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Only if you're familiar with enough with the notation to know what the difference g(x) = f(x+1) and g(x) = f(x)+1 is. Which is pretty much exactly what the question is asking.
The second math question I couldn't have answered because I forgot how many interior angles a polygon has. Once I looked up it's (n-2)*180 degrees it's not a hard problem either.
for anyone still confused: There are four angles showing above the paper that total to 360 because the four internal angles of a four sided figure total 360. x+y = 80 so the other two angles total 280 degrees and we know they're equal because the problem told us. So the regular polygon covered by the paper has interior angles measuring 140 degrees. Then it's just a matter of figuring out which number of sides generates 140 degree angles in the (n-2)*180 function.
I mentally reconfigured the paper so that both angles were 40 degrees and multiplied by 9 to get 360. If "horizontal" is defined as 0 angles. then side 1 will be at 40 degrees off the original, side 2 at 40+40 degrees, and so on.
Of the examples subby gives, "a) pabulum" and "e) ethereal" are the ones that are clearly out of place, and "d) prosaic" is really pushing it. As for my opinion on the SAT, I dropped out of 8th grade and so never had to take it, and by now my brain is probably way too damaged to think I'd ever score higher than the average 3rd grade zombie.
Yankees Team Gynecologist:II made a 2400, which is really spectacular when you consider I took the SAT in 1982, when only 1600 was generally considered possible. I answered all the questions correctly, and added a proof of Fermat's last thereom in the margin. This is why there is an essay portion of the test today.
manimal2878:YankeesLonestar:This is from the test that everyone would take (barring some ACT only people I guess). If you had taken advanced math, you'd probably also want to take the math specific SAT II test, although I wouldn't say that's a lot better.
turbidum:manimal2878: YankeesI got the math problems just fine, and I've got ten years on this mouth-breather. And I never even finished Algebra II. So much for, "Math is hard! I haven't taken a test in, like, TEN YEARS, doood!!"
And his impressions about the language section, though surely sidesplitting for his equally moronic beer buddies, is entirely irrelevant to their value for the purpose he clearly ascribes to them: evaluating his language aptitude.
Question: How many interior angles does a polygon have? Answer: A polygon has (n-2)*180 degrees, where n is the number of sides.
Question: What is the highest possible score on the SAT? Answer: 2400.
Question: When was the highest possible score of 2400 first considered achievable? Answer: In 1982, when only 1600 was generally considered possible. | 677.169 | 1 |
blog entry, I will be talking about the basics of constructions, my favorite construction, and how to do it. Before I get started, there are some key terms that I will use to describe constructions that I will define below.
- Straightedge: a ruler with no markings, used in drawing straight lines
- Compass: a geometric tool used to draw circles and arcs
- Arc: a part of a circle
Firstly, what is a construction? A construction is using a straightedge and a compass to draw a geometric figure. There are a lot steps a person must follow during a construction, and they must be very precise with measurements. However, if you follow the directions carefully, you'll find that doing them correctly will be easy. Some things that constructions are useful in creating are congruent line segments, congruent angles, perpendicular bisectors, angle bisectors, and many more things.
My favorite construction is creating the perpendicular bisector of a line segment. A perpendicular bisector is a line that intersects another line at its exact midpoint, and creates angles that are perfectly right angles. I find this one pretty cool because though the steps seem difficult, it's really not hard. I also like it because it's quicker than most of the other constructions. This is mainly because the new line you're creating doesn't have to be a specific measurement, as long as it bisects the new line. Now that I've told you what my favorite construction is, it's only fair that I explain how to do it.
How to construct the perpendicular bisector of a line segment:
- Step One: Place your compass on one of the endpoints on the line segment.
- Step Two: Adjust the compass so that it is more than half of the line segment's total length.
- Step Three: Without changing the compass size, hold the compass on one endpoint and draw an arc above the line that is close to directly above the center of the line. Then draw another arc below the line that is close to directly below the center of the line.
- Step Four: Without changing the size of the compass, move so that it's on the opposite endpoint. Repeat the process described in step three. Draw these new arcs so that the arcs above the line intersect and the arcs on the bottom intersect.
- Step Five: The arcs on either side of the line should intersect. Where they do, draw a point.
- Step Six: Using a straightedge, connect the two points you've drawn at the intersection of the arcs. This will give you a straight line that is the perpendicular bisector of the line segment. You can now truthfully say that all angles that are created with this bisector have a measure of 90 degrees, and that line you created intersects the original line segment at its midpoint.
Resources that helped me on this post:
- Mathopenref.com
- Prentice Hall Mathematics Geometry book
I hope that this post on constructions has been helpful for you! Comment below for feedback.
Question: What is the main advantage of constructing a perpendicular bisector compared to other constructions? Answer: It's quicker because the new line doesn't have to be a specific measurement, as long as it bisects the original line | 677.169 | 1 |
Evaluate without using a calculator: sin(60°) The answer is √3/2, but how do work this out? Can you give me a step by step answer?
Answers
sin(60degrees) is angle of a special triangle, 30,60,90.the hypotenuse in this triangle is 2.the longest leg which is opposite of 60 degrees is √3.now its asking for sin(60):imagine a triangle and and place the 30,60, 90 degrees in their respective places. and plug in the special triangle values. when you look at it sin(60) its opposite is √3 and the hypotenuse is 2. opposite over hypotenuse, so √3/2the shortest which is opposite of 30 degrees is 1.
Question: What is the formula used to calculate sin(60°) using the special triangle? Answer: Opposite side (√3) / Hypotenuse (2) | 677.169 | 1 |
given the two conics C1 and C2 consider the pencil of conics given by their linear combination λC1 + μC2
identify the homogeneous parameters (λ,μ) which corresponds to the degenerate conic of the pencil. This can be done by imposing that det(λC1 + μC2) = 0, which turns out to be the solution to a third degree equation.
given the degenerate cone C0, identify the two, possibly coincident, lines constituting it
intersects each identified line with one of the two original conic; this step can be done efficiently using the dual conic representation of C0
the points of intersection will represent the solution to the initial equation system
Dandelin spheres
See Dandelin spheres for a short elementary argument showing that the characterization of these curves as intersections of a plane with a cone is equivalent to the characterization in terms of foci, or of a focus and a directrix. In Geometry, a nondegenerate Conic section formed by a plane intersecting a cone has one or two Dandelin spheres characterized thus Each
See also
Focus (geometry), an overview of properties of conic sections related to the foci
References
^MathWorld: Cylindric section. In Geometry, the foci (singular focus) are a pair of special points used in describing Conic sections The four types of conic sections are the CircleA Lambert conformal conic projection ( LCC) is a conic Map projection, which is often used for Aeronautical charts In essence the projectionIn Mathematics, the matrix representation of conic sections is one way of studying a Conic section, its axis, vertices, foci, In mathematics a quadric, or quadric surface, is any D -dimensional Hypersurface defined as the locus of zeros of a QuadraticA quadratic function, in Mathematics, is a Polynomial function of the form f(x=ax^2+bx+c \\! where a \ne 0 \\!Rotation of Axes is a form of Euclidean transformation in which the entire Xy-coordinate system is rotated in the Counter-clockwise direction with respectIn Geometry, a nondegenerate Conic section formed by a plane intersecting a cone has one or two Dandelin spheres characterized thus EachProjective geometry, a pair of harmonic conjugate points on the Real projective line is defined by the following harmonic construction "Given three collinear
^ E. J. Wilczynski, Some remarks on the historical development and the future prospects of the differential geometry of plane curves, Bull. Amer. Math. Soc. 22 (1916) pp. 317-329.
External links
Eric W. Weisstein, Conic Section at MathWorld. Eric W Weisstein (born March 18, 1969, in Bloomington Indiana) is an Encyclopedist who created and maintains MathWorld MathWorld is an online Mathematics reference work created and largely written by Eric W
Question: How can the points of intersection between the identified lines and the original conics be found efficiently? Answer: Using the dual conic representation of C0 | 677.169 | 1 |
The plane of rotation is the plane containing m and n, which must be distinct otherwise the reflections are the same and no rotation takes place. As either vector can be replaced by its negative the angle between them can always be acute, or at most π/2. The rotation is through twice the angle between the vectors, up to π or a half-turn. The sense of the rotation is to rotate from m towards n: the geometric product is not commutative so the product nm is the inverse rotation, with sense from n to m.
Conversely all simple rotations can be generated this way, with two reflections, by two unit vectors in the plane of rotation separated by half the desired angle of rotation. These can be composed to produce more general rotations, using up to n reflections if the dimension n is even, n − 2 if n is odd, by choosing pairs of reflections given by two vectors in each plane of rotation.[9][10]
Bivectors
Bivectors are quantities from geometric algebra, clifford algebra and the exterior algebra, which generalise the idea of vectors into two dimensions. As vectors are to lines, so are bivectors to planes. So every plane (in any dimension) can be associated with a bivector, and every simple bivector is associated with a plane. This makes them a good fit for describing planes of rotation.
Every rotation plane in a rotation has a simple bivector associated with it. This is parallel to the plane and has magnitude equal to the angle of rotation in the plane. These bivectors are summed to produce a single, generally non-simple, bivector for the whole rotation. This can generate a rotor through the exponential map, which can be used to rotate an object.
Bivectors are related to rotors through the exponential map (which applied to bivectors generates rotors and rotations using De Moivre's formula). In particular given any bivector B the rotor associated with it is
This is a simple rotation if the bivector is simple, a more general rotation otherwise. When squared,
it gives a rotor that rotates through twice the angle. If B is simple then this is the same rotation as is generated by two reflections, as the product mn gives a rotation through twice the angle between the vectors. These can be equated,
from which it follows that the bivector associated with the plane of rotation containing m and n that rotates m to n is
This is a simple bivector, associated with the simple rotation described. More general rotations in four or more dimensions are associated with sums of simple bivectors, one for each plane of rotation, calculated as above.
Examples include the two rotations in four dimensions given above. The simple rotation in the zw-plane by an angle θ has bivector e34θ, a simple bivector. The double rotation by α and β in the xy-plane and zw-planes has bivector e12α + e34β, the sum of two simple bivectors e12α and e34β which are parallel to the two planes of rotation and have magnitudes equal to the angles of rotation.
Question: What are the two conditions for a reflection to result in a rotation? Answer: The vectors m and n must be distinct, and the angle between them must be acute or at most π/2. | 677.169 | 1 |
Question 597742: Write a system of equations and solve. The sum of the measures of the two smaller angles in a triangle is 2 degrees less than the measure of the largest angle. If the middle angle measures 40 degrees less than the largest, find the measures of the angles of the triangle.
I realize that my equations will have an x, y and z for the variable but trying to get those equations is what I am having difficulty with. Answer by scott8148(6628) (Show Source):
Question: What is the value of x, the measure of the largest angle? Answer: Solving the equation, we find x = 70 degrees. | 677.169 | 1 |
In my booklet it asks me to "draw a 144.32' segment in a counter clock wise direction. then it asks to show the delta angle, the length, and the radius of the curve."
how do u draw it with the instructions its giving me? how do you find out the delta angle, the length, and the radius?
this is not the only problem im facing. as there are 2 more instances where it happens.
30304
please look at Page 16,17. the problems im having difficult with are 4,5, and 10.
also im still having trouble with interpolation. i called my teacher, and he gave me some pointers but im still slightly confused. i really wish i had a formula to make everything go quickly. do you guys have a solution?
my formula for the first 90 ft contour line was x/50=1.1/4.9 equals 11.22 from the bench mark. i check with the other thread about this and it was right. but i don't remember how to add it, to form the 100' contour.
thanks in advance for your help.
2 projects to go and i graduate!:D
ReMark
29th Sep 2011, 07:14 pm
That distance is measured from the benchmark in an easterly direction. It is the location of the beginning of your 90 foot contour not the 100 foot contour.
The calculations you are doing are for the precise placement of points corresponding to a particular contour interval. I believe this drawing uses a 10' contour interval thus you'll have a 90, 100, 110, etc. contour. The points are also placed, if I am not mistaken, along the horizontal and vertical grid lines that you also had to include in the drawing.
paisis123
29th Sep 2011, 11:15 pm
K thanks for clearing the interpolation up. but can anyone help with the first part of my problem? i need to draw those line segments with the required information. but i still don't know how.
ReMark
29th Sep 2011, 11:32 pm
That section of the instructions have you drawing the cul-de-sac for Ocean Avenue if I'm not mistaken.
30311
To draw the arc correctly you have to go back and look carefully at steps 1 and 2 on the previous page (15) for further information (hint: radius). Once both curves are constructed you can obtain most of the required information you are being asked to provide via Properties or the List command. You will have to calculate the delta angle from the info given.
Question: In which direction should the segment be drawn? Answer: Counter clock wise | 677.169 | 1 |
Question: How to find vertices of an equilateral triangle knowing coordinates of centroid?
I want to find a triangle with the vertices A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3) knowing that the point G(1,1,1) is centroid of the triagle ABC and x1, y1, z1, x2, y2, z2, x3, y3, z3 are integer numbers, but I can not find. How do I tell Maple to do that?
Question: What is the user's difficulty in solving the problem? Answer: They cannot find the coordinates of the vertices A, B, and C. | 677.169 | 1 |
A good answer might be:
No. In this case, it is clear that walking in a straight line
to the final destination is shorter.
Summing Displacements =/= Summing Lengths
Summing displacements (vectors) is not the same as summing their length.
In the previous example, the sum of u and v
yields a vector that is shorter than the length of u
plus the length of v.
For now, just observe this by looking at the previous diagram,
and remembering that "a straight line is the shortest distance
between two points."
(Later on this will be discussed using the Pythagorean Formula.)
This fact is called the triangle inequality:
length( u + v ) <= length( u ) + length( v )
Here is a case where the length of the sum is much shorter than
the sum of the lengths:
e = ( 5, 4 )Tg = ( -4.9, -3.9 )Te + g = ( .1, .1 )T
(For clarity g has been moved
slightly moved away from where it should be.)
QUESTION 18:
Can you think of a situation where
the length of the result is equal to the sum of the length of the
two input vectors?
Question: What will be discussed later to explain this phenomenon? Answer: The Pythagorean Formula. | 677.169 | 1 |
Geometry equations are numerous and each equations serves a masterful purpose in the bigger picture of geometry as a whole. These deal mainly with lines and angles, but go on the further deal with other items as well. The geometric equations are explicit to finding the perimeter and the area of various shapes, the surface are and even the volume density of solid material. It is a branch of pure mathematics dealing with the measurement, properties, and relationships of points, lines, angles, and two-dimensional figures and three-dimensional figures. Though dealing with the two dimensional figures and the three dimensional figures are fairly rare. The interactive geometry concepts deal with Euclid's elements, Pythagoras theorems. Length and areas concepts are dealing with circles, spheres, and cylinders. There are so many different equations and they are easily understandable or they can be very difficult to understand. The majority of the time and equations are in keeping on a simple, palatable level for the sake of the majority who wish to repeat a given experiment. There are those rare times when the equation is so very difficult that the student reads the proof attached over a few times before fully understanding what the task was reasoned.
There is a variety of geometry equations and though they are in relation on some level they are all very independently different to. Some of the geometry equations are the interior angles, the angles, the sine angles, the cosine angles, the tangent angles, the cotangent angles; the secant angles the cosecant angles.
The sum of the triangle is equal to 180 degrees, while quadrilateral is 360 degrees. The sine is the ratio of length, while the cosine is the ration of length of side. The tangent is the ratio of length of opposite side, while the cotangent is the ratio of length of side on the opposite side. The secant is the ratio of the length of side, while the cosecant is the ratio of the length of the hypotenuse. Geometry equations take some time to learn and to absorb and understand. It does take persistent training on the part of the teacher. It also takes persistent training of the mind on the part of the student. This concept continues for a reason.
Question: What is the sum of the interior angles in a triangle? Answer: 180 degrees.
Question: Which two basic shapes are mentioned as having their perimeter and area calculated using geometric equations? Answer: Lines and angles. | 677.169 | 1 |
First of all, I have no idea what math teachers would say about any of this. I was playing around with some stuff, and I stumbled upon it. I read a few things to confirm I was right.
There are ways of looking at it: rotating the coordinate system or rotating all the points. I think the second way is a bit more intuitive. That's also the way I found it.
This is the basic setup for rotating a point.
First, we need to describe the point you want to rotate relative to the point of rotation and in terms of an angle from standard position.
Let's say the original point has coordinates of (xi, yi) and the point of rotation has coordinates of (xcenter, ycenter).
Since unit circle trigonometry is based off a circle at the origin, let's temporarily translate everything to the origin. Since we are working based on how the points are relative to the point of rotation and not a "fixed" point, we can move the whole system as long as we move it back. To do this, all we have to do is subtract (xcenter, ycenter) from both points. This puts the point of rotation at the origin and the other point at (xi- xcenter, yi - ycenter).
When you rotate an object around a point, the points are always the same distance from each other. This means if you draw a circle with the point of rotation at the center and the point you want to rotate on the circle, the rotated point will also be on the circle. Since we know the coordinates of the two points, we can use the distance formula to find this distance.
√((xi - xcenter)2 + (yi - ycenter)2)
Now by dropping a right triangle down from the point, we can find the angle between the point and standard position.
tan(θ) = o / h
tan(θ) = (xi - xcenter)/(yi - ycenter)
θ = arctan(xi - xcenter)/(yi - ycenter)
Now we have everything needed to express the original point using angles.
(Distance*cos(θ), Distance*sin(θ))
To rotate the point, just add the desired angle of rotation.
Since we have to move everything back to their original position, we need to add (xcenter, ycenter)
This looks terrible and is a pain to type so let's simplify it. According to the linear combination of sine and cosine, a*sin(θ)+b*cos(θ)= √(a2+ b2)*cos(θ - arctan(b/a)). The second half shows a very near resemblance to our two equations.
The shifts on the ends of the equations wouldn't matter because if the above statement is true wouldn't a*sin(θ)+b*cos(θ) + 5= √(a2+ b2)*cos(θ - arctan(b/a)) +5?
Starting with the x equation, a would be yi - ycenter because it is on the bottom of the fraction in the arctan.
That means that b is xi - xcenter
Question: What is the first step in rotating a point? Answer: Describe the point you want to rotate relative to the point of rotation and in terms of an angle from standard position.
Question: What is the formula to rotate the point by a desired angle? Answer: Add the desired angle of rotation to θ.
Question: What is the final step to move the point back to its original position? Answer: Add (xcenter, ycenter) to the coordinates of the rotated point.
Question: What is the formula to find the distance between the point of rotation and the point to be rotated? Answer: √((xi - xcenter)² + (yi - ycenter)²) | 677.169 | 1 |
A-level Mathematics/MEI/C1/Co-ordinate Geometry
Co-ordinates are a way of describing position. In two dimensions, positions are given in two perpendicular directions, x and y.
Straight lines
A straight line has a fixed gradient. The gradient of a line and its y intercept are the two main pieces of information that distinguish one line from another.
Equations of a straight line
The most common form of a straight line is . The m is the gradient of the line, and the c is where the line intercepts the y-axis. When c is 0, the line passes through the origin.
Other forms of the equation are , used for vertical lines of infinte gradient, , used for horizontal lines with 0 gradient, and , which is often used for some lines as a neater way of writing the equation.
Finding the equation of a straight line
You may need to find the equation of a straight line, and only given the co-ordinates of one point on the line and the gradient of the line. The single point can be taken as , and the co-ordinates and the gradient can be substituted in the formula . Then it is simply a case of rearranging the formula into the form .
You may only be given two points, and . In this case, use the formula to find the gradient and then use the method above.
Gradient of a line
The steepness of a line can be measured by its gradient, which is the increase in the y direction divided by the increase in the x direction. The letter m is used to denote the gradient.
Parallel and perpendicular lines
With the gradients of two lines, you can tell if they are parallel, perpendicular, or neither. A pair of lines are parallel if their gradients are equal, . A pair of lines are perpendicular if the product of their gradients is -1,
Distance between two points
Using the co-ordinates of two points, it is possible to calculate the distance between them using Pythagoras' theorem.
The distance between any two points A and B is given by
Mid-point of a line
When the co-ordinate of two points are known, the mid-point is the point halfway between those points. For any two points A and B, the co-ordinates of the mid-point of AB can be found by .
Intersection of lines
Any two lines will meet at a point, as long as they are not parallel. You can find the point of intersection simply by solving the two equations simultaneously. The lines will intersect at one distinct point (if a solution to their equation exists) or will not intersect at all (if they are parallel). A curve can however intersect a line or another curve at multiple points.
Curves
To sketch a graph of a curve, all you need to know is the general shape of the curve and other important pieces of information such as the x and y intercepts and the points of any maxima and minima.
Curves in the form
Here are the graphs for the curves , , and :
(Need to draw those later, just simple b&w curve sketches for each curve)
Question: What are the two main pieces of information that distinguish one straight line from another? Answer: The gradient of the line and its y-intercept | 677.169 | 1 |
Notice how the odd powers of all share the same general shape, moving from bottom-left to top-right, and how all the even powers of share the same "bucket" shaped curve.
Curves in the form
Just like earlier, curves with an even powers of all have the same general shape, and those with odd powers of share another general shape.
(Images here)
All curves in this form do not have a value for , because is undefined. There are asymptotes on both the and axis, where the curve moves towards increasingly slowly but will never actually touch.
Intersection of lines and curves
When a line intersects with a curve, it is possible to find the points of intersection by substituting the equation of the line into the equation of the curve. If the line is in the form , then you can replace any instances of with , and then expand the equation out and then factorise the resulting quadratic.
Intersection of curves
The same method can be used as for a line and a curve. However, it will only work in simple cases. When an algebraic method fails, you will need to resort to a graphical or Numerical Method. In the exam, you will only be required to use algebraic methods.
The circle
The circle is defined as the path of all the points at a fixed distance from a single point. The single point is the centre of the circle and the fixed distance is it's radius. This definition is the basis of the equation of the circle.
Equation of the circle
The equation of the circle is for a circle center (0,0) and radius r, and for a circle centre (a,b) and radius r.
So, for example, a circle with the equation would have centre (-2,3) and radius 5.
Circle geometry
When presented with a problem, it may appear at first that there is not enough information given to you. However, there are some facts that will help you spot right angles in relation to a circle.
The angle in a semi-circle is a right angle
The perpendicular from the centre of a circle to a chord bisects the chord
The tangent to a circle at a point is perpendicular to the radius through that point
Question: How can you find the points of intersection when a line intersects with a curve? Answer: You can find the points of intersection by substituting the equation of the line into the equation of the curve, then expanding the equation out and factorising the resulting quadratic. | 677.169 | 1 |
Have you drawn the figure yet? What have you done to try to answer the problem? What are you stuck on and need help with? Do you know the properties of a cyclic quadrilateral? Do you understand how to draw the figure? We don't just do problems for you but can help you get started.
Question: What is the current challenge you're facing and need assistance with? Answer: The text asks "What are you stuck on and need help with?", indicating that you're currently stuck and need help. | 677.169 | 1 |
In an ellipse with major axis of 2a, minor axis of 2b, and foci c (on the
major axis), the relationship c squared = a squared - b squared holds
true... how do the three numbers fit into a Pythagorean relationship?
I want to have my students draw a scale model of the solar system that
shows the orbits of the planets. Assuming I have the apogee and
perigee of each planet's orbit about the sun, they need to construct 9
ellipses with some degree of accuracy. What's the best way to go about
this?
A man stands in the center of a circle. On the circumference is a
tiger that can only move around the circle. The tiger can run four
times as fast as the man. How can the man escape the circle without
being eaten by the tiger?
When considering the case when circle C has center at the origin and
radius 1, we need to show that the equation of the circle orthogonal to
circle C and with center (h,k) is given by: x^2-2hx+y^2-2ky+1=0.
I'm purchasing a curved piece of glass for some furniture. The curve
(arc) is 60 inches long. The height (the midpoint of the chord to the
center of the arc) is 11 inches. I need to know the radius of this
curve so the glass company can make my glass. Any thoughts?
Question: How many ellipses do the students need to draw for the scale model of the solar system? Answer: 9 ellipses.
Question: In the scenario with the man and the tiger, what is the only way for the man to escape the circle without being eaten? Answer: The man can stand still and let the tiger run around the circle, as the tiger's faster speed will cause it to eventually pass the man and the man can then run off the circle. | 677.169 | 1 |
A right triangle ABC has to be constructed in the xy-plane [#permalink]
08 Sep 2010, 12:21
1
This post received KUDOS
00:00
Question Stats:
86%(01:00) correct
13%(02:51) wrong based on 3 sessionsQuestion should specify that x- and y-coordinates of A, B, and C are to be integers.
We have the rectangle with dimensions 9*7 (9 horizontal dots and 7 vertical). AC is parallel to y-axis and AB is parallel to x-axis.Note: A and C share the same x coordinate. A and B share the same y coordinate.
The x coordinates have 9 options (x = -1,0,1,2,3,4,5,6,7) pick 2: 1 for point B, and 1 for point A and C. 9P2 The y coordinates have 7 options (y= 1,2,3,4,5,6,7) pick 2: 1 for point A and B, and 1 for point C. 7P2 Number of triangles w/ A as the right angle = 9P2 * 7P2 = 3024
Question: How many options are there for the y-coordinate of point A (and B)? Answer: 7 | 677.169 | 1 |
This very broad topic is one of the most ancient areas of study known to man. It may not be hugely inviting for many students, but it is not possible to avoid. Included in this vast area are points, angles, lines, triangles, parallelograms, etc. - the list is indeed long.
Euclid is the point where our study of modern day geometry really begins. No questions will be set on him directly but we will use many of his approaches. There are two full questions on the ordinary level paper and three on the higher-level paper.
Key topics
Below is a list of all the key topics within this section of the Junior Cert syllabus. Areas of interest to higher-level students only have been highlighted.
For an interactive explanation of difficult concepts and the opportunity to solve practical problems within this section, consult our Geometry lessons.
Points, Lines, Angles
Types of angles and properties of angles
Distance from a point to a line
Parallel lines
Perpendicular lines
Triangles and types of triangles
The sum of the angles in a triangle equals 180°
The exterior angle of a triangle equals the sum of the interior opposite angles
Quadrilateral and types of quadrilateral
Circle: centre, radius, diameter
Chord, secant, segment, tangent
The angle in a semicircle is a right angle
A tangent to a circle is perpendicular to a radius at the point of contact
1. Vertically opposite angles are equal in measure.
2. The measure of the three angles of a triangle sum to 180°.
3. The exterior angle in a triangle equals the sum of the interior opposite angles.
4. If two sides of a triangle are equal in measure, then the sides opposite these angles are also equal in measure.
5. Opposite sides and opposite angles of a parallelogram are equal in measure.
6. A diagonal bisects the area of a parallelogram.
7. The measure of the angle at the centre of a circle is twice the measure of the angle at the circumference, standing on the same arc.
Deduction one: all angles on the same arc are equal.
Deduction two: the angle in a semicircle is a right angle. Link to lesson
Deduction three: the sum of the opposite angles in a cyclic quadrilateral is 180°.
8. A line through the centre of a circle perpendicular to a chord bisects the chord.
9. If two triangles are equiangular, the lengths of corresponding sides are in proportion.
10. In a right-angled triangle, the square on the hypotenuse equals the sum of the squares on the other two sides.
Sample questions (from official D.E.S. sample exam papers)
Ordinary level
Solution
Triangle abd and triangle adc.
Higher level
Question (Paper 2, Q.4a)
(i) Only one of the four hands shown below on the
right could be the image of the hand on the left
under a central symmetry. Say whether it's
number 1, 2, 3 or 4.
Question: In the higher-level sample question, which hand is the image of the hand on the left under a central symmetry? Answer: Number 3
Question: What is the sum of the angles in a triangle? Answer: 180 degrees
Question: Which shape has opposite sides and opposite angles of equal measure? Answer: A parallelogram
Question: If two triangles are equiangular, what is the relationship between the lengths of their corresponding sides? Answer: The lengths of corresponding sides are in proportion. | 677.169 | 1 |
Notsurehowtodrawitherebutonpaperyouwillseetheoriginalsimplex (whichwillbeatriangle)dividedinto4sub-triangles0,0.5,0.5)$.Theorderingisrestored with the vertices(v1,v5,v6),(v2,v4,v5),(v3,v4,v6)and(v4,v5,v6)transformation,i.e.,$(2,0,0)$refersto$1,0,0$.
I believe it might be a simple question with simple solution but I am not able to figure it out.Ialsothinkthattheremightbesomeorderinwhichtheverticesaregeneratedandmaycorrespondwithverticesofthesimplicesinthepartition. Can you please point me to some reference.For example consider the case of three dimensions. A simplex in 3-D is given as a convex hull of the vertices (0,0,1), (0 1 0) and (1,0,0). An exemplary FT of this example is the simplices with vertices v1 = (0,0,1), v2 = (0,1,0), v3 = (1,0,0), v4 = (0.5,0.5,0), v5 = (0,0.5,0.5) and v6 = (0.5,0,0.5).
Not sure how to draw it here but on paper you will see the original simplex (which will be a triangle) divided into 4 sub-triangles with the vertices (v1,v2,v5),(v3,v2,v4),v1,v5,v6), (v1,v3,v6)v2,v4,v5),(v3,v4,v6) and (v4,v5,v6)
My concern is that given the vertices v1,..,v6, how do I arrive at those 4 sub-simplices. I need it as I will be dealing with higher dimensions and finding the sub-simplices may not be that obvious in those cases.
I believe it might be a simple question with simple solution but I am not able to figure it out. Can you please point me to some reference.
Thank You have a doubt on Freudenthal TriangularizationTri
Determining the simplices in freudenthal triangulation
I have a doubt on Freudenthal Triangularization
Question: What does the user think might be a simple solution to their problem? Answer: The user thinks there might be some order in which the vertices are generated that corresponds with the vertices of the simplices in the partition.
Question: How many dimensions is the given simplex in the example? Answer: 3 | 677.169 | 1 |
Heron's formula is distinguished from other formulas for the area of a triangle, such as half the base times the height or half the modulus of a cross product of two sides, by requiring no arbitrary choice of side as base or vertex as origin.
Contents
The formula is credited to Heron (or Hero) of Alexandria, and a proof can be found in his book, Metrica, written c.A.D. 60. It has been suggested that Archimedes knew the formula over two centuries earlier, and since Metrica is a collection of the mathematical knowledge available in the ancient world, it is possible that the formula predates the reference given in that work.[2]
A modern proof, which uses algebra and is quite unlike the one provided by Heron (in his book Metrica), follows. Let a, b, c be the sides of the triangle and A, B, C the angles opposite those sides. We have
Question: When was Metrica written? Answer: c.A.D. 60 | 677.169 | 1 |
Perpendicularity is described as two lines in the same plane
that intersect to form right (90º) anglesParallel and perpendicular lines are shown below:
PO 2. Justify which objects in a collection M04-S5C2-06. Summarize mathematical Example:
match a given geometric description. information, explain reasoning, and draw Identify which of these shapes have perpendicular or
conclusions. parallel sides and justify your selection.
Connections: M04-S4C1-01, M04-S4C1-
03, M04-S4C1-05, M04-S4C1-06, M04- M04-S5C2-07. Analyze and evaluate
S4C1-07, M04-S4C3-03 whether a solution is reasonable, is
mathematically correct, and answers the
question. A possible justification that students might give is:
The square has perpendicular lines because the sides
meet at a corner, forming right angles.
PO 3. Describe and classify triangles by M04-S5C2-07. Analyze and evaluate A triangle can be described in more than one way.
angles and sides. whether a solution is reasonable, is
mathematically correct, and answers the Examples:
Connections: M04-S4C1-01, M04-S4C1- question. A right triangle can be both scalene and isosceles.
02, M04-S4C1-06 A scalene triangle can be right, acute and obtuseTriangles can be classified by:
Angles
o Right: The triangle has one angle that
measures 90º.
o Acute: The triangle has exactly three angles
that measure between 0º and 90º.
o Obtuse: The triangle has exactly one angle
that measures greater than 90º and less than
180º.
Sides
o Equilateral: All sides of the triangle are the
same length.
o Isosceles: At least two sides of the triangle
are the same length.
o Scalene: No sides of the triangle are the same
length Recognize which attributes (such M04-S5C2-05. Represent a problem Examples:
as shape or area) change and which do situation using any combination of words, Students should recognize that the area of the triangle
not change when 2-dimensional figures numbers, pictures, physical objects, or and the area of the rectangle are equal.
are cut up or rearranged. symbols.
Connections: M04-S4C4-04 M04-S5C2-07. Analyze and evaluate
whether a solution is reasonable, is
mathematically correct, and answers the
question.
Students should recognize that when the shape is
rearranged, the number of sides and vertices change,
Question: What is the definition of perpendicular lines according to the text? Answer: Perpendicular lines are two lines in the same plane that intersect to form right (90º) angles.
Question: When a 2D figure is cut up or rearranged, which attributes change and which do not? Answer: The number of sides and vertices change, but the area does not. | 677.169 | 1 |
[Median is calculated based on the number of elements in a set, while mean is calculated based on the values of those elements. A single outlier does not have a big impact on a median because it is still only 1 element of the set. However, a single outlier does change the mean significantly because its value is very different from the other values.]
Assessment Options
Have students write a letter to the school administration presenting the height they determined, including details on how the measurement was performed and why they believe it to be accurate.
Have students write a journal entry about the process of measuring the height of objects using isosceles triangles.
Have students write a journal entry about why individual measurements may be different. Students should address how reporting the mean or median of all measurements can be more reliable by accounting for the variation in measurements.
Allow students to measure the height of another object on their own.
Extensions
Allow students to come up with other ways to measure building and try them. On a sunny day, they can measure their shadow and the shadow of the building, and use a proportion to find the height of the building.
Students could also measure the height of other objects using their clinometers and tangent ratios. For example, using tangent x = opposite/adjacent, students could measure a common angle (30°, 45°, or 60°) substituting it for x, their distance from the object (adjacent), and solve for the height (opposite).
Teacher Reflection
Did students have sufficient understanding of how to use the clinometer before going outside? If not, how can you provide better direction?
Did students work well in their pairs? How else could you group them?
Did students have sufficient prior knowledge of isosceles right triangles? Did students understand the role of the triangle in calculating the building height?
Did students understand that measurements will turn out the same no matter who measures an object
Question: Which of the following is a suggested assessment option for students? Answer: Have students write a letter to the school administration presenting the height they determined, including details on how the measurement was performed and why they believe it to be accurate | 677.169 | 1 |
Pages
Monday, April 18, 2011
Perspective: X Marks the Center
Perspective is merely a trick to help create realistic-looking three dimensional space on a two dimensional surface.One, two and three point perspectives are very good tools for creating this sense of depth.However, within the basic rules of perspective there are little tricks to help you calculate distance of specific objects that can't be measured by using a vanishing point.Little things like these can turn a good drawing into a realistic drawing.
Edgar Degas, The Dance Foyer at the Opera on the Rue le Peletier, 1872
Today we'll take a look how to find the center of a rectangle in perspective.Though this doesn't sound like an overly helpful tool, once we get to the practical applications you'll see a world of possibilities.
Begin with a simple square.Draw a line from the top right corner to the bottom left corner and visa versa.The spot where these two lines cross marks the exact center of the square.Simple enough.
Now let's put this same square into one and two point perspectives (fig. 2).
Draw the same lines from corner to opposite corner (fig. 3).Even though these lines are not of equal length, where they cross will mark the precise center of the square.
Before dismissing this little trick as insignificant, take a look at some of its practical applications (fig. 4).Finding the center of a rectangle can help with the placement of elements in a perspective drawing.
Not only can we find the center of the square or rectangle, but by extending a line vertically from the middle of the X, we can also find the center of the top and bottom lines of the shape.This becomes particularly handy when working with things like buildings and windows.
As you can see in the house on the right in figure 4, the X can be used to subdivide space as well.By finding the center of the side of the house with an X, we can split it into two equal rectangles.We can then find their centers by placing an X within both rectangles.This is the only way to properly place the windows.
The archway of the door in the Degas painting from above is a good example of this perspective trick.By outlining the rectangle in which it sits, we can make an X and extend a line upward from its center.The point where this line meets the top of the rectangle will be the center (and thus the apex) of the arch.
The use of an X to find the center of a rectangle has many more applications than those shown here.So the knowledge of this simple trick will prove to be very helpful any time you are trying to create an accurate illusion of perspective
Question: What is a practical application of finding the center of a rectangle in perspective? Answer: Placing elements accurately in a perspective drawing, such as windows in a building. | 677.169 | 1 |
With such a starting point in place, we still need to model the two parabolic mirrors. Given our purpose and the complexity of 3-D modeling, we choose to model the mirascope using a two-dimensional approach or rather, a cross section of the 3-D mirascope. Accordingly, we need two parabolas. While a parabola can be constructed geometrically, an exclusively geometric approach is not well suited for our modeling task since we want to align the two parabolas with high accuracy. Furthermore, in building a dynamic model, we wish to leave room for open-ended exploration, which in our case may include changing the curvature of the parabolas and/or their vertical alignment. This analysis points to an algebraic way for constructing the two parabolas. Because the specific locations of the parabolas are not essential in this case, we could use a simple algebraic form: \(f(x)=ax^2+c\), where a controls the orientation and curvature of a parabola, and \(c\) moves it vertically for alignment. To facilitate subsequent open-ended explorations, we choose to use sliders for the initial parameters, \(a\) and \(c\), of the two parabolas. These sliders are simply visual representations of the two parameters and are shown as horizontal or vertical lines on the screen. Note that the symmetric shape of the mirascope dictates that if \(a\) is positive for the parabola that opens up, \(-a\)would be used for the one that opens down.
The above analysis is an essential phase of problem solving [4]. Although it takes time and may take many iterative steps, it unveils the underlying structure of the mirascope, clarifies the interdependent relations, and ultimately helps us devise a plan of action. In my teaching experience, I found that this step could be challenging for some prospective and classroom teachers, who tend to take the problem at the surface level and thus create a mere visual replica of the given problem using a quadratic function \(f(x)=ax^2+bx+c\) without explicitly defining the mathematical relations among the integral components. Therefore, in spite of the technical utilities and representational resources of a dynamic environment such as GeoGebra, the problem solver still has to make sense of the problem before he/she could come up with a preliminary plan of action. With the analysis above, we now proceed to the construction stage, during which we may be able to learn more about the problem itself and revise our plan
Question: What is the challenge some teachers face in this step of problem-solving? Answer: They tend to create a mere visual replica of the given problem using a quadratic function without explicitly defining the mathematical relations among the integral components.
Question: What is the algebraic form used to construct the parabolas? Answer: \(f(x)=ax^2+c\), where 'a' controls the orientation and curvature, and 'c' moves the parabola vertically.
Question: What is the next stage after the analysis phase? Answer: The construction stage. | 677.169 | 1 |
3 Questions from Cal II Quiz
I missed the following 3 questions on my Cal II quiz this week.
Someone please show me how to work them so I can find my mistake.
Show your work for a good rating.
13) Find the rectangular coordinates of the point(s) of
intersection of the following polar curves. r=6sin? r=6cos?
15) Calculate the area of the given region: r=2cos?, r=2sin?,
the rays ?=0 and ?=(?/4)
20) Which of the following represents the area outside
r=6cos(2?) but inside r=6?
a.
b
c
e
Once again, please show all work and explain what you are doing
(if necessary) for good rating. IF YOU COPY AND PASTE FROM ANOTHER
SITE OR THIS ONE I WILL GIVE YOU 1 STAR REGARDLESS OF WHETHER IT IS
CORRECT. Obviously I know how to use the internet too- that is not
what I need. Thank you. :0)
Question: What is the region to be calculated in question 15? Answer: The region is the area between the polar curves r = 2cos(θ) and r = 2sin(θ), bounded by the rays θ = 0 and θ = π/4. | 677.169 | 1 |
meridians and the parallels (which can be different).
At 51 degrees the ration lat/long should be 1/cos(51) = 1.59
(for the oblate spheroid that the Earth really is the figure
is nearer 1.585). If the figure found is different, outwith
errors, then the map is stretched N-S or E-W.
A little trigonometry is used to find any rotation of the
scales.
angle of 1 degree meridian =
tan-1 ((x1degLongTop - xb) - (x1degLongBtm - xa) / (ya - yb))
If the angle is small it might be assumed to be zero, no
rotation.
angle of 51 degree parallel =
tan-1 ((y51degLatLt - ya) - (y51degLatRt - yd) / (xd - xa))
If the angle is small it might be assumed to be zero, no
rotation.
The two angles can be compared to see the shape of a lat long
'cell'. This might be a rectangle or sheared into a
parallelogram. If the map is a trapezoid this comparison is
void.
Question: What is the formula to calculate the angle of a 1-degree meridian? Answer: tan-1 ((x1degLongTop - xb) - (x1degLongBtm - xa) / (ya - yb)) | 677.169 | 1 |
Jeffrey Everett's Discussions…Continue
"Looks like you may have figured out a way. My process for that is:
1. Take the core line of the doughnut (the primary circle of a torus or basically the circle at the center) and analyze it for length.
2. make a line of that length.
3. make a spiral…" wrapped around a donut. Am I clear enough? It's somewhat hard to describe. Anyway, how should I go about it? See More
Question: What is the third step in Jeffrey Everett's process? Answer: Make a spiral wrapped around a donut. | 677.169 | 1 |
Syndicate
You are here
Calculating distance
By ockley on Tue, 03/06/2012 - 16:56
One of the formulas, that you will use all the time, is the formula that calculates the distance between two points on the screen. As far as I know it is an old invention from the old Greeks, more specific the guy called Pythagoras. He made a discovery, that ensured his fame long into this century - the Pythagorean Theorem.
The Pythagorean Theorem
Used in geometry, this theorem states:
In a right-angled triangle the area of the square on the hypotenuse is equal to the sum of the areas of the squares of the other two sides.
The odd word here is the hypotenuse. This is the name for the side that is opposite to the right angle. As you can read, the theorem deals with a bunch of squares - rectangles with equal width and height - and the relation between the hypotenuse, and the two others. So in plain English the theorem states:
If you start from the right angle, and make two squares extend with the length of of the sides that extends from it, the sum of these two areas combined is the same as one single square with a width of the long side on the opposite side of the right angle.
You see, I tried to explain it simpler that the original theorem, but failed miserably. Maybe both quotes combined with the formula and a diagram does the job:
a2 + b2 = c2
Reverse engineer it, please
The tricky part here is that we just want the side and not an area. Luckily it isn't even a tricky part, because there is an easy way to calculate the side of a square, based on the area - the square root - well thought name. The root for the definition of a square is to take a length, and multiply it with it self. So we end up with
c = sqr((a * a) + (b * b))
C is almost always the length, we wan't to know, and A and B is almost always the values w know, because from any given two points you can draw a horizontal and vertical line. Find where the lines intersect, and measure the length from the intersection and out to points.
Fundamental implementations
If you want to find a distance between two points, this is how it is achieved in:
Java
Other implementations
Some of the more visual oriented languages like ActionScript and Processing has specific methods for calculating distances. ActionScript has a Point class, that only deals with points and the relation between them. In the same manner, Processing has a dist() method that that calculates the distance based on the given x- and y- coordinates, like this:
Question: What are the inputs to the dist() method in Processing? Answer: x- and y- coordinates
Question: What is the formula that represents the Pythagorean Theorem? Answer: a² + b² = c²
Question: In a right-angled triangle, which side is the hypotenuse? Answer: The side opposite the right angle | 677.169 | 1 |
37. As shown in the standard (x,y) coordinate plane below, P(6,6) lies on the circle with center (2,3) and the radius 5 coordinate units. What are the coordinates of the image of P after the circle is rotated 90 degree clockwise about the center of the circle.
I'm assuming that your sketch is drawn like in the problem. So when you rotating the circle 90 degrees upon (2,3), it seems as though P is the fourth quadrant which has the coordinates (x, -y). C is the only one that's like that.
Also, A and B just don't make sense.
Please correct me if I'm wrong.
Yes C is the correct answer ,but the different between C and D are so close -1 and 0. I did not know whether 90 degree would pass the x-axis or not...It seem like an estimating problem???
Anyone else>???
Estimating would be the easiest way to go about doing this problem.
You could, however, try to make a right triangle and find the distances but that's just too much work. I don't know if its worth the time, or if it is even feasible.
I didnt wanna fully work it out but this is the process. Since when rotation an point, the distance from the center stays the same to the point right? SO you know that the distance from the center to the point equals the distance from the center to the new point
you use the center coordinates 2,3 the orig coors 6,6 and the new coords x,y to solve it.
It gets pretty hectic
You could solve this problem in an easy way like the following :
1-Draw a line from the point (6,6) to the other side of the circle witch will make a 180 degree.
2- Because the circle is rotated clockwise you will draw a line from the midpoint of the line"vertical line 90 degree" you draw in step one and will go in the other side of the circle witch is toward the clock rotate.
3-when you connect the second line with the circle you will find that the circle is mating with the line in the point (5,-1).
Question: What is the center of the circle? Answer: (2,3)
Question: What are the coordinates of point P after the 90-degree clockwise rotation? Answer: (5,-1) | 677.169 | 1 |
GMAT/MBA Expert
Start with the definition of a square: four equal sides and four right angles.
statement 1 tells you that you have 2 right angles but does not tell you anything about the sides. insufficient.
STatement 2 tells you that you have 3 equal sides, but if one side is much smaller then this isn't a square - insufficient.
together - you have two right angles which means that you have three equal sides and at least 2 right angles, which will make for at least one set of parallel sides - with the third equal side it forces the fourth side to be equal and you have a square so the answer is C.
Question: What is the final answer according to the text? Answer: The answer is C. | 677.169 | 1 |
A student familiar with the definition of z-scores wonders why we use standard
deviations to calculate them. Illustrating two ways, Doctor Peterson explains the
concept of scaling that motivates this statistical measure.
As far as I know, a trapezoid is defined as a quadrilateral with exactly one set of parallel sides. However, a very highly regarded educator and textbook author recently argued that this definition is incorrect. His definition of a trapezoid is that it is a quadrilateral that has at least one pair of parallel sides. A square, therefore, would be considered a trapezoid. Is he correct or are thousands of books going to be published with the wrong definition?
Question: What is the alternative definition of a trapezoid proposed by a highly regarded educator? Answer: The alternative definition is that a trapezoid is a quadrilateral that has at least one pair of parallel sides, which would include a square. | 677.169 | 1 |
The length of the string is "c". "a" is the distance from the motor down vertically to where you want to be (the Y coordinate). "b" is the distance from the motor to where you want to be (the X coordinate). Doing this calculation for both of the motors gets you the distance you want each of the two string to be to land on a certain point. You want your frame to be much bigger than the area you're drawing on, because certain places are difficult to reach if it's not. In the picture attached to this step you'll notice there are two triangles, one with A1, B1, C1 and the second with A2, B2, C2. So:
Question: Which coordinate does "a" represent in the context of the triangles? Answer: The vertical coordinate (Y) | 677.169 | 1 |
We can remember this by imagining that the minus sign flips the direction of the edge, so if uv=−1, then edge-u must be identified antiparallel with edge-v, and if uv=1, they must be identified parallel. However, here is a picture to show why this works:
a'aa'±1a+1a+1a1× a'a'aa'±1a+1a'±11a'
On the left hand side, we show three adjacent faces, with edge numbers. One of the faces is face 1, and adjacent to that is face a. The edge between them must be numbered a in the numbering of face 1 (because on face 1, edge labels match face numbers), while on the numbering of face a, we've determined in the discussion above that it must be the projective multiplicative inverse of a, which we denote by a′. So a⋅a′=±1.
Let's say we've chosen to show the right-hand side of face 1. That means that the edge numbering will increase clockwise, so the edge that is one step clockwise of edge-a must be edge-(a+1), as shown, which in turn means that the third face shown must be face (a+1).
Now, depending on whether the right side of face 1 is identified with the left or the right side of face a, the edge numbering of face a will be either increasing or decreasing anticlockwise (respectively), so the edge one step anticlockwise of edge-a′ will be edge-a′±1 (respectively).
Now we multiply faces numbers by a′.
We send 1→a′
We send a→1
We send (a+1)→(a+1)⋅a′=(±1+a′)
where the sign in the last equation depends on whether a⋅a′=1 or a⋅a′=−1.
But the edge numbering is unchanged, so now that face 1 is on the left, the third face must match the edge label, and hence must be a′±1 depending on whether the right side of face 1 was identified with the left or the right side of face a (respectively). But the sign is also determined by the sign of a⋅a′.
Hence if a⋅a′=+1, then the right side of face 1 must be identified with the left side of face a (and hence they are joined parallel).
And if a⋅a′=−1, then the right side of face 1 must be identified with the right side of face a (and hence they are joined antiparallel).
QED
Question: What is the edge between face 1 and face 'a' numbered on face 1? Answer: 'a'
Question: What is the third face shown in the picture, if the right side of face 1 is identified with the left side of face 'a'? Answer: Face (a+1) | 677.169 | 1 |
William Wallace's proof of the "butterfly theorem"
Problem
AB is the diameter of a Circle, CD a chord cutting it at right angles in K, EF and HG two other chords drawn anyhow through the point K, and HF, EG chords joining the extremes of EF, HG. It is required to prove that MK is equal to KL.
Demonstration
Through L draw PQ parallel to GE, meeting KF in P and KH in Q.
Because of the parallels the angle HQP is equal to HGE, but HGE is equal to HFE, or to HFP, for they are in the same segment, therefore the angles HQP, HFP are equal, and hence the points H, Q, F, P are in the circumference of a circle, wherefore PL × LQ = FL × LH.
The above proposition is a particular case of a more general one extending to all the Conic Sections, which may be expressed thus.
If AB is any diameter of a Conic Section, and CD any right line cutting it in K, and parallel to a tangent at its vertices; also EF and HG two other lines drawn anyhow through K, to meet the conic section, the one in the points E, F, and the other in the points G, H, the straight lines EG, FH which join the extremities of these lines shall intercept upon CD the segments KM, KL which are equal to each other.
Question: What is the ratio of PL × LQ to FL × LH? Answer: They are equal, i.e., PL × LQ = FL × LH. | 677.169 | 1 |
tail to tip. The lengths of the vector arrows are drawn to scale, and the angles
are drawn accurately (with a protractor). Then, the length of the arrow
representing the resultant vector is measured with a ruler. This length is
converted into the magnitude of the resultant vector by using the scale
factor with which the drawing was constructed.
Ex: A car moves with a displacement A
of 275 m, due west and then with a
R displacement B of 125 m in a direction
B 55.0° north of west.
55°
A
The two displacement vectors A and
Vector A = 275 m; W B are neither colinear nor perpendicular
but add to give the resultant vector R.
Vector B = 125 m; @ 55° N of W
Ex: I ran 10.0 km due east but was pushed 5.0 km, north by a crosswind.
How far did I actually run? What was my angle from the start line?
R
5.0 km; north
Θ
10.0 km; east
- The resultant, R can be found by using
a ruler to measure the magnitude, and a
protractor to measure, the angle Θ.
C. Parallelogram Method to Vector Addition
Two vectors are drawn to scale and joined at their tails. Parallel lines
are drawn to complete the parallelogram.
The resultant is drawn from where the tails touch to the opposite corner.
Ex: A boat travels 2 km north, then 4 km north of east at an angle of 30°.
Determine the magnitude and direction of the resultant.
Two ways to solve for R
- Use a ruler and a protractor
R -Use vector resolution
2 km; n 4 km, e (math: Pythagorean's Theorem;
30° Trig functions)
D. Vector Component Method (Vector Resolution)
1. Resolve each vector into x and y components.
2. Assign positive or negative values to each component based on
which quadrant it falls in.
3. Reduce the problem to the sum of the two vectors.
Σ x = v1x + v2x + any others
Σ y = v1y + v2y + any others
4. Determine the magnitude and direction of the resultant. Since Σ x and Σ y
are at right angles, the Pythagorean Theorem can be used to determine the
magnitude of the resultant.
The definition of the tangent of the angle can be used to determine the
direction of the resultant.
tan Θ = vy Θ = tan-1
vx (v)
v
y
x
4 km, e 4 km, e
28° 28°
R R
2 km; n 2 km; n
Vector 1 4 km, e Vector 2
(2 km; n) v2y (4 km; e @ 28°)
28°
Question: In the given boat travel example, what is the magnitude of the resultant vector using the vector component method? Answer: 2√2 km (using Pythagorean Theorem)
Question: How can the magnitude and direction of the resultant vector be found in the parallelogram method? Answer: By measuring the length of the resultant with a ruler and its angle with a protractor.
Question: What mathematical theorem is used to determine the magnitude of the resultant in the vector component method? Answer: The Pythagorean Theorem | 677.169 | 1 |
Shapes
I built castles made out of shapes at school. I built big castles. I built little castles. I used cones. I used squares. I used circles. I used rectangles. I used semi-circles. There are attributes on shapes. Shapes roll. A triangle top is pointy. Some shapes have flat or straight sides. A few shapes have corners. Circles are curvy. A baby dragon went in my shaped castle. Some shapes have smooth ends. Some shapes have flat ends. A cone or triangle has 3 sides. A cube has 6 sides. A rectangle has 6 sides. A ball has 0 sides. A pentagon has 5 sides. A hexagon has 6 sides. A octagon has 8 sides. A square has 4 sides. A semi-circle has 0 sides. A semi circle is also called a half circle.
Question: How many sides does a cube have? Answer: 6 | 677.169 | 1 |
Oil Line Puzzle Instructions
Congruent triangles can be used to find the shortest
distance.
Consider the case of an oil company that needs to pump
oil from the main line to two different stations, A and B.
A single pumping station can tap into the line and pump oil
to these locations (A and B). Where is the best place to
put the station so you lay the least amount of pipe?
Load
OilLine.html in Netscape (NOT version 4 of MS Internet Explorer..there are Java bugs in this version)
Locate location A at (125,25) and location B at (235,50)
Put the Station half way between 125 and 235. What location is the Pumping
Station? What is the length of pipe needed? Do you think this is the
least possible?
Click "Show Hint" (if you can't see the hint, press
the "Hide Map" button too). Move the Pumping Station so that it is in line with
the hint. Where is the Pumping Station location now? What is the length
of pipe needed now?
Move the Station a little to the left and then to the right,
noting what happens to the length of pipe needed. Where is the best
place to put the station? Why?
Change Location B to (35,35). Without the hint showing, try
to find the best place to put the Pumping Station. Where should it be placed?
Now move A to (25,10) and B to (200,50). Where should the Pumping Station
be placed now?
Check your answer with the hint. Do you see any relation between the angles
in this situation and the billiards?
If the angle between the pumping station and location A is 35 degrees, what is the angle from location A to location B?
Question: What is the length of the pipe needed when the station is placed halfway between A and B? Answer: 110 units.
Question: If location A is moved to (25,10) and B to (200,50), where should the pumping station be placed? Answer: At the midpoint of A and B, which is (112.5,25). | 677.169 | 1 |
Saturday, October 1, 2011 at 12:59pm by kelvyn
Geomatry What is the value of the variable and BC if B is between A and C. AB=4x BC=5x AB=16
Tuesday, September 13, 2011 at 7:36pm by Chase
geometry Trapezoid ABCD has height 4, BC=5, and AD and BC are perpendicular. Find the area of the trapezoid.
Sunday, February 13, 2011 at 10:13pm by sally
geometry Trapezoid ABCD has height 4, BC=5, and AD and BC are perpendicular. Find the area of the trapezoid.
Sunday, February 13, 2011 at 10:12pm by sally
Math ABCD is a parallelogram. E is a point on DC extended, such that D and E are on opposite sides of BC. Let AE intersect BC and BD at F and G, respectively. If AG=180 and FG=108, what is EF?
Monday, May 20, 2013 at 3:00am by John
Trigonometry Make a sketch. I have a triangle ABC, where BC is the ground, AC is the tower. Angle B = 60°, angle C = 84.5°, making angle A = 35.5° by Sine Law: BC/sin 35.5° = 179/sin 60° I get BC = 120.026 So the shadow of the tower is 120 ft long
Tuesday, August 30, 2011 at 10:55pm by Reiny
math b the vertices of triangle ABC are A(-2,3), B(0,-3), and C(4,1). prove, by means of coordinate geometry, that the median to side BC is also the altitude to side BC.
Monday, May 25, 2009 at 1:44pm by jojo
geometry In triangle ABC, if line BC is one inch longer than line AB, line AC is 10 inches shorter than the sum of line BC and line AB , and the perimeter of triangle ABC is 72 inches, find the length of line BC?
Wednesday, November 3, 2010 at 10:14pm by meghan
maths --plse help me.. without using set square or protacter construct 1. triangle A BC WITH AB = 5.5 CM , BC = 3.2 CM CA = 4.8 CM 2. DRAW LOCUS OF A POINT WHICH MOVES SO THAT IT IS ALWAYS A DISTANCE OF 2.5 CM FROM B. 3. DRAW LOCUS OF POINT SO THAT IT IS EQUIDISTANT FROM BC & CA . 4. MARK POINT ...
Question: In the triangle ABC, if line BC is one inch longer than line AB, line AC is 10 inches shorter than the sum of line BC and line AB, and the perimeter is 72 inches, what is the length of BC? Answer: 24 inches | 677.169 | 1 |
Math sorry i dont see a picture. But, I'll explain what should be seen. Make a right triangle. From the angle <B, where B is 90 degrees, draw an altitude down to CA. Call that point D. We see that ABC and BCD are similar triangles. BC/CD=AC/BC, so BC^2=AC*CD. Likewise, ...
Monday, November 19, 2012 at 11:26am by mathtaculator
Question: What is the relationship between the areas of triangles ABC and BCD? Answer: The area of triangle ABC is twice the area of triangle BCD | 677.169 | 1 |
parabola
pa·rab·o·la
a plane curve formed by the intersection of a right circular cone with a plane parallel to a generator of the cone; the set of points in a plane that are equidistant from a fixed line and a fixed point in the same plane or in a parallel plane. Equation: y2 = 2 px or x2 = 2 py.
Gk. parabole "parabola, application" (see parable), so called by Apollonius of Perga c.210 B.C.E. because it is produced by "application" of a given area to a given straight line. It had a different sense in Pythagorean geometry. Related: Parabolic.
Question: Is a parabola a three-dimensional shape? Answer: No, a parabola is a two-dimensional shape. | 677.169 | 1 |
Two-dimensional space
From Wikipedia, the free encyclopedia
Bi-dimensional space is a geometric model of the planar projection of the physical universe in which we live. The two dimensions are commonly called length and width. Both directions lie in the same plane.
In physics and mathematics, a sequence of nnumbers can be understood as a location in n-dimensional space. When n = 2, the set of all such locations is called 2-dimensional Euclidean space or bi-dimensional Euclidean space.
In physics, our bi-dimensional space is viewed as a planar representation of the space in which we move, described as bi-dimensional space or two-dimensional space.
Non-convex
There exist infinitely many non-convex regular polytopes in two dimensions, whose Schläfli symbols consist of rational numbers {n/m}. They are called star polygons and share the same vertex arrangements of the convex regular polygons.
In general, for any natural number n, there are n-pointed non-convex regular polygonal stars with Schläfli symbols {n/m} for all m such that m < n/2 (strictly speaking {n/m}={n/(n-m)}) and m and n are coprime.
Question: What is the set of all locations in two-dimensional Euclidean space? Answer: The set of all sequences of two numbers. | 677.169 | 1 |
Special Features Of Isosceles Triangles
posted on: 18 Apr, 2012 | updated on: 18 May, 2012
Triangles are the polygons with three sides. We say that the triangle is the closed figure with three sides. Now we need to know that the Triangles are classified as per the measure of their lengths of their sides. We say that the triangles which have all the sides of the same measure, then the triangles are called equilateral triangles.
Now we will talk about isosceles triangles: An Isosceles Triangle has two equal sides and the third side is unequal, which works as the base of the triangle. If we look at the special properties of the isosceles triangle, we say:
As isosceles triangle have two lines of same length
As the two sides of the triangle are same, we say that the angles corresponding to the equal sides are also equal.
Also the Median of the isosceles triangle, drawn from non equal side, is also the perpendicular bisector of the non equal side.
As we know that the sum of the three angles of the triangle are equal, thus if one angle of the triangle is known, we are able to find the remaining two angles of the isosceles triangle. Let us see how: If the triangle has one angle = 70 degree, then if we want to find the measure of two angles of the triangle, which are unequal. Let the measure of those angles = x degrees. Now we say that 70 + x + x = 180,
2x + 70 = 180,
2x = 180- 70,
2x = 110,
X = 55, so other two angles of the triangle are 55 and 55 degrees
Similarly, if we know the two equal angles of the triangle, then we double the measure and then subtract it from 180 degrees to get the third angle of the triangle.
Points Shown above are the Special Features of Isosceles Triangles.
Special Features of Isoceles Triangles.
Topics Covered in Special Features Of Isosceles Triangles
Triangles are classified according to their length of the line segments and as per their angles. We know that if we have the Triangles classified as per the length of their line segments then the triangles are of following types: 1. Equilateral Triangle 2. Isosceles triangle 3. Scalene triangle. Here we are going to study about an Iso...Read More
A Median of a triangle is defined as the line segment which is used to join a vertex of a triangle to the midpoint of the opposite sides of a triangle. There are three vertices present in a triangle, so three medians are present in a triangle. Median value depends on the vertices. There are some properties of An Isosceles Triangle with a median which ...Read More
Question: What is the sum of the angles in any triangle? Answer: 180 degrees.
Question: What are the two main types of triangles based on the length of their sides? Answer: Equilateral and Isosceles triangles.
Question: In an isosceles triangle, is the median drawn from the unequal side also the perpendicular bisector of that side? Answer: Yes. | 677.169 | 1 |
The perimeter of any triangle is simply the sum of the lengths of the three sides; so, if p is the perimeter and the three sides are a, b, and c, then
p = a + b + c
Then, by the Pythagorean theorem, if c is the hypoteneuse (the side opposite the right angle), then
a^2 + b^2 = c^2
So, just plug in the squares of the two lengths you already have, solve for the remaining side, and take the square root of that. For example, if you already know the hypoteneuse is 5, and that one other side is 3, then
To find the perimeter of a right triangle when you have the length of two of the sides, you can use the Pythagorean theorem to first find the length of the third side. The Pythagorean theorem is a^2 + b^2 = c^2, in which c is the hypotenuse, the longest side, or the side opposite the right angle. Once you find what c is equal to, you can add the lengths of all three of the sides, which will equal the perimeter, or the total of all the lengths of the sides.
Question: If the hypotenuse (c) of a right triangle is 5 and one other side (a) is 3, what is the length of the third side (b)? Answer: To find b, you would use the Pythagorean theorem: a^2 + b^2 = c^2. Plugging in the known values, we get 3^2 + b^2 = 5^2, which simplifies to 9 + b^2 = 25. Solving for b^2 gives 16, so b = √16 = 4. | 677.169 | 1 |
The Knewton Blog
Nate is a content developer at Knewton, and he loves thinking up ways to help students with their GMAT prep.
–
Geometry is an important part of any GMAT test-taker's conceptual toolkit. On Data Sufficiency geometry questions, it's especially key to have an intuitive feel for what is and is not solvable given certain bits of information. Consider the following difficult problem:
A circle having center O is inscribed in triangle ABC. What is the measure of angle BAC?
The radius of the circle is 2.
Segment OA has length 4.
(A) Statement (1)Â ALONE is sufficient, but statement (2)Â alone is not sufficient to answer the question asked.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
(C) BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
(D) EACH statement ALONE is sufficient to answer the question asked.
(E) Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.
There are two ways to attack a problem like this. At the end of this article is an extremely well-thought out, coldly-reasoned, logical, academic explanation. While interesting, in terms of the GMAT it is an absolutely useless way to think about the question. It just takes too much time! Instead, you should train yourself to approach problems like these intuitively. Here's how:
First, check out this simple geometry tool. It's a handy JAVA applet that lets you see how the problem works visually. If you take a few moments to play with the applet, you may be able to get a better intuitive feel for the restrictions created by statements (1) and (2).
Try keeping the radius the same while changing the size and shape of the triangle. Notice the angle BAC changing? Next try keeping the length of segment OA the same while moving the circle around to change its radius. See the angle changing again?
By thinking intuitively, you can tell that neither statement is sufficient on its own. When the radius and segment length are fixed, though, it's another story. Once you know both these pieces of information, you can tell you're dealing with a 30-60-90 right triangle (more on this below), so finding the measure of angle BAC is a matter working with triangle properties.
That's why the answer here is C — statements (1) and (2) together are sufficient, but neither is sufficient alone.
The moral here is to avoid wordy reasoning in geometry whenever possible. Practice the art of visualization. You can't use a nifty applet on test day, but you can draw pictures of extreme cases and move the segments around in your head. This kind of intuitive reasoning is essential on Data Sufficiency geometry questions —where time is short and diagrams are seldom drawn to scale.
Question: What is the recommended approach to solve such problems on the GMAT? Answer: The recommended approach is to practice intuitive reasoning and visualization, as diagrams are seldom drawn to scale and time is short. | 677.169 | 1 |
In geometry, there are names for all polygons up to ten sides. Pass ten, and there are accepted names for the even sided polygons. Currently there are two proposed names for an 11 sided polygon, undecagon, and hendecagon For more information...
Polygons are named by number of Internal angles in Greek prefix. Greek prefix for eleven is Hendeca. So the eleven sided polygon will be called Hendecagon. Look here for more information:
Polygons are named by the Greek prefix of their total number of internal angles. Greek prefix for eleven is hendeca. So it would have name Hendecagon. Look here for more information:
A seven sided shape is called a heptagon. A polygon is named by the number of sides it has using a prefix that denotes the specific number; hence pentagon (5 sides), hexagon (6 sides), heptagon (7 sides), octagon (8 sides), etc.
A five sided shape is called a pentagon. The well known military headquarters in Washington D.C. has a pentagonal shape. All of its sides are the same length but a pentagon only must have five sides. They can be of different lengths.
A geometric shape that has twelve sides is called a dodecagon. Dodecagon shapes are not commonly used in everyday life items. However, there are a few countries that use the dodecagon shape for some of their coins.
A 12-sided shape or polygon is called a dodecagon. An interior angle of a dodecagon is equal to 150 degrees and the sum of all the interior angles is equal to 1800 degrees. You can find more information here:
A 5 sided shape is called a pentagon. The interior angles of a pentagon add up to 540 degrees. A pentagon is a polygon. The most famous pentagon is The Pentagon in Washington, D.C. It houses the US Department of Defense.
Question: What is the name of an eleven-sided polygon? Answer: Hendecagon
Question: What is the Greek prefix for eleven? Answer: Hendeca | 677.169 | 1 |
In the picture, some common angles, measured in radians, are given. Measurements in the counterclockwise direction are positive angles and measurements in the clockwise direction are negative angles.
Let a line through the origin, making an angle of θ with the positive half of the x-axis, intersect the unit circle. The x- and y-coordinates of this point of intersection are equal to cos θ and sin θ, respectively.
The triangle in the graphic enforces the formula; the radius is equal to the hypotenuse and has length 1, so we have sin θ = y/1 and cos θ = x/1. The unit circle can be thought of as a way of looking at an infinite number of triangles by varying the lengths of their legs but keeping the lengths of their hypotenuses equal to 1.
Note that these values (sin 0°, sin 30°, sin 45°, sin 60° and sin 90°) can easily be memorized in the form
but the angles are not equally spaced.
The values for 15°, 18º, 36º, 54°, 72º, and 75° are derived as follows:
From these, the values for all multiples of 3º can be analytically computed. For example:
The sine and cosine functions graphed on the Cartesian plane.
For angles greater than 2π or less than −2π, simply continue to rotate around the circle; sine and cosine are periodic functions with period 2π:
Its θ-intercepts correspond to those of sin(θ) while its undefined values correspond to the θ-intercepts of cos(θ).
The function changes slowly around angles of kπ, but changes rapidly at angles close to (k + 1/2)π.
The graph of the tangent function also has a vertical asymptote at θ = (k + 1/2)π, the θ-intercepts of the cosine function, because the function approaches infinity as θ approaches (k + 1/2)π from the left and minus infinity as it approaches (k + 1/2)π from the right.
All of the trigonometric functions of the angle θ can be constructed geometrically in terms of a unit circle centered at O.
Alternatively, all of the basic trigonometric functions can be defined in terms of a unit circle centered at O (as shown in the picture to the right), and similar such geometric definitions were used historically.
In particular, for a chord AB of the circle, where θ is half of the subtended angle, sin(θ) is AC (half of the chord), a definition introduced in India[5] (see history).
tan(θ) is the length of the segment AE of the tangent line through A, hence the word tangent for this function. cot(θ) is another tangent segment, AF.
sec(θ) = OE and csc(θ) = OF are segments of secant lines (intersecting the circle at two points), and can also be viewed as projections of OA along the tangent at A to the horizontal and vertical axes, respectively.
Question: In which direction are positive angles measured from the unit circle?
Answer: Positive angles are measured in the counterclockwise direction.
Question: Which trigonometric function has vertical asymptotes at angles of kπ, where k is an integer?
Answer: The tangent function | 677.169 | 1 |
14.0 Students prove the Pythagorean theorem. This theorem can be proved initially by using similar triangles formed by the altitude on the hypotenuse of a right triangle. Once the concept of area is introduced (Standard 8.0), students can prove the Pythagorean theorem in at least two more ways by using the familiar picture of four congruent right triangles with legs a and b nestled inside a square of side a +b.
8.0 Students know, derive, and solve problems involving the perimeter, circumference, area, volume, lateral area, and surface area of common geometric figures.
For rectilinear figures in the plane, the concept of area is simple because everything reduces to a union of triangles. However, the course must deal with circles, and here limits must be used and the number p defined. The concept of limit can be employed intuitively without proofs. If the area or length of a circle is defined as the limit of approximating, inscribing, or circumscribing regular polygons, then p is either the area of a disk of unit radius or the ratio of circumference to diameter, and heuristic arguments (see the glossary) for the equivalence of these two definitions would be given. The concept of volume, in contrast with that of area, is not simple even for polyhedra and should be touched on only lightly and intuitively. However, the formulas for volumes and surface areas of prisms, pyramids, cylinders, cones, and spheres (Standard 9.0) should be memorized. An important aspect of teaching three-dimensional geometry is to cultivate students' spatial intuition. Most students find spatial visualization difficult, which is all the more reason to make the teaching of this topic a high priority. The basic mensuration formulas for area and volume are among the main applications of geometry. However, the Pythagorean theorem and the concept of similarity give rise to even more applications through the introduction of trigonometric functions. The basic trigonometric functions in the following standards should be presented in a geometry course:
18.0 Students know the definitions of the basic trigonometric functions defined by the angles of a right triangle. They also know and are able to use elementary relationships between them. For example, tan (x) = sin (x)/cos (x), (sin (x))2 + (cos (x))2 = 1.
19.0 Students use trigonometric functions to solve for an unknown length of a side of a right triangle, given an angle and a length of a side.
Finally, the Pythagorean theorem leads naturally to the introduction of rectangular coordinates and coordinate geometry in general. A significant portion of the curriculum can be devoted to the teaching of topics embodied in the next two standards:
17.0 Students prove theorems by using coordinate geometry, including the midpoint of a line segment, the distance formula, and various forms of equations of lines and circles.
22.0 Students know the effect of rigid motions on figures in the coordinate plane and space, including rotations, translations, and reflections.
Question: What is the relationship between the sine and cosine functions of an angle in a right triangle? Answer: (sin(x))^2 + (cos(x))^2 = 1 | 677.169 | 1 |
Nine point circle
In geometry, the nine point circle is a circle that can be constructed for any given triangle. It is named so because it passes through nine significant points, with six of them lying on the triangle itself:
... the circle which passes through the feet of the altitudes of a triangle is tangent to all four circles which in turn are tangent to the three sides of the triangle...
The following image illustrates this theorem:
The point at which the incircle and the nine point circle touch is often called the Feuerbach point.
Feuerbach was not the first to discover the circle. At a slightly earlier date, Charles Brianchon and Jean-Victor Poncelet had stated and proven the same theorem. Soon after Feuerbach, mathematician Olry Terquem also proved what Feuerbach did and added the three points that are the midpoints of the altitude between the vertices and the orthocenter. Terquem was the first to use the name nine point circle (as he was the first to associate nine special points with the circle).
Other facts of interest:
The radius of the nine point circle is half the length of the radius of the circumcircle of the triangle.
The nine point circle bisects any line from the orthocenter to a point on the circumcircle.
The center of the nine point circle (the nine point center) lies on the triangle's Euler line, at the midpoint between the triangle's orthocenter and circumcenter.
If an orthocentric system of four points is given, any three of them define a triangle, and these four triangles all have the same nine point circle.
The centers of the incircle and excircles of a triangle form an orthocentric system. The nine point circle created for that orthocentric system is the circumcircle of the original triangle
Question: Where does the center of the nine point circle (the nine point center) lie on a triangle? Answer: The center of the nine point circle lies on the triangle's Euler line, at the midpoint between the triangle's orthocenter and circumcenter. | 677.169 | 1 |
C
Cause-Effect Diagram
A popular diagram used to analyze the causes of problems which provides an overview of all the possible causes. One starts at the right and lists the problem, and then extends a straight line to the left. From the line, one draws tangential lines and lists causes of the problems at the end of those lines. Lines can be drawn to the subsidiary lines as more discrete causes are considered, and so forth.
Question: How are causes of the problems listed in a Cause-Effect Diagram? Answer: At the end of tangential lines drawn from the initial line. | 677.169 | 1 |
sin (s + t) = sin s
cos t + cos s sin t
sin (s – t) = sin s
cos t – cos s sin t
cos (s – t) = cos s
cos t + sin s sin t
cos (s + t) = cos s
cos t – sin s sin t
you can find the sine and cosine for 3° (from 30° and 27°)
and then fill in the tables for sine and cosine for angles from 0°
though 90° in increments of 3°.
Again, using half-angle formulas, you could produce a table with
increments of 1.5° (that is, 1° 30'), then 0.75° (which is 45'), or even of
0.375° (which is 22' 30"). But how do you get a table with 1° increments?
Ptolemy recognized that there was no Euclidean construction to trisect
an angle of 3° to get an angle of 1°, but since the sine
function is almost linear for small angles, you could approximate
sin 1° just by taking 1/3 of sin 3°. With that
step, we can construct trig tables for trig functions with increments of 1°.
Better trig tables have been created throughout the centuries. For instance,
Ulugh Beg (15th century) constructed sine and tangent tables for every minute of arc to about
nine digits of accuracy!
Incidentally, if you have a table of sines, you can read it in reverse to
compute arcsine, so only one table is needed for both.
After computers: power series
Although computers and calculators could just store trig tables in their
memories, they can also compute trig functions directly, which is what they
usually do.
In the late 17th century, Newton and other mathematicians developed
power series. A power series is like a polynomial of unbounded degree.
For the various trig functions, these mathematicians found power series.
Here are the power series for sine and cosine (where x is an
angle measured in radians):
sin x =
x — x3/3! + x5/5! —
x7/7! +...
cos x =
1 — x2/2! + x4/4! —
x6/6! +...
The three dots ... mean that the expression is to go on forever, adding another
term, then subtracting a term, etc. The exclamation point ! is to be read
"factorial", and it means you multiply together the whole numbers from 1 up through
the given number. For example, 5!, "five factorial", equals 1 times 2 times 3 times 4 times 5,
which is 120, and so, 6! = 720.
These power series have infinitely many terms, but they get small so very fast that only the first
Question: What is the sine function of 1° approximated as? Answer: 1/3 of sin 3°
Question: Can you compute arcsine using a table of sines? Answer: Yes, by reading it in reverse
Question: What is the formula for the power series of sine (x) in radians? Answer: sin x = x - x³/3! + x⁵/5! - x⁷/7! +... | 677.169 | 1 |
Question 331516) )}}}
<pre><font size = 4 color = "indigo"><b>
Locate point E so that triangle EAB is congruent to triangle DAC))
)}}}
Draw ED))
)}}}
Extend BC to F so that CF = CD. Draw DF)),
green(line(2*cos(80*pi/180),0,1,0),line(.6736481777,.5652579374,1,0),
locate(1,0,F))
)}}}
I won't go through every step. I'll just tell you enough so
you can write it out like your teacher wants you to.
Using the fact that isosceles triangles have equal
base angles, that interior angles of a triangle have sum 180°,
that supplementary angles have sum 180°, and that vertical angles
are equal, you can now write the number of degrees in every
angle in the figure. That would be a good idea.
Therefore it is easy to show that
angle EAD = 100° = angle ADC = angle AEB.
Then by SAS, triangles ADC, AEB and AED are all congruent. So
AB = ED.
It is easy to show that angle DCF = 60° and that triangle CFD is
equilateral. You then show that BFDE is a parallelogram because
angle F = 60° = angle BED , angle EBF = 120° = angle EDF.
Then AB = ED = BF = BC + CF and since CF = CD, then AB = BC + CD.
If you have any questions as to why anything is true, you can ask
me in your thank-you note and I'll answer.
Edwin</pre>
Question: What shape is BFDE? Answer: A parallelogram
Question: What is the measure of angle EAD? Answer: 100° | 677.169 | 1 |
(This is a cognitive tuter program done via the internet. It will not let us go any further until we complete this blank. When we ask for a hint we get this: Please write a formula using the variable A and B to express the distances the pigeons flew to return home. In a right triangle, if the lengths of two legs are A and B, the hypotonuse is equal to the square root of (A*A+B*B)
PLEASE PLEASE HELP!! THANKS 1 solutions Answer 20900 by Earlsdon(6291) on 2006-04-23 10:14:52 (Show Source):
You can put this solution on YOUR website! You can draw the right triangle as follows:
The total north-south distance, A, is one leg of this triangle.
The total west-east distance, B, is the other leg and this, of course, is perpedicular the first leg, A.
The straight-line distance from the start of the trip to the finish is the hypotenuse of the right triangle.
You can express distances A and B as a function of the hypotenuse, which we'll call C, using a form of the well-known Pythagorean theorem:
Now, assuming that the pigeon took the direct route home...as the crow flies, so to speak, its route would be traced by the hypotenuse of the right triangle, or C.
The pigeon flew a distance of:
You can put this solution on YOUR website! Solve for x: Your first step is correct. Multiply both sides by (x+1) Simplify...here's where you erred! Subtract 3 from both sides. Solve this quadratic equation using the quadratic formula:.
You can put this solution on YOUR website! 1) A square whose area is 16 has sides of 4.
2) If partitioned into four congruent squares, the smaller squares will have sides of 2.
3) The center of the smaller squares will be at a distance of from the center of the parent square.
4) The radius of the circle whose circumference contains the centers of the four smaller squares will be.
5) The area of the circle is:
If statement 3) is not obvious, here's how I determined that.
From the center of one of the smaller squares, drop a perpendicular to one of its sides thus bisecting that side. You will have formed a right triangle whose legs are 1 each and whose hypotenuse is (See Pythagorean theorem).
The hypotenuse is the radius of the circle.
You can put this solution on YOUR website! Try this: Let x = the numerator and y = the denominator.
Rewrite this as: and substitute below. Solve for x. Divide both sides buy 2. Take the square root of both sides.
x = +or-6
y = 2x
y = +or-12
There are two answers:
Question: What is the relationship between the total north-south distance (A) and the total west-east distance (B) in the right triangle? Answer: In the right triangle, A and B are the lengths of the two legs, and they are perpendicular to each other.
Question: What is the radius of the circle whose circumference contains the centers of the four smaller squares? Answer: The radius of the circle is the hypotenuse of the right triangle formed by the legs of 1 unit each, which is √(1^2 + 1^2) = √2 units.
Question: What is the formula to express the distances the pigeons flew to return home in a right triangle? Answer: If the lengths of the two legs of the right triangle are A and B, the hypotenuse (the straight-line distance from the start to the finish) is equal to the square root of (A^2 + B^2). | 677.169 | 1 |
Question 351960: (1)Quadrilateral PQRS is a rectangle with diagonals PR and QS.If angle 1=18, find the measures of angle 2,angle 3 and angle 4
So I think angle 4 is also =18 but I have no idea what the other angles are or how I would find out???
Please help me with this... Click here to see answer by jrfrunner(365)
Question 353048: how do i convert a fraction to degree, min, sec? my book tells me how with a $100+ calculator, but my teacher told me to get a casio fx 115(or 300) es, which he uses. an example problem is Pi/9 rad. i do not understand how they get 20 deg, 0',0". how do i punch it into my calculator and how can i do it "manually"? Click here to see answer by jim_thompson5910(28598)
Question: What is the cost of the calculator mentioned in the text? Answer: $100+ | 677.169 | 1 |
The shaded triangle has one right angle (its a "right triangle") and two other angles, ∠x and
∠?. Complete the triangle to a rectangle. In the upper corner, we see two angles, ∠x and ∠y,
that add to 90◦ . But if we cut along the diagonal, we can rotate and slide the unshaded part to
exactly match the shaded part. That shows that ∠? = ∠y.
x
y
Conclusion: the two angles ∠x and ∠? in the original triangle add to 90◦ .
x
Fact 3
If a triangle contains a right angle, then
y the other two angles add up to 90◦
∠x + ∠y = 90◦
A triangle of any shape can be divided into two right triangles: slide a plastic triangle with
a right angle along the longest side of the triangle until its edge passes through the opposite
vertex, then draw the line. Then divides ∠y into two parts; call then ∠a and ∠b. Then
x y x a
b
z z
∠x + ∠a = 90◦ (the shaded triangle is a right triangle)
+ ∠b + ∠ z = 90◦ (the unshaded triangle is a right triangle)
∠x + ∠y + ∠z = 180◦ (adding).
Conclusion:
y
x
Fact 4
z The angles of any triangle add up to 180◦
∠x + ∠y + ∠z = 180◦
Equilateral and Isoceles Triangles
Equilateral – All 3 sides of equal length. If we rotate by 1/3 of a turn, the triangle matches
itself. Thus all 3 angles are equal. Since their sum is 180◦ , each must be 60◦ .
60
Fact 5
60 60 Each angle in an equilateral triangle is 60◦
Isoceles – At least 2 sides of equal length. The third side is called the base, and the angles
along the base are the base angles. If we fold along the line from the middle of the base to the
opposite vertex, the two sides match. Hence the base angles are equal.
Fact 6
The base angles of an isosceles triangle are equal
Base
Base
Question: What is the third side of an isosceles triangle called? Answer: The base
Question: If a triangle contains a right angle, what is the sum of the other two angles? Answer: 90° | 677.169 | 1 |
Pages
Friday, May 27, 2011
(This is a continuation of the previous post, talking about a previous GeoGebra project we did in Precalc.) Here are some samples of circles and associated waves that students did from scratch in GeoGebra: Part 1, where kids had to create a slowly counterclockwise rotating point along the unit circle; Part 2, where they had to make the circular path bigger; and Part 3, where they had to make the circle smaller and make the point rotate clockwise.
Here are some excerpts of student explanations as they reflect on how coefficients affect wave graphs and circular rotations. I pasted a different student response here for each question so you can read a variety of student "voices" as they attempt to articulate the concepts.
Student A - Question #1 asks the student to analyze the effects of coefficients on cosine wave and on the circle.
1. For the function f(x)=A*cos(Bx), A determines the radius of the circle and it therefore changes the amplitude of the cosine graph. In the circle, if A is greater than 1, for example 2, then the radius would be bigger because it would be 2 and the wave would be taller because it would go from 2 to -2. If it is less than 1, like 0.5, then the radius would be 0.5 and the wave amplitude would be smaller because it would go from 0.5 to -0.5. If the A happens to be negative then the graph would flip across the x-axis. When you flip the wave graph vertically along the x-axis the graph would start at -1 instead of 1. A horizontal flip across the y-axis wouldn't matter for cosine.
If B is greater than 1 then the point takes less time to make a period because it is moving faster. For example, if B would be 2 then you would complete 2 periods in the time it normally takes to make 1. It B is less than 1, i.e.: 0.5, you are slower, thus it takes you twice the time to make a period. For B in the cosine wave, it doesn't matter if it is negative or positive because it will be the same x-value regardless if the point in the circle is moving clockwise or counterclockwise.
Student B - Question #2 asks the student to analyze the effects of coefficients on sine wave and on the circle. Here, some of the explanations may duplicate part of their answer to #1, but I'm looking for the student to articulate that the sine wave flips if you make B negative, because it matters greatly whether your point rotating around the circle is going clockwise or counterclockwise, if you're tracking its height over time. This kid's response isn't perfect, but she's getting most of the ideas.
2. g(x)=A*sin(Bx)
Question: What happens to the cosine graph if A is negative? Answer: If A is negative, the cosine graph would flip across the x-axis, starting at -A instead of A.
Question: If A is greater than 1, how does it affect the radius of the circle and the amplitude of the cosine graph? Answer: If A is greater than 1, the radius of the circle would be bigger and the amplitude of the cosine graph would be taller, going from A to -A. | 677.169 | 1 |
You can put this solution on YOUR website! The locus of points in a plane that are equidistant from points A and B in the plane is a line that is the perpendicular bisector to line that connects the two points.
So the solution is which means that you need 2 cubic feet of soil containing 45% sand.
Probability-and-statistics/1988 1 solutions Answer 149329 by jim_thompson5910(28595) on 2009-06-03 16:42:19 (Show Source):
Numbers_Word_Problems/198849: An express train and a local train both leave Gray's Lake at 12:00 noon and head for Chicago 60 miles away. The express travels twice as fast as the local and arrives 2 hours ahead of it.Find the speed of each train.
60m/2h= t + 12n
I cannot get the right answer. 1 solutions Answer 149328 by jim_thompson5910(28595) on 2009-06-03 16:40:14 (Show Source):
Surface-area/198847: This question is from textbook
Directions state: Find the perimeters and areas of the squares and rectangles in questions 7-10.
The figure shown in the text has a picture of a square which is dissected in the middle to form two right triangles. The triangle has two 45 degree and one 90 degree angles. All that is given is six square root of two (6 square root 2) for the hypotenuse of the adjacent right triangles. My apoogies but I do not have a square root symbol key on my computer keyboard. As you can tell I need to solve for the perimeter and area, but I do not even know how to without data given for either side one or side two. 1 solutions Answer 149325 by jim_thompson5910(28595) on 2009-06-03 16:11:37 (Show Source):
x-intercept
To find the x-intercept, plug in and solve for x
Start with the given equation.
Plug in .
Simplify.
Divide both sides by to isolate .
Reduce.
So the x-intercept is .
Linear-systems/198842: Please help me solve:
Soybean meal is 14% protein; cornmeal is 7% protein. How many pounds of each should be mixed together in order to get 280lbs mixture that is 12% protein?
How many pounds of the cornmeal should be in the misture?
How many pounds of the soybean should be in the mixture?
Question: How many pounds of soybean meal and cornmeal should be mixed to get a 280-pound mixture that is 12% protein?
Answer: To solve this, we can use the following equations based on the problem statement:
Let x be the pounds of soybean meal and y be the pounds of cornmeal.
x + y = 280 (total weight)
0.14x + 0.07y = 0.12 * 280 (total protein content)
Solving these equations, we get:
x = 140 pounds of soybean meal
y = 140 pounds of cornmeal
Question: What is the locus of points in a plane that are equidistant from points A and B?
Answer: A line that is the perpendicular bisector of the line connecting points A and B.
Question: What is the given information in the problem about the square and rectangles?
Answer: The figure shown is a square dissected in the middle to form two right triangles, with the hypotenuse of the adjacent right triangles being 6√2. | 677.169 | 1 |
Spherical coordinates are given by ρ, ϴ, and Φ
ρ is the distance from the origin
ϴ is the angle from the positive x-axis to the positive y-axis
Φ is the angle between the positive z-axis towards the negative z-axis
NOTE: by convention, we put the following bounds on ρ:
ρ≥ 0
0 ≤ Φ
≤
∏
(15.8) Conversions between rectangular and spherical coordinates
x = ρsinϴcosΦ z = ρcosΦ
y = ρcosϴsinΦ ρ2 = x2 + y2 + z2,thus: ρ = [x + y + z]1/2
(15.7) What do we do when converting an integral to spherical coordinates?
Add the bonus term ρsin2Φ after converting to ρ, Φ, and ϴ coordinates
Question: In which direction does the angle ϴ measure from?
Answer: The positive x-axis to the positive y-axis | 677.169 | 1 |
optimization: square inscribed in a square
optimization: square inscribed in a square
1. The problem statement, all variables and given/known data
Each edge of a square has length L. Prove that among all squares inscribed in the given square, the one of minimum area has edges of length [tex]\frac{1}{2}L\sqrt{2}[/tex]
2. Relevant equations
3. The attempt at a solution
I started by drawing a square of sides L. Then labeled the vertices: (0,0) (L,0) (0,L) (L,L) then drew an inscribed square with variable x and the vertices were: (x,0) (L,x) (L-x,L) (0,L-x)
Then from this I set the distance of all the lines between each of these inscribed triangles vertices equal to one anotehr, to determine what values of X would work for the equation.
I determined that x=x. Is this true? Could you rotate a square 360 degrees while still being inscribed within a square?
I was stumped at this part, but if it's true. Then area of the inscribed square is going to be the distance of one of the inscribed squares length squared.
such as.. [tex]((x-L)^{2}+(x)^{2})^{2}[/tex]
Then i would take a derivative of this and determine the minimum point?
Is this right?
"Could you rotate a square 360 degrees while still being inscribed within a square?"
No, you cannot, if the four corners are still touching the sides. You must have made some small mistake.
Just find the area of the inscribed square as the sum of two triangles, by considering the vertices in proper order. Or you can find the dist between the parallel pairs of lines to get the area in terms of x. This one'd be easier. Whatever you do, you should be able to show the given result.
(The result shows that the least square is obtained by joining the mid-points of the bigger square.)
Question: How can the minimum area of the inscribed square be proven? Answer: By showing that the area of the inscribed square is the sum of two triangles, or by finding the distance between the parallel pairs of lines to get the area in terms of x, and then finding the minimum value of this function.
Question: What is the minimum area of the inscribed square? (Multiple Choice) A) (L^2/2) sqrt(2) B) L^2 C) (L^2/2) D) 2L^2 Answer: A) (L^2/2) sqrt(2) | 677.169 | 1 |
RNM27
RNM27
Visitor Messages
Hey Al, Do I just choose the degree and just do the work. Or does the job dimensions draw a right triangle for you. And you solve it plugging in your measurements? Which part of the job represents which side of the triangle. For example, Is the line of Kicks the hypotneuse? Is the spacing in the ceiling the opposite side? This is the part I am trying to figure out. Or do I just pick a degree tight enough to make the kicks in the panel and apply it?
Thanks for getting back to me so quickly. I saw that other site.It was helpful.But,they tell you to look at the triangle for references. But,I see no math except for spacing times cosecant. But, how did they determine 30 degrees? The parallel kick job they do on the 30 degree example is almost a carbon copy of the job I was assigned.My first kick is 19.5,24.25,29,33.75 for the last. I only have about 25" inches of space to work with.This is why I am using 4.75" center to center into the panel. I have other pipes to land in the panel.My spacing in the ceiling is 5.5". Some other people ran the pipes over head at 5.5" If I use 30 degrees an the ctc. is 11" apart I will not have enough room .Can I choose a higher degree to make the bends in the confines of 25" with 4.75" spacing?The box isn't knocked out yet.
Hello RN. To do the kicks as I've done you must first figure out the ammount that each pipe needs to be kicked. Then you take the C to C spacing of the conduits which in your case is 5.5. You must multiply that by the Multiplier of what ever degree you are kicking the pipes. Example 30* kick multiplier of 2. 2x5.5 = 11" That is the C to C spacing of the panel your landing onto. However is your panel already knocked out?
Question: What is the user's first kick measurement? Answer: 19.5 inches
Question: What is the center-to-center (CTC) spacing of the conduits in the text? Answer: 5.5 inches
Question: Has the box been knocked out yet? Answer: No, it hasn't been knocked out yet. | 677.169 | 1 |
A reference mark (of the plan or space) is the data of a base and a point of reference, in general noted O . We will suppose here that the base used for the vectors is the same one as that used for the reference mark. If the coordinates of the point has are ( xA, yA, zA) and those of the point B are ( xB, yB, zB), then the vector \ overrightarrow {AB} has as components:
x_A \\ y_A \\ z_A
\end{pmatrix}
By using the other notation, the coordinates of has are ( has 1, has 2, has 3) and those of the point B are ( B 1, B 2, B 3), and the vector U = AB has as components:
ui = bi - ai
Transformations
The transformations of a base are always the reverse of those of the functions which are described there. For example, the rotation of a reference mark in the trigonometrical direction is equivalent to the rotation of the vectors clockwise.
Question: What are the components of the vector U = AB in the other notation? Answer: The components of the vector U = AB in the other notation are (b1 - a1, b2 - a2, b3 - a3). | 677.169 | 1 |
I am a Java Developer.
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Post navigation
Four Points Determine A Square (Java)
Consider a list of four points on a plane; the points have integral coordinates, and their order is irrelevant. The four points determine a square if the distances between them are all equal, and the lengths of the two diagonals are also equal. For instance, the following lists are all squares:
Question: Are the coordinates of the points required to be positive? Answer: No, the coordinates can be any integers, including negative numbers. | 677.169 | 1 |
Question 12604: I have had very little geometry and don't even know where to begin with this problem.
If one-half of the complement of an angle plus three-fourths of the supplement of the angle equals 110 degrees, find the measure of the angle.
I appreciate the help tons!!!
capesch Click here to see answer by AdolphousC(70)
Question 12605: Please help me with the following word problem:
In triangle ABC, andle B is 8 degrees less than one-half of angle A and angle C is 28 degrees larger than angle A. Find the measures of the three angles of the triangle.
I have to admit that I have no idea of where to start here. Please walk me through step-by-step so that I can learn how to solve these on my own.
Thanks!!
Capesch Click here to see answer by AdolphousC(70)
Question 13131: I have a picture of a outdoor running track
The inside lane of the track will enclose an area that is the shape of a rectangle with a semicircle at each end. The length of the rectanjular part of the track is 2W. The outdoor running track is 400 meters around the inside lane.
Each lane measures 1.25 meters in width. Find the perimeter of the outer edgeof the inside Lane? Click here to see answer by glabow(165)
Question 13185: the perimeter of a rectangle is 120ft. the length of the rectangle is twice than the width. find the length and the width of the rectangle.
120=2L + 2W
120-2(2W +2)+2W
120=4W+4+2W
120=4W+2W+4
116=W-- LOST Click here to see answer by akmb1215(68)
Question 14264: 2 solutions are to be mixed to make 50 ml. of a solution that is 16% Bromine. One solution is 10% Bromine and the other solution is 40% Bromine. How much of each solution should be used?
Tried it but don't have the correct answer. Thanks for your help. Click here to see answer by Alwayscheerful(414)
Question 14514: Use the following information to solve questions 1-3. Right triangle ABC has one leg of length 3 in. and a hypotenuse of length 5 in. Find the length of the other leg
Question 14620: If a regular polygon has an ecterior angle of 18 degrees, how many sides would that polygon have?
and
If a regular polygon has an interior angle of 135 degrees, how many sides would that polygon have?
Question: How much of the 10% Bromine solution should be used in the fourth problem? Answer: 40 ml of the 10% Bromine solution should be used.
Question: Which two angles in a triangle are supplementary? Answer: The largest angle (angle C) and the smallest angle (angle A).
Question: What is the total perimeter of the rectangle in the third problem? Answer: The total perimeter of the rectangle is 120 feet.
Question: What is the relationship between angle A and angle B in the second problem? Answer: Angle B is 4 degrees less than angle A. | 677.169 | 1 |
A plane which passes through the cube's center produces a cross section
in form of a square (the leftmost cube in the illustration). A plane which
passes through the three corners of the cube only produces a cross section
in form of a regular triangle (the rightmost cube in the illustration).
The objective is to find the way how the plane should pass through the
cube in order to produce a cross section that is a regular hexagon. If the
cube's side is one unit, what is the side of the hexagon?
file size
161 KB
To print out PDF Print
'n' Play version of this puzzle, please download the latest version of the Adobe Reader.
Question: Which of the following is NOT a cross-section shape mentioned in the text? A) Square B) Triangle C) Circle D) Hexagon Answer: C) Circle | 677.169 | 1 |
This will always result in the correct angle. In our example, we have numerator = 0.8138 and denominator = 0.040009, with the denominator being positive. Then we compute arctan(0.8138/0.040009) = arctan(20.340) = 87.19°. Since the denominator was positive, there is nothing further to do, and 87.19° is the final result. (Had the denominator been negative, we would have computed 87.19° + 180° = 267.19°, which is the same as -92.81°, the other one of our two possible results found above.)
A final note: the equations we used assume that azimuth=0° corresponds to South, which is the usual convention in astronomy. If you prefer to work with a convention assuming that azimuth=0° corresponds to North, simply add 180° to the result obtained from the equations.
Bye,
Thomas
forrest noble
2011-Jan-20, 01:48 AM
I've also taught for awhile and would first ask, do you really understand where the equations come from in the first place that you are using and why they were set up that way and why you are using this or that equation for a particular application? If you cannot derive the equations in the first place then I think something is missing from your full understanding. Once you've derived these equations yourself from nothing then you have a pretty good understanding of why you are using them. After you can derive them once, then you do not have to do it again, you can simply remember them. My experience concerning math and physics has to do with students not understanding where the equations come from. There are a number of exceptions to this general idea but I don't know of any exceptions relating to an improved understanding of the subject via an understanding of where the equations came from in the first place. The equations of Quantum Chromodynamics for instance, according to what I have read, came from a long history of experimental results that enabled their equational derivation. GR came from many recorded observations that enabled a mathematical analog to be developed and tested against those observational results. What you are looking at in your example equations are relatively simply trigonometric relationships, whereby you could read the derivation of these equations from a textbook on trigonometry. Once reading of it and understanding it, you should be able to repeat the derivation without referring back to the text.
Question: Why is it important to understand where the equations come from? Answer: To have a full understanding of why and how they are used
Question: What is the value of the arctan function for 20.340? Answer: 87.19° | 677.169 | 1 |
The above techniques + your favorite computer algebra system make it easy to check that neither 1 degree nor 2 degrees are constructable angles. If Tn(x) denotes the Chebyshev polynomial of degree n (that is, cos(nx)=Tn(cos(x))), then cos(1) is a root of T60(x)-1/2, which has no irreducible factors of degree a power of two. cos(2) is a root of T30(x)-1/2, which also has no irreducible factors of degree a power of two. So, neither cos(1) nor cos(2) are constructable.
In particular, this means that the constructable whole-degree angles are precisely multiples of three degrees.
Yes, this is inelegant, but I'm too lazy to try to come up with a better solution.
Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?its pretty obvious that any odd number must have at least 1 prime factor that is odd. of course you can get arbitrarily close to an n-sect in the same way that a binary string can get arbitrarily close to representing an real number
Not really, since you don't know >OBC or point D. Given point D you can prove that it defines the trisection of >BOE but that doesn't help you find it.
You cannot "draw CA" until you have point A. Without knowing D, A could be anywhere on the line OE between E and some point to the right such that OE = EA. Basically you're asking us to take every point on the arc BE, find a point on the line OE which is 1 radius distant from the current point, and then pick the one which allows us to draw a line through some unknown point D. It can't be done.
Try it; remove points A and D and the lines that depend on them, then reconstruct the diagram.
Well, it can, it just uses a different set of laws for what you're allowed to do in a construction.
if you allow two ends of a compass to slide along the circle and line (at fixed distance OB) until they point at C, you're set. Yes, this involves picking something out of an infinite set, but so does taking the intersection between two lines.
Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?
Question: What is the degree of the Chebyshev polynomial of which cos(1) is a root? Answer: 60 | 677.169 | 1 |
Geometry Review by Leonard Blackburn, Parkland College
General Information
This web site is to serve as a brief and general review of basic facts of high school
geometry. The lessons are intended to be read by someone who has completed some such geometry
course and needs a refresher. It is not intended to be comprehensive and fully instructive.
There are examples
worked out in detail in the lessons and there are quiz problems for you to try.
So, in addition to recalling facts and formulas, you
will get to practice your problem solving skills.
There are nine lessons and quizzes given in PowerPoint format. Left-click on a link and open the
file. (Note: these instructions were developed for a PC utilizing Windows and Internet
Explorer.) The file will open in your browser and you will see the title page.
Once open, I recommend that you right-click on the screen and choose "Full Screen."
Once in full-screen mode, move the mouse around and you will see some navigation buttons
in the bottom-left corner of your screen. Left-click on the forward and back arrows as
necessary (you may also left-click anywhere else on the screen to move forward).
When you want to leave the lesson or quiz, right-click and choose "End Show." This will bring
you out of full-screen mode. Then use the back-button on your browser to return to the
Geometry Review home page.
The lessons are not written in a printer-friendly way. The material on some slides includes
animations and information that appears and disappears. Printing some slides will cause
some writing to appear on top of other writing. The quizzes may safely be printed out
without a loss of information.
The best way to learn from these files is to have paper and pencil ready so that you can work
out any details left out in the lessons and so that you can try the quiz problems. You may want
to take a look at a quiz first to gauge whether it is necessary for you to read a lesson, or
you may just want to read all the lessons in order, trying the quizzes after each one. Click
on the Table of Contents Link to see the contents of each lesson in more detail, to find out how
long a lesson is (measured by number of PowerPoint slides), and to find out how long a quiz is.
Instructors: please feel free to use, modify, and distribute this material as you see fit, while
giving credit to the author.
Question: How many lessons and quizzes are there in total? Answer: There are nine lessons and quizzes. | 677.169 | 1 |
This post is about alternate coordinate systems, in addition to the familiar Cartesian system, where each axis is at right angles to every other axis. The terminology I use is xMy where x is the number of axes and y the number of dimensions being measured. Cartesian coordinates are 1M1, 2M2, and 3M3 in 1, 2, or 3 dimensions. There is a 2M1 system, a 3M2 system and a 4M3 system in which all the coordinates add to zero for any point in space. 2M1 is simply a standard Cartesian single dimension paired with its mirror opposite. 3M2 has 3 axes oriented on a plane from the center of a triangle towards its three corners (or towards the centers of its three sides). 4M3 has 4 axes oriented in 3D space from the center of a tetrahedron to its four vertices (or towards the centers of its four faces). Here is a graphic representing the 3M2 system.
3M2 Coordinate System
Coordinates are given in the diagram for points p, q, and y. Point x is precisely between p and q – hence, its coordinates can be determined by averaging the coordinates of p (0,3,-3) and q (2,-3,1). So x is at (1,0,-1).
There are three different versions of 6M3, which I identify using a subscript to the 6, and which I will write here as 6(0)M3, 6(1)M3 and 6(2)M3. The first of these, 6(0)M3 is a doubling of 3M3, each axis pairing with its mirror image. 6(1)M3 has axes oriented from the center of a cuboctahedron (Bucky Fuller calls it a Vector Equilibrium) to its vertices. The 3M3, 4M3, 6(0)M3 and 6(1)M3 can all be easily lined up with the tetrahedron, octahedron, and cube. There are more complex systems, 6(2)M3, 10M3 and 15M3 that line up with the icosahedron and dodecahedron.
I add the (trademark) because the phrase is claimed (along with "prana calendar," "inner tuning" and other phrases) by Shyam Bhatnagar, who along with his one-time close companion Harish Johari were important influences in my life. I will not go into detail, but have been disappointed in how Shyamji has led his life, and have broken ties I once had with him. I still hold in high value the ideas to which he introduced me, some experiences for which he was a catalyst.
Question: Which shape do the axes of the 6(1)M3 system line up with? Answer: Cuboctahedron
Question: What is the number of axes and dimensions measured in the 2M1 coordinate system? Answer: 2 axes, 1 dimension
Question: What is the term used for the familiar coordinate system where each axis is at right angles to every other axis? Answer: Cartesian system | 677.169 | 1 |
7th grade geometry I need help desperately with geometry. the geometry is Area: Parallelograms, Triangles, and Trapezoids. It's really hard for me. Only if you get the concept. Anyway, Pleaswe help me!!!!!!!!!!
Monday, January 28, 2008 at 6:52pm by Janelle-Marie
precalc Suppose that z1=6-8i. Find: A. The Trig Form of the complex number z1, where your theta is in degrees. B. The Trig form of z1*z2, where z2=5[cos(60degrees)+isin(60degrees)] C. The Trig Form of (z1)^4
Monday, December 5, 2011 at 1:34pm by Joshua
trig Getting ready for trig semester finals. Can't figure out how they are simplifying these equations. (these are the review not actual test) tanx-sqr3=2tanx and 2sin^2+sinx=0
Sunday, December 11, 2011 at 4:49pm by Josh
Physics 2. you know the dispersion angle. YOu will have to use trig to find width at trig bottom. Draw the figure, label the paths, and I assume you know how thick the glass is. 3. Same principle as 2.
Friday, January 22, 2010 at 6:56pm by bobpursley
Trig Find values of all six trig functions if sin(theta)= 4/5 and theta is in the second quadrant.
Tuesday, August 23, 2011 at 2:47pm by Lucy
trig Solving Trig Functions: How Do You Solve For Sin 7.5 Using Half-Angle and Double-Angle Identities?
Wednesday, September 22, 2010 at 10:23pm by Spy[c]
trig 2sin(4x)[(cos(3/2)x)(cos(5x/2)-sin((15/4)x)]+x can someone simplify this using trig identities?
Sunday, May 13, 2012 at 9:08pm by Brianna
trig 405 I THINK I *'S 15 BY 27 AND GOT 405. NOW THAT IVE HELPED YOU WITH TRIG CAN YOU HELP ME WITH INTRODUCTORY ALGEBRA
Saturday, October 16, 2010 at 4:00pm by AliciaTrig HELP!!!! I dont know how to do the trig identity with this problem csc^4 x-cot^4x= Csx^2 x + cot^2x
Thursday, March 8, 2012 at 9:04am by Quetzally
trig find the exact values of the six trig functions of theta equals six pi divided by 8
Question: Which trigonometric identities is Josh struggling with in the fourth post? Answer: tanx - √3 = 2tanx and 2sin²x + sinx = 0
Question: What is the main topic of the first post? Answer: 7th grade geometry, specifically Area: Parallelograms, Triangles, and Trapezoids.
Question: What is the topic that Alicia wants help with after assisting with trigonometry? Answer: Introductory Algebra
Question: What is the complex number z1 in the third post? Answer: z1 = 6 - 8i | 677.169 | 1 |
IT says "if we delete one of the n+1 vertices of an n-simplex $[v_0, ...,v_n]$ then the remaining n vertices span an (n-1)-simplex, called a face ..."
So, is this the same "span" in linear algebra. Suppose we take a 1-simplex, or also known as a triangle, remove one of its vertices, we have two vertices and it spans the entire Euclidean space since they are linearly independent
Question: Do the remaining two vertices span anything? Answer: Yes, they span a line segment. | 677.169 | 1 |
Area
Area is a quantity expressing the two-dimensional size of a defined part of a surface, typically a region bounded by a closed curve. The term surface area refers to the total area of the exposed surface of a 3-dimensional solid, such as the sum of the areas of the exposed sides of a polyhedron.
Right Triangles
A right triangle has one 90° internal angle (a right angle). The side opposite to the right angle is the hypotenuse; it is the longest side in the right triangle.
Triangles
A triangle is one of the basic shapes of geometry: a polygon with three corners or vertices and three sides or edges which are line segments.
In Euclidean geometry any three non-collinear points determine a unique triangle and a unique plane.
Question: How many vertices does a triangle have? Answer: Three | 677.169 | 1 |
y1 = R1 - COS (A) * R1
y2 = R2 - COS (A) * R2
Now, to find angle (A), we must construct two right triangles. For one triangle, draw a line from the center of R2 horizontally to the vertical line that intersects the center of R1. The length of this line, which is the height of our triangle, is W. Draw another line from the center of R1 to the center of R2. This is the Hypoteneus of the triangle. The third leg (base) of the triangle is from the center of R1 to the point where the horizontal leg of the triangle intersects the vertical axis. This leg is H-R1-R2 in length. From there it is simple to calcualate the length of the hypoteneus, and the angle that the hypoteneus makes with the horizontal axis, which we'll call A1.
Draw a line from the center of R2, parallel to the line you plan to cut between the two radii. Now, draw a line from the center of R1 to intersect that line at a 90 degree angle. The base of this second triangle is R1 + R2, and the hypoteneuse is the same as the hypoteneus (D) of the previous triangle (above). What we're after here is the ANGLE between the hypoteneuse and the line you drew parallel to the cutting path. We'll call this angle A2.
The formula for that angle (A2) is: A2 = ARCSIN ( (R1 + R2) / D)
Now, for my next magical trick: The angle (A) that we were looking for at the start of this adventure is the sum of A1 and A2, which we just calculated. This is also the COMPLIMENT of the angle of the straight line you want to cut, which is 90 - A
Now that you have that angle (A), you can just plug A into the formulas given at the top to calculate x1, x2, y1 and y2. From there, you can do the rest.
Drawing
Here's a drawing of the two triangles I was talking about last night.
Before you calculate angle A using A1 and A2, check to see if the distance H - R1 - R2 is positive or negative. If the two radii are large and this distance is negative, you will have to SUBTRACT A1 from A2 to get angle A instead of adding them together to get A.
Unless Personal Designer can open DesignView "DV suffix" files, it won't be of much help.
Then again, the folks at Pro E bought the DesignView code and maybe it will work/import. Supposedly the DesignView code was incorporated into a later version of ProE or one of it's lower ended sketch pads.
Will check it out (if it's free as I already have DV2, DV3 and the latest DV5).
Dan
I enjoyed working through your solution. (more stimulating than SUDOKU)
Question: What is the formula to calculate angle A1? Answer: It is not explicitly given in the text, but it can be calculated using the lengths of the sides of the first triangle.
Question: What is the angle of the straight line you want to cut? Answer: 90 - A
Question: What is the length of the base of the first triangle? Answer: H - R1 - R2
Question: What is the formula to calculate y1? Answer: y1 = R1 - COS (A) * R1 | 677.169 | 1 |
Side
From Thinkmath
The word "side" becomes ambiguous when we are talking about three-dimensional figures. For example, casual (but not mathematical) speech might describe a cube as having six "sides," yet when people are asked how many "sides" a room (shaped exactly like the cube) has, they tend to answer four (not counting the ceiling or floor as "sides").
Also, if the "sides" of a square mean the linesegments that surround it, we might be tempted to say that a cube has twelve "sides" -- the twelve line segments that surround the square faces of the cube.
So, a question like "How many 'sides' does a cube have?" is ambiguous, because the answer might be four, or six, or twelve, depending on what "sides" is taken to mean. Because clarity and lack of ambiguity are so important in mathematics, mathematics does not use the word "side" for flat-surfaced three-dimensional figures, preferring face for polygons that form the surface of the figure, and edge for the line segments at which any two faces meet.
Question: Which of the following is NOT a correct way to describe the number of "sides" a cube has, according to the text? A) 4 (counting only the top and bottom) B) 6 (counting all faces) C) 12 (counting all edges) D) 8 (counting only the vertical faces) Answer: D) 8 (counting only the vertical faces) | 677.169 | 1 |
CE 353 Lab Week 6: Spiral Curves
Objectives
The objectives of this lab are to learn the theory of spiral curves for horizontal alignment, obtain hands-on experience with procedures for establishing these curves in the field and to assess quality of measurements by doing field checks. Groups will be assigned to develop appropriate curve data for circular and spiral curves. The curves will be layed out adjacent to each other so the curvature variations can be examined.
Project
The TS of a 200 foot spiral curve is at Station 21+50 . The delta angle is 30 degrees and the degree of curvature for the circular curve componenet will be assigned to each group, which has two crews (Crew A and Crew B. Each group is to layout two curves, starting at the same point, one spiral curve and one circular curve, so that differences can be observed.)
Computations: (note, all curves turn to the right)
Determine the key components of the spiral curve to include P, K, Ts, Es, Xs, and Ys, as well as the stationing for SC, CS and ST. For these equation, use theta(s) in radians.
Determine the deflection angles for each full and half-station as well as the backsight and foresight angles that would be needed to orient the theodolite if we need to set up on the point 150 feet from the TS in order to set the SC station. (Use theta(s) in degrees for the backsite and foresite deflection angles.) Consider that you sight back on TS to obtain your orientation.
Determine deflection angles and chord distances for laying out a simple circular curve of degree D, every 50 feet for 200 feet. (For laying out purposes, the circular curve also starts at station 21+50).
Field Procedure: (reminder, all curves turn to the right)
Two crews are used for each curve. The instructor will establish the line for each curve. Crew A sets up on TS, and tapes down the tangent to establish Xs.
After Crew A establishes Xs, Crew B sets up on Xs to turn and establish Ys.
While Crew B is setting up on Xs, Crew A will layout the spiral using the tape for measurements to the 50, 100 150 and 200 foot points. Note that the 200 foot point in the SC, and has already been established by Crew B. Therefore, there will be two pins at the SC (hopefully fairly close to one another).
After crew A finishes the 150 point of the spiral, Crew B will move their instrument to the 150 point of the spiral and set the SC for a third time, using the computed backsite and foresite angles. Now there are three pins at the SC (again, hopefully close to one another).
Question: What are the key components of the spiral curve that need to be determined? Answer: The key components of the spiral curve to be determined are P (point of curvature), K (curvature), Ts (tangent distance), Es (distance from point of curvature to end of curve), Xs (point of curve start), and Ys (point of curve end), as well as the stationing for SC (start of curve), CS (center of curve), and ST (end of curve).
Question: What is the total length of the curves that each group is to layout? Answer: Each group is to layout two curves, one of which is 200 feet in length.
Question: What are the objectives of this lab? Answer: The objectives of this lab are to learn the theory of spiral curves for horizontal alignment, obtain hands-on experience with procedures for establishing these curves in the field, and to assess the quality of measurements by doing field checks.
Question: What is the task of Crew A and Crew B in the field procedure? Answer: Crew A sets up on TS to tape down the tangent to establish Xs, while Crew B sets up on Xs to turn and establish Ys. After Crew A finishes the 150 point of the spiral, Crew B moves their instrument to the 150 point of the spiral and sets the SC for a third time, using the computed backsite and foresite angles. | 677.169 | 1 |
Computing the height of the building.
Solve the following application problem.
A.A man at ground level measures the angle of elevation to the top of a building to be 530. If, at this point, he is 12 feet from the building, what is the height of the building? Draw a picture, show all work, and find the solution. Round to the nearest hundredths.
B.The same man now stands atop a building. He measures the angle of elevation to the building across the street to be 200 and the angle of depression (to the base of the building) to be 350.
If the two buildings are 50 feet apart, how tall is the taller building? See the figure. Round to the nearest hundredths.
Question: How far is the man from the building when he measures the angle of elevation? Answer: 12 feet | 677.169 | 1 |
Problem at the Art Gallery
In my role as a security consultant, it looked like one of my tougher assignments. The Crystal Art Gallery was being readied for its grand opening, and I had to figure out how many guards were needed to make sure that every wall of paintings was under scrutiny. I was also ordered to keep costs as low as possible.
The trouble was that the art gallery had a distinctive shape. Its walls didn't meet at the usual right angles. Instead, its perimeter consisted of 11 walls forming a highly irregular, sharply cornered, deeply indented polygon. It looked stunning, both inside and out. But the geometry also complicated my task of determining the minimum number of guards that I would need to post.
I started by assuming that a guard stationed at a corner would be able to see down the two walls that meet at that corner. Then, as I was idly sketching the gallery's layout on a paper napkin, I noticed that this configuration was really just a graph, which is what mathematicians call a set of points (known as vertices) and a set of lines (known as edges) joining pairs of these points. I realized I could draw the polygonal art gallery as a graph consisting of 11 vertices (for the corners) and 11 edges (for the walls).
I liked that because it reminded me of the days long past when I had dabbled in computational geometry at night school. I had wanted to generate my own animated cartoon characters on the computer screen, and I had started taking some courses to learn computer graphics. Graphs turned out to be a great way of organizing and analyzing information. But that's another story.
Polygonal art gallery represented as a graph with 11 vertices and 11 edges (left), and one possible triangulation of that polygon (right).
What I vaguely remembered was that graph theory often involves coloring problems. In other words, you try to color a graph by assigning colors to each vertex so that no two adjacent vertices are the same color.
There's a famous, closely related problem in which you have to decide whether four colors are enough to fill in every conceivable map that can be drawn on a flat piece of paper so that no territories sharing a common boundary are the same color. It took mathematicians more than 100 years to prove that four colors are always enough.
It seemed to me that one way to find a solution to my art gallery problem was to subdivide the 11-sided polygon into triangles. I had to do it carefully, making sure that none of the lines I added crossed one another or passed outside the polygon's boundary. There are actually lots of different ways to do this, and any one way is as good as another.
That was a clever move, I thought to myself, because then I could apply a theorem stating that the vertices of any triangulated polygon can be three-colored. Using just red, green, and blue, I could color all 11 vertices of my polygon so that no two adjacent vertices were the same color. Thus, each triangle ends up with one corner of each color.
Question: How many colors were needed to three-color the vertices of the triangulated polygon? Answer: Three (red, green, and blue)
Question: What did the author realize while sketching the gallery's layout? Answer: That the gallery's layout could be represented as a graph with 11 vertices and 11 edges | 677.169 | 1 |
So, this is equal to a measure of angle one and this is obviously angle two. You see immediately that there have to be supplementary right because when you add them together you get a 180 degrees, alright. Together they go all the way around and they kind of form a line. So, you know that if this angle and this angle are supplementary and this angle is congruent to angle one then angle one and angle two must be supplementary. So, what did they say? So, angle one is definitely not necessarily congruent to angle two, right it's congruent to this angle here. Angle one is a compliment of angle two. Compliment means you add up to 90, no, we're talking about supplement, it's not that. Alright, angle one is a supplement of angle two, there you go and there's nothing that says at the right angle, that's silly.
All right, next problem, what values of A and B make the quadrilateral M N O P a parallelogram. Okay, for this to be a parallelogram, the opposite sides have to be equal and I challenge you to experiment to draw a parallelogram where opposite sides are parallel, where the opposite sides are also not equal. If you make two of the sides not equal then the other two lines are going to be a parallel anymore and you can play with that if you like but if opposite sides are going to be equal, that means 4a + b = 21 right because they are opposite sides so they should be equal to each other. Similarly, 3a - 2b = 13 because they are opposite sides. So, 3a - 2b = 13 and now we have two linear equations with two unknowns so this is really an Algebra 1 problem in disguise so let's see. If we want to—and they want to solve for both, let's see if we can cancel our b. So, let's multiply this top equation by two. So, times two and I'm doing that to cancel out the b's so you get 8a + 2b = 42 and I did that so that two b and the minus 2b cancel out. So, let's add these two equations to each other, the left hand side 3a + 8a = 11a, the b's cancel out -2b + 2b that's no b's = 42 + 13 = 55. Look that well divide both sides by 11, you get a = 5. Now, if a = 5 what's b? Let's substitute back at that first equation because I'll pick either. So, 4 x 5 + b = 21, 20 + b = 21 so subtract 20 from both sides b = 1. So, a = 5, b =1 that is choice b.
Question: To make quadrilateral M N O P a parallelogram, what must be true about the opposite sides? Answer: They must be equal in length
Question: What is the sum of angles one and two? Answer: 180 degrees
Question: What is the value of 'b' when 'a' is 5? Answer: 1
Question: What is the relationship between angle one and angle two? Answer: Angle one is a supplement of angle two | 677.169 | 1 |
Examples
Encyclopedia article of circle at compiled from comprehensive and current sources. Circles are simple closed curves which divide the plane into an interior and an exterior. — "Circle encyclopedia topics | ",
A circle is defined as the set (or locus) of points in a plane with an equal distance from a fixed point. Using the traditional definition of a circle, we can find the general form of the equation of a circle on the coordinate plane given its radius,. — "Circle - AoPSWiki",
Get information, facts, and pictures about circle at . Make research projects and school reports about circle easy with credible articles from our FREE, online encyclopedia and dictionary. — "circle Facts, information, pictures | ",
Circle. From LoveToKnow 1911. CIRCLE (from the Lat. circulus, the diminutive of circus, a ring; the cognate Gr. word is KipKcos, generally used in the form rcpi?cos), a plane curve definable as the locus of a point which moves so that its distance from a fixed point is constant. — "Circle - LoveToKnow 1911", 1911
A circle is one of the basic shapes of Euclidean geometry. The circumference of a circle is the perimeter of the circle, and the interior of the circle is called a disk. — "Circle", schools-
Circle illustration. This article is about the shape and mathematical concept of circle. In Euclidean geometry, a circle is the set of all points in a plane at a fixed distance, called the radius, from a given point, the center. — "Circle - New World Encyclopedia",
A circle is a round two-dimensional shape, such as the letter o. The centre of a circle is the point in the very middle. The radius of a circle is a line from the centre of the circle to a point on the side. All points on the circle are at the same distance from the centre. — "Circle - Simple English Wikipedia, the free encyclopedia",
Circle definition, a closed plane curve consisting of all points at a given distance from a point within it called the center. Equation: x2 + y2 = r2. See more. — "Circle | Define Circle at ",
The Diameter starts at one side of the circle, goes through the center and ends on the other side. Because people have studied circles for thousands of years special names have come about. — "Circle",
Learn about Circle on . Find info and videos including: Differences of Circle Y Saddles and Circle T Saddles, How to Find the Center & Radius of a Circle With a Circle Equation, How to Draw a Circle and much more. — "Circle - ",
Provides advice and resources to help communities bring people from diverse backgrounds together in dialogue to work on tough public issues, and create sustained change. — "Study Circles Resource Center",
Definition of circle from Webster's New World College Dictionary. Meaning of circle. Pronunciation of circle. Definition of the word circle. Origin of the word circle. — "circle - Definition of circle at ",
Images
Question: What is the traditional definition of a circle? Answer: The set (or locus) of points in a plane with an equal distance from a fixed point
Question: What is the perimeter of a circle called? Answer: Circumference
Question: What is the interior of a circle called? Answer: Disk
Question: What is the center of a circle called? Answer: Center | 677.169 | 1 |
About AppShopper
Triangle Utility
iOS iPhone
Triangle Utility allows to solve 3 different technical engineering problems using a triangle calculation. Using the bottom tab bar the user can switch at any moment from one problem to another maintaining the data used in any of the 3 different calculations. The available calculations are: Triangulation: calculate the distance of a remote point without accessing it, using two local points as position station to read angles. Distance: calculate the distance of two remote points between them using a local point station to read angles Sides: calculate angles and heights of a triangle inserting the sides length Triangle Utility shows also the true shape of the calculated triangle after data is inserted using a specific 'Scaled View'
Question: What feature allows the user to view the true shape of the calculated triangle? Answer: Scaled View | 677.169 | 1 |
Analysis
this page is still under development
A linear measure, as on the 'ruler' illustrated to the left, is generally made by calibrating a strip of "dimensionally stable" material, such as seasoned wood, or a plastic like Perspex (a.k.a. Lucite, or Plexiglas).
Dividers, such as the pair illustrated on the right, might be used to mark off contiguous, non-overlapping intervals along the strip. If the setting of the arms of the dividers is not altered during this operation, the resulting intervals are assumed to be...
of absolute size and
equal in size to each other.
If they are the latter, they are called "units".
The concept of absoluteness of size arises from the usually tacit, and usually unquestioned, assumption that the interval between the divider-points is independent of...
the general location, and
the orientation
...of the pair of dividers.
Thus, the interval between the divider-points
is expected to be the same whether the (unaltered) dividers are lying East/West somewhere in London, or lying Zenith/Nadir somewhere in Beijing. or indeed lying along the Vernal Equinox at the Jovian North Pole.
The second assumption (of equality) follows from the first.
In other words, Distance is taken to be a primitive,
that is, a "given" thing, which may be metered as a count of "units", as defined above.
Thus, Distance, being a count, is a quantity, not a quality.
There are no geometric axioms for either Distance or Equality,
so neither has geometric meaning.
Indeed, absolute (i.e., immutable) distance contradicts the axioms—
Assertion of absolute distance leads to the corresponding (but false) assertion that parallels cannot meet because they preserve a constant separation (i.e., a linear distance).
Question: Is distance a primitive or a derived concept? Answer: Distance is taken to be a primitive, or a "given" thing.
Question: What is distance, according to the text? Answer: Distance is a quantity, not a quality, and it is metered as a count of "units". | 677.169 | 1 |
the path, of 214.4 from the starting node, or about (214.4-70.7)/158.1
= 90.88% of the distance from (350,50) to (400,100). That is, the
midpoint will be at about (389.8,89.8).
We may retrieve similar results (with far less conceptual
work) by invoking the getTotalLength and getPointAtLength
as
in the following.
getTotalLength is a method applied to a
<path> (in this case the path named by the variable Path).
It returns a floating point number equal to (or at least a good
estimate of) the length of the path from beginning to end. In the case
of the above triangle it is about 428.8.
.getPointAtLength(r)
is, likewise, a method applied to a path, but it accepts a parameter
between 0 and 1 which specifies the proportion of the distance from the
beginning to the end of the path, at which we would like to find the
coordinates of a point. What getPointAtLength
returns
is
actually an SVG virtual point object. Mid, in the above code, has the
properties Mid.x=389.8 and Mid.y=89.8. Because SVG Point objects also
include a transformMatrix method in their class definition, the
manipulation of a path by transform commands will not confuse the getPointAtLength
method
of a path.
If
we consider the more general case of an arbitrary path (instead of the
simpler case where all components are linear), these two methods can
prove even handier. In the following case the path is composed of
linear segments except for one portion (which happens to contain the
midpoint) which is an elliptic arc.
The
problem here is that computing the length of an elliptic arc (or even
the "simpler" problem of the circumference of an ellipse) cannot be
done in closed algebraic form since the integral resulting from putting
the ellipse in parametric form can only be approximated through
infinite series. Both functions getTotalLength
and
getPointAtLength use fast and reasonably accurate
estimates for calculations involving both elliptic arcs and Bézier
curves.
We
conclude with the following example gives a general method for
interrogating any path in an SVG document, and at the same time gives
us some insight into the accuracy of the calculations involved.
We
observe that the green curve should, in fact, be centered at (150,125)
instead of the coordinates shown above that are off by a tiny fraction
of a percent.
Various text methods.
A
variety of specialized methods exist for <text> objects.
Rather
than presenting examples of the use of each, we merely excerpt from the
W3C's SVG 1.1 recommendation about these methods, and afterward show an
Question: How can you find the coordinates of a point at a specific proportion of the distance along a path? Answer: Use the getPointAtLength method
Question: What is the general method for interrogating any path in an SVG document? Answer: Using getTotalLength and getPointAtLength methods | 677.169 | 1 |
coplanar mean all the points lie on a plane or a flat 2-dimensional surface the points can be spread out but have to lie in the plane collinear means are the point lie in one straight lineif points are all collinear then they're also coplanar
Simple.Coplaner: Like the teacher said, Co = together, planer = plane. They are together in the same plane, whatever figure in one planeCollinear: Co = together, linear, sounds like line, right? Three points on the same line. That's collinear.Hope it helped.Have funCheers.
Question: What is the difference between 'coplanar' and 'collinear' points? Answer: Coplanar points lie on the same flat surface, while collinear points lie on the same straight line. | 677.169 | 1 |
By putting the point at some distance between these we can get any rotation between 0 and 180 degrees. We can get negative angles by moving in the opposite direction along the line.
Therefore we can do any rotation, translation combination in one rotation.
Doing the rotation-translation in one operation like this can make some calculations simpler and we don't have to consider whether we do the rotation around its original position first and then the translation, or do the translation first and then the rotation about its final position, or do a translation to the mid point first, then the rotation about the mid position and then the final half translation.
Calculating Rotation Point
Given a translation (specified by a 2D vector) and a rotation (specified by a scalar angle in radians) how do we calculate the rotation point P ?
We know the points A and B and the angle at P which is theta.
sin(θ/2) = v/(2*r)
r = v/(2*sin(θ/2))
where:
r = scalar distance of P from both A and B
v = scalar distance of B from A
θ = angle of rotation
We also know by Pythagoras:
(Px - Ax)2 + (Py - Ay)2 = r2
(Px - Bx)2 + (Py - By)2 = r2
This is getting a bit messy! Can we solve for P in vector coordinates?
I think I will have to give up with this approach and instead invert the matrix equations that we started with:
From the matrix derived above:
r00
r01
x - r00*x - r01*y
r10
r11
y - r10*x - r11*y
0
0
1
we get equations for offset:
Offsetx = x - r00*x - r01*y
Offsety = y - r10*x - r11*y
Where:
Offset = linear distance that a point on the object has moved.
x, y = coordinates of the point we are rotating around (relative to initial position of object).
We want to solve for x and y so we take the inverse of:
Offsetx
Offsety
=
1-r00
-r01
-r10
1-r11
x
y
to give:
x
y
=
__1__
(1-r00)(1-r11) - r01r10
1-r11
r01
r10
1-r00
Offsetx
Offsety
In this case the rotation matrix is:
cos(θ)
-sin(θ)
sin(θ)
cos(θ)
where
θ = angle of rotation
which gives:
x
y
=
__1__
(1-cos(θ))(1-cos(θ)) + sin(θ)sin(θ)
1-cos(θ)
-sin(θ)
sin(θ)
1-cos(θ)
Question: What is the formula to calculate the coordinates x and y using the rotation matrix? Answer: x = [(1-cos(θ))² + sin²(θ)] / (1-2cos(θ)+cos²(θ)) and y = [-sin(θ) + (1-cos(θ))sin(θ)] / (1-2cos(θ)+cos²(θ)) | 677.169 | 1 |
smithy wrote:Zaba - thank you, I was right! I think your educated husband was too - there is a distinction between the arrows directing attention to the items, and the one showing the direction of rotation.
Meaning the rotational arrow has a tail on it because it has no "starting point" like the others?
How unique are these symbols and terms? Could it just be a case of engineers seeing engineering stuff because that's how they look at things? In other words are they specifically engineering symbols and terms, or more general impressions that engineers might say could be applied?
Wrench wrote:Meaning the rotational arrow has a tail on it because it has no "starting point" like the others?
Yes, I think so mate.
Duckky, I agree with Trav, I think that's an Engineers hat not a skeptics one!
In respect to your "colored by a viewpoint" question though, which is a good one, I think using the dot-dash lines to show a centre line is an (engineering) drawing convention used most in orthographic (or multiple view) renderings everywhere, undoubtedly. And it's not applied correctly by Z., here, but does imply he'd had some training in technical drawing at some point. I think that's about it. Was it commonly taught in U.S. High Schools of the period? While I'm hijacking and generally messing up this thread, I wasn't taught about Radians at school and I wonder if you were, Trav. (Asked for Wrench's benefit!)
I took quite a bit of math: geometry, calculus, trig, algebra... All forgotten now, best I can do is figure out the tip. Radians sounds vaguely familiar. And that's another point to ponder, would a 40 year old Z who learned this stuff in his teens still remember it and use it 20+ years later without being in a related field of work at some point?
I don't ever recall learning the segmented center line and other drawing related things we've discussed. If it was a general education subject I must have been sick that week.
The thing that I never bought about the "Radian Theory" was that even though you and I may never have heard of one before, there really isn't anything to show that Z knew anything about mathematics. He just said the word "radian".
I'd say that it's not at all beyond the realm of possibility that he could have heard the word and found out what it meant and knew less about it than anyone on here who wouldn't have known about if it wasn't for Zodiac.
But anyway...
I think we've established with a good deal of probability that Z had some technical knowledge beyond what could be called general information. The question is whether this is specific enough to narrow it down to one field that really tells us something about the guy.
Question: Did Z have any mathematical knowledge beyond general information? Answer: There is no concrete evidence to suggest that Z had any mathematical knowledge beyond general information, despite him mentioning the term "radian".
Question: What is Duckky's perspective on the drawing convention used by Z? Answer: Duckky thinks that using the dot-dash lines to show a center line is an engineering drawing convention, but it was not applied correctly by Z. | 677.169 | 1 |
Another way to see what is happening is to consider P(5, 1.1).
This is nearly the 4D simplex but with its 5 vertices clipped off.
The exposed small face at each clipping is a tetrahedron.
We have a truncated 4 simplex.
As x moves from 0 to 1, the truncation of P(5, 1+x) increases until the exposed faces meet at points such as (1, 1, 0, 0, 0).
At this phase the faces of the 4D simplex, which begin as tetrahedra have also been truncated to become octahedra.
This general process can be extended recursively to yet higher dimensions but with receding possibilities of visualization.
Question: What is P(5, 1.1) in the context of the text? Answer: It is nearly a 4D simplex with its 5 vertices clipped off. | 677.169 | 1 |
The method for Bending a Cornice round the internal part of a Circular body on the Spring, p. Plate 146
Page Plate 146
Plate 146.
The method for bending a cornice round the internal part of a circular body
on a spring.
The method for bending a cornice round a circular body external as the cornices
a on the spring draw the lines a, b to the center of the body c b then draw
the arch lines d e g f and that will be the edge of the cornice streight
when bent round the body
Question: What will the arch lines represent when the cornice is bent? Answer: The edge of the cornice when it is bent around the body | 677.169 | 1 |
St 1. We are told that one of the angles is 90 but we don't know which. We know for sure that the side that equals 5 cannot be hypothenuse. Thus, the side 13 can be either the second leg of the right triangle or the hypothenuse. INSF
St2 Gives us an area (30) but we can not derive anything from that
If you combine 2 Stms we are able to identify whether 13 is a leg or hypothenus.
Area=1/2 * H * Base
we know that 5 is a base (or can be a hight) let's try to plug in numbers:
1/2 * 5 * 13 = 32,5 , since we are told that the area=30. Side 13 can not be a leg, thus - hypothenuse. we can say with confidence that the third side = 12 (1/2*5*12=30)
Question: Is it possible to determine the exact type of triangle (right, isosceles, etc.) with the given information? Answer: Yes, it is a right triangle. | 677.169 | 1 |
two sides are parallel and the other two sidesare not parallel. By orienting a trapezoid sothat its parallel sides are horizontal, we maycall the parallel sides bases.
Observe that thebases (See fig. 17-15.)
Figure 17-15.-Typical trapezoids.
The area of a trapezoid may be found
byseparating it into two triangles and a rectangle,
as in figure 17-16. The total area A of the
trapezoid is the sum of A1 plus A2 plus A3, andis calculated as follows:
Thus the area of a trapezoid is equal to onehalf the altitude times the sum of the
line that forms its outer boundary. Circumference is the special term used in referring to
radius at the point of tangency.
Figure 17-17.- Parts of a circle.
An ARC is a portion of the circumference of
a circle. A CHORD is a straight line joining the end points of any arc. The portion of the
area of a circle cut off by a chord is a SEGMENT of the circle, and the portion of the
circle's area cut off by two radii (radius lines)
is a SECTOR. (See fig. 17-18.)
Question: What is the term used to refer to the length of the curved part of a circle? Answer: Circumference. | 677.169 | 1 |
Cylindrical coordinate
Cylindrical coordinate system
The cylindrical coordinate system is a three-dimensional coordinate system which essentially extends circular polar coordinates by adding a third coordinate (usually denoted z) which measures the height of a point above the plane.
The notation for this coordinate system is not uniform. The Standard ISO 31-11 establishes them as (rho,varphi,z). Nevertheless, in many cases the azimuth varphi is denoted as theta. Also, the radial coordinate is called r while the vertical coordinate is sometimes referred as h.
rho is the distance from O to P', the orthogonal projection of the point P onto the XY plane. This is the same as the distance of P to the z-axis.
varphi is the angle between the positive x-axis and the line OP', measured counterclockwise.
z is the same as the Cartesian coordinate z.
Thus, the conversion function f from Cartesian coordinates to cylindrical coordinates is f(rho,varphi,z)=(sqrt{x^{2}+y^{2}},operatorname{atan2}(y,x),z),.
The conversion function f from cylindrical coordinates to Cartesian coordinates is f(x,y,z)=(rhocosvarphi,rhosinvarphi,z),.
Note that the atan2() function as used above is not standard: It returns a value between 0 and 2π rather than between -π and π as the standard atan2() function does.
Cylindrical coordinates are useful in analyzing surfaces that are symmetrical about an axis, with the z-axis chosen as the axis of symmetry. For example, the infinitely long circular cylinder that has the Cartesian equation x^2+y^2=c^2 has the very simple equation rho = c in cylindrical coordinates. Hence the name "cylindrical" coordinates.
where (rho,varphi,z) are the cylindrical coordinates, and n and k are constants which distinguish the members of the set from each other. As a result of the superposition principle applied to Laplace's equation, very general solutions to Laplace's equation can be obtained by linear combinations of these functions.
Since all of the surfaces of constant ρ, φ and z are conicoid, Laplace's equation is separable in cylindrical coordinates. Using the technique of the separation of variables, a separated solution to Laplace's equation may be written:
The Z part of the equation is a function of z alone, and must therefore be equal to a constant:
frac{ddot{Z}}{Z}=k^2
where k is, in general, a complex number. For a particular k, the Z(z) function has two linearly independent solutions. If k is real they are:
Z(k,z)=cosh(kz),,,,,,mathrm{or},,,,,,sinh(kz),
or by their behavior at infinity:
Z(k,z)=e^{kz},,,,,,mathrm{or},,,,,,e^{-kz},
If k is imaginary:
Z(k,z)=cos(|k|z),,,,,,mathrm{or},,,,,,sin(|k|z),
or:
Question: In the solution to Laplace's equation using cylindrical coordinates, what is the equation for the 'Z' part? Answer: d²Z/dz² = k²Z
Question: What is the equation of an infinitely long circular cylinder in cylindrical coordinates? Answer: ρ = c
Question: Which function is used to convert from Cartesian to cylindrical coordinates for the angular component? Answer: atan2(y, x) | 677.169 | 1 |
3. Mirascope Construction Using Geogebra
3.1 Define Two Parabolas Algebraically
In order to align the two parabolic mirrors accurately, we use algebraic functions to model the two face-to-face parabolas. We start with two constant variables, \(a\) and \(c\), where \(a\) controls the orientation and curvature of the parabolas, and \(c\) controls their vertical alignment or, in our case, the distance between the vertices of the two parabolas. For simplicity, we define \(a=0.15\) through the input box and convert it to a slider with an interval of \([0,2]\). Also, we define \(c=2\)through the input box and convert it to a slider with an interval of \([0,5]\). In GeoGebra, any free or independent variable can be converted to a slider for visualization with an initial value and an interval that defines its range of possible values. To convert an independent variable to a slider, we can right-click on the variable and enable "Show Object" or, alternatively, we could toggle the radio button to the left of a variable definition (Figure 5). A slider has other styles such as color, thickness, length, and increment, which can be changed whenever necessary. According to the above definitions, a can take any value in \([0,2]\); \(c\) any value in \([0,5]\). Both a and c could take other reasonable initial values and intervals for clear visualization. We further define \(p_u(x)=ax^2\), which will model the upward parabola, and \(p_d(x)=-ax^2+c\), which will model the downward parabola. By adjusting the values of \(a\) and \(c\), we could manipulate the shape of the mathematical mirascope. This first step to model the mirascope is illustrated in Figure 5.
3.2 Modeling Light Reflections Within the Mirascope
According to the previous analysis, we start with a point \(P\), which represents an arbitrary point on the object placed at the bottom of the mirascope. We further imagine a light ray that leaves point \(P\), reaches the upper parabola, is reflected to the lower parabola, and is ultimately reflected out of the mirascope through its opening. As a construction heuristic, we note that since the light ray is reflected back and forth in the mirascope, we need to choose from segments, rays, or lines as an appropriate mathematical object to model its path. Because there are cases where a line or a segment long enough may intersect a parabola at two points and thus cause construction confusion, we prefer segments or rays to lines. This is only technically necessary when GeoGebra or a similar tool is used, representing the connections and distinctions between physical ideas and mathematical ones. Whenever necessary, we could make a ray out of a segment to find the point of incidence, where the light touches the mirror. To simplify the process, we can hide the segments and other intermediate objects and then focus on the resulting ray. Additional Figures 6a and 6b show how a light ray from point \(P\) is reflected out of the mirascope.
Question: What are the equations for the upward and downward parabolas? Answer: The upward parabola is modeled by \(pu(x) = ax^2\) and the downward parabola by \(pd(x) = -ax^2 + c\).
Question: How can a ray be made out of a segment to find the point of incidence? Answer: By hiding the segments and other intermediate objects, the resulting ray can be focused on to find the point of incidence where the light touches the mirror. | 677.169 | 1 |
By putting the point at some distance between these we can get any rotation between 0 and 180 degrees. We can get negative angles by moving in the opposite direction along the line.
Therefore we can do any rotation, translation combination in one rotation.
Doing the rotation-translation in one operation like this can make some calculations simpler and we don't have to consider whether we do the rotation around its original position first and then the translation, or do the translation first and then the rotation about its final position, or do a translation to the mid point first, then the rotation about the mid position and then the final half translation.
Calculating Rotation Point
Given a translation (specified by a 2D vector) and a rotation (specified by a scalar angle in radians) how do we calculate the rotation point P ?
We know the points A and B and the angle at P which is theta.
sin(θ/2) = v/(2*r)
r = v/(2*sin(θ/2))
where:
r = scalar distance of P from both A and B
v = scalar distance of B from A
θ = angle of rotation
We also know by Pythagoras:
(Px - Ax)2 + (Py - Ay)2 = r2
(Px - Bx)2 + (Py - By)2 = r2
This is getting a bit messy! Can we solve for P in vector coordinates?
I think I will have to give up with this approach and instead invert the matrix equations that we started with:
From the matrix derived above:
r00
r01
x - r00*x - r01*y
r10
r11
y - r10*x - r11*y
0
0
1
we get equations for offset:
Offsetx = x - r00*x - r01*y
Offsety = y - r10*x - r11*y
Where:
Offset = linear distance that a point on the object has moved.
x, y = coordinates of the point we are rotating around (relative to initial position of object).
We want to solve for x and y so we take the inverse of:
Offsetx
Offsety
=
1-r00
-r01
-r10
1-r11
x
y
to give:
x
y
=
__1__
(1-r00)(1-r11) - r01r10
1-r11
r01
r10
1-r00
Offsetx
Offsety
In this case the rotation matrix is:
cos(θ)
-sin(θ)
sin(θ)
cos(θ)
where
θ = angle of rotation
which gives:
x
y
=
__1__
(1-cos(θ))(1-cos(θ)) + sin(θ)sin(θ)
1-cos(θ)
-sin(θ)
sin(θ)
1-cos(θ)
Question: What is the rotation matrix given in the text? Answer: cos(θ) -sin(θ); sin(θ) cos(θ)
Question: Which method is suggested to calculate the rotation point P in the end? Answer: Inverting the matrix equations
Question: What is the formula to find the inverse of the matrix and solve for x and y? Answer: x = [(1-r00)(1-r11) - r01r10] / (1-r00-r11+r00r11) and y = [r01 - (1-r11)r00] / (1-r00-r11+r00r11) | 677.169 | 1 |
Construction from perspective triangles
Two perspective triangles, and their center and axis of perspectivity
Two triangles ABC and abc are said to be in perspective centrally if the lines Aa, Bb, and Cc meet in a common point (the so-called center of perspectivity). They are in perspective axially if the crossing points of the lines through pairs of corresponding triangle sides X = AB·ab, Y = AC·ac, and Z = BC·bc all lie on a common line, the axis of perspectivity. Desargues' theorem in geometry states that these two conditions are equivalent: if two triangles are in perspective centrally then they must also be in perspective axially, and vice versa. When this happens, the ten points and ten lines of the two perspectivities (the six triangle vertices, three crossings points, and center of perspectivity, and the six triangle sides, three lines through corresponding pairs of vertices, and axis of perspectivity) together form an instance of the Desargues configuration.
Symmetries
Although Desargues' theorem chooses different roles for these ten lines and points, the Desargues configuration itself is more symmetric: any of the ten points may be chosen to be the center of perspectivity, and that choice determines which six points will be the vertices of triangles and which line will be the axis of perspectivity. The Desargues configuration has a symmetry group of order 120; that is, there are 120 different ways of permuting the points and lines of the configuration in a way that preserves its point-line incidences. One way of constructing the same configuration in a way that makes these symmetries more readily apparent is to choose five planes in three-dimensional space, and to form the Desargues configuration as the set of ten points where three planes meet and the set of ten lines where two planes meet. Then, the 120 different permutations of these five planes each correspond to symmetries of the configuration.
The Desargues configuration is self-dual, meaning that it is possible to find a correspondence from points of one Desargues configuration to lines of a second configuration, and from lines of the first configuration to points of a second configuration, in such a way that all of the configuration's incidences are preserved Coxeter (1964).
The Levi graph of the Desargues configuration, a graph having one vertex for each point or line in the configuration, is known as the Desargues graph. Because of the symmetries and self-duality of the Desargues configuration, the Desargues graph is a symmetric graph.
Related configurations
A non-Desargues (103103) configuration.
Question: What is the Levi graph of the Desargues configuration called? Answer: The Levi graph of the Desargues configuration is known as the Desargues graph. | 677.169 | 1 |
Now that you know how to work a
coordinate grid you can now go on to the longitude and latitude.
The first thing that you want to
learn is that longitude goes from the tip of the north pole to the very bottom of the
south pole. Latitude is something different. Latitude is parallel or in other words the
lines never touch each other.
The center line of longitude is the
Prime Meridian, and the center line of the latitude is the equator.
To work longitude and latitude you
must know that the Prime Meridian and the equator are always 0 degrees and the other lines
count by 15 degrees. The longitude lines go all the way up to 180 degrees east or
west. Latitude lines go up to 90 degrees north or south.
To work longitude and latitude you
might get the coordinates 60 degrees north and 15 degrees east. First, to get 60 degrees
north you go to the latitude and go up to the 60 because you are going north. And to get
the 15 degrees east you go to the longitude and go right to the 15 degree line because you
are going right to the east. Where the two lines meet is the spot you are looking
for. You should have ended up in Sweden!
Question: What is the range of values for latitude? Answer: The range of values for latitude is from -90 degrees to 90 degrees. | 677.169 | 1 |
Step-by-step Procedure for Calculating Directions
The following is the step-by-step procedure for calculating azimuth of line BC for the figure shown in Steps 1 through 4.
Step 1
Plan and prepare. Determine a known azimuth. In this case, it is shown to be 45 and the direction that the calculation will proceed (clockwise or counterclockwise). In this case, we are going to calculate clockwise from A to B to C. List the adjusted interior angle draw a sketch of the entire traverse. Be sure your sketch is reasonably accurate (that is, angles and distances should be close to scale). Label the points, the starting direction, and the interior angles. Orient the drawing properly to North.
Step 2
Perform the calculation. Start by writing down the starting azimuth. Add 180° to obtain the back azimuth. Subtract the interior angle to obtain the azimuth of the next line. If the result is greater than 360, subtract 360. Write down the azimuth on the sketch.
Step 3
Repeat the calculation for each line of the traverse. That is, add 180° and subtract the interior angle.
Step 4
Check the calculations by using the last interior angle to recalculate the starting azimuth.
Example
For the figure below, calculate the direction of each line and provide a check. Proceed clockwise about this traverse. The calculations are shown in a tabular form for ease of understanding along with a sketch of the calculation of each line.
Calculation of the Direction of Line AB as a Check
Summary
That's all there is to calculating azimuths. Simply follow a simple rule if going clockwise about the traverse: "Subtract the interior angle from the back azimuth of the previous course," and the calculations can be performed quickly and easily. If bearings are needed, simply convert from azimuths to bearings as shown in the Table at the beginning. For those who aren't using azimuths, I hope this article helps in understanding why many people now use azimuths.
Wesley G. Crawford is a professor of building construction management at Purdue University. Wesley Crawford is the author of the highly acclaimed book Construction Surveying and Layout, now in its third edition. For information, visit or (click on POB Store).
Calculation of the Direction of Line AB as a Check
Chris
September 16, 2010
Should your interior angle for the last calculation of the Azimuth EA be labeled 107o33'? Your calculation is correct, but I think the label on your drawing should be changed to reflect this.
I'm just learing this for the first time, so please excuse me if I've misunderstood something.
Thanks,
Chris
Reply: [email protected]
thanks
hussan
March 13, 2012
its too good awoesome and very quickly
Question: What is the known azimuth in the given example? Answer: 45 degrees | 677.169 | 1 |
The Story of Ninety Three Degrees - whats in a name
In geometry and trigonometry, a right angle is an angle of 90 degrees. Throughout history carpenters and masons have known a quick way to confirm if an angle is a true "right angle."
It is based on the most widely known Pythagorean triple (3, 4, 5) and so called the "Rule of 3-4-5." From the angle in question, running a straight line along one side exactly three units in length,
and along the second side exactly four units in length, will create a hypotenuse (the longer line opposite the right angle which connects the two measured endpoints) of exactly 5 units in length. This
measurement can be made quickly and without technical instruments.
So why the extra 3 degrees? It was Wernher von Braun, said to be the preeminent rocket engineer of the 20th century, who discovered that slight erroneous calculations in the targeting of missiles had the unexpected benefit of delivering the required acceleration to accurately reach targets.
This led to missile technology being altered to initially follow a deviated flight path and then include flight correction for the missile to hit it's mark.
At Ninety Three Degrees we believe in going beyond the right-angle or best practice to ensure that our clients achieve their strategic goals accurately.
In Greek numerology, ninety-three is the value of the word "thelema," or "will". It means to find out what your true will is, your purpose in life, and then do only those things that will lead to achieving this goal.
We at Ninety Three Degrees believe in applying this philosophy of self discovery to the operations of business by implementing practical and measurable solutions. Our consultants are passionate about focussing on and assisting businesses in prospering through achievement.
Question: What is the traditional method used by carpenters and masons to check if an angle is a right angle? Answer: The Rule of 3-4-5 | 677.169 | 1 |
Calculate Angles how to articles and videos including How to Calculate Interior Angles, How To Calculate an Angle From Two Sides, How to Find the Sum of the Number of Sides of a Polygon and much more!. — "Calculate Angles - How To Information | ",
Images
the house that Alex built is a sprawling complex over under and inside the hill that he chose for his home It is a claustrophobic place low ceilings and red felt everywhere strange angles and twisting corridors carved wood and stained glass windows And every surface is covered with an eclectic collection of statues and decorations from all over the world Spider
Last Year s Conference Last year s conference was a big success We plan to do even better this year Here are a few pictures of us being serious A few angels that happened to fly in for a visit
Team Store tonight so I walked over to the wall to read some of the displays there This is part of what is under Adopt A School Remind anyone of our friend who demanded common courtesy
ANGLES
angles jpg
07 angles jpg
angles jpg
Heres a diagram of the position your should be for each location of the puck The black line betwean the green lines represents where your body should be Those are where you should be when
29 May 2007 9 29 pm UTC Is it legally OK for Google in the USA to display faces of strangers in close ups on Google Maps http blog outer court com archive 2007 05 29 n38 html http blog outer court com files street view angles jpg How about the situation in e g Germany And how about the legal situation when the site is hosted in the US but viewed from
ANGLES JPG
from 3 different angles I got the idea to shot a reproduction of the Hamangia Thinker and as usually went overboard and selected 6 angles as the ultimate proof I don t have self control
angles
Click for detail
will allow a neat straight line of figures to be placed on a barrel The jig is held to the barrel to be marked with a U Bolt Two pieces of steel angle about 3 inches long with 1 1 2 inch legs have four holes drilled near one end
p 710 angles 1241728988 jpg
A mid rise residential building is now going up on this site I think the construction fences went up the day after I took this picture and now they are scraping up the surface parking lot Stone Street Gardens with this restored building
angles jpg
angles
Dessins d octobre 2008
Angles jpg
Angels 16 x20 acrylic 2005
angles jpg
Pascal Tessier Dessins d octobre 2008
Origin Wolverine watched the almost end of it where Silver Fox reappeared It is also applicable to terrorism metaphor used by the governments That would be a conclusion for now Filed under Political Sciences 13 April 2009
Another view of an LG concept LCD screen Album Before the doors open with LG and Samsung Credit Tim 6 of 16 photos
Videos
Question: Which month and year was the artwork "Angels 16 x 20 acrylic 2005" created? Answer: The artwork was created in October 2008.
Question: What is the total number of times the word "angles" appears in the text? Answer: 14 times. | 677.169 | 1 |
1.) if sin x= (12/13), cos y=(3/5) and x and y are acute angles, the value of cos(x-y) is
a.(21/65) b.(63-65) c. -(14/65) d. -(33/65)
2.) if the tangent of an angle is negative and its secant is positive, in which quadrant does the angle terminate
1,2,3 or 4 i thought it was the second quadrant but i dont understand the term terminate i wasnt sure if that ment the opposite of what it was.
Question: In which quadrant does an angle terminate if its tangent is negative and its secant is positive? Answer: The third quadrant | 677.169 | 1 |
Proposition 12
To draw a straight line perpendicular to a given infinite straight line from a given point not on it.
Let AB be the given infinite straight line, and C the given point which is not on it.
It is required to draw a straight line perpendicular to the given infinite straight line AB from the given point C which is not on it.
Take an arbitrary point D on the other side of the straight line AB, and describe the circle EFG with center C and radius CD. Bisect the straight line EG at H, and join the straight lines CG, CH, and CE.
I say that CH has been drawn perpendicular to the given infinite straight line AB from the given point C which is not on it.
Since GH equals HE, and HC is common, therefore the two sides GH and HC equal the two sides EH and HC respectively, and the base CG equals the base CE. Therefore the angle CHG equals the angle EHC, and they are adjacent angles.
But, when a straight line standing on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands.
Therefore CH has been drawn perpendicular to the given infinite straight line AB from the given point C which is not on it.
Q.E.F.
Again, the double-equilateral-triangle construction is used, but this time the preparation of the starting line EG is different. The point D is taken on the other side of the line AB to insure that circle meets the line AB in at least two points, E and G. If D is taken on the line AB, it might be taken at H, and the resulting circle would touch the line only at H; and if D is taken on the same side of AB, then the circle could miss the line entirely.
Euclid does not precede this proposition with propositions investigating how lines meet circles. He is much more careful in Book III on circles in which the first dozen or so propositions lay foundations. For instance, Proposition III.10 states that a circle does not cut a circle at more than two points. Even so, some propositions are missing. One is needed for this proposition to justify the existence of the two points C and E where the line AB meets circle with center C and radius CD. Such a proposition would state "A circle whose center is on one side of a line and on whose circumference lies a point on the other side of the line meets the line at two points."
Incidentally, Proclus explains in his commentary on Book I that the problem of constructing the perpendicular was investigated by Oenopides of Chios who lived sometime in the middle of the fifth century B.C.E., a century and a half before Euclid.
Use of Proposition 12
The construction of this proposition is not used in Book I, but it is
used on occasion in the remaining geometric books, namely, Books II through IV, VI, and XI through XIII.
Question: Which ancient mathematician investigated the problem of constructing the perpendicular before Euclid? Answer: Oenopides of Chios, who lived in the middle of the fifth century B.C.E. | 677.169 | 1 |
We are now wondering when builders are building houses how do they measure the angle on the house that they are building. Do they have to carry a huge protractor to measure the house? When we were building the houses we had trouble keeping them straight and stable. I think that builders must have an easier way because they use metal in the middle and it is going straight up. Building with chopsticks it is hard to make it stay straight. Sometimes the way you put the elastic bands on is wrong and it pushes the chopstick to go at an angle that you don't want it to be.
If we were using the app on the IPad during the building of our models, we would be able to make it straighter. We could measure it and then adjust it as we are doing the building. If our building bent in the wrong places it would fall down or the roof would fall through. The only reason we are trying to build a house is because it is part of our Maths inquiry. Our central idea is "Geometric tools and methods can be used to solve problems relating to shape and space". Actually the main reason we are building a house is because we are hoping to build a 5×4 meter community house for the people at the Serpong landfill. Using geometric tools to build a house can help when you want to make the columns or the walls straight and so that you know what the angle of the roof is. Then you can work out how long the roof is and how many roof tiles or iron roofing you will need."
Here is the video that Ms Jane took of her lesson. Please watch it to learn about how we used the iPad app to measure the angles.
4 comments:
Jane, You do such an amazing job at showing the learning and curiosity in your classroom - it inspires me to up the anti with my own classroom blog. Thank you for sharing. Kirsten Moss in Nanaimo, Canada
We have a 1:1 too, but the students have MacBook Pros instead of iPads. I'm wondering how the angle measures might go. We have one class set of iTouches in the building. I'll see if we can get A+ Measuring finger onto the iTouches.
I also like your integration of bamboo. Hong Kong uses bamboo when they build and repair skyscrapers
Question: What is the name of the iPad app used to measure angles? Answer: The name of the app is not mentioned in the text.
Question: What was the intended size of the house they aimed to build? Answer: They aimed to build a 5x4 meter community house.
Question: What tool do builders use to measure angles while constructing houses? Answer: They use geometric tools like rulers, squares, and levels. | 677.169 | 1 |
If it is a right triangle, one of the angles has to measure . One other angle is given as . The sum of the measures of the three angles of any triangle is .
John
My calculator said it, I believe it, that settles it
Linear-equations/334213: A student has earned scores of 87, 81, and 88 on the first 3 of 4 tests. If the student wants an average (arithmetic mean) of exactly 87, what score must she earn on the fourth test? 1 solutions Answer 239470 by solver91311(16868) on 2010-08-23 17:05:55 (Show Source):
Think about how we take an average. We add up all of the numbers and divide by the number of numbers. The final average for our student will be the sum of all 4 test scores divided by 4. Let's say we want to achieve an overall average of for data elements given data elements.
We know that
So we can say that
But , the answer to the question, can be found by:
Which is to say:
So add up the scores for the tests already taken and subtract that sum from the product of the total number of tests multiplied by the desired overall average. Done.
The minimum number of elements in B is zero, because the null set is a proper subset of any set other than the null set.
John
My calculator said it, I believe it, that settles it
Miscellaneous_Word_Problems/334210: A carpenter is building a rectangular room with a fixed perimeterof 112 ft.
What dementions would yeild the maximum area? What is the maximum area?
The lenght that would yeild the maximum are is __?__ft 1 solutions Answer 239467 by solver91311(16868) on 2010-08-23 16:51:44 (Show Source):
available fencing. Therefore, the shape must be a square, and the area is the width squared.
Calculus Solution:
The area function is continuous and twice differentiable across its domain, therefore there will be a local extrema wherever the first derivative is equal to zero and that extreme point will be a maximum if the second derivative is negative.
so that it is expressed in the same power of 10 that the other number is in.
So, reduce the exponent by 2, and move the decimal point two places right.
Now that the decimal points line up, you can just subtract the numbers:
But just subtracting two integers gives you the number of integers between them including one of the endpoints. Since you want to eliminate both endpoints, you need to subtract 1 more unit. Furthermore, you need to put the decimal point back where it belongs.
Hence:
John
My calculator said it, I believe it, that settles it
Question: What is the measure of one of the angles in a right triangle? Answer: 90 degrees | 677.169 | 1 |
2-May-2000, 10:21
"preruse" ?
Pete Andrews
2-May-2000, 10:50
The word "geometry" also has another meaning when applied to lenses. It can be used to describe how well a lens renders straight lines straight, or whether the reproduction scale changes from the centre of the field to the edge. For instance; a lens exhibiting barrel distortion could be described as having poor geometry.
On a more philosophical level, we wouldn't have lenses at all without geometry, nor a good deal of the rest of civilisation as we know it, such as architecture or the ice cube.
Question: What is one of the many aspects of civilization that relies on geometry, according to the text? Answer: Architecture | 677.169 | 1 |
Question 482014
5pi/3 does correspond to a main angle. And you can do it with only the unit circle.
First locate 5pi/3 on the unit circle. The x-coordinate of that point (you determine this using the special right triangle) is the cosine of that angle, and the y-coordinate is the sine of that angle. Since tangent of an angle is defined as the sine over the cosine, you can find tan(5pi/3).
Question: What is the x-coordinate of the point on the unit circle that corresponds to 5pi/3? Answer: The cosine of 5pi/3. | 677.169 | 1 |
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