. We have to figure out the perimeter of this inner triangle right over here. We already figured out one length is 2. We figured out another length is 2 square roots of 3.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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So we're doing pretty good. We have to figure out the perimeter of this inner triangle right over here. We already figured out one length is 2. We figured out another length is 2 square roots of 3. And then all we have to really figure out is what ED is. And we can do that because we know that AD is going to be the same thing as BC. We know that this entire length, because we're dealing with a rectangle, is the square root of 3.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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We figured out another length is 2 square roots of 3. And then all we have to really figure out is what ED is. And we can do that because we know that AD is going to be the same thing as BC. We know that this entire length, because we're dealing with a rectangle, is the square root of 3. This entire length. If that entire length is square root of 3, if this part, this AE, is 1 over the square root of 3, then this length right over here, ED, is going to be square root of 3 minus 1 over the square root of 3. That length minus that length right over there.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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We know that this entire length, because we're dealing with a rectangle, is the square root of 3. This entire length. If that entire length is square root of 3, if this part, this AE, is 1 over the square root of 3, then this length right over here, ED, is going to be square root of 3 minus 1 over the square root of 3. That length minus that length right over there. And now to find the perimeter is pretty straightforward. We just have to add these things up and simplify it. So it's going to be 2.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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That length minus that length right over there. And now to find the perimeter is pretty straightforward. We just have to add these things up and simplify it. So it's going to be 2. So let me write this. Perimeter of triangle BED is equal to, this is short for perimeter. I just didn't feel like writing the whole word.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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So it's going to be 2. So let me write this. Perimeter of triangle BED is equal to, this is short for perimeter. I just didn't feel like writing the whole word. Is equal to 2 over the square root of 3 plus square root of 3 minus 1 over the square root of 3 plus 2. And now this just boils down to simplifying radicals. You could take a calculator out and get some type of decimal or approximation for it.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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I just didn't feel like writing the whole word. Is equal to 2 over the square root of 3 plus square root of 3 minus 1 over the square root of 3 plus 2. And now this just boils down to simplifying radicals. You could take a calculator out and get some type of decimal or approximation for it. Let's see, if we have 2 square roots of 3 minus 1 square root of 3, that'll leave us with 1 over the square root of 3. 2 square root of thirds, I should say. 2 over the square root of 3 minus 1 over the square root of 3 is 1 over the square root of 3.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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You could take a calculator out and get some type of decimal or approximation for it. Let's see, if we have 2 square roots of 3 minus 1 square root of 3, that'll leave us with 1 over the square root of 3. 2 square root of thirds, I should say. 2 over the square root of 3 minus 1 over the square root of 3 is 1 over the square root of 3. And then you have plus the square root of 3 plus 2. And let's see, I can rationalize this. If I multiply the numerator and the denominator by the square root of 3, this gives me the square root of 3 over 3 plus the square root of 3, which I can rewrite that as plus 3 square roots of 3 over 3.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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2 over the square root of 3 minus 1 over the square root of 3 is 1 over the square root of 3. And then you have plus the square root of 3 plus 2. And let's see, I can rationalize this. If I multiply the numerator and the denominator by the square root of 3, this gives me the square root of 3 over 3 plus the square root of 3, which I can rewrite that as plus 3 square roots of 3 over 3. I just multiplied this times 3 over 3 plus 2. And so this gives us, this is the drum roll part now. So 1 square root of 3 plus 3 square roots of 3, and all of that over 3 gives us 4 square roots of 3 over 3 plus 2.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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If I multiply the numerator and the denominator by the square root of 3, this gives me the square root of 3 over 3 plus the square root of 3, which I can rewrite that as plus 3 square roots of 3 over 3. I just multiplied this times 3 over 3 plus 2. And so this gives us, this is the drum roll part now. So 1 square root of 3 plus 3 square roots of 3, and all of that over 3 gives us 4 square roots of 3 over 3 plus 2. Or you could have put the 2 first. Some people like to write the non-irrational part before the irrational part. But we're done.
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30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3
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Sort the expressions according to their values. You can put any number of cards in a category or leave a category empty. And so we have this diagram right over here. And then we have these cards that have these expressions. And we're supposed to sort these into different buckets. So we're trying to say, well, what is the length of segment AC over the length of segment BC equal to? Which of these expressions is it equal to?
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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And then we have these cards that have these expressions. And we're supposed to sort these into different buckets. So we're trying to say, well, what is the length of segment AC over the length of segment BC equal to? Which of these expressions is it equal to? And then we should drag it into the appropriate buckets. So to figure these out, I've actually already redrawn this problem on my little, I guess you'd call it, scratch pad or blackboard, whatever you want to call it. This right over here is that same diagram blown up a little bit.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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Which of these expressions is it equal to? And then we should drag it into the appropriate buckets. So to figure these out, I've actually already redrawn this problem on my little, I guess you'd call it, scratch pad or blackboard, whatever you want to call it. This right over here is that same diagram blown up a little bit. Here are the expressions that we need to drag into things. And here are the buckets that we need to see which of these expressions are equal to which of these expressions. So let's first look at this.
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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 that same diagram blown up a little bit. Here are the expressions that we need to drag into things. And here are the buckets that we need to see which of these expressions are equal to which of these expressions. So let's first look at this. The length of segment AC over the length of segment BC. So let's think about what AC is. The length of segment AC.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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So let's first look at this. The length of segment AC over the length of segment BC. So let's think about what AC is. The length of segment AC. AC is this right over here. So it's this length right over here in purple over the length of segment BC. Over this length right over here.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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The length of segment AC. AC is this right over here. So it's this length right over here in purple over the length of segment BC. Over this length right over here. So it's the ratio of the lengths of two sides of a right triangle. This is clearly a right triangle. Triangle ABC.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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Over this length right over here. So it's the ratio of the lengths of two sides of a right triangle. This is clearly a right triangle. Triangle ABC. And I could color that in just so you know what triangle I'm talking about. Triangle ABC is this entire triangle that we could focus on. So you could imagine that it's reasonable that the ratio of two sides of a right triangle are going to be the sine of one of its angles.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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Triangle ABC. And I could color that in just so you know what triangle I'm talking about. Triangle ABC is this entire triangle that we could focus on. So you could imagine that it's reasonable that the ratio of two sides of a right triangle are going to be the sine of one of its angles. And they give us one of the angles right over here. They give us this angle right over here. You say, well, no, all they did is mark that angle.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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So you could imagine that it's reasonable that the ratio of two sides of a right triangle are going to be the sine of one of its angles. And they give us one of the angles right over here. They give us this angle right over here. You say, well, no, all they did is mark that angle. But notice, one arc is here. One arc is here. So anywhere we see only one arc, that's going to be 30 degrees.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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You say, well, no, all they did is mark that angle. But notice, one arc is here. One arc is here. So anywhere we see only one arc, that's going to be 30 degrees. So this is 30 degrees as well. You have two arcs here. That's 41 degrees.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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So anywhere we see only one arc, that's going to be 30 degrees. So this is 30 degrees as well. You have two arcs here. That's 41 degrees. Two arcs here. This is going to be congruent to that. This over here is going to be 41 degrees.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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That's 41 degrees. Two arcs here. This is going to be congruent to that. This over here is going to be 41 degrees. This is three arcs. They don't tell us how many degrees that is. But this angle with the three arcs is congruent to this angle with the three arcs right over there.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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This over here is going to be 41 degrees. This is three arcs. They don't tell us how many degrees that is. But this angle with the three arcs is congruent to this angle with the three arcs right over there. So anyway, this yellow triangle, triangle ABC, we know the measure of this angle is 30 degrees. And then they give us these two sides. So how do these sides relate to this 30 degree angle?
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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But this angle with the three arcs is congruent to this angle with the three arcs right over there. So anyway, this yellow triangle, triangle ABC, we know the measure of this angle is 30 degrees. And then they give us these two sides. So how do these sides relate to this 30 degree angle? Well, side AC is adjacent to it. It's literally one of the sides of the angle that is not the hypotenuse. So let me write that down.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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So how do these sides relate to this 30 degree angle? Well, side AC is adjacent to it. It's literally one of the sides of the angle that is not the hypotenuse. So let me write that down. This is adjacent. And what is BC? Well, BC is the hypotenuse of this right triangle.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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So let me write that down. This is adjacent. And what is BC? Well, BC is the hypotenuse of this right triangle. It's the side opposite the 90 degrees. So this is the hypotenuse. So what trig function, when applied to 30 degrees, is equal to 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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Well, BC is the hypotenuse of this right triangle. It's the side opposite the 90 degrees. So this is the hypotenuse. So what trig function, when applied to 30 degrees, is equal to the adjacent side over the hypotenuse? Let's write down SOH CAH TOA just to remind ourselves. So SOH CAH TOA, sine of an angle is opposite over a hypotenuse. Cosine of an angle is adjacent over a hypotenuse.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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So what trig function, when applied to 30 degrees, is equal to the adjacent side over the hypotenuse? Let's write down SOH CAH TOA just to remind ourselves. So SOH CAH TOA, sine of an angle is opposite over a hypotenuse. Cosine of an angle is adjacent over a hypotenuse. So cosine, let's write this down, cosine of 30 degrees is going to be equal to the length of the adjacent side, so that is AC, over the length of the hypotenuse, which is equal to BC. So this right over here is the same thing as the cosine of 30 degrees. So let's drag it in there.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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Cosine of an angle is adjacent over a hypotenuse. So cosine, let's write this down, cosine of 30 degrees is going to be equal to the length of the adjacent side, so that is AC, over the length of the hypotenuse, which is equal to BC. So this right over here is the same thing as the cosine of 30 degrees. So let's drag it in there. This is equal to the cosine of 30 degrees. Now let's look at the next one. Cosine of angle DEC. Where is DEC?
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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So let's drag it in there. This is equal to the cosine of 30 degrees. Now let's look at the next one. Cosine of angle DEC. Where is DEC? So DEC. So that's this angle right over here. I'll put four arcs here so we don't get it confused.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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Cosine of angle DEC. Where is DEC? So DEC. So that's this angle right over here. I'll put four arcs here so we don't get it confused. So this is angle DEC. So what is the cosine of DEC? Well, once again, cosine is adjacent over a hypotenuse.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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I'll put four arcs here so we don't get it confused. So this is angle DEC. So what is the cosine of DEC? Well, once again, cosine is adjacent over a hypotenuse. So cosine of angle DEC, the adjacent side to this, well, that's this right over here. You might say, well, isn't this side adjacent? Well, that side, side DE, that is the actual hypotenuse.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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Well, once again, cosine is adjacent over a hypotenuse. So cosine of angle DEC, the adjacent side to this, well, that's this right over here. You might say, well, isn't this side adjacent? Well, that side, side DE, that is the actual hypotenuse. So that's not going to be the adjacent side. So the adjacent side is E. The adjacent side is, I could call it EC. It's the length of segment EC.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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Well, that side, side DE, that is the actual hypotenuse. So that's not going to be the adjacent side. So the adjacent side is E. The adjacent side is, I could call it EC. It's the length of segment EC. And then the hypotenuse is this right over here. It's the length of the hypotenuse. The hypotenuse is side DE or ED, however you want to call it.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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It's the length of segment EC. And then the hypotenuse is this right over here. It's the length of the hypotenuse. The hypotenuse is side DE or ED, however you want to call it. And so the length of it is, we could just write it as DE. Now what is this also equal to? We don't see this choice over here.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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The hypotenuse is side DE or ED, however you want to call it. And so the length of it is, we could just write it as DE. Now what is this also equal to? We don't see this choice over here. We don't have the ratio EC over DE as one of these choices here. But what we do have is we do get one of the angles here. They give us this 41 degrees.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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We don't see this choice over here. We don't have the ratio EC over DE as one of these choices here. But what we do have is we do get one of the angles here. They give us this 41 degrees. And the ratio of this green side, the length of this green side over this orange side, what would that be in terms of if we wanted to apply a trig function to this angle? Well, relative to this angle, the green side is the opposite side. And the orange side is still the hypotenuse.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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They give us this 41 degrees. And the ratio of this green side, the length of this green side over this orange side, what would that be in terms of if we wanted to apply a trig function to this angle? Well, relative to this angle, the green side is the opposite side. And the orange side is still the hypotenuse. So relative to 41 degrees, so let's write this down. Relative to 41 degrees, this ratio is the opposite over the hypotenuse. It's 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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And the orange side is still the hypotenuse. So relative to 41 degrees, so let's write this down. Relative to 41 degrees, this ratio is the opposite over the hypotenuse. It's the cosine of this angle. But it's the sine of this angle right over here. So sine is opposite over hypotenuse. So this is equal to the sine of this angle right over here.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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It's the cosine of this angle. But it's the sine of this angle right over here. So sine is opposite over hypotenuse. So this is equal to the sine of this angle right over here. It's equal to the sine of 41 degrees. So that is this one right over here, the sine of 41 degrees. So let's drag that into the appropriate bucket.
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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 sine of this angle right over here. It's equal to the sine of 41 degrees. So that is this one right over here, the sine of 41 degrees. So let's drag that into the appropriate bucket. So let's sine of 41 degrees is the same thing as the cosine of angle DEC. Only have two left. So now we have to figure out what the sine of angle CDA is. So let's see, where is CDA?
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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So let's drag that into the appropriate bucket. So let's sine of 41 degrees is the same thing as the cosine of angle DEC. Only have two left. So now we have to figure out what the sine of angle CDA is. 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.
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Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3
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