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On orthogonality, that is implicit in the definition usually adopted - but you will note that the formula for the cosine then involves dividing by zero. It would be possible to adopt the idea that orthogonal vectors were non-zero, but that would mean that every time you used the concept it would be necessary to consider special cases. It turns out that the usual definition works OK - the special cases don't cause a problem (like you might expect with dividing by zero).
In fact you will find that there are times when you will use this property of the zero vector (it is the only one which has zero dot product with every other vector) to prove other useful facts about vectors and their arithmetic - so having the word "orthogonal" to hand is convenient.
The angle does not have to be in any range. This formula does not give the angle at all. It only gives a cosine of the angle. It so happens that if $0 \leq \theta \leq \pi$ then $\cos \theta$ could have any value from $1$ to $-1$. This sort of mapping just makes sense.
I assume by "standard position" you mean "start at the origin".
The thing is that vectors have no positions at all. They are only lengths and directions. A vector is very often associated with some point, but that is not part of the vector. When finding the angle, you only draw vectors from the origin for clarity.
The only definition of orthogonality of $\vec{x}$ and $\vec{y}$ is that $\vec{x} \cdot \vec{y} = 0$. Naturally $\vec{0}$ always satisfies this and is thus orthogonal to every other vector.
Given two points $x$ and $y$ on the unit sphere $S^{n-1}\subset{\mathbb R}^n$ the spherical distance between them is the length of the shortest arc on $S^{n-1}$ connecting $x$ and $y$. The shortest arc obviously lies in the plane spanned by $x$ and $y$, and drawing a figure of this plane one sees that the length $\phi$ of the arc in question can be computed by means of the scalar product as $$\phi =\arccos(x\cdot y)\ \ \in[0,\pi]\ .$$ This length is then also called the angle between $x$ and $y$.
When $u$ and $v$ are arbitrary nonzero vectors in ${\mathbb R}^n$ then $u':={u\over |u|}$ and $v':={v\over |v|}$ lie on $S^{n-1}$. Geometrical intuition tells us that $\angle(u,v)=\angle(u',v')$. Therefore one defines the angle $\phi\in[0,\pi]$ between $u$ and $v$ as
$$\phi:=\arccos{u\cdot v\over|u|\>|v|}\ .$$
Question: Is it possible for two vectors to be orthogonal if one of them is the zero vector? Answer: Yes, any vector is orthogonal to the zero vector.
Question: What is the definition of orthogonality between two vectors x and y? Answer: x · y = 0 | 677.169 | 1 |
Remember that a^2 +b^2 = c^2. That should read ( a squared) plus (b squared) equals (c squared)the hypotenuse is c.So to answer your question, a right traingle could have many different length sides as long as it satifies this equation. However, in order to get the exact length for a particular triangle, you need more information. For example if you knew the length of one of the sides that would help. Or even if you did not know the length of one of the sides, but you knew that they were the same length, that would help. So to find the sides of this right triangle, you really need more information either the length of one of the sides or what is the measure of another inside angle beside the already mentioned right-angle.
technically even though this is a right triangle you need two sides to determine the 3rd side; to answer your question however sides a and b that satisfy pythagorean will work ie a^2+b^2=36(or 6 squared)
Question: Can a right triangle have sides of any length? Answer: No, the sides of a right triangle must satisfy the Pythagorean theorem, so they cannot be of any length. | 677.169 | 1 |
Proving Segment Relationship
In this lesson our instructor talks about proving segment relationship. She talks about five essential parts to a good proof. She also discusses reasons used in proof such as undefined, definitions, postulates, and previously proven theorems. She talks about congruence of segments and the theorem. She does some proof example for setting up proofs. Four complete extra example videos round up this lesson.
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Proving Segment Relationship
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
Question: What is the main topic of this lesson? Answer: Proving segment relationship | 677.169 | 1 |
Thanks! that formula worked like a charm for automating my radius calculations along a railroad track. It gave pretty much the same results as least squares circel fitting.
Anyone know what the name of the formula is and/or a scientific paper i can refer to it?
I have never seen a name given to it. It is included in most math reference books. In the CRC Standard Mathematical Tables, it is covered in the Geometry section, under sector and segment of a circle. (Page numbers vary in different editions)
Question: Is the formula's name and origin discussed in the text? Answer: No, the text only mentions that it is included in most math reference books but does not provide a name or origin. | 677.169 | 1 |
Identify the 3D Shapes PowerPoint is a great introduction to learning the names of 3D shapes. Students are asked to look at a shape and decide whether it is a cube, cylinder, cone, rectangular prism, pyramid, or sphere. Each shape is shown twice. This would also go well with my Super Shapes 2D and 3D packet!
Be sure that you have an application to open this file type before downloading and/or purchasing.
474.6 KB | 13
Question: How many slides are there in the PowerPoint? Answer: There are 13 slides in the PowerPoint. | 677.169 | 1 |
CBSE Board Class 10 Math Sample Papers 2010
CBSE Board Sample Papers 2010 for Class 10 Math
Sample Paper – 2010
Class – X
Subject –Mathematics
General Instructions:
1.All questions are compulsory.
2.The question paper consists of 30 questions divided into four sections – A, B, C and D. Section A contains 10 questions of 1 mark each, Section B is of 5 questions of 2 marks each, Section C is of 10 questions of 3 marks each and section D is of 5 questions of 6 marks each.
3.In question on construction, the drawing should be neat and exactly as per the given measurements.
4. In question on construction, the drawing should be neat and exactly as per the given measurements.
5.Use of calculator is not permitted. .
SECTION – A
1.Is number 5 x 3 x 17 + 17 a prime number?
2.The sum and product of the zeroes of the quadratic polynomial are – ½ and – 3. What is the quadratic polynomial?
20.Find the area of the minor and major segment of a circle, given that the central angle is 600 and radius of the circle is 14 cm.
21.Which term of an A.P 3, 10, 17…………. will be 84 more than its 13th term?
22.Two circles of radii 2 cm and 3 cm with their centres 7 cm apart. Draw the tangents from the centre of each circle to the other circle.
23.Find the ratio in which the point P( m, 6 ) divides the join of A( - 4, 3 ) and B ( 2, 8 ). Also find the value of m.
24.A two digit number is such that product of its digits is 18, when 63 is subtracted from the number, the digits interchange their places. Find the number.
25.In ABC, right angle at B. AD and CF are the two medians drawn from A and C, if AC = 5 cm and AD = ( 3√5) / 2 cm, find the length of CE.
SECTION –D
26.In a flight of 600 km an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/h and the time of flight increased by 30 minutes. Find the duration of the flight.
27.A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream, also find the silver paper required to cover each ice cream cone.
Question: What is the total number of questions in the paper? Answer: There are 30 questions in total.
Question: What is the maximum marks for section D? Answer: Each question in section D is worth 6 marks, so the maximum marks for section D is 5 * 6 = 30 marks.
Question: What is the use of a calculator not permitted for? Answer: The use of a calculator is not permitted for any of the questions in the paper. | 677.169 | 1 |
Angles/15274: two angles are complementary.The mesure of one angle is twice the mesure of the other angle. what is the smaller of the two angles? 1 solutions Answer 7573 by rapaljer(4667) on 2005-10-13 20:45:04 (Show Source):
Linear_Algebra/15263: Sarah is visting a friend about two miles away from her house. she sees a hot-air balloon directly above her house. Sara estimates that the angel of elevation formed by the hot-air balloon is about 15 degrees. about how high is the hot-air balloon?
You can put this solution on YOUR website! This figure is a right triangle with the balloon at a height of x miles above Sara's house. The base of the triangle represents the distance from Sara's house to where Sara is at her friend's house. The angle from Sara's friend's house to the balloon is 15 degrees, so trigonometry is required.
You can put this solution on YOUR website! First, notice that Bob's money is given in terms of Anthony, and Anthony's money is given in terms of Lisa. So, let x = the quantity at the end of all of this, which is Lisa.
You can put this solution on YOUR website! Express the fraction 2/5 as a decimal by dividing 5 into 2.00 (you get .40), and then convert the decimal to a percent, by moving the decimal two places to the right. This means that .40 = 40%.
Another way to do this particular problem is to start with the fraction 2/5 and multiply the numerator and denominator by 2. This gives the very special fraction 4/10, which as a decimal would be .4 or .40. Then convert to a percent as above .40 = 40%.
Angles/15255: if two angles of a triangle are complementary find the number of degrees in the third angle of the triangle 1 solutions Answer 7563 by rapaljer(4667) on 2005-10-13 20:08:46 (Show Source):
You can put this solution on YOUR website! If two angles are complmentary, then the sum of those angles is 90 degrees. Since the sum of all three angles must be 180 degrees, the third angle must be 180-90 = 90 degrees.
You can put this solution on YOUR website! Let x = width of the rectangle
x+3 = length of the rectangle
Area = L*W
x(x+3)= 70
x=-10
Reject, since width of a rectangle can't be negative
x= 7 cm Width
x+ 3 = 10 cm Length
Check:
Area= 7*10= 70 sq.cm.
R^2 at SCC
Question: What is the measure of the smaller angle in the complementary angles scenario? Answer: 75 degrees
Question: What is the length of the rectangle in the given problem? Answer: 10 cm
Question: What is the width of the rectangle in the given problem? Answer: 7 cm | 677.169 | 1 |
Parts of a Cone
Date: 04/18/2001 at 12:59:14
From: Brian McCormick
Subject: Parts of a solid cone
Hello,
I am a second grade teacher and we are currently teaching a unit on
shapes. The question came up as to whether or not a solid cone has any
edges. My contention is that the definition of an edge is where two
planes intersect, and therefore a cone cannot have an edge. Another
teacher says that the curved surface of a cone represents an infinite
number of planes, and therefore represents an infinite number of
edges.
I would very much appreciate your response, and don't be afraid to get
technical. This is as much to satisfy my own curiosity as to let the
kids know the proper answer.
Brian McCormick
Date: 04/18/2001 at 14:25:21
From: Doctor Peterson
Subject: Re: Parts of a solid cone
Hi, Brian.
We get this question from time to time, and can never really give a
definite answer. The word "edge" is used in different ways; often
people get in trouble by introducing the concept of "edge" in the
context of polyhedra (where it does mean the intersection of two flat
faces), but then talking about curved surfaces like cones without
additional comment.
Here's the definition in the Academic Press Dictionary of Science
and Technology:
1. in graph theory, a member of one of two (usually finite) sets
of elements that determine a graph; i.e., an element of the edge
set. The other set is called the vertex set; each element of the
edge set is determined by a pair of elements of the vertex set...
2. a straight line that is the intersection of two faces of a
solid figure.
3. a boundary of a plane geometric figure.
In the latter sense (which I think is appropriate in discussing a
cone, even though the dictionary only mentioned plane figures and not
curved surfaces), the cone has one edge. I definitely would not bring
in the idea of "an infinite number of edges"; that kind of reasoning
generally leads to trouble! I would simply say that we can extend the
concept of edge either from the world of polyhedra (definition 2) or
from the world of plane geometry (definition 3) to apply to possibly
curved boundaries of possibly curved surfaces, as long as we say that
we are doing so. This also agrees with definition 1, which likewise
does not require straightness (indeed, there is no such concept in
graph theory), and which relates to boundaries when we consider planar
graphs (as in Euler's polyhedral formula).
What definition you use depends on what you are going to do with it.
If you are just describing objects, my loose definition is fine. If
you are going to prove theorems involving planes and angles, you'll
want to restrict yourself to the polygonal definition, but then you
Question: What is the difference between the two definitions of an edge mentioned in the text? Answer: The first definition is specific to graph theory and involves sets of elements, while the second definition is more general and can apply to boundaries of both plane and curved surfaces. | 677.169 | 1 |
Lines
Perspective
can cause you to see lines differently depending on how they are presented.
If you drew three equal lines, but drew slanting lines on the end of each
as in the figure below, the added lines would change the way you see them.
The top line looks bigger and the bottom line looks the smallest.
But
this illusion can be destroyed if you change the order of the lines. Below
the lines now appear to be the same length.
Sometimes
we see things differently because of what its surroundings. If you look
at the figures below, one of the circles looks bigger in each figure,
but they are really the same size. Can you figure out why?
You
can also use straight lines to make a circle appear top bend, and circle
to make straight lines appear to bend. If you put a circle around a square,
the circle appears to pinch in at the point where it touches the square.
If
you put a series of circles inside that same circle, the lines of the
square appear to bend. Even the direction you look at something may make
something look different to you. The two squares below are the same size,
but the one on the left looks bigger because it is turned to be a diamond.
So how you look at something and what surrounds it adds up to the perspective
and that adds up to optical illusions.
Question: Are the two squares in the last example the same size? Answer: Yes. | 677.169 | 1 |
"You are a professional, after all", Greyskin added.
"Allright," Smith surrendered, "never mind the color. But there's also something about the straight lines being perpendicular?"
Mrs. Redroot shook her head, trying to get rid of the old ghost of her middle school education.
Lehare slammed his fist on the table: "Smith, let's try to avoid this 'sixth grade, sixth grade'. Let's keep this civil. Let's not make suggestions or even insult each other. Let's maintain a constructive dialogue. The people gathered here are not a bunch of idiots, are they?"
"I agree", Greyskin said.
Smith took out a piece of paper.
"All right then," he said. "Let me draw this out for you. This is a line, right?"
Mrs. Redroot nodded in agreement.
"Let's draw another one," said Smith. "Is it perpendicular to the first one?"
"well…"
"Yes, it is perpendicular."
"Well, you see?" Mrs. Redroot exclaimed happily.
"Wait, there is more. Now we draw a third line. It is perpendicular to the first line?"
A silence followed, the other attendees deeply in thought. Smith then answered his own question:
"Yes, it is perpendicular with the first line. However, it does not cross the second line, it's parallel to it."
Another silence. Then, Mrs. Redroot rose and walked around the table to view Smith's work from behind his shoulder.
"Well…" she said, "yes, I suppose."
"That's exactly the problem," Smith quickly said, trying to maintain the achieved success. "As long as there are two straight lines, they can be perpendicular. But when there are more…"
"Can I try please?" Mrs. Redroot asked.
Smith handed her over the pen. Mrs. Redroot cautiously drew several lines. "And what about this?"
Smith sighed. "This is called a triangle. No, these are not perpendicular lines. Plus, there are three, not seven."
Mrs. Redroot looked disgruntled.
"But why are they blue?", Lehare asked suddenly.
Greyskin supported this question: "Yes, that's what I wanted to ask as well."
Smith blinked a couple of times, staring at the drawing. "My pen is blue," he said in the end. "I just tried to show you…"
"Well, maybe that's the problem?", Lehare quickly interrupted him, with the air of someone who just solved a major problem and is anxious to share his solution with others before he forgets it. "Your lines are blue. Draw them in red, and let's see what happens then."
"I don't have a red pen on me", Smith admitted. "But I can certainly…"
"I can tell you with absolute certainty," Smith desperately stated, "that in red the result will be exactly the same."
Question: What was Smith's final statement about drawing the lines in red? Answer: He stated that drawing the lines in red would result in the same outcome.
Question: Was Smith initially focused on the color of the lines? Answer: Yes, he was. | 677.169 | 1 |
April 19, 2011
This is the second of an occasional series of posts about perspective. Many people believe that geometrical perspective, single point, two point and three point are actually an accurate representation of what we see. Cameras see in this way after all so it must be right mustn't it? Well actually no, it is a compromise as are all methods of making our very three dimensional world fit conveniently on a flat surface. Here I am going to deal with a very old problem that perplexed Renaissance artists as they struggled to find solutions to the problems of illusory painting. Vredeman de Vries and other artists published learned books full of geometrical construction but certain problems seemed impossible to resolve. Columns were a big feature of architecture of the time and they often occurred in long arcades, perfect fodder for the perspective method you would think. However it turns out that round columns are exactly the type of object that causes the neat geometrical rules to fall apart. What I intend to do here is highlight the issues as clearly as I can and then point the way to the various solutions that later artists arrived at. It is a sad fact that all modern books on perspective that I have seen do not even seem to realise that the issues are there, let alone giving any practical advice to overcome them.
.
Here we have a plan view of a simple set up consisting of a row of columns, a cube and a green triangle which marks where our viewer is standing.
Below is what we get if we construct using one point perspective what that viewer would see.
.
Well here we are, at first glance it seems sort of OK. Look more closely though and we have some problems. In our plan we can clearly see that all the
columns are the same size. That does not however seem to be the case in our perspective projection. The column on the far left is a lot wider than the one
straight in front of us. Worse when we look at the plan the far left column it is actually further away from us and should appear smaller not larger. Something
is plainly awry. Looking more closely still the base of the far left column seems oddly tilted. This is exactly the result a camera would give on a fairly wide
angle lens giving a viewing angle of about 70 degrees. So when you wonder why you looked so fat in that group photo this is the cause and in future you
would be best to make sure you are in the middle of the line! Lets take another case.
.
Here is the plan of a simple set up, as before the green triangle marks our viewpoint. We often get rows of things receding from us, looking down a romanesque
church nave would be an example. Below is how traditional two point perspective renders the scene.
.
Once again at a quick glance it all seems well but a closer one shows that the left column again shows problems. These are clearer still when we isolate
that part as in the green circle. There is a weird tilt which is plainly not how we would really see the base of such a column. If you then tried to add capitols
Question: What is the text's intended goal? Answer: The text aims to highlight the issues with traditional perspective methods when dealing with round columns and point the way to the various solutions that later artists arrived at.
Question: According to the text, why do round columns cause problems in perspective? Answer: Round columns cause problems because they make the neat geometrical rules of perspective fall apart, resulting in columns appearing wider or tilted when they should not be.
Question: What is the cause of the distorted appearance of columns in one point perspective? Answer: The distortion is due to the wide viewing angle, similar to what a camera with a wide-angle lens would capture. | 677.169 | 1 |
Homework 10/5/99
Chapter 3:
Q6. When two vectors are added, the magnitude of the sum will be the greatest
when the vectors point in the same direction. In this case the magnitude will be 7.5
km. When the vectors point in different directions the sum will be smaller.
The sum will be the smallest when the vectors point in exactly opposite directions.
In this case the magnitude of the sum will be 0.5 km.
Q7. Two vectors must have the same magnitude and point in opposite directions to
exactly cancel, that is, to add up to zero. Three unequal vectors can add up to zero
if the two shorter vectors are opposite the longest vector and the sum is zero.
Three unequal vectors can also add up to zero when they form a triangle and the
head of the third vector meets the tail of the first vector
Q8. The magnitude of a vector can equal one of its components when the vector
lies along the x or y axis. If a vector is at an angle to the x or y axis, the
component is smaller than the magnitude of the vector. The component can never be
greater than the magnitude of the vector.
Q9. According to the Pythagorean theorem, the sum of the squares of the
components equals the magnitude squared. When the magnitude squared is zero, all of
the components must be zero since they are all positive when squared and cannot cancel one
another out.
Question: Which of the following is NOT a condition for three unequal vectors to add up to zero? A) The two shorter vectors are opposite the longest vector B) The vectors form a triangle C) The vectors have the same magnitude Answer: C) The vectors have the same magnitude
Question: Can a component of a vector ever be greater than its magnitude? Answer: No | 677.169 | 1 |
Question 188646: This is a triganometry problem using Vectors length and direcion. In a regatta, a sailboat sailed 33 km west and 73 km south from the port, and then took the shortest route back. Find the distance and southern most angle in the triangle.
I did find the distance which is 80.1 km. Click here to see answer by Alan3354(30993)
for each of the following points, find the coordinates of the image point under a half-turn about the origin:
a) (4,0) b) (0,3)
c) (2,4) d) (-2,5)
e) (-2,-4) f) (a,b)
g) (-a,-b) Click here to see answer by Alan3354(30993)
Question 193337: r is the midpoint of line segment pt, and q is the midpoint of line segment pr. If s is a point between r and t such that the length of segment qs is 10 and the length of segment ps is 19, what is the length of segment st? Click here to see answer by RAY100(1637)
Question 194691: I need help in explaining this to my 5th grade daughter. Please illustrate the following for me: A 3-D right prism shape has these dimensions: H=6 cm, L=3 cm, and W=4 cm.
1) What is the perimeter if we know all 3 dimensions?
2) Is it possible to solve: H= 6 cm and W = 4cm, but L=x? How do I solve for "x"?
3) Is it possible to solve: H=6 cm, L= 3 cm, but W=x? How do I solve for "x"?
I appreciate any help I can as soon as possible! Thanks. Click here to see answer by solver91311(16897)
Question 198827: Hi, this is my question as found in the text book:
"A rectangular classroom is 8m wide, 15 m long, and 3 m high. Determine the inclination of the main (longest) diagonal of the classroom."
I am not sure how to determine the inclination of the diagonal, please help!
Thank you! :) Click here to see answer by nerdybill(6962)
Question 199295: Create a graph with precisely two odd vertices and two even vertices. State which vertices are even and which are odd.Use a dot and letter to mark each vertex.
this question is not from the textbook, it's from a handout thanks. Click here to see answer by Edwin McCravy(8909 solver91311(16897 rfer(12662 RAY100(1637)
Question 200701: If
Question: In a graph with precisely two odd vertices and two even vertices, which vertices are even and which are odd?
Answer: Not possible to determine from the text, as the graph's structure and vertices are not described.
Question: What is the southernmost angle in the triangle formed by the sailboat's path?
Answer: Not provided in the text
Question: Is it possible to solve for L when given H=6 cm and W=4 cm?
Answer: No, L is not given and cannot be solved for with the provided information. | 677.169 | 1 |
1 line = no intersection
2 lines =1 intersection
3line - 3 intersections
4 line = 6 intersection,then how many intersections will 10 lines have? please any one solve it as soon as possible. i have exam on it please urgent!!!!!!!!!!! stanbon(57361) Theo(3464)
Question 206927: I donot know whether to put this question on this department or in the others. If you know where to submit the question please let me know. Anyways the question is the following;
Kindly, help me solve this question it is urgent!
I have never solved this type of question.
A wheel of radius 40cm is turning about a fixed axis through the centre of the wheel at acentre rate of 2 revolutions per minute. Calculate the
(a) Angle through which the wheel turns in 10 seconds.
(b) Distance moved by a point on the rim in 10 seconds.
I hope someone will solve it. I shall be highly obliged to you.
Thanks,
Question: What is the urgency level of the question? Answer: High urgency, as the user mentions "urgent!!!!!!!!!!!" and has an exam on the topic. | 677.169 | 1 |
An identity is an equation that is true for all of the possible values of its variables. Trig identities are important, they involve the sums or differences of angles.
What are the Trigonometric Identities?
The above identities can be used to determine that other trigonometric equations are also identities. To do so, you will need to use your algebraic background to show that the expression on one side of the equals sign can be changed into the expression on the other side of the equals sign.
Trigonometry(Cliff's Quick Review)If you require some additional review to assist you to grasp trigonometric identities and their applications, this resource will amply supply you with the tools you need to reinforce trigonometric concepts.All the concise and easy to follow tutorials in this selection will assist the struggling trig. student to understand identities, functions, polar coordinates, triangles, vectors and inverse functions and equations. Cliffs notes tend to be the favored among students needing some additional work in introductory levels.
Schaum's Outline of TrigonometryChapter 8 deals with the trigonometric basic relationships and identities. Overall, this resource focuses on all concepts related to Plane Trigonometry.Detailed explanations, step by step solutions make this trigonometry resource one of the best to assist you to solve all types of trigonometric problems. Whether you're looking to brush up on concepts before taking your tests or if you simply want to try to solve a variety of problems, this book is sure to help you understand and extend your knowledge in trigonometry.
Question: What are some of the topics covered in this resource? Answer: identities, functions, polar coordinates, triangles, vectors, and inverse functions and equations.
Question: Which type of identities are specifically mentioned in the text? Answer: Trig identities (trigonometric identities) | 677.169 | 1 |
college cornerstone what time and financial constraints have you face since starting college? How did you deal with them
math The measure of the supplement of an angle is 20 degrees more than three times the measure of the original angle. Find the measures of the angles.
trig Two men on the same side of a tall building notice the angle of elevation to the top of the building to be 30o and 60o respectively. If the height of the building is known to be h =120 m, find the distance (in meters) between the two men.
Question: What is the height of the building mentioned in the trigonometry problem? Answer: 120 meters | 677.169 | 1 |
Microsoft Excel
regular polygon, all of the sides are equal, and all of the
interior angles ? are equal. The
perimeter P of a regular polygon is given by the following
formula:
P=a*n
and the sum S of the interior angles (in degrees) is given by
the following formula:
S=(n-2)*180
where n is the number of sides in the regular polygon.
Suppose that data regarding 20 regular polygons has been placed
in a spreadsheet, such thatcells A1:A20 and B1:B20 contain,
respectively, the values of the length of the side of thepolygon (a
in inches) and the corresponding number of sides of the polygon
(n).Design the remaining part of the spreadsheet to satisfy the
requirements described below. Ineach case, describe clearly how you
would complete the spreadsheet. Provide the exact syntaxfor each
formula and tell what cell it goes in. If a formula is to be
copied, give the formula andtell what range it is copied through.
If you want to introduce additional columns, just indicatewhich
ones and what formulas you will put in them.
(a). What formula would you write in cells C1 and copy down to
determine theperimeter of each of the polygons?
(b). What formulas would you write in cells D1 and copy down to
determine thevalue of S for each polygon?
Question: What is the formula to calculate the sum of the interior angles of a regular polygon? Answer: S = (n - 2) * 180
Question: If you wanted to calculate the area of each regular polygon, which formula would you use? Answer: Area = (a^2 n) / (4 tan(π/n)) | 677.169 | 1 |
Search community
turns
This unit uses one of the digital learning objects, Angles, to support students as they investigate measuring and drawing angles using other angles as units of measurement. It is suitable for students working at level 2 of the curriculum because they estimate and measure the size of other angles using other angles and not with compasses and protractors. This unit includes background information on teaching about angles, a sequence that can be used by the teacher when working with a group of students on the learning object, and ideas for independent student work.
In this unit, students will learn to investigate polygons using an instrument of their own construction. They should be able to prove the general formula for the number of degrees in any polygon (including a triangle). Finally they will investigate an interesting locus problem related to polygons.
This unit is about estimating and measuring angles. Initially, the students will be required to describe an amount of turn from a particular position to another using the 'circular' benchmarks of 0,1/4, 1/2, 3/4, and full turn. Estimation language such as 'just about', 'between', 'not quite', 'just over', and similar terms may be used to qualify their descriptions. As the unit progresses, the students will be exposed to the more formal unit of measurement, the 'degree'. They will be involved in constructing a link between the fractions of the circle with which they are familiar and the corresponding degree values, where 360 degrees is used as the measure of a complete turn.
Only towards the end of the unit are students introduced to the protractor as a tool that can be used to measure an angle. Estimation will continue to play an important role, however.
In this unit students explore movement and direction concepts in the context of programming a robot to move. They will be developing sets of instructions to accomplish tasks, focusing on the use of right, left, forward, backwards and quarter turns.
In this unit we develop the concept of angle to see that an angle may be constructed in a clockwise or anticlockwise direction. We see the effect of clockwise and anticlockwise turns on objects. We also think about corners of objects that are equivalent to quarter turns. We also think about whether a corner can be a half turn. These ideas are explored physically.
In this unit we look at the beginning of the concept of angle. Here we are interested in students understanding quarter and half turns and to begin to see that 'angle' is something involving 'an amount of turn'. These ideas are explored by using students' bodies, toys, games and art.
Question: What is the initial concept of an angle that students are introduced to? Answer: Quarter and half turns, and the idea that 'angle' involves 'an amount of turn'.
Question: What is the measure of a complete turn in degrees? Answer: 360 degrees. | 677.169 | 1 |
Vivian Kerr is a regular contributor to the Veritas Prep blog, providing tips and tricks to help students better prepare for the GMAT and the SAT.
]]> of Geometry - Part II
20 May 2013 15:59:46 +0000Karishma week, we discussed how drawing extreme diagrams can help solve Geometry questions. Today we will see how to solve another Geometry question by making diagrams. The diagram can help you understand exactly what it is that you need to do; doing it will be quite straightforward.
Question: If 10, 12 and 'x' are sides of an acute angled triangle, how many integer values of 'x' are possible?
(A) 7
(B) 12
(C) 9
(D) 13
(E) 11
Solution: The question is very interesting. It asks you for an acute triangle i.e. a triangle with all angles less than 90 degrees. It's a little hard to wrap your head around it, isn't it? We know that the third side of a triangle can take many values. Right from a little more than the difference of the other two sides to a little less than the sum of the other two sides (Since we know that the sum of any two sides of a triangle is always greater than the third side). So x can be anything from a little more than 2 to a little less than 22. But how do we find out the values for which all the angles will be less than 90?
We want no obtuse or right angles. An obtuse angled triangle has one angle more than 90. So the thought here is that before one of the angles reaches 90, find out all the values that x can take.
Look at the figure given above. The value of x in the first figure is very small – slightly more than 2 – minimum required to make a triangle. There is an obtuse angle in that triangle. We keep making x bigger and bigger and the angle keeps becoming smaller till it reaches 90 (Fig III). We use Pythagorean theorem to get the value of x in that case:
x = √(12^2 – 10^2)
x = √44 which is 6.something
x should be greater than 6.something because the angle cannot be 90.
We further keep increasing x and all the angles are acute now. We reach Fig V where we hit another right triangle. We use Pythagorean theorem again to get the value of x (the hypotenuse) in this case:
x = √(12^2 + 10^2)
x = √244 which is 15.something
x should be less than 15.something so that the angle is not 90.
Further on, in Fig VI, we obtain an obtuse angle again.
We only need integral values of x so values that x can take range from 7 to 15 which is 9 values.
Answer (C).
Question: What is the topic of the blog post? Answer: Solving Geometry questions using diagrams
Question: Who is the author of the blog post? Answer: Vivian Kerr | 677.169 | 1 |
.......and to convert the tangent of the angle to a percentage, simply multiply by 100.
Example: 1 in 3 hill = 1/3 = 0.3333 = (0.3333 x 100) = 33.33%.
Or take the same slope downwards, dropping 1 foot in height for every
3 feet traversed horizontally = 1 / -3 = -0.3333 = (-0.3333 x 100) =
-33.33%.
Thus any gradient expressed as "y in x" can
very easily be converted to a percentage gradient. NB In this example, the value 0.3333 is
the tangent of the angle.
A fourth way? Instead of dividing the height risen by the horizontal
distance along the base (the tangent of the angle of the slope), sometimes the ratio is taken as
the height risen by the length of the slope (height / length of the hypotenuse), in
trigonometry this is referred to as the sine of the angle. For obvious
reasons it is easier to work out the length of a slope (the
hypotenuse) than the base horizontal distance. The values of the
sine and tangent of a slope are often quite similar however to our knowledge the
convention is ALWAYS to express a road gradient by reference to the
tangent of the angle and never its sine!
Can I work out the angle (q) of the slope from its percentage gradient?
Answer: Yes, but to do so involves delving into some tricky
trigonometry. (Don't worry too much, we will cheat and use an on-line
scientific calculator). NB: this only works for gradients up to 100% (45° angle). Step 1: convert the % gradient back to the tangent (TAN) of the angle by dividing by 100, that bits easy! Step 2:
Next convert the value of TAN to the angle of the slope using the inverse function of TAN. The
inverse function of TAN is referred to as ARCTAN, or sometimes ACTAN or
ATAN, and in mathematical annotation is written as:
tan-1(x) where x is the value of TAN
Unfortunately working out the
arctangent by hand is fiendishly difficult, involves something called
Gregory's formula and should only be attempted by higher mathematical
types. Step 3: so cheat! Click on this scientific calculator
then key in the TANgent number (eg for a 100% gradient, key in "1"), now
switch on the inverse function of the calculator by clicking the "INV" button, then click
the "TAN" button (which in the inverse function gives tan-1 ).
Hey presto! The numerical answer given by the calculator is the ANGLE of the slope; so
for example if you keyed in "1" you got back the answer q= 45°.
Question: Can you calculate the angle of a slope from its percentage gradient? Answer: Yes, but it involves using the inverse tangent function (arctan). | 677.169 | 1 |
Using
colored electrical tape, draw a set of coordinate axes (on the floor) down
the center of your classroom, dividing the students' desks into four
quadrants. (Avoid using masking tape as it is difficult to remove
from the floor once students have walked on it.) Establish where the
positive x-axis and the positive y-axis are located.
The
students can now be asked to identify "who" is their:
1.
reflection over the x-axis.
2.
reflection over the y-axis.
3.
point reflection in the origin.
4.
translation under a given pattern.
Questions
can be posed in various ways:
a
straightforward manner, such as: "Which student is your reflection over the
x-axis?"
an
indirect manner, such as: "Susan is someone's point reflection in the origin. Who is
that person?"
an
inquisitive manner, such as: "Pick any student in the room. Using your knowledge of
transformations, tell me how you could be "transformed" to
that student's position." (Teacher demonstrates by giving an
example.)
Be
sure to call on all students during the activity. Questions can be
directed at specific students or the entire class. It is possible to engage
all students without embarrassing any one pupil. Directing questions
to the entire class at the beginning of the activity may give the slower
student time to process how the activity will work and allow him/her to
orient himself/herself with the grid of desks. Students should,
however, realize that they WILL be expected to participate in the activity.
Written
responses are also possible - but the oral aspect of this activity is its
charm.
Question: What is the purpose of the activity? Answer: To help students identify and understand reflections and translations using their positions in the classroom
Question: How many quadrants are created by the coordinate axes? Answer: Four | 677.169 | 1 |
You can put this solution on YOUR website! Since you must pick a blue from both bags to be "successful," you must multiply the individual probabilities...so
1/3 * 1/4 = 1/12, your answer.
You can put this solution on YOUR website! It is a bit difficult to explain the entire procedure, but in essence what you do is to eliminate the same variable from pairs of equations, via the linear combination method. You want to "boil" three equations and three unknowns down to two equations and two unknowns...for example, from
x-2y+z =7
2x-5y+2z =17
-3x+7y+5z =-32
we can double the first equation and subtract it from the second one, eliminating x...repeat that procedure with the first and third equation, etc...
You will find the solution
x = 2, y = -3, and z = -1
You can put this solution on YOUR website! What we do is multiply each term in the first expression times each term in the second expression...so we have
(3x + 2)(x^2 - 5x + 2) =
3x^3 - 15x^2 + 6x + 2x^2 - 10x + 4
Now collect like terms...
= 3x^3 - 13x^2 - 4x + 4
and we're done...
Angles/38259: if one angle measures 140 degrees, what is the probability that if you draw a line inbetween that both angles will be acute? My first answer was 89:140, but i'm not sure if i worked it correctly. 1 solutions Answer 23856 by fractalier(2101) on 2006-05-18 10:38:39 (Show Source):
You can put this solution on YOUR website! The way I looked at is this...
The random line can be drawn in the middle 39.999999 degrees of the large angle and still split the 140 degree angle into two acute angles, so the odds are roughly 40 out of 140, or 2/7.
You can put this solution on YOUR website! The word linear means line. A linear relationship between two variables, when graphed, is a straight line. It also means that both variables are expressed to the first power only, no roots, no higher exponents.
Question: What does the word "linear" mean in the context of relationships between variables? Answer: A linear relationship means that both variables are expressed to the first power only, and when graphed, it results in a straight line.
Question: If you were to draw a line in the middle of the 140-degree angle, what would be the measure of each of the new angles? Answer: Both new angles would measure 70 degrees.
Question: What is the result of multiplying (3x + 2)(x^2 - 5x + 2)? Answer: 3x^3 - 13x^2 - 4x + 4 | 677.169 | 1 |
Ellipse
The semi-minor axis of an ellipse is one half of the minor axis, running from the center, halfway between and perpendicular to the line running between the foci, and to the edge of the ellipse. The minor axis is the longest line segment that runs perpendicular to the major axis.
The semi-minor axis of an ellipse is the geometric mean of the maximum and minimum distances and of the ellipse from a focus — that is, of the distances from a focus to the endpoints of the major axis:
A parabola can be obtained as the limit of a sequence of ellipses where one focus is kept fixed as the other is allowed to move arbitrarily far away in one direction, keeping l fixed. Thus a and b tend to infinity, a faster than b.
Hyperbola
In a hyperbola, a conjugate axis or minor axis of length 2b, corresponding to the minor axis of an ellipse, can be drawn perpendicular to the transverse axis or major axis, the latter connecting the two vertices (turning points) of the hyperbola, with the two axes intersecting at the center of the hyperbola. The endpoints (0, ±b) of the minor axis lie at the height of the asymptotes over/under the hyperbola's vertices. Either half of the minor axis is called the semi-minor axis, of length b. Denoting the semi-major axis length (distance from the center to a vertex) as a, the semi-minor and semi-major axes' lengths appear in the equation of the hyperbola relative to these axes as follows:
The semi-minor axis and the semi-major axis are related through the eccentricity, as follows:
Question: How is the semi-minor axis of an ellipse defined in terms of distances from a focus? Answer: It is the geometric mean of the maximum and minimum distances from a focus to the endpoints of the major axis. | 677.169 | 1 |
Now, the very last thing I want to tell you about these functions is - well, the values at certain famous points. A lot of times you want to find the values at some big point - let me tell you the values that you should know for certain. You should know - I'll write them first in degrees. You should the trig functions at zero degrees, at 30 degrees, at 45 degrees, at 60 degrees, and at 90 degrees. You should know all the trig value functions just off the top of your head for those values, at least. Now let me convert them to the language of Calculus. Let me convert those to radians. This would be zero radian, and you could try this on your own by just looking at conversion that pi equals 180. This is going to give you pi over six radian. This will give you pi over four radian, like we saw before. This will give you pi over three radian. This will give you pi over two; these are all radian. And if you don't believe me here for example, take pi over three and now multiply that by 180 over pi. If you imagine putting in a 180 over pi here - let's just do that one for fun, so you can see that. I put in a 180 over pi. Well, you see the pi's cancel and three goes into 180, 60 times. So in fact you see 60 degrees.
Question: What is the conversion of 60 degrees to radians?
Answer: Pi over three radians (π/3).
Question: True or False: The text provides the trigonometric values for 45 degrees.
Answer: True. | 677.169 | 1 |
I don't know if you do the sorts of proofs where students make errors like saying that triangles are congruent by SAS when really they meant to say ASA, but if you do, I may have a trick for you. When my students get sloppy I have them start writing 3 column proofs. The first two columns are the ones you're used to, but you number each line, and the third column lists the line numbers that have the facts you're using, so if you are saying triangles are congruent by SAS, then column 3 will have generally have 3 line numbers in it: one that cites congruent sides, one that cites congruent angles, and another for congruent sides: SAS. It really cuts down on some of the sloppy mistakes.
I'm guessing, and this is pure speculation, that if I were doing this, I'd try the post-its with the plan to go through and check students' errors. Then I"d get home, realize that was a stupid, time-consuming idea, and lesson plan for every other class instead. I'd still have to remove all the post-its and I wouldn't be doing anything with them.
vlorbik, I know, but 2-column Euclidian geometry proofs don't seem to lend themselves very effectively to that goal. ThereLsquared, it's on my list to upload it. The biggest obstacle will be that many of my materials were inherited and I only have paper copies. It's going to be a lot of work unless I just scan to pdf and upload those.
I've see the third column used before and I can see where it would prevent sloppiness; thanks for sharing. Our honors-level teachers do it that way. I'm not sure it's appropriate for the level of students I have, but maybe I should try it out next year and see.
Therethe point *is* to keep it simple. "two-column" proofs *eliminate* the need for prose style altogether.
let the english department grapple with getting these fine young people to know when they've written a decent english sentence. this is important work... far more than writing proofs of any kind... but it's *too hard*.
is this "statement" good code? (Y/N) does this "reason" justify it? (Y/N)
lather, rinse, repeat.
math is here reduced to an uncannily-computerlike *binary logic*.
the "justifications" for the second column will depend on *definitions* and *theorems* (loosely; any previously proven piece of code is here glorified with the name of "theorem"); nothing else. again, this is "keep it simple" with a vengeance.
laying everything out with this absurd-once-you-know-how rigidity is precisely what makes it *possible*... i say... for your students to "grade" each other's work with confidence; keep it if you have any say in curriculum (say i that never has had so far).
Question: Why does the author think that two-column Euclidean geometry proofs don't lend themselves effectively to checking students' errors? Answer: The author thinks that two-column proofs don't lend themselves effectively to checking students' errors because they don't provide a clear reference to the facts used in the proof.
Question: What does the author think about the use of prose style in two-column proofs? Answer: The author thinks that two-column proofs eliminate the need for prose style altogether.
Question: What is the author's speculation about using post-it notes to check students' errors? Answer: The author speculates that using post-it notes to check students' errors would be a time-consuming and ineffective idea. | 677.169 | 1 |
Several curves are related to the Cycloid. When we relax the requirement that the fixed point be on the edge of the circle, we get the curtate cycloid and the prolate cycloid. In the former case, the point tracing out the curve is inside the circle, and, in the latter case, it is outside. A trochoid refers to any of the cycloid, the curtate cycloid and the prolate cycloid. If we further allow the line on which the circle rolls to be an arbitrary circle then we get the epicycloid (circle rolling on outside of another circle, point on the rim of the rolling circle), the hypocycloid (circle on the inside, point on the rim), the epitrochoid (circle on the outside, point anywhere on circle), and the hypotrochoid (circle on the inside, point anywhere on circle).
All these curves are roulettes with a circle rolled along a uniform curvature. The cycloid, epicycloids, and hypocycloids have the property that each is similar to its evolute. If q is the product of that curvature with the circle's radius, signed positive for epi- and negative for hypo-, then the curve:evolute similitude ratio is 1+2q.
Question: What is the property that the cycloid, epicycloids, and hypocycloids share? Answer: Each is similar to its evolute. | 677.169 | 1 |
In this book I have attempted to follow a middle course between the treatise which fully proves the propositions of elementary geometry and the syllabus which contains no proofs whalever. The early propositions are proved at length in order to make clear the form of geometrical demonstration and the details of proof are gradually removed in order to throw the pupil on his own resources. It hardly needs argument to show the wisdom of retaining and intensifying the pupils interest in the study, and many teachers are convinced that this is best done by expecting easy original work very early and by making the exercises an integral part of the course. How far I have succeeded in giving just the right amount of assistance of course depends on the character of the class. The teacher should be ready to supplement and expand the hints to meet the need of students. This requires judgment, as all teaching does. I shall be especially grateful for criticism and suggestion on this feature of the book. The order of propositions in Plane Geometry is nearly the conventional order of American text-books, but 1 have placed in Book I the elementary relations of rectilinear figures and of the circle; in Book II proportional line segments including similar and regular figures; in Book III the relations of areas, and measurement. The constructions are established before they are used in demonstration.
Modern Geometry is to present to the more advanced students in public schools and to candidates for mathematical honours in the Universities a concise statement of those propositions which I consider, to be of fundamental importance, and to supply numerous examples illustrative of them. Results immediately suggested by the propositions, whether as particular cases or generalized statements, are appended to them as Corollaries. The Examples are printed in smaller type, and are classified under the Articles containing the principal theorems required in their solution. The more difficult ones are fully worked out, and in most cases hints are given to the others. The reader who is familiar with the first six books of Euclid with easy deductions and the elementary formulae in Plane Trigonometry will thus experience little difficulty in mastering the following pages. I have dwelt at length in Chap. II. on the Theory of Maximum and Minimum. Chap. III. is devoted to the more recent developments of the geometry of the triangle, initiated 21 1873 by Lemoine spaper entitled Sur quelques propride sd un point remarquable du triangle.
Question: What type of font is used for the examples in 'Modern Geometry'? Answer: The examples are printed in smaller type.
Question: Who initiated the more recent developments of the geometry of the triangle, as mentioned in the text? Answer: Lemoine
Question: What is the order of propositions in Plane Geometry according to the author? Answer: The author follows a conventional order of American textbooks, with Book I focusing on the elementary relations of rectilinear figures and the circle, Book II on proportional line segments, and Book III on the relations of areas and measurement. | 677.169 | 1 |
One of the main purposes in writing this book has been to try to present the subject of Geometry so that the pupil shall understand it not merely as a series of correct deductions, but shall realize the value and meaning of its principles as well. This aspect of the subject has ben directly presented in some places, and it is hoped that it pervades and shapes the presentation in all places. Again, teachers of Geometry generally agree that the most diflBcult part of their work lies in developing in pupils the power to work original exercises. The second main purpose of the book is to aid in the solution of this difficulty by arranging original exercises in groups, each of the earlier groups to be worked by a distinct method. The pupil is to be kept working at each of these groups till he masters the method involved in it. Later, groups of mixed exercises to be worked by various methods are given. In the current exercises at the bottom of the page, only such exercises are used as can readily be solved in connection with the daily work. All difficult originals are included in the groups of exercises as indicated above.
The pedagogy of Geometry is divisible into three parts, the Science of Geometry (the facts), the Logic of Geometry (the framework), and the Art or Technique of Geometry. The Science and Logic of Geometry are presented in the current text-books. The Art of Geometry is given in the following pages. One of the great causes of complaint among students of Geometry is the lack of a systematic course of procedure. The student has no guide as to what to do next, and is generally unable to give any reason for the different steps he takes, except that they are logically correct and that they produce an answer. On the other hand, he ought to know why each step in the proof is taken. Only thus can he proceed with real understanding and be prepared to attack intelligently a new and unknown problem. Thus in the problem. To compute the length of the bisector of the angle of a triangle, the first step given by all the text-books is, draw the circumscribing circle. But why? What suggests such a step? What preparation has been given to the student to lead him to such a construction? No text-book has yet answered the question. The following pages will show not only what suggests the circle, but will provide the student with such a course of procedure (technique) that he could not miss it if he would. The current text-books supply the student with propositions and a logical proof of their correctness, but they do not put him in a position to deduce his own propositions nor to build a proof independently for himself.
Question: What is one of the main complaints of students studying geometry? Answer: The lack of a systematic course of procedure, or a guide on what to do next in solving problems. | 677.169 | 1 |
(Such a transformation indeed exists. It belongs to the class of
inversions and its existence is established in Inversive
Geometry after just a few definitions.)
For concentric circles, i.e., circles with the same center, the
statement is quite obvious. The radii of the given circles and n must stand
in a certain relationship for the chain to close on itself. Whether it does
or not does not depend on the position of the starting point.
There's a proof by induction. For n = 3, the theorem is obvious. Two
pairs of parallel sides form equal inscribed angles that are bound to
subtend equal chords — the third sides.
Let n = 4, and two polygons A1A2A3A4 and B1B2B3B4 be given such that , A1A2 || B1B2, A2A3 || B2B3, and A3A4 || B3B4. The diagonals A1A3 and B1B3 are equal by the previous paragraph. Therefore, inscribed angles A3A4A1 and B3B4B1 are equal and have parallel sides: A3A4 || B3B4. Therefore, their other sides are also parallel: A4A1 || B4B1.
The last paragraph also demonstrates one half of the inductive step: if the statement holds for an odd n = k, it also holds for n = k+1.
Assume now that the statement holds for n = k, where k is even. Let two polygons A1A2...AkAk+1 and B1B2...BkBk+1 have the first k sides parallel. Then by the just established first half of the statement, A1Ak || B1Bk. The angles A1AkAk+1 and B1BkBk+1 have parallel sides and are therefore equal. Being inscribed into the same circle, they subtend equal chords: Ak+1A1 = Bk+1B1. Q.E.D.
Question: What is the final step in the proof by induction? Answer: Q.E.D. | 677.169 | 1 |
Is the point over my ray? Confusing...
This is a discussion on Is the point over my ray? Confusing... within the C++ Programming forums, part of the General Programming Boards category; I am trying to detect if my point (x & y) are over my ray (rayrot).
This is the best ...
I don't see why you're multiplying the cos and sin values by the same coordinates you wish to compare to. You should multiply those with the radius to get the correct X and Y values.
Also, you will get an inverse result if the coordinates are negative.
Just a guess, and it might not be the most efficient way to solve this, but couldn't you check if the inverse tangent of y/x is greater than rayrot? Essentially, if the angle that the position vector (x,y) makes with the x axis is greater than the angle the ray makes with the x axis, then the point must be above the ray.
Of course, you'd have to be careful to make adjustments if the point or ray is not in the first quadrant.
I am trying to detect if my point (x & y) are over my ray (rayrot).
This is the best I could come up with, but it's not working. How should I do this?
Code:
if ( x > cos ( rayrot ) * y && y > sin ( rayrot ) * x )
Ray - what ray?
A Ray consists of a point and a direction, where the direction would normally be a vector. However this 'rayrot' seems to be just a single value, therefore it is just an angle.
The problem would however make sense, if you state that the ray happens to cross through (0, 0) for example (which I'll assume). However since you're using angles instead of vectors, you also need to state which direction an angle of zero points in. (I'll assume that increasing angle turns clockwise, and that zero is straight up, for now).
This would mean that a point (x, y) is on the ray if:
Code:
(x * cos(rayrot) == y * sin(rayrot))
However since floating point math isn't exact you'll need to use something like this:
Question: What is the condition provided by the second responder for a point (x, y) to be on the ray? Answer: (x cos(rayrot) == y sin(rayrot))
Question: What is the issue with the user's initial code snippet? Answer: The user is multiplying the cos and sin values by the same coordinates (x and y) they wish to compare, instead of multiplying them with the radius to get the correct X and Y values. | 677.169 | 1 |
egthomas asked
I understand the rules of horizontal, reflection, stretch and vertical translations however not seeing a graph and just given this information in the practice problem is very confusing. I understand to show this problem you will have to do on paper and I thank you for effort. Plz show steps so I can fully understand this concept.
Question: What is the student's tone towards the assistant? Answer: The student's tone is polite and respectful, using "please" and thanking the assistant for their effort. | 677.169 | 1 |
If on the coordinate plane (6,2) and (0,6) are the endpoints of the diagonal of a square, what is the distance between point (0,0) and the closest vertex of the square?
a) 1/sqrt(2) b) 1 c) sqrt(2) d) sqrt(3) e) 2*sqrt(3)
if (0,6) and (6,2) are end points of a diagonal of a square - shouldnt they be two of the vertices of the square too? also, if it were the vertices, the diagonal length shows that the quadrilateral isnt a square....which is where im most confused.... can anybody help?? thanks in advance.
this is what OA says:
First, we have to "guess" the coordinates of the vertex of the square closest to the origin. We can do it by making a sketch. The coordinates of the vertex are (1,1) and it is units away from the origin.
I cannot understand why do we have to do such complex calculations. If the vertices of the diagonal are 0,6 and 6,2 then the closest vertex will be (0,2). So the shortest distance from 0,0 should be 2Some questions can be solved over 2 minutes. There are 75 minutes and 37 questions. Some can be done under 1 minute even under 30 seconds. Today I scored 51q and although 2 problems got my 6-7 minutes I finished exam in 64 minutes That will give help you I think. So the easier under 1 that will give you a chance to solve hards in 5-6 minutes.
You don't need to use the distance formula here; instead we can use slopes. I posted this solution a while back to another forum:
We have two endpoints of a diagonal of a square. We can use the following:
-the midpoint of one diagonal is the midpoint of the other diagonal; -the diagonals are perpendicular.
If (0,6) and (6,2) are endpoints, (3,4) is the midpoint.
From (0,6) to (3,4), we go right 3 and down 2; that is, we increase x by 3 and decrease y by 2: the slope is -2/3. Consider the perpendicular diagonal- its slope is the negative reciprocal, i.e. 3/2. From (3,4), on a perpendicular line, to find a point the same distance from (3,4) as (0,6) is, we can decrease x by 2 and decrease y by 3, or we can increase x by 2 and increase y by 3. The endpoints of the other diagonal are (1,1) and (5,7).
Question: What is the midpoint of the diagonal with endpoints (0,6) and (6,2)? Answer: The midpoint is (3,4). | 677.169 | 1 |
a = xi + yj + zk
The coefficients of the i, j, and k parts of the equation are the vectors components. These are how long each vector is in each of the 3 axis.
For example, the vector equation pointing to the point ( 3, 2, 5 ) from the origin ( 0, 0, 0 ) in 3D space would be:
a = 2i + 3j + 5k
The second way I will represent vectors is as column vectors. These are vectors written in the following form:
Where x, y, and z are the components of that vector in the respective directions. These are exactly the same as the respective components of the vector equation. Thus in column vector form, the previous example could be written as:
There are various advantages to both of the above forms, although column vectors will continue to be used. Various mathematic texts may use the vector equation form.
Vector Mathematics
There are many ways in which you can operate on vectors, including scalar multiplication, addition, scalar product, vector product and modulus.
Modulus
The modulus or magnitude of a vector is simply its length. This can easily be found using Pythagorean Theorem with the vector components. The modulus is written like so:
a = |a|
Given:
Then,
Where x, y and z are the components of the vector in the respective axis.
Addition
Vector addition is rather simple. You just add the individual components together. For instance, given:
The addition of these vectors would be:
This can be represented very easily in a diagram, for example: The individual components are simply subtracted from each other. The geometric representation however is quite different from addition. For example:
The visual representation
It may be easier to think of this as a vector addition, where instead of having:
c = a – b
We have:
c = -b + a
Which according to what was said about the addition of vectors would produce:
You can see that putting a on the end of –b has the same result.
Scalar Multiplication
This is another simple operation; all you need to do is multiply each component by that scalar. For example, let us suggest that you have a vector a and a scalar k. To perform a scalar multiplication you would multiply each component of the vector by that scalar, thus
The Scalar Product (Dot Product)
The scalar product, also known as the dot product, is very useful in 3D graphics applications. The scalar product is written:
This is read "a dot b".
The definition of the scalar product is:
Θ is the angle between the two vectors a and b. This produces a scalar result, hence the name scalar product. This operation has the result of giving the length of the projection of a on b. For example:
The length of the thick gray horizontal line segment would be the dot product.
The scalar product can also be written in terms of Cartesian components as:
We can put the two dot product equations equal to each other to yield:
With this, we can find angles between vectors.
Question: What is the addition of the vectors 'a = 2i + 3j' and 'b = 1i + 2j'? Answer: The addition of the vectors 'a = 2i + 3j' and 'b = 1i + 2j' is 'c = 3i + 5j'.
Question: Which of the following is NOT a way to operate on vectors, according to the text? A) Scalar multiplication B) Addition C) Division D) Scalar product Answer: D) Division
Question: What does the scalar product (dot product) operation produce? Answer: The scalar product (dot product) operation produces a scalar result, giving the length of the projection of one vector onto another. | 677.169 | 1 |
This lesson explains the basics and concepts about the quadrilaterals. You'll learn it
starting from your earlier learning about polygons. All this you'll learn in the
contents presented by the instructor in own handwriting, using video and with the
help of several examples with solution.
Quadrilateral: in geometry, a plane closed figure formed by four line segments is called
a quadrilateral. It is a polygon with four sides, four vertices and with four
angles.
The interior angles of a quadrilateral add up to
360.
Quadrilaterals are simple i.e. not self-intersecting or complex i.e. self-intersecting.
Simple quadrilaterals may be either convex or concave.
(More text below video...)
(Continued from above)
There are three topological types of quadrilaterals: convex quadrilaterals,
concave quadrilaterals, and crossed quadrilaterals or butterflies type
(shown in the figure below):
Now you'll explore more on conceptual understanding about important quadrilaterals (Figure below).
Parallelogram- is a quadrilateral with two pairs of parallel sides. Equivalent conditions are that
opposite sides are of equal length; the opposite angles are equal and the diagonals bisect
each other. Notice: parallelograms also include the rectangle, square, rhombus and rhomboid. Square: is a regular quadrilateral having all four sides of equal length, and all four angles are
right angles. An equivalent condition is that opposite sides are parallel; diagonals are of
equal length and bisect each other at right angles. A quadrilateral is a square 'if and only'
if it is a rhombus and a rectangle both. Rectangle- is a quadrilateral in which all the four angles are right angles. An equivalent
condition is that the diagonals bisect each other and are equal in length. Trapezoid: is a quadrilateral that has exactly two sides parallel, but it's a type of
quadrilateral that is not a parallelogram. One of the parallel sides is the base and the
non-parallel sides are legs. Rhombus: is a quadrilateral in which all the four sides are of equal length. Equivalent conditions
are that opposite sides are parallel and opposite angles are equal. The diagonals perpendicularly
bisect each other. Kite: is a quadrilateral in which two adjacent sides are of equal length and the other two
sides are also equal. Thus the angles between the two pairs of equal sides are equal,
and the diagonals are perpendicular.
Furthermore, you'll learn the relationship with the help of the 'Venn Diagram'
i.e. the position and overlap of the circles indicating the relationships between the quadrilaterals
and that helps you better understand to put together various types of quadrilaterals:
Notice in the figure above, how different quadrilaterals relate to each
other i.e. the relationship amongst important quadrilateral group types i.e.
Question: What is the defining characteristic of a parallelogram? Answer: A quadrilateral with two pairs of parallel sides | 677.169 | 1 |
Applications of Trigonometry Lesson 3: Dot Product of Vectors lesson includes dot products of vectors, projection of vectors, work, and various applications in real-world situations. There is an eight-page "Bound-Book" style foldable to accompany the lesson, along with a *.pdf file of the completed set of notes.
Compressed Zip File
Be sure that you have an application to open this file type before downloading and/or purchasing.
1938.34
Question: What is the size of the compressed zip file containing the lesson materials? Answer: The size of the compressed zip file is 1938.34. | 677.169 | 1 |
Ptolemy's Ptools
Taking the Ptools into Cyberspace!
Using the properties of triangles,
we can investigate 3D games in some very interesting ways. In these games
the objects get larger as your view gets closer and smaller as your view
gets farther away. In order to look realistic game programmers must follow
the same rules of perspective that we are investigating (angles of separation).
All we need to do to make use of this fact is ...........
1) Choose a beginning item of known or estimated size and
2) Find what "scale" we are seeing things in on our particular
computer system while running a particular game
We will show you three examples. Since the competition that we are
doing this page for is partly judged on how much response we get from YOU
(and because it's much more FUN than just watching!) we would like to get
everyone involved by sending us the results of your cyberspace surveys
and we will post them on this site, making it easier for others to find
starting points on these games, or to just measure their way "farther
out into cyberspace".
Question: What is the ultimate goal of measuring one's way "farther out into cyberspace"? Answer: The ultimate goal is not explicitly stated in the text. | 677.169 | 1 |
Tilted Squares
Activity time:35 minutes Level / prior knowledge: area of a square and triangle,KS3/KS4 Subject / curriculum links / skills: problem solving, generalising a solution, Pythagoras' Theorem, Preparation time: none (unless you want to make sure you can solve the problem yourself) Extra resources: projector/IWB Commentary
I used this problem with a bottom set Year 10 class studying for higher GCSE. I didn't follow the teachers' notes to this activity having already learnt Pythagoras' Theorem. Also, the solution is quite messy and hard to read since it's handwritten so I created my own. My class has 10 students in it so we worked through the activity as a whole class and I got them to lead the discussion and the problem solving. We worked out a general solution to the problem in half an hour but they already knew the theorem so it would take much more time starting from scratch. The students liked the fact they were using what they had learnt to solve a different problem without my help.
By Bridget Keely
Question: What is the estimated time required to complete this activity? Answer: 35 minutes | 677.169 | 1 |
When two lines in a plane meet one another, we call Angle the inclination of the lines to one another.
We denote angles either by Greek letters (α, β, γ, and so on)or by the three point indicating the lines meeting (writing the point where they meet in the middle). For example, in the picture on the right, the angle would be called ABC, because it is the inclination between AB and BC, which meet at B.
An angle could also be formed by two lines which are not straight, and we call Rectilinear Angle one which is formed by two straight-lines. However, all angles in this course will be assumed to be rectilinear unless otherwise specified.
If a straight-line meets another straight-line at a point in the middle of it, and the two angles it makes are equal to each other, we call those angles Right-Angles, and we call the two straight-lines Perpendicular. For example, in the picture below, the lines AB and CD are perpendicular, and the angles ADC and BDC are right-angles.
We call an angle Obtuse if it is greater than a right-angle, and we call it Acuteif it is less than a right-angle.
Besides straight-lines, the other most useful and basic geometric figure is the Circle. A circle is a plane figure enclosed by one single line, called the Circumference. It also has a point, called the Center, such that all straight-lines from the center to the points on the circumference are equal to each other, in the sense that they have the same shape and length. We call each of those straight-lines a Radius.
A straight-line between two points on the circumference that passes through the center is called a Diameter. In the picture on the side, the straight-line AB is a diameter. Each diameter cuts the circle exactly in half, and each of the two halves is called a Semi-Circle.
A Figurethat is bounded by straight-lines is called a Rectilinear Figure. If it is contained by three straight-lines, it is called a Triangle, if it is contained by four, it is called a Quadrilateral, and if it is contained by more than four sides, it is called a Multilateral. Here is a picture of several rectilinear figures:
A triangle is called Equilateral if it has three equal sides, Isosceles if it has only two equal sides, and Scalene if all three sides are unequal. Also, a triangle is called Right-angled if it has a right-angle, Obtuse-angled if it has an obtuse angle, and Acute-angled if all three angles are acute.
Question: If a straight line meets another straight line at a point in the middle and the two angles it makes are equal, what are those angles called? Answer: Right-Angles | 677.169 | 1 |
Proposition 45
To construct a parallelogram equal to a given rectilinear figure in a given rectilinear angle.
Let ABCD be the given rectilinear figure and E the given rectilinear angle.
It is required to construct a parallelogram equal to the rectilinear figure ABCD in the given angle E.
Join DB. Construct the parallelogram FH equal to the triangle ABD in the angle HKF which equals E. Apply the parallelogram GM equal to the triangle DBC to the straight line GH in the angle GHM which equals E.
Thus, with a straight line GH, and at the point H on it, two straight lines KH and HM not lying on the same side make the adjacent angles together equal to two right angles, therefore KH is in a straight line with HM.
Since FK is equal and parallel to HG, and HG equal and parallel to ML also, therefore KF is also equal and parallel to ML, and the straight lines KM and FL join them at their ends. Therefore KM and FL are also equal and parallel. Therefore KFLM is a parallelogram.
Therefore the parallelogram KFLM has been constructed equal to the given rectilinear figure ABCD in the angle FKM which equals the given angle E.
Q.E.F.
With this construction any rectilinear area can be applied to a line in an angle, that is, it can be transformed into a parallelogram with whatever angle you want and with one side whatever you want. That is a satisfactory solution to the question "what's the area of this figure?"
But the question "what's the area of a circle?" is not answered in the Elements. See the note on squaring the circle after proposition II.14 for more discussion of this question.
Use of Proposition 45
This construction is used in propositions II.14, VI.25, and XI.32. Like many of the other constructions in Book I, it is used to make constructions in different planes as is done in XI.32.
Question: What is the main objective of Proposition 45? Answer: To construct a parallelogram equal to a given rectilinear figure in a given rectilinear angle.
Question: What is the significance of the construction in Proposition 45? Answer: It allows any rectilinear area to be applied to a line in an angle, transforming it into a parallelogram with a desired angle and side length. | 677.169 | 1 |
Fire hydrant : what shape is at the very top of a fire hydrant?
This activity begins an exploration of geometric shapes by asking students why the five-sided (pentagonal) water control valve of a fire hydrant cannot be opened by a common household wrench. The activity explains how geometric shape contributes to the usefulness of many objects. A hint calls students' attention to the shape of a normal household wrench, which has two parallel sides. Answers to questions and links to resources are included.
Diagonals to Quadrilaterals
Instead of considering the diagonals within a quadrilateral, this lesson provides a unique opportunity: Students start with the diagonals and deduce the type of quadrilateral that surrounds them. Using an applet, students explore characteristics of diagonals and the quadrilaterals that are associated with them.
Image:Geometric Solids and their Properties
A five-part lesson plan has students investigate several polyhedra through an applet. Students can revolve each shape, color each face, and mark each edge or vertex. They can even see the figure without the faces colored in — a skeletal view of the "bones" forming the shape. The lesson leads to Euler's formula connecting the number of edges, vertices, and faces, and ends with creating nets to form polyhedra. An excellent introduction to three-dimensional figures!
Slicing solids (grades 6-8)
So what happens when a plane intersects a Platonic solid? This virtual manipulative opens two windows on the same screen: one showing exactly where the intersection occurred and the other showing the cross-section of the solid created in the collision. Students decide which solid to view, and where the plane will slice it.
Studying Polyhedra
What is a polyhedron? This lesson defines the word. Students explore online the five regular polyhedra, called the Platonic solids, to find how many faces and vertices each has, and what polygons make up the faces. An excellent applet! From this page, click on Polyhedra in the Classroom. Here you have classroom activities to pursue with a computer. Developed by a teacher; the lessons use interactive applets and other activities to investigate polyhedra.
Analyzing Data Sets
(NCTM, 2006, p.20).
As reflected in this set of resources, the emphasis here is on understanding descriptive statistics; in particular, measures of center. You will find tutorials, lesson ideas, problems, and applets for teaching these topics, and even full projects that can involve your middle school students in worldwide data collection.
Describing Data Using Statistics
Investigate the mean, median, mode, and range of a data set through its graph. Manipulate the data and watch how these statistics change (or, in some cases, how they don't change).
Understanding Averages
Written for the student, this tutorial on mean, median, and mode includes fact sheets on the most basic concepts, plus practice sheets and a quiz. Key ideas are clearly defined at the student level through graphics as well as text.
Plop It!
Question: What can students do with the applet in the "Geometric Solids and their Properties" lesson? Answer: They can revolve, color faces, mark edges or vertices, and view a skeletal view of the polyhedra.
Question: What is the purpose of the activity that begins with exploring the shape of a fire hydrant's valve? Answer: To explain how geometric shape contributes to the usefulness of many objects. | 677.169 | 1 |
In mathematics, a spiral is a curve which emanates from a central point, getting progressively farther away as it revolves around the point.-Spiral or helix:...
using polar coordinates, an extra full turn gives rise to a quite different point on the curve.
Theta is the eighth letter of the Greek alphabet, derived from the Phoenician letter Teth...
, a circular arc centered at the vertex of the angle is drawn, e.g. with a pair of compasses. The length of the arc s is then divided by the radius of the circle r, and possibly multiplied by a scaling constant k (which depends on the units of measurement that are chosen):
The value of θ thus defined is independent of the size of the circle: if the length of the radius is changed then the arc length changes in the same proportion, so the ratio s/r is unaltered.
Units
In physics and all science, dimensional analysis is a tool to find or check relations among physical quantities by using their dimensions. The dimension of a physical quantity is the combination of the basic physical dimensions which describe it; for example, speed has the dimension length per...
, angles are considered to be dimensionless. There are several units used to measure angles. Of these units, listed below according to magnitude, the 1 quad. = 90° = π/2 rad = 1/4 turn = 100 grad. In German the symbol ∟ has been used to denote a quadrant, and is especially easy to construct with ruler and compasses. The degree, minute of arc and second of arc are sexagesimal subunits of the Babylonian unit. 1 Babylonian unit = 60° = π/3 rad ≈ 1.047197551 rad.is the angle subtended by an arc of a circle that has the same length as the circle's radius (k = 1 in the formula given earlier). Oneis 2π radians, and one radian is 180/π degrees, or about 57.2958 degrees. The radian is abbreviated rad, though this symbol is often omitted in mathematical texts, where radians are assumed unless specified otherwise. When radians are used angles are considered as dimensionless. The radian is used in virtually all mathematical work beyond simple practical geometry, due, for example, to the pleasing and "natural" properties that the trigonometric function
Trigonometric function
In mathematics, the trigonometric functions are functions of an angle. They are used to relate the angles of a triangle to the lengths of the sides of a triangle...
s display when their arguments are in radians. The radian is the (derived) unit of angular measurement in the SI
Si
Si, si, or SI may refer to :- Measurement, mathematics and science :* International System of Units , the modern international standard version of the metric system...
A turn is an angle equal to a 360° or 2 radians or \tau radians. A turn is also referred to as a revolution or complete rotation or full circle or cycle or rev or rot....
Question: What is the formula to calculate the value of θ for a circular arc? Answer: The length of the arc (s) divided by the radius of the circle (r), possibly multiplied by a scaling constant (k). | 677.169 | 1 |
A reference angle is the acute version of any angle determined by repeatedly subtracting or adding 180 degrees, and subtracting the result from 180 degrees if necessary, until a value between 0 degrees and 90 degrees is obtained. For example, an angle of 30 degrees has a reference angle of 30 degrees, and an angle of 150 degrees also has a reference angle of 30 degrees (180-150). An angle of 750 degrees has a reference angle of 30 degrees (750-720).
Using trigonometric functions
A Euclidean angle is completely determined by the corresponding. In particular, if is a Euclidean angle, then using trigonometric functions we find,
and
for two numbers x and y. So an angle in the Euclidean plane can be legitimately given by two numbers x and y.
To the ratio y/x there correspond two angles in the geometric range 0 < θ < 2π, since
If infinities are permitted for the quotient y/x one can define the angle θ as a function of x and y using the inverse tangent function for all points except the origin, assuming the inverse tangent varies from -π/2 to π/2,
The result will vary from -π to π. The values of x and y determine which quadrant the angle is in. Alternatively one can use the inverse cosine function assuming the result for the inverse cosine varies from 0 to π,
Angles between curves
In mathematics, a curve is, generally speaking, an object similar to a line but which is not required to be straight...
(mixed angle) or between two intersecting curves (curvilinear angle) is defined to be the angle between the at the point of intersection. Various names (now rarely, if ever, used) have been given to particular cases:—amphicyrtic (Gr. , on both sides, κυρτός, convex) or cissoidal (Gr. κισσός, ivy), biconvex; xystroidal or sistroidal (Gr. ξυστρίς, a tool for scraping), concavo-convex; amphicoelic (Gr. κοίλη, a hollow) or angulus lunularis, biconcave.
Dot product and generalisation
In mathematics, Euclidean space is the Euclidean plane and three-dimensional space of Euclidean geometry, as well as the generalizations of these notions to higher dimensions...
, the angle θ between two vectors u and v is related to their dot product
Dot product
In mathematics, the dot product or scalar product is an algebraic operation that takes two equal-length sequences of numbers and returns a single number obtained by multiplying corresponding entries and then summing those products...
and their lengths by the formula
This formula supplies an easy method to find the angle between two planes (or curved surfaces) from their normal vectors and between skew lines
Skew lines
Question: What is the formula to find the angle θ between two vectors u and v using their dot product and lengths? Answer: θ = arccos((u · v) / (|u||v|))
Question: Which Greek word is the origin of the term 'concave' in mathematics? Answer: κοίλη (koilē)
Question: What is the reference angle of an angle of 390 degrees? Answer: 30 degrees (390 - 360) | 677.169 | 1 |
Welcome back! I think that we are on the verge of snow! Although it has been fairly warm lately, we are getting very close to Thanksgiving, which means snow is around the corner!
On a sadder note, my favorite spot, baseball, is done for the year. The Rangers lost, and the Mets didn't even make the Playoffs, but I guess there is next year!
One thing that hasn't stopped is Geometry Class! The First Quarter is over! Here is a brief summary of what we're doing in class:
1. Well, we just finished learning about supplementary, complementary, vertical and adjacent angle pairs. To start this quarter, we are learning about angle pairs formed by two lines and a transversal. A transferal is a line that intersects 2 other lines. When this happens, many different angles are created, and by using new theorems and postulates, we can decide which are congruent, which are supplementary, and even the measures of some pairs!
2. Sometimes we encounter a problem where they tell us some simple information, and then tell us to answer a question based off of that information. Usually, these problems are easy, but now , they require much more planning and thinking. As an addition to the new angles pairs, we have started Proofs. A Proof is a series of steps used to prove a concept or statement true. We have learned how to do Proofs in Paragraph Form and Column Form. We have been practicing Proofs on concepts we have already learned. This is difficult because when we are told that something is true, we never learn how to prove that it is true. We never even stop to think about the steps used to achieve the statement. I now have to do this, and although it is tough, I think I am getting the hang of it!
On another note, we have a few more units to go in this section, in which we will cover parallel line proofs, the angles in triangles and polygons, lines in the coordinate plane, parallel and perpendicular line slopes, and constructing parallel and perpendicular lines.
Well, that is what we have been doing in class lately!
Before I go, I have one thing to say to all the Patriots Fans out there who watched last Sunday's game:
Question: What are the two forms of proofs the speaker mentioned learning? Answer: Paragraph Form and Column Form
Question: What is the next topic the speaker will cover in Geometry Class after parallel line proofs? Answer: The angles in triangles and polygons
Question: What is the speaker currently learning in Geometry Class? Answer: Angle pairs formed by two lines and a transversal, and how to do proofs. | 677.169 | 1 |
Have students write in their own words why they think perpendicular bisectors and circumcenters are helpful in finding the delivery regions for the pizzerias.
Extensions
Instead of points A, B, C, D, and E being pizzerias, let them represent rain gauges holding different amounts of rain measured in inches. Tell students that each block on the grid represents one square mile, and ask them to estimate the total volume of rainfall in cubic inches that fell on Squaresville.
Set up Squaresville on a coordinate plane with the bottom left corner being the origin with four locations identified. Rather than the points A, B, C, and D being pizzerias, let them represent obstacles in a square room (with a side length of 8 feet) around which a robot needs to maneuver. Assuming the robot should stay the same distance from the two nearest obstacles, ask students to find the equations of the lines that the robot can travel (and their domains).
Teacher Reflection
Were students able to accurately construct the perpendicular bisectors and circumcenters using your method of choice (Mira, patty paper, or straight edge and compass)? Why or why not? Would another method have worked better?
Was students' level of enthusiasm/involvement high or low? Explain why.
How did students demonstrate understanding of the materials presented?
Were concepts presented too abstractly? too concretely? How would you change them?
Did you set clear expectations so that students knew what was expected of them? If not, how can you make them clearer?
Did you find it necessary to make adjustments while teaching the lesson? If so, what adjustments, and were these adjustments effective
Question: What is one way to change the level of abstraction of the concepts presented? Answer: Make them more concrete or more abstract, depending on the current level. | 677.169 | 1 |
All squares are special cases of rhombuses and rectangles (and trapezoids), and all rhombuses and rectangles are special cases of paralleograms which are in turn special cases of trapezoids which are in turn special cases of quadrilaterals.
_________________
1) AC(imagine: diagonal of a quadrilateral) bisect BD (imagine: another diagonal of a quadrilateral) bisect each other in point E 2) Angle ABC = 90 degrees
As I see opinions differ.
The key here lies in the word bisect, which is to divide in halfs.
From the first statement the figure could be a rombus and a rectangle. Insufficient.
From the second statement it could be anything with a 90 degree angle. Imagine an ugly figure with two relatively short sides and a 90 degree angle between them and with third and forth sides realy-realy long ones, ending up in the stratosphere. Insufficient.
Question: What is the difference between a rectangle and a rhombus? Answer: A rectangle has all angles equal to 90 degrees, while a rhombus has all sides equal in length | 677.169 | 1 |
Triangles/456884: The sides of a length of a triangle are x, x+4, and 20, where 20 is the longest side. For which range of values is x an acute triangle 1 solutions Answer 313550 by robertb(4012) on 2011-06-02 22:12:03 (Show Source):
You can put this solution on YOUR website! For the triangle to be acute, , where c is the longest side of the triangle. Note that it is given that the longest side has length 20.
Hence, we must have , or , after simplifying, , or (x+16)(x - 12) >0. Since x must have positive values only, x > 12.
From the triangle inequality, we get the relation x +(x+4) > 20, or x > 8. Also, since 20 is the longest side, we must have x + 4 < 20, or x < 16. Hence from the initial conditions, we must have 8 < x < 16.
Intersect the preceding interval with the interval x > 12.
Therefore, for the triangle to be acute, we must have 12 < x < 16.
Surface-area/456855: The area of a regualr Octogon is 50 Cm2. How do you find the area of a regular octogon with sides 3 times as large? 1 solutions Answer 313510 by robertb(4012) on 2011-06-02 20:33:53 (Show Source):
You can put this solution on YOUR website! The equation of the parabola has the form
From the given, the x-coordinate of the vertex must be -1. The y-coordinate must be midway between y = -3 and y = -4. Hence, k = -7/2.
Also, a = -4 - (-7/2) = -1/2, the directed distance from the focus to the vertex.
==> , or
You can put this solution on YOUR website! The form of the equation is , where (h,k) is the vertex.
==>
a = directed distance from the focus to the vertex (which in absolute value is equal to the distance between the vertex and the directrix.) Hence a = -1, and therefore, the equation is
.
Permutations/456128: A poker hand consists of five cards from a standard deck of 52. How many poker hands are there that don't have any spades in them? 1 solutions Answer 313110 by robertb(4012) on 2011-06-01 02:23:44 (Show Source):
Question: Is the longest side of the triangle given in the text? Answer: Yes
Question: What is the form of the equation of the parabola with the vertex at (-1, -7/2)? Answer: y = -(1/2)(x + 1)² - 7/2
Question: What is the triangle inequality relation given in the text? Answer: x + (x+4) > 20
Question: What is the condition for x based on the triangle inequality? Answer: 8 < x < 16 | 677.169 | 1 |
Re: A few questions
Hi;
Not necessary Agnishom found the link belowHello Bobbym, Would you mind if I go to bed nowOf course you can get some sleepOh you found it, thanks. I could not remember where I copied it from. I cleaned it up a little in my notes Agnishom!
There are other triangles other than right triangles that work. I was just pointing out that there are only two that are right triangles. The proof of that is probably much easier than the proof for triangles in general. It will be interesting to see the general proof.
P.S. I tried your link and found the proof. Thanks!
Last edited by noelevans (2012-12-12 08:26:34Hi noelevans! Will you explain the proof! This is Dan's proof of 4 years ago from your link.
Let a,b,c denote the sides of triangle ∆ABC and P and A its perimeter and area respectively. Note that Heron's formula states that the area of ∆ABC is: A = √[s(s-a)(s-b)(s-c)] Where s denotes the semi-perimeter, s = (a+b+c)/2. To find all such triangles such that P=A we must have √(a+b+c)(b+c-a)(a+c-b)(a+b-c) = 4(a+b+c). Put x=b+c-a, y=a+c-b and z=a+b-c, where x,y,z are positive integers (true by the triangle inequality). We see that Heron reduces to the Diophantine equation 16(x+y+z)=xyz.
We see from above that x,y,z must all be even integers. So, we may put x=2m, y=2n, z=2k to get,
mnk=4(m+n+k) ⇒ mnk-4m = 4(n+k) ⇒ m = 4(n+k)/(nk-4)
Without loss of generality assume m≥n≥k, then we have 2m≥2n≥n+k ⇒ nk≤12. Since 4<nk≤12 we may test integral values of n and k to find all such triangles. Finally, we see that there is only 5 such triangles: (a,b,c) = (5,12,13), (6,8,10), (6,25,29), (7,15,20), (9,10,17) ..................................................................................................................................
I followed pretty easily to the three little lines starting with mnk=4(m+n+k) but had a little difficulty figuring out why 4<nk≤12. So I went back to the line mnk=4(m+n+k) and got it from there. First rewrite mnk=4(m+n+k) as nk = 4(m+n+k)/m.
Question: What is Heron's formula for the area of a triangle? Answer: A = √[s(s-a)(s-b)(s-c)], where s is the semi-perimeter
Question: What is the relationship between the perimeter (P) and the area (A) of the triangles being discussed? Answer: P = A
Question: How many such triangles are there that satisfy the given conditions? Answer: 5 | 677.169 | 1 |
Trigonometry and Advanced Math
People sometimes describe trigonometry as the science of circles
and angles. It's trigonometry that helps you calculate the hypotenuse
of a triangle or the diameter of a circle. However, when you use
trigonometry in Excel, you probably won't be worrying about shapes;
instead, you'll be using some type of formula from a scientific field
that requires common trigonometric calculations like the cosine or
tangent.
Students of the sciences know that trigonometry turns up
anywhere you need to think about space, including geography,
astronomy, kinematics, and optics. Less direct applications of
trigonometry turn up in just about every other scientific field—from
chemistry to social science.
Question: True or False: Trigonometry is only used in the fields of mathematics and physics. Answer: False. | 677.169 | 1 |
Therefore if a straight line falling on two straight lines makes the alternate angles equal to one another, then the straight lines are parallel to one another.
Q.E.D.
There is implicitly assumed an ambient plane. The term "alternate angles" doesn't have a meaning unless the lines all lie in a plane.
Note that Euclid does not consider two other possible ways that the two lines could meet, namely, in the directions A and D or toward B and C.
About logical inverses
Although this is the first proposition about parallel lines, it does not require the parallel postulate Post.5 as an assumption. This proposition I.27 and the parallel postulate can be made to look more similar if they are reworded (with the help of I.13).
Proposition 1.27.
If a straight line falls on two straight lines, then if the alternate angles are equal, then the straight lines do not meet.
Post.5.
If a straight line falls on two straight lines, then if the alternate angles are not equal, then the straight lines meet [on a certain side of the line].
If the remark about the side is dropped, then the conclusions are logical inverses of each other, and the logical inverse of a statement is logically equivalent to the converse.
Although the contrapositive is logically equivalent to the statement, Euclid always proves the contrapositive separately using a proof by contradiction and the original statement. Similarly, the inverse is proved using the converse. Sometimes all four statements appear in separate propositions as in propositions X.5 through X.8. Other times the four appear as four statements in one proposition as in X.9. More often than not, however, the contrapositive and inverse make no appearance, and, of course, the converse only appears when it can be proved.
Use of Proposition 27
At this point, parallel lines have yet to be constructed. That occurs in proposition I.31 which uses this proposition to verify that lines constructed there are parallel. This proposition is also used in the next one and in I.33.
Question: What does "Q.E.D." stand for? Answer: "Quod Erat Demonstrandum" which means "that which was to be shown". | 677.169 | 1 |
Thanks
I had 4 as answer. Can someone please elaborate on the answer explanation.
KhurramEach side of the square must have a length of 10. If each side were to be 6, 7, 8, or most other numbers, there could only be four possible squares drawn, because each side, in order to have integer coordinates, would have to be drawn on the x- or y-axis. What makes a length of 10 different is that it could be the hypotenuse of a Pythagorean triple, meaning the vertices could have integer coordinates without lying on the x- or y-axis.
For example, a square could be drawn with the coordinates (0,0), (6,8), (-2, 14) and (-8, 6). (It is tedious and unnecessary to figure out all four coordinates for each square).
If we label the square abcd, with a at the origin and the letters representing points in a clockwise direction, we can get the number of possible squares by figuring out the number of unique ways ab can be drawn.
a has coordinates (0,0) and b could have the following coordinates, as shown in the picture:-BM-
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why we have to take points 6, 8 .( we know that side of square is 10 )please correct me
bluementor wrote:
mjjking wrote:
can somebody explain how and why we find 8,6 and 6,8?
I think its best to use a diagram to illustrate this.
A couple of notes on this one. I think everyone is comfortable with the fact that the square must have sides of length 10, so I will ignore that.
Good Guessing Strategy
- I can quickly sketch the 4 "easy" squares that sit on the axes, so 4 feels too easy - eliminate A.
- I can also see that anything I can draw in one coordinate will be mirrored in the other 3, so the answer must be a multiple of 4 - eliminate B and D.
- This just leaves C and E - not a bad guess if stuck (and that would only have taken a minute or so to do).
Actually Solving
The problem gives us a square in a coordinate plane so step 1 should be to sketch something. I would draw at least one of the "easy" squares (the ones that sit on the axes), but then start to think about rotating it. The side length 10 becomes a distance from the origin = This should be sending off a beacon for you to use Pythagorean Theorem rules. Add to that the fact that the vertices have to be integers = which right triangles have integer side lengths?? AH thats right, the Pythagorean triples.
Question: What is the length of each side of the square in the given problem? Answer: 10 units
Question: What are the coordinates of one possible square with sides of 10 units? Answer: (0,0), (6,8), (-2, 14), and (-8, 6)
Question: Why is it not necessary to find all four coordinates for each square? Answer: The number of possible squares can be determined by figuring out the number of unique ways the first two points (a and b) can be drawn, as the other two points will be determined by the side length of 10. | 677.169 | 1 |
(designed for MSIE, requires JavaScript, best view 1024 x 768, auxiliary window needed) Let Sbe the ground plane (desert, floor tiles)
and B an image plane (the computer screen) perpendicular to S.
The observer's eye A is at height h above S and distance d in front of B.
The observer is looking straight at B, so we have the principal ray
AH, orthogonal to B, which intersects B at the principal point H (see Fig.1).
Now imagine a real scene (e. g. the wall) built up on S.
Let P be a point in the scene (e.g. the top of the corner of the wall).
The position of P is given by three coordinates
(xP|yP|zP)
in an orthogonal x,y,z coordinate system:
The origin is at H, the x and y axes lie in the plane B with the x axis pointing to the right and the y axis upward, while the z axis points away from A to H. The ray AP runs from A to P.
Where it intersects B is the image point P*
of P; in the x,y coordinate system of the image plane B it has coordinates
P*(xP*|yP*).
Measurements in the image plane (with explanation of terms)
By inverting our mapping equations we can now very easily determine the proportions of various distances in the real scene, knowing where they map to in the image.
How*) the inset works:
Each cursor position corresponds to a particular image point P* derived from a real point
P (see above).
Line 2 shows the coordinates of P* in pixels (origin of coordinates is the principal point H).
Line 3 shows the coordinates of P in metres (origin of coordinates is the eye point A).
Line 5. Radio switch up: if you click on a point in the image, the corresponding real point V becomes the reference point for measuring a length.
Line 6. Radio switch down: the observer's eye A is the reference point.
Line 7 shows the real coordinates of the reference point in metres.
Line 8 shows the distance in metres from the reference point to the real point represented by the cursor.
(Hint: use the numerical arrow keys to move the cursor.) *) The calculation window can be moved by drag and drop (blue title bar) to a convenient part of the screen. If coordinates are displayed incorrectly, click the principal point on the horizon 12 pixels away from the right edge of the levelling staff.
If no auxiliary window has opened, allow your browser to do so, or click here: 'no popup'-version
Question: How can you move the calculation window? Answer: The calculation window can be moved by drag and drop (blue title bar) to a convenient part of the screen.
Question: What are the coordinates of the image point P in the image plane B? Answer: The coordinates of the image point P in the image plane B are (xP|yP). | 677.169 | 1 |
From Higher Dimensions Database
The pyrochoron, also known as the pentachoron and the 5-cell, is the four-dimensional simplex, and has the lowest possible element count of any flat, non-degenerate four-dimensional shape. It consists of 5 regular tetrahedra joined at their faces, folded into 4D to form a 4D volume. It is a special case of the pyramid where the base is a tetrahedron. There are 3 tetrahedra surrounding every edge.
Equations
Net
The net of a pentachoron is a tetrahedron surrounded by 4 more tetrahedra.
Projection
Cell-first / vertex-first projection
The following diagram shows a perspective projection of the pentachoron.
The dotted line shows the far edge of the outer tetrahedron. The blue lines are inside the outer tetrahedron in this projection. The center of the projection where these blue lines meet is actually the apex of the pentachoron pointing away from us in the 4th direction.
All 5 tetrahedral cells of the pentachoron are present in this diagram: the outer tetrahedron, and the 4 "inner" tetrahedra outlined by one triangular face of the outer tetrahedron and 3 of the blue lines each. Although they appear as slightly flattened tetrahedra, this is only because they are being viewed at from an angle. In actuality, they are perfectly regular tetrahedra.
The following diagrams illustrate 3 of these cells.
Note that if a 4D being were to look at an actual pentachoron, it would not be able to see all 5 cells at once. Rather, it would either see only the outer tetrahedron (if it looks at the "base" of the pentachoron), or the 4 inner cells (if it looks at the apex of the pentachoron).
Edge-first / face-first projection
The next diagram shows the pentachoron viewed at from another angle.
In this diagram, three of the pentachoron's cells are arranged around the central axis indicated by the blue line. The other two cells are the upper and lower halves of the outer shape, which is called a trigonal bipyramid. All the cells appear somewhat deformed from a regular tetrahedron, because they are all being viewed at from an angle. Some of these cells are shown below:
Again, this projection represents two possible views of the pentachoron. From one view, the 4D being would see only the upper and lower tetrahedra. From the opposite view, it would see the 3 inner tetrahedra. It cannot see all 5 cells at once unless the pentachoron is transparent.
Question: How many regular tetrahedra are joined to form a pyrochoron? Answer: 5
Question: What is the name of the four-dimensional simplex with the lowest possible element count? Answer: Pyrochoron | 677.169 | 1 |
Geometry ABC is a triangle with area equal to 20 . The incircle of triangle ABC has radius equal to 2 and the circumcircle of triangle ABC has radius equal to 6 . If sinA+sinB+sinC=a/b , where a and b are coprime positive integers, what is the value of a+b ?
Geometry ƒ¡ is a circle with center O . A and B are points on ƒ¡ such that the sector AOB has perimeter 40 . What is the measure of ÚAOB (in radians) when the area of the sector AOB is maximized
Trigonometry Let è be the angle between the x-axis and the line connecting the origin O(0,0) and the point P(−3,−4) , where 180<è<270. Given that sinè+cosè+tanè=−a/b, where a and b are coprime positive integers. What is the value o...
MEchanics M-1 (Velovity time graphs) A dragonfly, at rest on a bullrush, decides to fly to a second bullrush 18 m away. It accelerates uniformly to a speed of. 5 m s-1, then immediately decelerates uniformly to rest on the second bulrish. Howlong the journey took?
Physics We use a wrench to turn nuts on bolts because they require less force. Consider a hexagonal nut 1 cm in diameter. We can tighten this nut with one of two wrenches, wrench A with lever arm 10 cm and wrench B with lever arm 20 cm. Both wrenches have a very small mass, so you may...
Trigonometry Let S be the sum of all the possible values of sin x that satisfy the following equation: 5-2cos^2(x)-7sin(x) = 0 S can be written as a/b, where a and b are coprime positive integers. What is the value of a + b?
Calculus The location of a dot P at a given time t in the x-y plane is given by (x,y) = (t - sin t, 1 - cos t). What is the distance traveled by P in the interval 0 <= t <= 2pi
Question: What is the perimeter of the sector AOB in circle ƒ¡? Answer: 40
Question: What is the radius of the incircle of triangle ABC? Answer: 2
Question: What is the value of a + b in the equation sinA + sinB + sinC = a/b for triangle ABC? Answer: 10 | 677.169 | 1 |
Angles are often measured in a common unit of degrees. One full
circle equals 360 degrees. But, angles can also be measured in a
scientific unit called radians. One full circle equals 2π
radians, which equals 360 degrees. Since 360 degrees = 2π
radians, π in radians equals 180 degrees and π/
2 equals 90 degrees.
It is
understood when given equations such as y = Sin(X) that X is in radians.
If not, then they would give a degrees sign.
Plotting
the Graph of Y = Sin(X)
In this equation, Y ranges from -1 to +1 since the Sine of any angle cannot be
greater than 1 and cannot be less than -1.
Each time the angle X is increased by 2π, the
trigonometric functions of X repeat. So, the graph of Y = Sin(X) is a
repeating function with a period of 2π.
The amplitude of equations when graphed is distance from the X-axis to the
highest point on the curve. The amplitude is the number in front of the
function either Sin(x) or Cos(x). In this case, the amplitude is 1.
The frequency of a graph is the number of times the graph repeats itself in the
interval 0 to 2π. The frequency can be found
from the equation. It is the number in front of the angle X. In this
case, the frequency is also 1. If the equation given is Y = 2Sin(1/2X),
then the amplitude is 2, and the frequency is 1/2.
To find the period just by looking at the equation, you find the frequency and
plug it into the formula: Period = 2π/frequency.
The
Graph of Y = Sin(X) with an interval of 0≤X≤2π
Plotting Y =
Cos(X)
The
amplitude is 1, the frequency is 1, and the period is 2π.
The
Graph of Y = Cos(X) with an interval of 0≤X≤2π
Plotting the Graph
of Y = Tan(X) with the interval of 0≤X≤2π
Since Tan(X) = Sin(X)/Cos(X), when Cos(X) is zero, the tangent is
undefined. This happens when X equals π/2 and 3π/2.
"Near" these values, the function is extremely large and approaches
infinity. The lines X = π/2 and X = 3π/2
are called asymptotes.
The graph
Cot(X) is similar to the graph of Tan(X). The only differences are that
the asymptotes are at X=0 and X= π and the graph is
horizontally flipped.
Question: What is the frequency of the graph y = Sin(X)? Answer: 1 | 677.169 | 1 |
In geometry, topology, and related branches of mathematics, a closed set is a set whose complement is an open set. In a topological space, a closed set can be defined as a set which contains all its limit pointsleft|150px|thumb|A [[spheroid]].A great ellipse is an ellipse passing through two points on a spheroid and having the same center as that of the spheroid. Equivalently,it is an ellipse on the surface of a cylinder centered at the origin....
Arc length
Determining the length of an irregular arc segment is also called rectification of a curve. Historically, many methods were used for specific curves...
of an arc of a circle with radius and subtending an angle (measured inA central angle is an angle which vertex is the center of a circle, and whose sides pass through a pair of points on the circle, thereby subtending an arc between those two points whose angle is equal to the central angle itself...
— equals . This is because
Substituting in the circumference
and solving for arc length, , in terms of yields
An angle of degrees has a size in radians given by
and so the arc length equals
A practical way to determine the length of an arc in a circle is to plot two lines from the arc's endpoints to the center of the circle, measure the angle where the two lines meet the center, then solve for L by cross-multiplying the statement:/360 = L/Circumference.
For example, if the measure of the angle is 60 degrees and the Circumference is 24", then
60/360 = L/24
360L=1440
L = 4".
This is so because the circumference of a circle and the degrees of a circle, of which there are always 360, are directly proportionate.
Arc area
The area between an arc and the center of a circle is:
The area has the same proportion to the circle area as the angle to a full circle:
We can get rid of a on both sides:
By multiplying both sides by , we get the final result:
Using the conversion described above, we find that the area of the sector for a central angle measured in degrees is:
Arc segment area
The area of the shape limited by the arc and a straight line between the two end points is:
To get the area of the arc segment, we need to subtract the area of the triangle made up by the circle's center and the two end points of the arc from the area . See Circular segment
Circular segment
In geometry, a circular segment is an area of a circle informally defined as an area which is "cut off" from the rest of the circle by a secant or a chord. The circle segment constitutes the part between the secant and an arc, excluding the circle's center...
for details.
Arc radius
Using the equality in the intersecting chords theorem (also known as power of a point
Power of a point
Question: Is a closed set in topology defined by its limit points? Answer: Yes, a closed set in topology is defined by containing all its limit points.
Question: What is the definition of a circular segment in geometry? Answer: A circular segment is an area of a circle that is "cut off" from the rest of the circle by a secant or a chord, excluding the circle's center. | 677.169 | 1 |
QuoteAgain, pretty much lost on me. Looking at that web page, it doesn't say anything about absolute angle, it just shows one solution. Does absolute angle mean it's always the smaller of the two possible angles?Actually, I'm not sure whether you should bother explaining any of that. I'm willing to take your word for it and just try using the calculations you provide. I just need to know how to find the smallest of the two possible angles, in degrees.Using the dot product as above will always get you the smallest angle between two angles, but obviously theres a bigger angle that exists:
1 2 3 4 5 6
| / |a/ | / |/
b
The smallest angle is clearly 'a', yet the largest angle 'b' loops all the way around the other way. So b = 360 - a (or b = 2PI - a in radians).
Question: What is the relationship between the larger angle 'b' and the smaller angle 'a'? Answer: b = 360 - a (or b = 2PI - a in radians) | 677.169 | 1 |
1st 2 sides are 16 each and they form angle = 60 degrees
So If u draw a diagonal the triangle that contains diagonal and these 2 sides will be equilateral, with sides 16.
Therefore area of Triangle = (sqrt3/4)*16*16 =64*sqrt3
For the other triangles sides are=16,14,10
Hence its area = 40sqrt3
Area of quad = sum of 2 triangles = 104 sqrt3!!
Question: What is the total area of the quadrilateral formed by these triangles? Answer: 104√3 | 677.169 | 1 |
In graphical perspective, a vanishing point is a point in the picture plane Π that is determined by a line in space. Given the oculus or eye point O and a line L not parallel to Π, let M be the line through O and parallel to L. Then the vanishing point of L is the intersection of M and Π. Traditional linear drawings use one to three vanishing points.
A curvilinear perspective is a drawing with either 4 or 5 vanishing points, in 5-point perspective the vanishing points are mapped into a circle with 4 vanishing points at the cardinal headings N,W,S,E and one at the circle origin.
A reverse perspective is a drawing with vanishing points that are placed outside the painting with the illusion that they are "in front of" the painting.
Vanishing points can also refer to the point in the distance where the two verges of a road appear to converge. This is often used to help assess the upcoming curves in the road; to judge the radius and therefore the entry speed and optimum line. If the vanishing point moves towards you or to your sides, the curve is tightening. If the vanishing point moves away from you or comes to center, the curve is straightening.
Question: What is a vanishing point in graphical perspective? Answer: A vanishing point is a point in the picture plane that is determined by a line in space. | 677.169 | 1 |
Hello: These are my boxes in one point perspective, and various boxes. Please comment on them. Question: I was looking at a small rectangular table today. It was below my eyeline and about 5 feet away. The dimensions were about 12" x 18", and as I looked at it, the side edges appeared to be wider apart at the back side than the near side, so that they looked like \ and / rather than / /. To the point I went over and measured the table. When I returned to my seat, I still had that impression--that the edges were \ and /. What causes that distortion, and how can I believe my eyes with that? That phenonema would explain some of the problem others are having, too. Logically, and realistically, as I measured them, the sides are parallel. Wisually, they did not appear to do se, even as I studied them, measured, and looked at them again.
Thanks
Crafor
crafor
12-06-2011, 08:47 PM
lesson 2 perspective
Hello,
This is the table for exercise 4. Thank you for checking it.
Crafor
arnoud3272
12-07-2011, 03:48 AM
Diane - Very good job :clap::clap:. All correctly done.
:thumbsup:
arnoud3272
12-07-2011, 04:13 AM
crafor - Very well done too :clap::clap:.
the side edges appeared to be wider apart at the back side than the near side,.... Yes, I know, it is a common phenomenon. It's your brain playing tricks, like those optical illusions. ( But if you hold a knitting needle or so parallel to one side and move it to the other side without changing the angle - possibly this will need some practice -, you will see they converge.
So you undersand why for centuries after its discovery, artists were very enthusiastic about perspective, it freed them from the need to measure and re-measure everything.
BTW, this skill to move while keeping the angle constant is very useful for transfering angles from observation to drawing :thumbsup:.
astropaint
12-08-2011, 09:02 AM
Hi Arnoud,
I have corrected some boxes per your critique. I also re-did the table and drew a chair. Please provide your feedback. Thanks in advance for all your help!
astropaint
12-08-2011, 09:05 AM
Hi Arnoud,
jrl11528
12-08-2011, 09:19 AM
Updated boxes and tissue box. Didn't mean for this to take so long but work has a way of interfering. I hope to have the last two parts of this lesson completed this evening.
Question: How far away was Crafor from the table? Answer: About 5 feet
Question: What caused the distortion Crafor perceived in the table's side edges? Answer: An optical illusion, where the brain tricks the eyes into perceiving the edges as wider apart at the back side than the near side
Question: How can one verify that the table's side edges are parallel, despite the visual distortion? Answer: By measuring the table or using a straight object, like a knitting needle, to compare the angles | 677.169 | 1 |
1) Is insufficient - This length doesn't help us except to tell us that BCD is not a right isosceles triangle. 2) x = 60. This means that Angle BCD = 180-60 =120. The remaining angle CBD = 180-120-30 = 30. This tells us that the triangle is isosceles. BC = CD = 6.
As it's DS question no need for actually finding the value of BC, rather than to determine that it's possible to determine it with either of statements:
(1) BD=6\sqrt{3}, we know CD, BD and the angle between them the opposite side BC is fixed and has single value, meaning that you can not draw two or more triangles with given two sides and the angle between them. Sufficient.
(2) x=60. Again we know x, hence we know all the angles in triangle BCD, plus we know one of the sides CD=6, again only one such triangle exists, hence BC value can be determined. Sufficient.
So just for my knowledge, if we know 2 sides of a triangle, and the angle in between we can safely determine that the information is sufficient?
When answering I did think about what you guys said, but thought that we a) couldn't assume that if we have 2 fixed sides and an angle we can derive the 3rd (althgouh I know we don't need to actaully derive it) and b) that trigonometry wasn't really required knowledge for the GMAT?
So just for my knowledge, if we know 2 sides of a triangle, and the angle in between we can safely determine that the information is sufficient?
This is a very good question. Well, I think everybody agrees that knowing such tips is very important for GMAT. Especially in DS as it helps to avoid time wasting by not calculating an exact numerical values.
When can we say that information given is sufficient to calculate some unknown value in triangle? Think it's the same as determining congruency. If we are given some data and we can conclude that ONLY one triangle with given measurements exists, it should mean also that with given data we can calculate anything regarding this triangle.
Determining congruency:
1. SAS (Side-Angle-Side): If two pairs of sides of two triangles are equal in length, and the included angles are equal in measurement, then the triangles are congruent.
2. SSS (Side-Side-Side): If three pairs of sides of two triangles are equal in length, then the triangles are congruent.
3. ASA (Angle-Side-Angle): If two pairs of angles of two triangles are equal in measurement, and the included sides are equal in length, then the triangles are congruent.
So, knowing SAS or ASA is sufficient to determine unknown angles or sides.
Question: Is knowing two sides and the angle between them sufficient to determine the third side of a triangle? Answer: Yes.
Question: If we can conclude that only one triangle with given measurements exists, can we also calculate any unknown value regarding that triangle? Answer: Yes. | 677.169 | 1 |
Name some phenomena that may be characterized by graphs of the shape you produced. With each phenomenon, describe in what units the horizontal and vertical measurements might be made. [Sample answers: Height of a rider on a Ferris wheel: vertical measurement of height, horizontal measurement of time. Wave in the water: vertical and horizontal measurement might both be distances, or, if they think of a buoy, the horizontal component of the graph might be time. Length of daylight, in hours, on the vertical axis might be graphed against date, in days, on the horizontal axis. Sound waves might have air pressure on the vertical axis and time on the horizontal axis.]
Teacher Reflection
Was assigning specific roles to students in the groups effective, allowing for efficient data collection?
Was there adequate time for discussion of the key points in the lesson?
Do your students understand that dividing measurements on a circle by the radius of the circle converts the measurements into radian measure?
Were your students comfortable describing points on a circle in terms of arc length instead of angle?
This lesson was careful to avoid introducing a formula for translating between radians and degrees. Are you comfortable with that approach, or would you rather have students learn the formulas first
Question: Which of the following is a reason given for not introducing a formula for translating between radians and degrees in the lesson? Answer: The lesson was careful to avoid introducing a formula | 677.169 | 1 |
Of course, you don't have to -- I know you have better things to do than to teach mathematics to some Canadian who really should have been paying more attention in school. ;-)
(You can probably safely assume that I have an upper level of understanding of math, however -- I did take entry-level Calculus in school, and I didn't do too bad (relatively speaking to my normal suckiness) either.)Well, the subtraction and addition parts should be simple enough for starters--basically I changed the coordinates so that they were relative to the pivot point, by subtracting the pivot, and then after rotation I added that back in.
The rotation part is a formula I can never fully remember, so I have to work it out every time. But basically it goes like this:
For the sake of discussion I'll call the rotated coordinates x' and y'.
Consider a right triangle. One side is x, and it goes all the way from the origin to (x,0). The next side is y, and it goes from (x,0) to (x,y). The hypotenuse goes from the origin to (x,y). Now, tip that triangle clockwise, pivoting it around the origin'd be a lot easier to explain with diagrams.
The way I figure it out is: x' will be the same as x if no rotation is used, so that's the cosine term. If you rotate 90° clockwise, it's y, and that's the sine term. Therefore x'=x*cos(angle)+y*sin(angle). Similar reasoning goes into y'; it's normally y unless you rotate, so that's the cosine term, while y'=-x if you rotate 90° clockwise, so y'=y*cos(angle)-x*sin(angle). I can never keep the signs straight without doing this each time because coordinate systems differ is probably one of the few times in my life that I've actually understood a concept like this -- the former line being a hypoteneuse of a new triangle made it easy to understand.
Thanks a bunch.
(Incidentally, I'm taking grade 11 mathematics* through night school, so I'll build up a geometric foundation within the next few months. =))
* It's too lengthy to explain the B.C. math curriculum here -- suffice it to say that it's a comprehensive course designed to teach people upper high-school level mathematics in all fields, ranging from geometry to algebra. Grade 12 mathematics is top level non-calculus stuff that you take before college.
Question: What is the user's current level of mathematical understanding? Answer: The user has an upper level of understanding of math, having taken entry-level Calculus in school. | 677.169 | 1 |
Hi! My son has been working independently on the Geometric Approach so I am somewhat lost when he comes to me with a question (even after I try to review the earlier lessons leading up to the problem). I am hopeful you will be able to catch me up so I can help him with the chart on Worksheet 9-1. I see that the chart's first division multiplied by the second division equals the number of pieces. Are the answers interchangable and may other combinations work? For instance, for 18 pieces, may he use 6 as the first division and 3 as the second division -or- may he use 2 as the first division and 9 as the second division? Thank you for helping this momma!
You are correct. You are basically looking for 2 numbers that multiply together to equal the number of pieces you are splitting the triangle into. If you are trying to split a triangle into 18 pieces, you can split the triangle into 9 pieces and then split each of the 9 pieces in half. (9 x 2 is 18). Another option, you can also split the triangle into 6 pieces and then each of the 6 pieces into 3 more pieces. (6 x 3 is 18). You can see a sample of the triangle divided into 6's on worksheet 5.
I hope that has helped, but if you have further questions, you can feel free to call RightStart's help desk at (888)272-3291 or email them at [email protected].
Question: If you want to divide a triangle into 24 pieces, what is one possible combination of divisions? Answer: 12 and 2.
Question: Can you divide a triangle into 18 pieces by first dividing it into 9 pieces and then dividing each of those into 4 pieces? Answer: No, that would result in 36 pieces. | 677.169 | 1 |
Archimedes had reduced the problem of finding a regular hexagon to that of finding two points that divided a line segment into two mean proportionals. He then used a construction somewhat like that of the painting to find a line segment divided as desired. Crockett Johnson's papers include not only photocopies of the relevant portion of Heath, but his own diagrams.
The painting is #104 in the series. It is in acrylic or oil on masonite., and has purple, yellow, green and blue sections. There is a black wooden frame. The painting is unsigned and undated. Relevant correspondence in the Crockett Johnson papers dates from 1974.
This painting is part of Crockett Johnson's exploration of the properties of the heptagon, extended to include a 14-sided regular polygon. The design of the painting is shown in his figure, which includes many of the line segments in the painting. Here Crockett Johnson argues that the triangle ABF in the figure is the one he sought, with angle FAB being one seventh of pi. Segment CD in the figure, which appears in the painting, is the length of the edge of a regular 14-sided figure inscribed in a portion of the larger circle shown.
The painting, of oil or acrylic on masonite, is number 105 in the series. It is drawn in shades of cream, blue, and purple on a light purple background. It has a metal frame and is unsigned.
This whimsical painting is part of Crockett Johnson's exploration of ways to represent the sides and angles of a regular heptagon using line segments of equal length. In its mathematics, it follows closely the construction from isosceles triangles within a rhombus used in the painting Heptagon from Ten Equal Lines (#104 in the series - 1979.1093.71). However, both the line segments shown and the appearance of the paintings are quite different.
Here three pairs of carefully selected equal lines at appropriate equal angles combine with a seventh line of equal length to give a construction of three sides and two angles of a regular heptagon. All but one of the endpoints of the lines lie on a parallelogram (the rhombus mentioned previously), hence the title. The segment of the heptagon is on the right side of the painting. In Crockett Johnson's figure for the work, the segment is lettered BCPE.
The painting, in oil or acrylic on masonite, is #106 in the series. It has a dark purple background. The pairs of line segments are in turquoise, green, and lavender, with the vertical one in white. This increases the drama of the painting, but obscures the heptagon. There is a wooden frame. The painting is signed on the back: HEPTAGON STATED BY (/) SEVEN TOOTHPICKS (/) (BETWEEN PARALLELS) (/) Crockett Johnson 1973.
Question: What is the title of painting #106? Answer: Heptagon Stated by (/) Seven Toothpicks (/) (Between Parallels) (/)
Question: In which year was painting #106 signed? Answer: 1973
Question: Which of the following is NOT a color used in painting #106? A) Turquoise B) Green C) Red Answer: C) Red | 677.169 | 1 |
Toward the end of his life, Crockett Johnson took up the problem of constructing a regular seven-sided polygon or heptagon. This construction, as Gauss had demonstrated, requires more than a straight edge and compass. Crockett Johnson used compass and a straight edge with a unit length marked on it. Archimedes and Newton had suggested that constructions of this sort could be used to trisect the angle and to find a cube with twice the volume of a given cube, and Crockett Johnson followed their example.
One may construct a heptagon given an angle of pi divided by seven. If an isosceles triangle with this vertex angle is inscribed in a circle, the base of the triangle will have the length of one side of a regular heptagon inscribed in that circle. According to Crockett Johnson's later account, in the fall of 1973, while having lunch in the city of Syracuse on Sicily during a tour of the Mediterranean, he toyed with seven toothpicks, arranging them in various patterns. Eventually he created an angle with his menu and wine list and arranged the seven toothpicks within the angle in crisscross patterns until his arrangement appeared as is shown in the painting.
Crockett Johnson realized that the vertex angle of the large isosceles triangle shown is exactly π/7 radians, as desired. The argument suggested by his diagram is more complex than what he later published. The numerical results shown in the figure suggest his willingness to carry out detailed calculations.
Heptagon from its Seven Sides, painted in 1973 and #107 in the series, shows a triangle with purple and white sections on a navy blue background. This oil or acrylic painting on masonite is signed on its back : HEPTAGON FROM (/) ITS SEVEN SIDES (/) (Color sketch for larger painting) (/) Crockett Johnson 1973. No larger painting on this pattern is at the Smithsonian.
This is one of a series of paintings in which Crockett Johnson explored ways of constructing the regular heptagon. The construction is his own, and a drawing for it is attached to the back of the painting. By an arrangement of ten equal line segments, he produced three sides and two angles of a regular heptagon. Two sides and one angle are actually shown in the painting.
Crockett Johnson supposed that four equal isosceles triangles, constructed with six equal line segments, were arranged as shown in his figure to form sides of a rhombus and of a parallelogram within it. Two adjacent sides of the rhombus also served as the long sides of equal triangles oriented in the opposite direction. Finally, a line parallel to one of these sides passed through points of intersection of the sides of triangles.
Question: Who was the artist of the painting "Heptagon from its Seven Sides"? Answer: Crockett Johnson | 677.169 | 1 |
Three very similar paintings in the Crockett Johnson collection are closely related to the Crockett Johnson described the construction of his isosceles triangle in the diagram reproduced. The horizontal line segment below the circle on the painting corresponds to unit length BF in the figure, and the triangle is ABF. Three of the four light-colored sections of the painting highlight important points in the construction. The critical steps are drawing a perpendicular bisector to the line segment BF, marking off an arc of radius equal to the √(2) with center F, and measuring the unit length AO along a marked straightedge that passes through B and intersects the perpendicular bisector at A. Finally, one finds the side of the regular inscribed heptagon.
Construction of Heptagon is #117 in the series. The oil painting on masonite is in shades of purple, cream, turquoise, and black. It has a black wood and metal frame. The work is unsigned. The surface appears damaged, perhaps from water. See also #115 (1979.1093.77) and #108 (335571).
Three very similar paintings in the Crockett Johnson collection are closely related to7 in the image. The horizontal line segment below the circle on the painting corresponds to unit length BF in the figure, and the triangle is ABF. The light colors of the painting highlight important points in the construction - marking off an arc of radius equal to the square root of 2 with center F, measuring the unit length AO along a marked straight edge that passes through B and ends at point A on the perpendicular bisector, and finding the side of the regular inscribed heptagon.
This version of the construction of a heptagon is #108 in the series. The oil painting on masonite with chrome frame was completed in 1975 and is unsigned. It is marked on the back: Construction of the Heptagon (/) Crockett Johnson 1975. See also paintings #115 (1979.1093.77) and #117 (1979.1093.79) in the series.
Question: What shape is being constructed in the paintings? Answer: A heptagon
Question: What year was painting #108 completed? Answer: 1975 | 677.169 | 1 |
Examples of Pythagorean triples include (3, 4, 5) and (5, 12, 13). There are infinitely many such triples,[2] and methods for generating such triples have been studied in many cultures, beginning with the Babylonians[3] and later ancient Greek, Chinese and Indian mathematicians.[4] The traditional interest in Pythagorean triples connects with the Pythagorean theorem;[5] in its converse form, it states that a triangle with sides of lengths a, b and c has a right angle between the a and b legs when the numbers are a Pythagorean triple. Right angles have various practical applications, such as surveying, carpentry, masonry and construction. Fermat's Last Theorem is an extension of this problem to higher powers, stating that no solution exists when the exponent 2 is replaced by any larger integer.
[edit] Diophantine equations
Main article: Diophantine equation
Fermat's equation xn + yn = zn is an example of a Diophantine equation.[6] A Diophantine equation is a polynomial equation in which the solutions must be integers.[7] Their name derives from the 3rd-century Alexandrian mathematician, Diophantus, who developed methods for their solution. A typical Diophantine problem is to find two integers x and y such that their sum, and the sum of their squares, equal two given numbers A and B, respectively:
A = x + y\
B = x^2 + y^2\
Diophantus' major work is the Arithmetica, of which only a portion has survived.[8] Fermat's conjecture of his Last Theorem was inspired while reading a new edition of the Arithmetica,[9] which was translated into Latin and published in 1621 by Claude Bachet.[10]
Diophantine equations have been studied for thousands of years. For example, the solutions to the quadratic Diophantine equation x2 + y2 = z2 are given by the Pythagorean triples, originally solved by the Babylonians (c. 1800 BC).[11] Solutions to linear Diophantine equations, such as 26x + 65y = 13, may be found using the Euclidean algorithm (c. 5th century BC).[12] Many Diophantine equations have a form similar to the equation of Fermat's Last Theorem from the point of view of algebra, in that they have no cross terms mixing two letters, without sharing its particular properties. For example, it is known that there are infinitely many positive integers x, y, and z such that xn + yn = zm where n and m are relatively prime natural numbers.[note 1]
[edit] Fermat's conjecture
Problem II.8 in the 1621 edition of the Arithmetica of Diophantus. On the right is the famous margin which was too small to contain Fermat's alleged proof of his "last theorem".
Question: What is the name of the theorem that extends the problem of Pythagorean triples to higher powers, stating that no solution exists when the exponent 2 is replaced by any larger integer?
Answer: Fermat's Last Theorem
Question: Who was the 3rd-century Alexandrian mathematician who developed methods for solving Diophantine equations?
Answer: Diophantus
Question: What is the converse form of the Pythagorean theorem stating about a triangle with sides of lengths a, b, and c?
Answer: It states that such a triangle has a right angle between the a and b legs when the numbers are a Pythagorean triple. | 677.169 | 1 |
First, it takes an x coordinate for the center of the shape, not the edge. Second, it takes a y coordinate for the center of the shape. The third parameter is the radius of the shape. The fourth parameter describes the beginning angle in radians. The fifth parameter describes the ending angle in radians, and the sixth and final parameter describes the direction the arc will be drawn. Here are those parameters again as a list.
x coordinate of the shape's center point
y coordinate of the shape's center point
radius
beginning angle in radians
ending angle in radians
drawing direction (called anticlockwise parameter)
Here is some example code that will produce a circle.
context.arc(275,275,200,0, Math.PI*2,true);
context.arc(275, 275, 200, 0, Math.PI*2, true);
Notice that the circle's center point will be at 275, 275 or approximately the middle of the canvas I have been using in this tutorial.
The radius, which is the third parameter, is set to 200 pixels. This means that the complete circle will be 400 pixels in diameter.
The beginning angle is at zero radians along the circle's circumference. The ending angle is set to π times 2 or 2π.
Recall a radian is a ratio between the radius and the arc's length. Interestingly, when you are measuring in radians, there is one spot on the circumference of the circle that has two values. That spot is at zero or 2π. I think of it as a spot on the right side of the circle. You get to this spot by following the x axis out to the right the length of the radius. By setting the beginning and ending angles at the same point on the circumference, we get a complete circle.
Wikipedia provides a helpful illustration of how radians relate to degrees on a circle.
Finally, the so-called anticlockwise parameter determines which way the shape will be drawn. Setting it to "true" means the arc will move in a counterclockwise fashion.
In canvas, a circle is drawn using an arc that has the same beginning and ending angle.
To help explain the anticlockwide parameter, I am going to reset the ending angle of my circle to π rather than 2π, drawing a half circle. Because I am drawing in the counterclockwise direction, I will be keeping the top half of the circle.
context.arc(275,275,200,0, Math.PI,true);
context.arc(275, 275, 200, 0, Math.PI, true);
Drawing a half circle in the counterclockwise direction places the shape at the top of what was the circle.
If I draw the arc in the clockwise direction, by setting the anticlockwise parameter to "false", the half circle is flat on top or on the bottom half of the original circle.
Question: What are the units of the beginning and ending angles? Answer: Radians. | 677.169 | 1 |
For convex two-dimensional shapes, the centroid can be found by balancing the shape on a smaller shape, such as the top of a narrow cylinder. The centroid occurs somewhere within the range of contact between the two shapes. In principle, progressively narrower cylinders can be used to find the centroid to arbitrary accuracy. In practice air currents make this unfeasible. However, by marking the overlap range from multiple balances, one can achieve a considerable level of accuracy.
Of a finite set of points
The centroid of a finite set of points in is
By geometric decomposition
The centroid of a plane figure can be computed by dividing it into a finite number of simpler figures , computing the centroid and area of each part, and then computing
Holes in the figure , overlaps between the parts, or parts that extend outside the figure can all be handled using negative areas . Namely, the measures should be taken with positive and negative signs in such a way that the sum of the signs of for all parts that enclose a given point is 1 if belongs to , and 0 otherwise.
For example, the figure below (a) is easily divided into a square and a triangle, both with positive area; and a circular hole, with negative area (b).
(a)
(b)
(c)
The centroid of each part can be found in any list of centroids of simple shapes (c). Then the centroid of the figure is the weighted average of the three points. The horizontal position of the centroid, from the left edge of the figure is
The same formula holds for any three-dimensional objects, except that each should be the volume of , rather than its area. It also holds for any subset of , for any dimension , with the areas replaced by the -dimensionalBy integral formula
Integration is an important concept in mathematics and, together with its inverse, differentiation, is one of the two main operations in calculus...
where the integral is taken over the whole space , and g is the characteristic function of the subset, which is 1 inside X and 0 outside it. Note that the denominator is simply theof the set X (However, this formula cannot be applied if the set X has zero measure, or if either integral diverges.)
Another formula for the centroid is
where Ck is the kth coordinate of C, and Sk(z) is the measure of the intersection of X with the hyperplane defined by the equation xk = z. Again, the denominator is simply the measure of X.
For a plane figure, in particular, the barycenter coordinates are
where A is the area of the figure X; Sy(x) is the length of the intersection of X with the vertical line at abscissa
Abscissa
In mathematics, abscissa refers to that element of an ordered pair which is plotted on the horizontal axis of a two-dimensional Cartesian coordinate system, as opposed to the ordinate...
x; and Sx(y) is the analogous quantity for the swapped axes.
Of an L-shaped object
This is a method of determining the centroid of an L-shaped object.
Question: What is the range within which the centroid occurs when using a narrow cylinder? Answer: The range of contact between the two shapes.
Question: What are the barycenter coordinates for a plane figure? Answer: The barycenter coordinates are given by the formula where A is the area of the figure, Sy(x) is the length of the intersection of X with the vertical line at abscissa x, and Sx(y) is the analogous quantity for the swapped axes.
Question: How can the centroid of a finite set of points in be calculated? Answer: By dividing the set into simpler figures, computing the centroid and area of each part, and then computing the weighted average.
Question: Can the centroid of a convex two-dimensional shape be found using a narrow cylinder? Answer: Yes, in principle. | 677.169 | 1 |
If you don't know what a cross product is try this alternative method for evaluating the area of the triangle. Lets say we have a triangle whose vertices are lattice points. As shown in the picture below draw a vertical line from A so that this line intersects at .
Use coordinated of and to write down an equation of the line . Find coordinates of using the equation of this line and the -coordinate of . Now using this information evaluate the area of as the sum of two areas of triangles and and show this area is at least .
This shows that for any polygon , .
Now assume for a triangle , we have . Using two vertices of this triangle you can draw a rectangle whose sides are parallel to the axes and contains this triangle and two of its vertices are the vertices of this triangle as shown in the following picture.
As shown in the above picture, divide this rectangle into four triangles. For each one of this triangles we know that , but for the rectangle we already showed that which shows for the triangle the equality should hold also.
Some parts of this proof needs a bit of work that is left to the reader.
I discussed FLT in an earlier post and stated a problem that if solved FLT for will be deduced from there. Below is a solution to that problem.
Assume is a "non-trivial" triple of integers that satisfies
(1) .
By "non-trivial" we mean "". Obviously we can assume that and are positive. If we show for any such triple, there is another triple satisfying the same equation with a smaller , then using the method of "infinite descent" we get a contradiction, and therefore we can deduce that there is no "non-trivial" solutions for .
Now one can show that and are relatively prime. Otherwise you can find a "smaller" solution for the equation (1). Therefore is a Pythagorean triple and we have two relatively prime integers and such that:
Using the fact that is odd (why?) and the equation we can deduce that is odd and is even (why?). Now since and are relatively prime and is a perfect square we can deduce that and are perfect squares (why?). Therefore there are two relatively prime integers and such that and .
Now look at the equation and use the formula for Pythagorean triples. So there are relatively prime integers and such that:
Now since is a perfect square, should be a perfect square. Therefore both and are perfect squares. Therefore there are integers and such that and . Combine this with the equations and and you get . So, the triple is a triple satisfying (1) which is what we were looking for. We only need to show this triple is "non-trivial" and that . I leave this to the reader, but it is fairly easy.
Question: What is the formula used to find another triple satisfying the same equation with a smaller 'c'? Answer: The triple (x, y, z) = (x^2 - 2xy + y^2, 2xy, x^2 + 2xy + y^2) where x and y are relatively prime integers such that x^2 + y^2 = c.
Question: How many triangles are formed when a rectangle containing a triangle is divided into four parts? Answer: Four triangles. | 677.169 | 1 |
Notation
The following notations hold for all six trigonometric functions: sine (sin), cosine (cos), tangent (tan), cotangent (cot), secant (sec), and cosecant (csc). For brevity, only the sine case is given in the table.
Definitions
Periodicity, symmetry, and shifts
These are most easily shown from the unit circle:
For some purposes it is important to know that any linear combination of sine waves of the same period but different phase shifts is also a sine wave with the same period, but a different phase shift. In other words, we have
Pythagorean identities
Note that the second equation is obtained from the first by dividing both sides by cos²(x). To get the third equation, divide the first by sin²(x) instead.
Angle sum and difference identities
These are also known as the addition and subtraction theorems or formulas.
The quickest way to prove these is Euler's formula. The tangent formula follows from the other two. A geometric proof of the sin(x + y) identity is given at the end of this article.
Double-angle formulas
These can be shown by substituting x = y in the addition theorems, and using the Pythagorean formula for the latter two. Or use de Moivre's formula with n = 2.
The double-angle formulas can also be used to find Pythagorean triples. If (a, b, c) are the lengths of the sides of a right triangle, then (a2 − b2, 2ab, c2) also form a right triangle, where angle B is the angle being doubled. If a2 − b2 is negative, take its opposite and use the supplement of B.
This substitution of t for tan(x/2), with the consequent replacement of sin(x) by 2t/(1 + t2) and cos(x) by (1 − t2)/(1 + t2) is useful in calculus for converting rational functions in sin(x) and cos(x) to functions of t in order to find their antiderivatives. For more information see tangent half-angle formula.
Products-to-sum identities
These can be proven by expanding their right-hand-sides using the addition theorems.
Sum-to-product identities
Inverse trigonometric functions
Every trigonometric function can be related directly to every other trigonometric function. Such relations can be expressed by means of inverse trigonometric functions as follows: let φ and ψ represent a pair of trigonometric functions, and let arcψ be the inverse of ψ, such that ψ(arcψ(x))=x. Then φ(arcψ(x)) can be expressed as an algebraic formula in terms of x. Such formulas are shown in the table below: φ can be made equal to the head of one of the rows, and ψ can be equated to the head of a column:
Table of conversion formulas
φ / ψ
sin
cos
tan
csc
sec
cot
sin
cos
tan
csc
sec
cot
One procedure that can be used to obtain the elements of this table is as follows:
Question: If you want to express sine in terms of arctangent, what is the formula? Answer: sin(arc tan(x)) = 2x / (1 + x²)
Question: Which trigonometric function is used as a reference for brevity in the table? Answer: Sine (sin)
Question: What is the relationship between the sum of two angles and their sine, as per the angle sum identity? Answer: sin(x + y) = sin(x)cos(y) + cos(x)sin(y) | 677.169 | 1 |
(OMK 2008, Bongsu) Ali(OMK 2008, Bongsu) An equilateral triangle with sides 2 units is inscribed in a circle. Find the area of the circle.
(OMK 2008, Muda) Given that ABC is an isosceles triangle with angle ABC = angle ACB = 80 degrees. Two points E and F are on AB and AC, respectively, such that angle ABF = 10 degrees and angle ACE = 20 degrees. Let line AM where M is the midpoint of BC intersects CE at the point N. Show that FN is parallel to AB.Share this entry:
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34 Responses to Olimpiad Matematik Kebangsaan (Bahagian 3)
1. An old man has a total of 258 children,grandchildren and great geandchildren.None of his great grandchildren has any children.All the others have an equal number of children.How many children does the old man have .
Afiqah,
We have to make the problems hard because even at the current standard many students get more than 20 points (out of 30) on the paper. If we make it easier then we will have like 200 perfect scorers out of the 9000 contestants, which will make it impossible to select the winner
Question,
I have the past year papers but I am not authorized to share them. PERSAMA holds the copyright and they usually compile a book with past year problems and solutions.
You can buy my books on OMK at several centers (UKM and UiTM for sure). If not, you can order from us at the "Books" link on top.
Depends on the level. Bongsu SK tends to be in high teens to low 20s. Muda and Sulung SK is around 12-20 range, but again, it depends on which year. I heard that in some years, some SK winners only scored 9.
Alimay i know how to solve OMK 2008 bongsu, question 3? in triangle ABC, the lines AD, BE and CF meet at O, given that area of triangleAOF = 84, area triangle BOD = 40, area triangle COD = 30 and area triangle COE = 35, find the area of triangle ABC.
may i know how to get the olympiad question booklet..?
i entered this thing once in f2 and manage to get SK..
but in f4, some teachers are very x adil.. they chose those from ex-mrsm only.. =.=" of course.. they would trust them more then the 'ordinary' ex government school student….
Question: How many children does the old man have? Answer: The number of children the old man has is not provided in the text.
Question: What is the measure of angle ABF? Answer: 10 degrees | 677.169 | 1 |
Attached is a chart I made for connecting beams at right angles for the various orientations of the two beams.
For each of the three ways that two beams can meet at a right angle, the bold numbers indicate which connectors might be used for a rigid connection. The non-bold numbers can also make the connection but it won't be rigid.
Question: What does the chart provide for each way the beams can meet? Answer: It provides numbers indicating which connectors might be used for a rigid connection, with bold numbers being the ones that ensure a rigid connection. | 677.169 | 1 |
CIRCLE: Draw a large circle from top to bottom with your two index fingers (each finger draws half of the circle.
CENTER: Point both fingers forward together in the center of the circle you just drew.
RADIUS: From the "center" gesture, move one finger from the center to the edge of the circle (the other stays put).
DIAMETER: From the "center" gesture, move both fingers out to the edge.
"Ordered pairs is like learning to run" (run in place) "before you can fly ( flap your arms like wings) Because you have to do x( cross your arms in the shape of an x) before you do Y ( make a Y as if you were doing the YMCA dance)
It is Eye-soscele Triangle, when you say "eye" you point to your eyes, then make a triangle with your hands, you have a pair of eyes just like an isosceles triangle has a pair of congruent sides...
Absolute Value:
"Absolute value (hold hand vertically and parallel to each other) is the distance (spread your arms far apart) from a number (on one hand hold up all 5 fingers andstart moving that hand towards the other) to zero (use the other hand to make zero)"
Question: What number is used to represent the starting point of the absolute value distance? Answer: Zero. | 677.169 | 1 |
A (convex) polytope is the convex hull of some finite set of points. Each polytope of dimensions d has as faces finitely many polytopes of dimensions 0 (vertices), 1 (edge), 2 (2-faces), · · · , d-1 (facets). Two-dimensional polytopes are usually called polygons, three-dimensional ones polyhedra. Two polytopes are said to be isomorphic, or of the same combinatorial type, provided there exists a one-to-one correspondence between their faces, such that two faces of the first polytope meet if and only if the corresponding faces of the second meet. The prism and the truncated pyramid of Figure 9 are isomorphic, the correspondence being indicated by the letters at the vertices. To classify the convex polygons by their combinatorial types, it is sufficient to determine the number of vertices υ; for each e + f = 2 for every convex polyhedron, where υ, e, and f are the numbers of vertices, edges, and faces of the polyhedron. Though this formula became one of the starting points of topology, Euler was not successful in his attempts to find a classification scheme for convex polytopes or to determine the number of different types for each υ. Despite efforts of many famous mathematicians since Euler (Steiner, Kirkman, Cayley, Hermes, Brückner, to mention only a few from the 19th century), the problem is still open for polyhedra with more than 19 vertices. The numbers of different types with four, five, six, seven, or eight vertices are 1, 2, 7, 34, and 257, respectively. It was established by American mathematician P.J. Federico in 1969 that there are 2,606 different combinatorial types of convex polyhedra with nine vertices. The number of different types for 18 vertices is more than 107 trillion.
The theory of convex polytopes has been successful in developments in other directions. The regular polytopes have been under investigation since 1880 in dimensions higher than three, together with extensions of Euler's relation to the higher dimensions. (The Swiss geometer Ludwig Schläfli made many of these discoveries some 30 years earlier, but his work was published only posthumously in 1901.) The interest in regular polyhedra and other special polyhedra goes back to ancient Greece, as indicated by the names Platonic solids and Archimedean solids.
Question: What are the faces of a polytope of dimension d? Answer: A polytope of dimensions d has as faces finitely many polytopes of dimensions 0 (vertices), 1 (edges), 2 (2-faces), up to d-1 (facets).
Question: How can we classify convex polygons by their combinatorial types? Answer: It is sufficient to determine the number of vertices (v) for each. | 677.169 | 1 |
No. |a + b| is not generally equal to |a| + |b|, although there are important relationships called the triangle inequality and the polarization identity relating the two expressions, and it is good to know the geometry of when they are equal.
You must first calculate the vector a + b by adding components, and then take the length |a + b| of that vector.
Question: Which of the following is NOT a valid statement according to the text? A) |a + b| = |a| + |b| B) |a + b| is generally not equal to |a| + |b| C) The triangle inequality relates |a + b| and |a| + |b| Answer: A) |a + b| = |a| + |b| | 677.169 | 1 |
Tip #17 to Get a Top ACT Math Score
You couldn't ask for easier points to boost your score. Every ACT has a midpoint and/or distance question. If you don't know the formulas, you don't stand much chance. But if you memorize them, right here and now, you will gain points, guaranteed! In fact, take a few minutes right now to cut out the flashcards from the back of this book. Bring them everywhere you go, school, sports, DMX concerts, parties, etc. Everyone loves math flash cards!
To find the midpoint between two points on a graph, use the midpoint formula. All this formula really says is, "Take halfway between the x numbers and halfway between the y numbers." Halfway between two numbers is the same as the average of the numbers, so the midpoint formula is like the average formula:
The distance formula also makes sense if you really look at it. It is based on the Pythagorean theorem (Skill 24). To find the distance between two points on a graph, use the formula
Cut out the flashcards at the end of this book to help you memorize these formulas.
Let's look at this question:
Solution: A midpoint is halfway between two points, really just the average. That's what the midpoint formula gives us, the average of the two points:
Correct answer: C
Example Problems
Easy
The endpoints of on the real number line are –14 and 2. What is the coordinate of the midpoint of ?
–8
–6
–2
0
8
A line in the standard (x, y) coordinate plane contains the points M( –2, 4) and N(8, 10). What point is the midpoint of ?
(–2, 4)
(3, 3)
(5, 3)
(5, 7)
(3, 7)
Medium
In the standard (x, y) coordinate plane, what is the distance between the points (–4, 2) and (1, –10) ?
3
5
9
11
13
A diameter of a circle has endpoints (–2, 10) and (6, –4) in the standard (x, y) coordinate plane. What point is the center of the circle?
(2, 3)
(4, 6)
(6, 4)
(12, 8)
Cannot be determined from the given information
Hard
A city's restaurants are laid out on a map in the standard (x, y) coordinate plane. How long, in units, is the straight-line path between Paul & Elizabeth's restaurant at (17, 15) and The Green Bean restaurant at (12, 5) ?
11
Answers
B A midpoint is halfway between the two points, really just the average. That's what the midpoint formula gives us:
K A midpoint is halfway between the two points, really just the average. That's what the midpoint formula gives us:
E Use the distance formula:
Question: What is the distance between the points (-4, 2) and (1, -10)? Answer: 13
Question: What is the distance formula based on? Answer: The distance formula is based on the Pythagorean theorem.
Question: Which point is the midpoint of the line containing points M( -2, 4) and N(8, 10)? Answer: (3, 3) | 677.169 | 1 |
Pre-Calc I made a sketch, and y = |9x| consists of two lines first: y = 9x starting at the origin, and the second: y = -9x also starting in the origin, resulting in a graph looking like a V If I understand the question correctly, you now want the area of the triangle in the second ...
Tuesday, December 7, 2010 at 11:19pm by Reiny
Help!!!!!!!!! Calculate Triangle G°, free energy change, that occurs when 12g of Shila react with excess And at 298k according to the following reaction: N2(g)+3H2(g)<======>2NH3(g) Triangle G°=-34KJ/mole
Wednesday, December 19, 2012 at 10:00am by Lost Girl!! 7:45pm 1:03amMonday, April 5, 2010 at 11:59pm by Abbey
math right triangle the hypotenuse of a right triangle is 1 inch longer than one leg and 8 inches longer than the other. Find the length of each side of the triangle. Let l1, l2 be the legs. l1^2 + l2^2 = h^2 but h= l1+1 and h= l2 + 8 solve these for l1, l2, put them in the first equation and ...
Tuesday, December 5, 2006 at 8:59pm by darren
PLEASE HELP TANGENTS AND CIRCLES PROBLEM GEOMETRY circle A at bottom circle B at upper right circle C at upper left side of triangle at upper right = x so (1/2) of the bottom is x/2, draw vertical altitude through top of triangle and center of circle A NOW draw line through centers of circles A and B extending beyond at both ...
Saturday, February 2, 2013 at 3:44pm by Damon
geometry In the figure shown (sorry no diagram, but description below), NK and ML are parallel, JN=10 feet, NK=14 feet, JL=22 feet, and ML=24. Find KL. Its a triangle with JN and NM on the left side and JK and KL on the right side. NK is a horizontal line in the middle of the triangle...
Wednesday, February 9, 2011 at 5:43pm by shb
Pre-Algebra An acute triangle has all acute angles (less than 90 degrees) An obtuse triangle has one obtuse angle, more than 90 deg. and less than 180 deg 28 + 58 = 86 180 - 86 = 94 therefore, this is an obtuse triangle
Tuesday, January 11, 2011 at 7:51pm by helper
Question: In the right triangle problem, what is the length of the hypotenuse (h) in terms of the legs (l1 and l2)? Answer: The length of the hypotenuse (h) is l1 + 1 and l2 + 8.
Question: What is the relationship between JN and NK in the geometry problem? Answer: JN is parallel to NK.
Question: What is the graph of the function y = |9x| like? Answer: The graph of the function y = |9x| looks like a V, consisting of two lines: y = 9x starting at the origin and y = -9x also starting at the origin. | 677.169 | 1 |
David Singmaster.Off the rails.The Weekend Telegraph (18 Feb 1989) xxiii&(25 Feb 1989) xxiii.Gives the Ripley and Always results and asks which is correct and whether the wrong one can be corrected -- cf Ripley above.
5 comments:
I got the same answer, but with a different approach. I put A at the origin and D on the positive x-axis. Then the equation (sin x)/x = 2640/2640.5 falls out with simple trig. WA can handle that (with FindRoot) quickly to get x = 0.0337.
I think you need to take the compression of the steel into account. A mile of steel rail would be pretty heavy and at that small angle (if it could be held perfectly in a line above the ground) it would probably compress itself until almost flat...
Thanks for this interesting geometry problem which shows how circular lengths can only be compared approximately to rectilinear lengths, except if the angle has a sine which we know precisely (60, 45, 30, ... deg). If we had arc = chord * pi/2 (rad), or 2*pi/(3*sqrt(3)), ... the height can be determined to the precision of which we know pi or square roots. I don't know about the compression of steel, but as it is a thin rail of steel, it is probable that it will bend. Maybe a physically more probable solution would be that the rail bends parabolically. Parabolic segments and heights can be calculated more precisely than circular segments and heights, so this would yield a more precise height.
Question: What is the main question raised in the article? Answer: Which result is correct and whether the wrong one can be corrected | 677.169 | 1 |
Parallel postulate
The Parallel postulate, also called Euclid's Fifth Postulate on account of it being the fifth postulate in Euclid's Elements. It states:
If a line segment intersects two straight lines forming two interior angles on the same side sum to less than two right angles then the two lines segments, if extended indefinitely, meet on that side on which are the angles less than the two right angles.
Several properties of Euclidean geometry are logically equivalent to Euclid's parallel postulate, meaning that they can be proved in a system where the parallel postulate is true, and that if they are assumed as axioms, then the parallel postulate can be proved.
One of the most important of these properties, and the one that is most often assumed today as an axiom, is Playfair's axiom, named after the Scottish mathematicianJohn Playfair. It states:
A line can be drawn through any point not on a given line parallel to the given line.
Many attempts were made to prove the parallel postulate in terms of Euclid's first four postulates.
The independent discovery of non-Euclidean spaces by Gauss and Lobachevsky finally demonstrated the independence of the parallel postulate.
(See "History" under non-Euclidean geometry for further discussion.)
Some of the other statements that are equivalent to the parallel postulate appear at first to be unrelated to parallelism.
Some even seem so self-evident that they were unconsciously assumed by people who claimed to have proved the parallel postulate from Euclid's other postulates.
Here are some of these results:
If three angles of a quadrilateral are right angles, then the fourth angle is also a right angle.
There exists a pair of straight lines that are at constant distance from each other.
Fact-index.com financially supports the Wikimedia Foundation. Displaying this page does not burden Wikipedia hardware resources. This article is from Wikipedia. All text is available under the terms of the GNU Free Documentation License.
Question: Which of the following is NOT a property equivalent to the Parallel Postulate? (A) If three angles of a quadrilateral are right angles, then the fourth angle is also a right angle. (B) There exists a pair of straight lines that are at constant distance from each other. (C) Two parallel lines never intersect. Answer: (C) Two parallel lines never intersect (This is the definition of parallel lines, not a property equivalent to the Parallel Postulate)
Question: Which of the following is a true statement about the Parallel Postulate? (A) It can be proven using Euclid's first four postulates. (B) It is a necessary truth. (C) It is independent of Euclid's first four postulates. Answer: (C) It is independent of Euclid's first four postulates | 677.169 | 1 |
above is not working properly, review the anatomy for
the
course. You may need to update your browser or download a free media
plug-in.
To record direction, use north, east, south, west, northeast
etc. until you learn to use azimuth. Azimuth is
measured in degrees around the horizon (0 degrees when facing due
North to 360 degrees after
you have rotated in a circle once. North (N) = 0 through E = 90, S
= 180, W = 270).
The graphic below is a different view of the same telescope,
this time representing a change in azimuth. In this case you are looking
down on the top of the telescope and as it rotates on its base, you
can see the view change on the right.
Question: How is azimuth measured? Answer: In degrees around the horizon, from 0 degrees (facing due North) to 360 degrees (after rotating in a circle once). | 677.169 | 1 |
A characteristic of horizontal lines, such as the edges of an overhang, in reference to a point of interest, such as a point on a façade or the interior of a space. The profile angle is the altitude from the point of interest to the intersection of the horizontal line with the normal to it plane passing through the point of interest.
Question: Is the profile angle measured from the point of interest to the horizontal line itself or to the intersection with the normal plane? Answer: To the intersection with the normal plane | 677.169 | 1 |
Math (Geometry) Square ABCD has M as the midpoint of AB, N as the midpoint of BC, P as the midpoint of CD and Q as the midpoint of MP. If the area of AMNPDQ is 20, what is the area of ABCD?
Wednesday, May 1, 2013 at 6:42am by Hale
Math (Geometry) Square ABCD has M as the midpoint of AB, N as the midpoint of BC, P as the midpoint of CD and Q as the midpoint of MP. If [AMNPDQ]=20, what is the value of [ABCD]? Details and assumptions [PQRS] refers to the area of figure PQRS.
Monday, April 29, 2013 at 11:28am by Bob ...
Wednesday, December 10, 2008 at 4:03pm by Celina
Physics You need to find distance from each speaker from the midpoint as a function of the angle. Then set the difference of distances equal to for a, multiples of wavelength of 277hz, and for b, multiples of half wavelengths. I would recommend use the law of cosines. Your known will ...
Saturday, December 12, 2009 at 12:06am by bobpursley
hialeah miami lakes Calculate the coordinates of the midpoints as the mean values of the vertex endpoints. The midpoint of PQ is (2,-8.5) The midpoint of PR is (-1,-5.5) The distance between those points is sqrt(3^2 + 3^2) = sqrt18 = 4.24
Tuesday, May 3, 2011 at 12:52pm by drwls
math It's all a matter of knowing the basic distance and midpoint formulas. The distance between two points with coordinates (x1,y1) and (x2,y2) is given by d=the square root of (x2-x1)^2 + (y2-y1)^2 If a line segment has endpoints at (x1,y1) and (x2,y2) the the midpoint of ...
Thursday, April 7, 2011 at 11:12am by Sarah
math you use the midpoint formula to find the midpoint, but since you already know the midpoint, and one of the endpoints, plug in what you know in to the formula, and solve for the other endpoint. ?
Monday, October 18, 2010 at 9:08pm by y912
Math: Distance and Midpoint Three points on the edge of a circle are (-220, 220), (0, 0), and (200, 40), where each unit represents 1 foot. What is the diameter of the circle to the nearest 10 feet? I know the answer is supposed to be 550 ft, but I keep coming up with the wrong answer. I found the ...
Question: What is the midpoint of BC in square ABCD? Answer: N
Question: What is the area of square ABCD given that the area of AMNPDQ is 20? Answer: The area of square ABCD is 80.
Question: What is the formula to calculate the distance between two points with coordinates (x1,y1) and (x2,y2)? Answer: d = √[(x2-x1)² + (y2-y1)²]
Question: What is the midpoint of AB in square ABCD? Answer: M | 677.169 | 1 |
Figure 3
In the right-angled triangle, we refer to the lengths of the three sides according to how they are placed in relation to the angle θθ. The side opposite to the right angle is labeled the hypotenuse, the side opposite θθ is labeled opposite, the side next to θθ is labeled adjacent. Note that the choice of non-90 degree internal angle is arbitrary. You can choose either internal angle and then define the adjacent and opposite sides accordingly. However, the hypotenuse remains the same regardless of which internal angle you are referring to (because it is ALWAYS opposite the right angle and ALWAYS the longest side).
We define the trigonometric functions, also known as trigonometric identities, as:
These functions relate the lengths of the sides of a right-angled triangle to its interior angles.
Note:
The trig ratios are independent of the lengths of the sides of a triangle and depend only on the angles, this is why we can consider them to be functions of the angles.
One way of remembering the definitions is to use the following mnemonic that is perhaps easier to remember:
Table 2
Silly Old Hens
S in =O pposite H ypotenuse S in =O pposite H ypotenuse
Cackle And Howl
C os =A djacent H ypotenuse C os =A djacent H ypotenuse
Till Old Age
T an =O pposite A djacent T an =O pposite A djacent
You may also hear people saying Soh Cah Toa. This is just another way to remember the trig functions.
Tip:
The definitions of opposite, adjacent and hypotenuse are only applicable when you are working with right-angled triangles! Always check to make sure your triangle has a right-angle before you use them, otherwise you will get the wrong answer. We will find ways of using our knowledge of right-angled triangles to deal with the trigonometry of non right-angled triangles in Grade 11.
Investigation : Definitions of Trigonometric Functions
In each of the following triangles, state whether aa, bb and cc are the hypotenuse, opposite or adjacent sides of the triangle with respect to the marked angle.
For most angles θθ, it is very difficult to calculate the values of sinθsinθ, cosθcosθ and tanθtanθ. One usually needs to use a calculator to do so. However, we saw in the above Activity that we could work these values out for some special angles. Some of these angles are listed in the table below, along with the values of the trigonometric functions at these angles. Remember that the lengths of the sides of a right angled triangle must obey Pythagoras' theorem. The square of the hypotenuse (side opposite the 90 degree angle) equals the sum of the squares of the two other sides.
Table 3sinθsinθ
0
1212
1212
3232
1
0
tanθtanθ
0
1313
1
33
--
0
Question: Which side of a right-angled triangle is always the longest? Answer: The hypotenuse
Question: In a right-angled triangle, which side is the adjacent side with respect to the angle θ? Answer: The side next to θ | 677.169 | 1 |
Parallelogram?
Start with any general quadrilateral, and connect the midpoints of
consecutive sides, making an inscribed quadrilateral as in the diagram. That
inscribed quadrilateral, in the diagram, seems to be a parallelogram. Let me
conjecture that this inscribed quadrilateral is a parallelogram with half the
area of the original quadrilateral. Can you prove or disprove either part of my
conjecture?
Question: What is the relationship between the area of the original quadrilateral and the inscribed parallelogram? Answer: The area of the inscribed parallelogram is half that of the original quadrilateral | 677.169 | 1 |
In fact there are fifteen triangles with the same shape as (1,2,3) (including (1,2,3) itself). They are: (1,2,3), (2,3,4), (3,4,5), (4,5,6), (5,6,7), (6,7,8), (7,8,9), (8,9,10), (9,10,11), (10,11,12), (11,12,13), (12,13,14), (13,14,15), (1,14,15) and (1,2,15). Each of these triangles can be obtained from (1,2,3) by rotating it around the circle. These fifteen triangles together form an equivalence class.
Likewise there are fifteen triangles with the same shape as (1,2,4). Said another way (1,2,4) has fifteen triangles in its equivalence class (but we need to be careful here to emphasize that we are considering (1,2,4) and say (1,3,4) to have different shapes: they are mirror images of each other, but not rotations of each other).
In all, the 455 triangles fall into 31 equivalence classes. 30 of these classes have fifteen triangles. The final equivalence class consists of five triangles. More on that in a moment.
The strategy is this: Pick an equivalence class. Rotate through all of the cheeseburgers in that class. Next move on to another equivalence class.
Let's list the equivalence classes by writing down a single triangle in each class. OK, I'm not actually going to write all of them out, but I will do the next best thing. You'll see.
First we have twelve equivalence classes with triangles of the form (1,2,n), where n=3,4,5,…,14. We don't allow n=15 (i.e. (1,2,15) because that is the same class as (1,2,3). That gives us 180 cheeseburgers.
Then there are nine classes of the form (1,3,n), where n=5,6,7,…,13. (n=15 would give the same class as (1,2,4) and n=14 the same class as (1,3,5)). That's another 135 cheeseburgers.
Next we get six classes of the form (1,4,n), with n=7,8,9,…,12. Another 90 cheeseburgers.
Similarly there are three classes of the form (1,5,n), with n=9,10,11. 45 more cheeseburgers.
Question: How many classes are there of the form (1,5,n), with n ranging from 9 to 11? Answer: 3 | 677.169 | 1 |
Proving Diagonals Perpendicular
Date: 07/28/2003 at 09:10:05
From: Dan16etta
Subject: Perpendicular diagonals
Given the points a(-4,1), b(2,3), c (4,9) and d (-2,7), show that
quadrilateral abcd is a parallelogram with perpendicular diagonals.
Date: 07/28/2003 at 13:35:48
From: Doctor Barrus
Subject: Re: perpendicular diagonals
First I'm going to draw a rough picture of the four points, so I can
see what we're talking about.
| * c
|
d * |
|
|
|
| * b
|
a * |
------------+-----------
We want to show that the two diagonals of the quadrilateral are
perpendicular. In other words, if we drew a line passing through a
and c, and another line passing through b and d, those two lines
would be perpendicular to each other.
Now how can you tell if two lines are perpendicular? One way, most
likely the way you'll want to use here, is to look at their slopes.
If the product of two lines' slopes is equal to -1, then the two
lines are perpendicular.
So let's find the slopes of the two diagonals.
rise
Slope ac = ------
run
9 - 1
= ----------
4 - (-4)
= 8/8
= 1
Similarly (you'll want to work this out for yourself), the slope of
line bd is -1. Now if we take the two slopes and multiply them, we get
slope AC * slope BD = 1 * -1 = -1
and since the product is -1, we know that these two lines are
perpendicular. Make sense?
I hope that helps. If you have further questions, feel free to write
back. Good luck!
- Doctor Barrus, The Math Forum
Question: What is the slope of the diagonal 'ac'? Answer: 1
Question: What is the slope of the diagonal 'bd'? Answer: -1 | 677.169 | 1 |
Stated this way, compass and straightedgeconstructions appear to be a parlor game, rather than a serious practical problem.
The set of ratios constructible using compass and straightedge from such a set of ratios is precisely the smallest field containing the original ratios and closed under taking complex conjugates and square roots.
I answered the second question by saying it was impossible to trisect an angle with a straightedge and a compass, and gave the person a reference to some modern algebra books as well as an article Evelyn Sander wrote about squaring the circle.
Mark on the straightedge the length between A and B. Take the straightedge and line it up so that one edge is fixed at the point B. Let D be the point of intersection between the line from A parallel to BC.
Move the marked straightedge until the line BD satisfies the condition AB = ED, that is adjust the marked straightedge until point E and point D coincide with the marks made on the straightedge.
The compass establishes equidistance, and the straightedge establishes collinearity.
The compass is anchored at a center point, and keeps the pencil at a fixed distance from that point.
In many commonly accepted constructions (e.g., congruent angles), the compass radius is set by the distance between two points, and then the compass is centered on some third point, elsewhere on the drawing.
Allow me to offer the more liberal argument that trisection of an arbitrary angle with straightedge and compass is indeed possible; it just takes a (countably) infinite number of steps.
It is possible to exactly bisect an arbitrary angle using only straightedge and compass.
Bisection, for example, requires an infinitely accurate straightedge and compass -- which I think can only be constructed in an uncountably infinite number of steps -- and nobody gets bent out of shape over assuming the existence of these ideal instruments.
It's tempting to try to use a straightedge and compass to lay out the hatchet right on the desired angle, but it also can't be done without trial and error.
You can't tighten nuts with a saw or cut a board with a wrench, and expecting a straightedge and compass to do something beyond their capabilities is equally futile.
Geometry with straightedge and compass creates a similar illusion; eventually we believe the points we can construct are all the points that exist.
/dutchs/PSEUDOSC/trisect.HTM (5079 words)
From the time of Euclid, geometric constructions were done solely with a straightedge and compass(Site not responding. Last check: 2007-10-19)
It is not possible to construct with a ruler and compass a line whose length is the numerical value of a root of a cubic equation with rational coefficients having no rational roots (Davis, 227).
Since traditional constructions are performed with the use of only a straightedge and compass there is no unit of measurement involved.
Question: What is the main illusion created by geometry with straightedge and compass? Answer: It creates the illusion that the points we can construct are all the points that exist.
Question: Who was the first known mathematician to perform geometric constructions solely with a straightedge and compass? Answer: Euclid
Question: Are traditional constructions with a straightedge and compass performed with a unit of measurement involved? Answer: No, there is no unit of measurement involved. | 677.169 | 1 |
Question 45424
The angle of depression from the top of a building to apoint on the groudn is 32 deg 30'. How far is the point on the ground from the top of the building if the buildlng is 252 m high?
Draw the picture.
You have a right triangle with an acute angle of 32 deg 30' and
a side opposite the acute angle of length 252 m.
Use the sine to find the hypotenuse, as follows:
sin(32 deg 30')=252/hyp.
hyp=252/sin(32 deg 30')
hyp=252/0.53729961...
hyp=469 m
Cheers,
Stan H.
Question: Which of the following is the length of the side of the triangle opposite the 32 degrees 30 minutes angle? A) 252 m B) 469 m C) 32.5 m Answer: A) 252 m | 677.169 | 1 |
In that moment a math student will saw the problem, figure out, and jump out of the window.
Avistew
05/14/2010, 07:17 am
That's not mathese at all. That's what I would have said too, if you hadn't used number and confused me. Something doesn't qualify as mathese if I can understand it :p
EDIT: by the way, my whole reasoning was "they only give us one number, and it's for something that doesn't even have a name. The thing kinda look like a circle. I'm gonna say it's one so the answer is 20".
So out of curiosity, how do you actually know that AB? and AC? are the same length? Is it because of the right angles?
GinnyN
05/14/2010, 07:42 am
So out of curiosity, how do you actually know that AB? and AC? are the same length? Is it because of the right angles?
It's a Rectangle, thanks to the right angles. If you separate them with AD, you get two Square Triangles. AB has the same lenght than CD, and AC has the same lenght than BD. Since you have an Square Triangle, you can use the Pythagorean Theorem to figure out AD, which is the hypotenuse, by using AB (or CD) and BD (Or AC). So, it's something like this:
=> (AD^2) = (AB^2) + (BD^2)
=> AD = sqrt(AB^2 + BD^2)
By the way, there's a way to figure out AC and AB if you suppose those are equal. Of course, by knowing AD = 20.
Psy
05/14/2010, 07:44 amThere are quite a few specific locations, but the answer you're looking for is the north pole
GinnyN
05/14/2010, 07:47 am
There are quite a few specific locations, but the answer you're looking for is the north pole
Now I'm curious for the other few specific locations!
Avistew
05/14/2010, 07:47taumel
05/14/2010, 07:56 am
Uhm, it's 20, as it's all about, and so the distance AD, the radius.
GinnyN
05/14/2010, 07:59In a equation, I can't think in a way to do it.
In fact, you can say there's no actual answer (For lack of information), because we're just assuming it's a quarter of a Circle. If it were the quarter of a Elipse, for example, AE =/= AF and we're screwed, unless they also give us AE.
Question: What is the date and time of Avistew's post? Answer: 05/14/2010, 07:17 am
Question: What is the relationship between AB and CD, as per GinnyN's explanation? Answer: AB has the same length as CD | 677.169 | 1 |
How to find Euler's angles?
How to find Euler's angles?
I have read about Euler's angles and matrices, including zxz,zyz, etc . I am not obliged to use a specific rotation but rather I want to figure out what angles I need to use for alpha,theta, gamma in the specific matrix. For example, I have a vector from the centre with vertex x,y,z of (0,0,15), I want to rotate it to (3,-12,15), what angles should I use or how can I obtain these angles ?
If let's say I choose zyz, i am finding difficulties finding the corresponding angles after the first x-y plane rotation.
Please disregard the scaling factor and just consider the orientation.
Question: Should the scaling factor be considered in this task? Answer: No, the user asks to disregard the scaling factor and only consider the orientation | 677.169 | 1 |
Perpendicular
Perpendicular is a geometric term that may be used as a noun or adjective. The fundamental meanings pertain to the position of straight lines relative to one another, in which two lines are positioned at an angle of ninety degrees, which is defined as a right angle. Any two lines in Euclidean space which are perpendicular to one another and which intersect automatically define a plane.
Naturally, if a line is given, then a perpendicular is any line at a ninety-degree angle to that line. This is an important property in geometry and trigonometry since important properties accrue to line systems containing right angles. When graphinging, the convention is to use either an X and Y axis, or to use an X, Y, and Z axis, which are defined as being mutually perpendicular. Right triangles, too, include two perpendicular lines and so have special properties, which are the foundation of trigonometry
Question: Which branch of mathematics is the foundation of which is based on the properties of right triangles? Answer: Trigonometry | 677.169 | 1 |
In this lesson we will look at the basic definition and properties of the right angled triangle.
Right Angled Triangle
'Right Angled Triangle' is a triangle with one internal angle equal to 90 degrees (right angle).
The side opposite to the right angle is called "hypotenuse" and hypotenuse is the longest
side of the right angled triangle. The other two sides adjacent to the right angle are called
legs or catheti.
Properties of Right Angled Triangle
1. If any two side lengths are given then we can find the third side length by famous
"Pythagorean theorem". i.e. If we let 'a' be the length of the hypotenuse and 'b' and 'c'
be the lengths of the other two sides, the theorem can be expressed as the equation.
2. Acute angles of the right angled triangle are complimentary. i.e. Sum of the two acute angles is 90 degrees.
3. If both acute angles are same then the both legs are of equal length and vice-versa. Hence both acute angles are 45 degrees.
Types of Right Angled Triangle
We can categorized the right angled triangle into three categories.
(a) 30-60-90 triangle
In 30-60-90 triangle, angles are 30 degree, 60 degrees and 90(b) Isosceles right angled triangle
In Isosceles right angled triangle, one right angle and acute angles are of 45Both legs are of equal length
(c) Scalene right angled triangle
In Scalene right angled triangle, one right angle and other two angles are not equal.
No two sides are equal in Scalene right angled triangle.
30-60-90 triangle is a particular case of Scalene right angled triangle.
Question: What is a 30-60-90 triangle? Answer: A type of right-angled triangle with angles 30, 60, and 90 degrees
Question: Which type of right-angled triangle has two sides of equal length? Answer: Isosceles right-angled triangle | 677.169 | 1 |
Since the sum on angles in triangle is 180 then the ratio of degrees is 180/9 = 20
2*20:3*20:4*20 = 40:60:80
sufficient
statement II
In order for all the angles of the triangle ABC be smaller than 90 degrees you cannot have a right angle. If you did then you could express the ratio of this triangle sides by the Pythagoras Theorem A^2+B^2 = C^2.
Question: If the angles of a triangle are in the ratio 2:3:4, what are the degrees of the angles? Answer: 40, 60, 80 degrees | 677.169 | 1 |
Then, since BC is to GH as GH is to CF, and, if three straight lines are proportional, then the first is to the third as the figure on the first is to the similar and similarly situated figure described on the second, therefore BC is to CF as the triangle ABC is to the triangle KGH.
Therefore also the triangle ABC is to the triangle KGH as the parallelogram BE is to the parallelogram EF. Therefore, alternately, the triangle ABC is to the parallelogram BE as the triangle KGH is to the parallelogram EF.
And KGH is also similar to ABC. Therefore this figure KGH has been constructed similar to the given rectilinear figure ABC and equal to the other given figure D.
Q.E.F.
Note that it isn't proposition V.14 being invoked near the end of the proof, but an alternate form of it. See the Guide to V.14.
This proposition solves a similar problem, to find a figure with the size of one figure but the shape of another, a problem reputedly solved by Pythagoras. It is used in the proofs of propositions VI.28 and VI.29
Question: Is the triangle ABC similar to the triangle KGH? Answer: Yes | 677.169 | 1 |
Vectors- How do you number to use to multiply by cos and sin? 257cos198? how do you get 198?
Johnny asked
An airplane flies 257 km in a straight line in a direction of 18.0 degrees North of West. It then changes direction and flies in a straight line for 233 km in a direction of 78.0 degrees South of West. Find the total displacement (distance and direction) of the airplane from its starting point.
Question: What is the angle of the displacement vector with respect to the positive x-axis? Answer: 96 degrees | 677.169 | 1 |
A circle is a special form of this... I won't show it again.
A straight line - the single body problem again.
Two straight lines - parallel this time. This is something new.
No intersection at all. The kōan of conics.
This is new, but whether it is something is a question I will not attempt to address.
Now for the proof that cylindrical intersection #1 really is an ellipse. Place two double-headed cones
coaxially with the cylinder. Arrange them so that:
The apexes of both cones are in the "up" direction, from anywhere on the curve in question.
The "top" cone intersects the highest point on the curve in question.
The "bottom" cone intersects the lowest point on the curve in question.
Diagram:
Notice that the entire curve lies in between the two cones. Consider the intersection of the plane that
determines the curve with each of the cones. We know both of these are ellipses; we know that the intersection
of the plane with the "upper" cone lies wholly outside our unknown curve; and we know that the intersection of
the plane with the "inner" cone lies wholly inside our unknown curve.
Diagram:
The final stage of the proof lies in pushing the apexes of both cones to infinity, which "squeezes" our unknown curve
in between two known ellipses. Since the horizontal distance between the cones goes to zero as 23/2 / tan(θ), our
unknown curve cannot help but be elliptical-ish... to any degree of precision... and it is, thus, an ellipse. QED #2.
For the final problem, we must make a "Keplerian observation." The observation in question is that our parametric point,
like the planets, sweeps out equal areas in equal times. What makes this interesting is that the parametric point does not move in the same way
that a planet does... so it's a little odd that such a similar result should hold... but it does.
Let's talk a little about Kepler's second law. A planet moves around the sun in an elliptical orbit. Fine.
The sun lies at one of the foci of the ellipse. An imaginary line between the planet and the sun will sweep out area at a constant rate.
That is to say, when the planet is far from the sun, that line will be long... and it will move correspondingly slowly.
When the planet is near to the sun, that line will be short... and it will move correspondingly quickly.
Conversely, our parametric point sweeps around the origin at constant angular velocity. So this is trivially
"equal areas in equal times". Right?
Not quite. It's true that our parametric point appears to have constant angular velocity, if you project
its path into the x-y plane... that is, if you view it from directly above, from a point on the z-axis.
But that's a silly way to look at things. To get an idea of the point's actual motion, it's far more natural
Question: What is the special form of a circle mentioned in the text? Answer: A circle is a special form of an ellipse.
Question: What is the proof shown in the text for the shape of the curve in a cylindrical intersection? Answer: The proof shows that the curve is an ellipse.
Question: What is the "Keplerian observation" mentioned in the text? Answer: The observation is that the parametric point, like the planets, sweeps out equal areas in equal times. | 677.169 | 1 |
I used two equations and two unknowns and came up with x =50 degrees and y = 40 degrees. that doesn't work when i use y=40 in the second picture with the angle DBC = (y-50)degrees. This would make that angle a negative number. Click here to see answer by [email protected](15624)
Question 213077: Another unit of angle measure, used primarily in engineering, is called a gradian. There are 100 gradians in a right angle. Is 1 gradian smaller or larger than 1 degree? Why?
I believe 1 gradian is smaller than 1 degree because we're comparing 100 (gradian) to 90 (degrees in right angle). It seems too simple so I'm asking for confirmation or correction.
Thank you. Click here to see answer by Alan3354(30966)
Question 213212: This is a question from a worksheet... One of two copmlemtentary angles added to one-half of the other yields 72degrees. Find the half the measure of the larger.
not quite sure how to work the problem........thank you for your time! i really appreciate it! Click here to see answer by [email protected](15624)
Question: In the worksheet problem, what are the two complementary angles? Answer: Let's call them A and B, where A + B = 90 degrees.
Question: What are the values of x and y that the user initially found? Answer: x = 50 degrees, y = 40 degrees | 677.169 | 1 |
In the figure above, s is the length of the side
opposite the 3, so s = 3.
• Square
A square is a rectangle with 4 equal sides.
If PQRS is a square, all sides are the same length
as QR. The perimeter of a square is equal to four
times the length of one side.
• Trapezoid
A trapezoid is a quadrilateral with one pair of
parallel sides and one pair of nonparallel sides.
In the quadrilateral above, sides EF and GH are
parallel, while sides EH and GF are not parallel.
EFGH is therefore a trapezoid.
87. AREAS OF SPECIAL QUADRILATERALS
Area of Rectangle = Length*Width
The area of a 7-by-3 rectangle is 7*3 = 21.
Area of Parallelogram = Base*Height
The area of a parallelogram with a height of 4
and a base of 6 is 4*6 = 24.
Area of Square = (Side)^2
The area of a square with sides of length 5 is 5^2 = 25.
Area of Trapezoid =
Think of it as the average of the bases (the two
parallel sides) times the height (the length of the
perpendicular altitude).
In the trapezoid ABCD above, you can use side AD
for the height. The average of the bases is so the area is 5*8, or 40.
88. INTERIOR ANGLES OF A POLYGON
The sum of the measures of the interior angles
of a polygon is (n – 2)*180, where n is the
number of sides.
Sum of the angles = (n – 2)*180 degrees
The eight angles of an octagon, for example, add
up to (8 – 2)*180 = 1,080.
To find one angle of a regular polygon, divide the
sum of the angles by the number of angles (which
is the same as the number of sides). The formula,
therefore, is:
Interior angle =
Angle A of the regular octagon
above measures 1,080/8 degrees.
Circles
89. CIRCUMFERENCE OF A CIRCLE
Circumference of a circle = 2πr
Here, the radius is 3, and so the circumference is 2π(3) =
6π.
90. LENGTH OF AN ARC
An arc is a piece of the circumference. If n is the
measure of the arc's central angle, then the formula
is:
Length of an Arc =
In the figure above, the radius is 5 and the measure
of the central angle is 72°. The arc length is
72/360 or 1/5 of the circumference:
91. AREA OF A CIRCLE
Question: If PQRS is a square, what is the length of each side? Answer: The same length as QR
Question: What is the area of a rectangle with dimensions 7 by 3? Answer: 21 | 677.169 | 1 |
Ellipse #9
This is the same method as Ellipse #6, and
shows that an entire family of ellipses can be drawn with a spirograph. Notice
that the endpoints of the line move in straight lines, and the midpoint moves
in a circle. The outer boundary of this family of ellipses is called an
Astroid.
The above Java animation was created with
Cinderella (a geometry program).
Question: How do the endpoints of the line move? Answer: In straight lines | 677.169 | 1 |
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