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These resources have been reviewed and selected by STEM Learning's team of education specialists for factual accuracy and relevance to teaching STEM subjects in UK schools.
Vectors
This interactive excel sheet deals with a variety of vector topics. It begins with sheets which require students to identify column vectors from line segments and then to find the magnitude and direction of vectors expressed in column form.
The next interactive sheets illustrate the fact that a+nb lies on a straight line and finding the point of intersection of two straight lines. Later sheets deal with some vector geometry topics including the mid-point of a line, the centroid of a triangle and parallelogram formed by the midpoints of a quadrilateral.
There are a further eight sheets of questions on a number of different vector topics which may be duplicated for classroom use | 677.169 | 1 |
Angle chasing
Angle chasing
Angle chasing is a technique to solve or help in solving a kind of geometric problems by studying the angles first. It is often advisable to start geometrical problem-solving with angle-chasing as it's a simple way to get further information about the problem.
Angle chasing stages
upplementary angles
Supplementary -those that make up a straight line- can be found quickly.
um of the angles in triangles and quadrilaterals
In every polygon the sum of the external angles is always 360°. There is a nice formula for the internal angle sum.
Uses for circumferences
Two different triangles having one side in common have all vertices on just one circle if their angles opposite the common sides are equal.List of mathematical jargon — The language of mathematics has a vast vocabulary of specialist and technical terms. It also has a certain amount of jargon: commonly used phrases which are part of the culture of mathematics, rather than of the subject. Jargon often appears in… … Wikipedia
Pirates of the Caribbean (theme park ride) — Pirates of the Caribbean is a dark ride at the Disneyland, Magic Kingdom, Tokyo Disneyland, and Disneyland Paris theme parks. The last attraction in which Walt Disney himself had participated in the design, but never experienced, since it openedDrifting (motorsport) — A Toyota Supra in drifting exhibition in Commerce, Georgia Drifting refers to a driving technique and to a motorsport where the driver intentionally over steers, causing loss of traction in the rear wheels through turns, while maintaining vehicle … Wikipedia | 677.169 | 1 |
Three Dimensional Geometry Class 12 Notes By Master notes the fundamental concepts and applications that make this subject a cornerstone of mathematical understanding.
Detail
Class : – 12th
Subject : – Maths
Introduction to 3D Space
In Class 12, Three-Dimensional Geometry introduces students to the richness of 3D space. We move beyond the familiar 2D plane, venturing into a world where objects possess depth, height, and width. Understanding the fundamental axes—x, y, and z—becomes the cornerstone of navigating this three-dimensional realm. Three Dimensional Geometry Class 12
Coordinate Systems and Vectors
Class 12 students delve into coordinate systems that extend into three dimensions. Vectors emerge as the language of movement in this space, guiding us through the intricate coordinates that define points, lines, and planes.
Equations and Representations: Bringing Geometry to Life
Equations of Lines in 3D
Class 12 introduces the elegance of expressing lines in 3D through equations. Students learn to decipher the significance of direction ratios and parallelism, capturing the essence of lines as they traverse through space. Three Dimensional Geometry Class 12
Intersection of Lines and Planes: Where Paths Converge
Line and Line Intersection
In the intricate dance of 3D space, lines intersect and diverge. Class 12 students explore the conditions for line intersection, unraveling the algebraic and geometric nuances that define these critical points. Three Dimensional Geometry Class 12
Plane and Line Intersection
As lines weave through planes, Class 12 brings forth the dynamics of their intersection. Through systematic analysis, students discern the scenarios where lines pierce through planes or remain confined within them.
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Skew Lines and Shortest Distances: The Geometric Drama Unfolds
Understanding Skew Lines
Class 12 navigates the concept of skew lines—lines that neither intersect nor lie on the same plane. Students delve into the conditions that define skewness, unraveling the intricacies of lines moving independently in 3D space. Three Dimensional Geometry Class 12
Shortest Distance Between Lines
The pursuit of the shortest distance between lines becomes a focal point. Class 12 introduces students to the mathematics that defines this distance, exploring the interplay of vectors and projections that govern the spatial relationships between lines.
Planes in Space: A Mathematical Canvas
Equations of Planes
Building upon the basics, Class 12 deepens the exploration of planes. Students grapple with the complexities of multiple planes, understanding how their equations encapsulate the geometric arrangements of points and vectors in three dimensions.
Angle Between Two Planes
The concept of angles takes a new dimension as Class 12 unveils the methodology for calculating the angles between planes. This mathematical lens provides insights into the spatial relationships that define the orientation of planes in 3D space.
3D Geometry in Physics and Engineering
The practicality of 3D geometry extends beyond mathematics. Class 12 students discover how physics and engineering leverage these principles to model and analyze real-world scenarios. From projectile motion to structural design, 3D geometry becomes a powerful tool in various disciplines.
Computer Graphics and Animation
In the digital realm, 3D geometry is the backbone of computer graphics and animation. Class 12 unveils the role of vectors, coordinates, and transformations in creating virtual worlds, emphasizing the interdisciplinary nature of mathematical concepts. 3d geometry class 12
Visualization Techniques
Given the spatial nature of the subject, visualization becomes a potent tool. Class 12 encourages students to develop a mental canvas, visualizing lines, planes, and intersections to enhance comprehension and problem-solving abilities.
In Conclusion: Navigating Mathematical Dimensions with Master Notes
As we conclude our expedition through "Three-Dimensional Geometry Class 12," remember that Master Notes is your dedicated guide. Whether you're unraveling the equations of planes or deciphering the shortest distance between lines, our study material is crafted to be your comprehensive companion.
So, embrace the depth, height, and width of mathematical exploration, and let your journey through 3D space be one of curiosity, discovery, and mastery. In the dynamic landscape of Class 12 education, Master Notes ensures that you not only learn but also thrive. Happy studying! 3d geometry class 12 | 677.169 | 1 |
What is a bearing in math?
NOVEMBER 02, 2022
Our cpmpany offers different What is a bearing in math? at Wholesale Price,Here, you can get high quality and high efficient What is a bearing in math?
Bearings | Definition, Examples & Uses | GCSE Maths StudyWhat are Bearings in Maths? A bearing measures the movement of an angle in a clockwise direction and always on the north line. The bearing of a point is the
Bearings Worksheets | Bearings Questions | Maths Made EasyBearings are a way of expressing the angle between two objects. There is a specific set of rules about how bearings should be calculated and expressed. 1Bearings tutorial - fully interactive | Vivax SolutionsMathematics. Bearings Tutorial. Since we are fairly familiar with the four main directions - North, South,
Definition of Bearing - Math is FunIllustrated definition of Bearing: Three-Figure Bearings: The angle in degrees measured clockwise from North. It is common to put extra
Bearings - Mathematics GCSE Revision - Revision MathsBearings. A bearing is an angle, measured clockwise from the north direction. Below, the bearing of B from How to solve Bearings Questions - Help with IGCSE GCSEWhat are Bearings. On your GCSE / IGCSE maths exam your knowledge on trigonometry might be tested by a question involving bearings. Bearings are used to
Bearings - Interactive Maths Series software (interactiveNote: The bearing of a point is the number of degrees in the angle measured in a clockwise direction from the north line to the line joining the centre Directions and Bearings - Interactive Maths Series softwareNote: N30ºE means the direction is 30º east of north. The bearing to a point is the angle measured in a clockwise direction from the north
bearing ~ A Maths Dictionary for Kids Quick Reference byBb. bearing. • the angle of direction in relation to a north-south line. • measured in degrees from the north in a clockwise direction. EXAMPLES: bearing Bearing - Word Problems | Brilliant Math & Science WikiMost bearing word problems involving trigonometry and angles can be reduced to finding relationships between angles and the measurements of the sides of a | 677.169 | 1 |
Project- Find the Celebrity who Looks like You using Computer Vision
Now, we would write a function that calculates the Euclidean distance between the face embeddings that we calculated in the previous steps. This is done to see how far apart the test image is from the known faces. It returns the distance that is minimum, and the path to that image.
What is Euclidean Distance?
The Euclidean distance (also known as the Pythagorean distance) between two points in Euclidean space is the length of a line segment between those two points. It can be calculated from the Cartesian coordinates of the points using the Pythagorean theorem. Given below is the formula to calculate the Euclidean distance between 2 points in n-dimensional space:
Here, the face_distance() function compares a list of face encodings to a known face encoding and get a Euclidean distance for each comparison face. The distance tells you how similar the faces are. | 677.169 | 1 |
GoGeometry Action 72!
Creation of this applet was inspired by a tweet from Antonio Gutierrez (GoGeometry).
For the two circles shown, both common external tangents are shown.
The segments that pass through the two pink points are tangents to these circles as well.
How can we formally prove what this applet informally illustrates? | 677.169 | 1 |
figure above, lines l1 and l2 are parallel. What is the value
[#permalink]
14 Sep 2022, 03 figure above, lines l1 and l2 are parallel. What is the value [#permalink] | 677.169 | 1 |
How To Find The Moment Of Inertia Of A Semicircle
In order to find the moment of inertia of a semicircle, we need to recall the concept of deriving the moment of inertia of a circle. The concept can be used to easily determine the moment of inertia of a semicircle.
1. We will first begin with recalling the expression for a full circle.
I = πr4 / 4
In order to find the moment of inertia, we have to take the results of a full circle and basically divide it by two to get the result for a semicircle.
Now, in a full circle because of complete symmetry and area distribution, the moment of inertia relative to the x-axis is the same as the y-axis.
Ix = Iy = ¼ πr4
M.O.I relative to the origin, Jo = Ix + Iy = ¼ πr4 + ¼ πr4 = ½ πr4
Now we need to pull out the area of a circle which gives us;
Jo = ½ (πr2) R2
Similarly, for a semicircle, the moment of inertia of the x-axis is equal to the y-axis. Here, the semi-circle rotating about an axis is symmetric and therefore we consider the values equal. Here the M.O.I will be half the moment of inertia of a full circle. Now this gives us;
Ix = Iy = ⅛ πr4 = ⅛ (Ao) R2 = ⅛ (πr2) R2
Now to determine the semicircle's moment of inertia
We will basically follow the polar coordinate method.
The moment of inertia of the semicircle about the x-axis is
y = r sin θ
dA = r drd θ
We will now determine the first moment of inertia about the x-axis. We get;
Ix = o∫ro∫π y2 ⋅ dA
= o∫ro∫π (r sin θ) 2 . r drd θ
= o∫ro∫π r3 [(1 – cos2θ)/2](1/2) d2θdr
We will now determine the first moment of inertia about the x-axis. We get; | 677.169 | 1 |
Past Classes
What is so special about a triangle? Are circles useful? Do proofs have to be so difficult and boring? And what does "power of a point" REALLY mean, anyway?
This class will address questions such as these by offering an interesting and easy-to-understand introduction to selected topics within plane geometry. It will emphasize basic principles and techniques of mathematical proof as a means of affirming visual intuition and uncovering surprising and beautiful truths. | 677.169 | 1 |
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Geometry at WorkGeometry in Learning Kristin P. Bennett Rensselaer Polytechnic Institute Erin J. Bredensteiner University of Evansville Goal: To classify objects into two classes based on specific measurements made on each object. Examples: tumors: benign or malignant patients: healthy or with heart disease Congressmen: Republicans or Democrats
• Data is collected for a large sample of individuals. • Individuals are assigned to one of two classes by experts. • A perceptron is created. A perceptron is a linear model that is used to classify points into two sets. • New individuals are then classified by a computer using the perceptron.
A plane with normal vector wRn is given by the vector equation xw= where is the Euclidean distance from the origin for R. If p is the position vector of a point on the plane, then (x – p) w =0 xw – p w= 0 xw = p w= xw = and g gives the location of the plane relative to the origin. This plane is g units away from the parallel plane through the origin, xw = .
Definition: Let x be a point in Rnto be classified as a member of class A or class B. A perceptron with weights wRn and threshold R assigns x to class A or to class B using the following rule: If xw – then x A. If xw – <then x B. By convention, if xw = then x B Linearly Inseparable Case We need some terminology before we can define the cases.
A set is convex if the segment connecting any two points in the set is also in the set. The convex hull of a set of points is the smallest convex set that contains the set. Let A1,A2, … , Am be the points in set A. Then u1 A1+u2 A2 +… + umAm is in the convex hull of A if u1 +u2 +… + um = 1 and ui > 0: There are many planes lying between the two sets. This happens when the convex hulls of the sets A and B are disjoint. Linearly Inseparable Case: There is no plane lying between the two sets. This happens when the convex hulls of the sets A and B intersect.
Solution of the Linearly Inseparable Case If there is no plane that separates the two sets, there are many different planes that could define a perceptron. Robust Linear Programming. Minimize the maximum distance from any misclassified point to the separating plane. This can give importance to one hard-to-classify point. Multisurface Method of Pattern Recognition. Find the plane that misclassifies the least number of points. Generalized Optimal Plane. Reduce the average distance of misclassified points from the separating plane and decrease the maximum classification error. | 677.169 | 1 |
Basic Characteristics of the Dot Product in Mathematics
Vector is known as the quantity that can be very much capable of providing people with a clear idea about the magnitude as well as direction. Several kinds of mathematical operations can be perfectly performed on the vector for example addition as well as multiplication. The multiplication of the vector can be done in two ways which will be the utilisation of Dot product as well as cross-product in the whole process.
The definition of the concept of Dot product can be given into different kinds of manners so that everybody will be able to have a good command over the entire thing.
People need to have a clear idea about the Dot product concept and the scalar product of two vectors A as well as B of magnitude will be taking place with the utilisation of the concept of Dot product very successfully without any kind of doubt.
The basic properties of the Dot product vector have been explained as follows:
The dot product of two vectors is very much commutative
If A.B is equal to 0 then it can be seen that either be or a zero which very well suggests that either of the vectors will be zero so they will be perpendicular to each other.
People need to have a clear idea about the scalar product of vectors in the whole process.
The Dot product of vector to itself will be the magnitude squared of vector
The dot product will always follow the distributive law as well over here
In terms of orthogonal coordinates, it is also important for people to have a clear idea about the vector system in the world of mathematics as well.
In very simple terms Dot product is known as the product of the magnitude of two vectors as well as the cosine of the angle between two vectors. The resultant of the Dot product will lie into the same plane of the two vectors and the Dot product might be positive real numbers or a negative real number in the whole process.
In the cases of geometry, this is perfectly undertaken by having an idea about the construction component of one factor in the direction and then multiplying it with the magnitude of the other factor in the whole process. The vector will help in representing the direction as bad as magnitude and the magnitude of the vector will be the square root of the sum of the squares of the individual constituents of the vector. Apart from this one will also need to be clear about the projection of vector so that direction of another one can be found very easily and there will be no chance of any kind of issue in the whole process.
It is also important for kids to have a clear idea about the angle between to better funding process with the utilisation of Dot product so that working rules are perfectly made available and there is no doubt at any point in time. The product of the force applied and the displacement will be known as work and application of the scalar product will also be perfectly undertaken in the calculation of the work. So, both of these products and methods are perfectly available in the world of physics to make sure that applications will be easily undertaken and there will be no chance of any kind of error at any point in time.
Hence, depending upon the official platforms like Cuemath or Brighterly for online math classes is important so that everybody will be on the right track of dealing with things from the comfort of their home place without any kind of issue. | 677.169 | 1 |
2021 Oregon Math Guidance: 3.GM.A.1
Cluster: 3.GM.A - Reason with shapes and their attributes.
STANDARD: 3.GM.A.1
Standards Statement (2021):
Understand that shapes in different categories may share attributes and that shared attributes can define a larger category1
4.GM.A.1, 5.GM.B.3, 5.GM.D.6
N/A
3.G.A.1
3.GM.A Crosswalk
Standards Guidance:
Clarifications
Recognize rhombuses, rectangles, and squares as examples of quadrilaterals, and draw examples of quadrilaterals that do not belong to any of these subcategories.
There should be a focus on the investigation of quadrilaterals, specifically, but other polygons should also be explored.
Progressions
Students should be able to analyze, compare, and classify two dimensional shapes by their properties. Because they have built a firm foundation of several shape categories, these categories can be the raw material for thinking about the relationships between the classes. (Please reference page 13 in the Progression document).
Examples
Compare and classify shapes by their sides and angles.
Recognize rhombi, rectangles, squares, and trapezoids as examples of quadrilaterals, and draw examples of quadrilaterals that do not belong to any of these subcategories. | 677.169 | 1 |
I want to create linestrings with a definite angle (e.g. 160°) and length (e.g. 2m) that are fixed to a number of points of another linestring. So, I want to use the ST_DumpPoints function to find the points and bind the created linestrings to them.
Is there a way to declare an angle (α) during the linestring creation?
Here is an example picture:
I want to create the blue linestrings.
EDIT
The angle (α2) in the picture isn't really exemplary. But rather an azimuth of 160° (like α1).
UPDATE
The answer from Evil Genius helped me calculating the maximum width of a polygon with a given aspect.
1 Answer
1
You could accomplish this a few different ways depending on what sort of output you are wanting, but the concept is the same. It's generally easier to do a simple rotation followed by a translation rather than trying to calculate the coordinates in a single step.
In this case, the basic steps are:
Create a line of the desired length at the origin (0,0). This line should run along the axis that you wish to measure your angle from and have it's center at the origin.
Rotate the line around the origin.
Translate the line by the coordinates of the point that you want it to be centered on.
The following PostGIS view creates the lines from your example scenario. A few things are assumed:
The geometry column is called shape
The angle is measured from the x axis. Your example drawing was a bit confusing since you first mentioned 40 degrees, drew a dotted vertical line, but then said it should be around 160 degrees. I've interpreted that to mean you actually want to measure from the x axis.
The data is projected in the same units that you want to measure with (ie. meters).
To break down what's actually going on with that last line, starting from the inner most:
ST_SetSRID(ST_Translate(ST_Rotate(ST_MakeLine(ST_MakePoint( 0,1.0),
ST_MakePoint(0,-1.0)),
radians(40)), ST_X(vertex), ST_Y(Vertex)),
ST_SRID(vertex)) AS newline
ST_MakePoint(1.0,0.0) and ST_MakePoint(-1.0,0.0): Create the endpoints for a horizontal line that is our desired length and centered on the origin.
ST_SetSRID(..., ST_SRID(vertex)): Give the new line the same SRID as the input geometry.
If you are using PostGIS 2.0 you can simplify this since you can specify a different origin for ST_Rotate.
If you want to rotate to an angle based on the slope of the line, you'll have to calculate that first and add it to the rotation angle.
If the data isn't projected in the same units that you want to measure in you can still do something similar, but you'll need an extra step:
Create a line (projected in something that uses what you want to measure in)
Rotate
Reproject to your target projection
Translate to the target point
Edit
I now understand what you mean by the angle. Essentially, you want a rotation clockwise from the Y axis (0 is up, 90 is right, 180 is down, etc.).
You do still need to use the radians function since ST_Rotate expect the angle in radians. You should be able to get the correct angle with two small changes:
Start with a vertical line (use ST_MakePoint(0.0,1.0) and ST_MakePoint(0.0,-1.0))
Multiply your angle by -1. This will make it negative, causing ST_rotate to rotate it in a clockwise direction. radians(<angle> * -1)
You're an evil genius! Thank You. It works perfect, but setting up the right angle. I work with angles (stored in a column), which are like an exposition or aspect (0 to 360 degrees). An aspect of 200° for example (like geo direction SSW): when I plot the linestrings they have not the right angle. Using degrees() the geo direction is like SSE and using radians() like SW. | 677.169 | 1 |
Breadcrumb
Geometry: Dilations
Dilations
Geometry
A dilation is a transformation that changes the length of all line segments by the same proportion. A dilation does not change the shape of a figure, but it can change the size. Because the size of the figure changes, dilations are not isometries. Because the shape is invariant under a dilation, the original figure and its image are similar. By "similar" I mean the mathematical notion of similarity: Angles are congruent and sides are proportional.
If the proportion involved in the dilation is
equal to one, then no noticeable change will
have occurred. That would be the identity
dilation. If the proportion is less than one,
the figure will shrink. If the proportion is
greater than one, the figure will grow.
Solid Facts
A dilation is a transformation that changes the length of all line segments by the same proportion.
Just as you saw with isometries, collinearity, between-ness and angle measures are invariant under a dilation. You saw that any two congruent triangles are related to each other by a sequence of isometries. It turns out that any two similar triangles are also related to each other by a sequence of isometries and a dilation. In other words, if two triangles are similar, then you can make one from the other using only isometries to move it around and a dilation to change the | 677.169 | 1 |
Quadrilaterals & their properties(anmol)
A rhombus is a type of parallelogram, and what distinguishes its
shape is that all four of its sides are Congruent. There are several
formulas for the rhombus that have been derived by the
formulae for the parallelogram.
A rectangle is a parallelogram with four right
angles. A rectangle is a parallelogram, its
opposite sides must be congruent and it must
satisfy all other properties of
parallelograms .
A square has the properties of a rhombus
and a rectangle. Its sides intersect at
90° and all four sides are congruent.
A kite is a 4-sided flat shape with straight sides that:
* has two pairs of sides.
* each pair is adjacent sides (they meet) that are equal in
length.
Also, the angles are equal where the pairs meet.
Diagonals (dashed lines) meet at a right angle, and one of
the diagonal bisects (cuts equally in half) the other.
ISOSCELES TRAPEZIUM
• An isosceles trapezoid is a
trapezoid where the legs have
equal length. It can also be
defined as a convex
quadrilateral with a line of
symmetry bisecting one pair of
opposite sides, making it
automatically a trapezoid.
PROPERTIES OF A QUADRILATERAL
• Interior Angle Sum Property:
According to this property, the
sum of the interior angles of the
quadrilateral is 360°
• Exterior Angle Sum Property:
According to this property, the
sum of the exterior angles of
General Trapezoid
Median of a
Trapezoid
The MEDIAN of a trapezoid is
the line segment that joins the
midpoints of the two non-parallel
sides of the trapezoid.
The length of the MEDIAN a
trapezoid is the line segment that
joins the midpoints of the two
non-parallel sides of the
trapezoid.
• Two opposite sides (the bases) are parallel.
• The two other sides (the legs) are of equal
length.
• The diagonals are also of equal length.
• The base angles of an isosceles trapezoid
are equal in measure | 677.169 | 1 |
Areas of Parallelograms and Triangles
Oct 13, 2011
90 likes | 578 Views
Areas of Parallelograms and Triangles. Lesson 9-4. Parallelogram. A parallelogram is a quadrilateral where the opposite sides are congruent and parallel. A rectangle is a type of parallelogram, but we often see parallelograms that are not rectangles (parallelograms without right angles).
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Areas of Parallelograms and Trianglesallelogram • A parallelogram is a quadrilateral where the opposite sides are congruent and parallel. • A rectangle is a type of parallelogram, but we often see parallelograms that are not rectangles (parallelograms without right angles).
Area of a Parallelogram • Any side of a parallelogram can be considered a base. The height of a parallelogram is the perpendicular distance between opposite bases. • The area formula is A=bh A=bh A=5(3) A=15m2
Area of a Triangle • A triangle is a three sided polygon. Any side can be the base of the triangle. The height of the triangle is the perpendicular length from a vertex to the opposite base. • A triangle (which can be formed by splitting a parallelogram in half) has a similar area formula: A = ½ bh. | 677.169 | 1 |
1 Math Buddies -Grade Lesson #13 Congruence, Symmetry and Transformations: Translations, Reflections, and Rotations Goal: Identify congruent and noncongruent figures Recognize the congruence of plane figures resulting from geometric transformations such as translation (slide), reflection (flip) and rotation (turn). Identify figures that are symmetric and lines of symmetry Vocabulary: Congruent figures have the same size and shape. The angles and line segments that make up the plane figure are exactly the same size and shape. Two shapes or solids are congruent if they are identical in every way except for their position; one can be turned into the other by rotation, reflection or translation. A figure or shape is Symmetrical when one-half of the figure is the mirror image of the other half A Line of Symmetry divides a symmetrical figure, object, or arrangement of objects into two parts that are congruent if one part is reflected (flipped) over the line of symmetry Transformation is an operation that creates an image from an original figure or pre-image. Translations, Reflections and Rotations are some of the transformations on the plane. Although there is a change in position for the original figure, there is no change to the shape or size of the original figure. Translation (Slide) is a transformation of an object that means to move the object without rotating or reflecting it. Every translation has a given direction and a given distance. Reflection (Flip) is a transformation of an object that means to produce its mirror image of the object on the opposite side of a line. Every reflection has a mirror line or a line of reflection. A reflection of an "R" is a backwards "R" Rotation (Turn) is a transformation of an object that means to turn it around a given point, called the center. Every rotation has a center of rotation, an angle of rotation, and a direction (counterclockwise and clockwise). Tessellations are patterns of shapes that cover a plane without gaps (holes) or overlaps are called tessellations. Related SOL: 4.17 The student will b) identify congruent and noncongruent shapes; and c) investigate congruence of plane figures after geometric transformations such as reflection (flip), translation (slide) and rotation (turn), using mirrors, paper folding, and tracing.
2 Math Buddies -Grade Materials: 2 Mira Sheets of Patty Paper 50 assorted Pattern Block Pieces 1 set of colored pencils 2 Sets of Tangrams (7 piece Chinese puzzle) Translation, Reflection and Rotation Concentration Cards (20) Goal 1: Recognize congruent and non-congruent plane figures Activity 1.1: Warm-Up: Congruent Object Search 1. Say: Look around the room. Can anyone identify two objects or figures that appear to be exactly alike? (Answers will vary) 2. Say: Lets look at these two objects (or figures). How many sides do the objects have? How many angles do the objects have? Are the shapes the same size? Do they have the same shape? 3. Say: Congruent figures have the same size and shape. Would you say these two objects (or figures) are congruent? 4. Say: To further explain congruence, think about going to your favorite mall and looking at dozens of copies of your favorite CD on sale. All of the CDs are exactly the same size and shape. In fact, you can probably think of many objects that are mass-produced to be exactly the same size and shape. Congruent objects are exactly the same- they are duplicates of one another. In Mathematics, if two figures are congruent and you cut one figure out with a pair of scissors, it would fit perfectly on top of the other figure. So, if two quadrilaterals (4 sided) are the same size and shape, they are congruent. If two pentagons (5 sided) are the same size and shape, they are congruent. 5. Say: Now, let s hear from you. Would you please describe what a pair of congruent objects or figures have in common? Students might suggest that congruent figures have the same size and shape. The angles and line segments that make up the congruent figures are exactly the same size and shape. Say: Yes, congruent figures have the same size and shape. 6. Say: Look around the room and see if you can identify two other objects in the classroom that are congruent. What did you find? Wait for answers. After the math buddies have chosen two objects, say: Can you explain to us why the objects are congruent? 7. Ask the following leading questions to guide the students in a discussion as to why the congruent figures are congruent. Depending on the objects, ask: How many sides do the shapes have? How many angles do the shapes have? How can you tell that the shapes the same size?
3 Math Buddies -Grade How do you know these are the same shapes? When we say two objects are congruent, does the color of the shape matter? (no) 8. Say: On paper, draw this symbol. Say: The mathematical symbol used to denote congruent is. The symbol is made up of two parts: ~ which means the same shape (similar) and = which means the same size (equal). Congruent Symbol Activity 1.2: Congruent or Not 1. Say: Open your book to Lesson #13: Student Activity Sheet #1:Congruent or Not. Look at the various shapes and determine whether they are congruent. Put a check under yes or no to indicate your answer. Then explain why they are or are not congruent. 2. Answers: 1. No, not the same size. 2. Yes, even though one is shaded. 3. No, different size and shaped triangles. 4. Yes, lines don t change the shape or size. 5. No, different size. 6. Yes, different position but the same shape and size. Activity 1.2: Tantalizing Triangles 1. Say: We can further refine our definition of congruent figure by saying that two shapes or solids are congruent if they are identical in every way except for their position; a figure can be moved by slides, flips or turns, and still be congruent. 2. Open the set of colored pencils for the Math Buddies to use and give each student one piece of patty paper to use as tracing paper. Say: Open your book to Lesson #13: Student Activity Sheet #2: Tantalizing Triangles. The objective of this activity is to find the tantalizing triangles that are congruent. To determine if they are congruent, carefully trace one of the triangles and then move the traced triangle around the page to find others that are congruent to it. Remember it does not matter what position the shape is in relative to another shape. Color any congruent triangles you find with the same color pencil. Then trace a second triangle and continue the same process. There are four different shaped triangles and all triangles should be colored. Good luck! Answers: Set #1: A, E, N, K are congruent Set #2: D, I, M, P are congruent Set #3: C, J, H, L are congruent Set #4: B, G, O, F are congruent Goal 2: Recognize the congruence of plane figures resulting from geometric transformations such as reflection (flip), translation (slide), and rotation (turn). Activity 2.1: Warm-Up: Transformations with Tangrams 1. Describing figures and visualizing what they look like when they are transformed through translations (slides), reflections (flips), and rotations (turns), or when they are put together or
4 Math Buddies -Grade taken apart in different ways are important aspects of the geometry program in elementary school. In this activity, students will use the seven tangram pieces to explore the transformation of shapes as they work to solve a few tangram puzzles. The potential for a high-quality spatial visualization experiences provided this activity that involves the use of manipulatives should enhance student understanding of transformations. The manipulative to be used is Tangrams, which are an ancient Chinese moving piece puzzle, consisting of 7 geometric shapes. 2. Give each student a set of tangrams and say: This is a set of seven tangram pieces from the ancient Chinese puzzle. The Tangram shapes were used for recreational activity in China thousands of years ago. The word Tangram is derived from tan, meaning Chinese, and gram, meaning diagram or arrangement. Spread them out on the table and point to the pieces as I say them: the square, two small triangles, one medium triangle, two large triangles, and one parallelogram. 3. Say: Let s examine each of the five different Tangram pieces, and determine the area of each piece, assuming that the small triangle has an area of one unit. Answers Small Triangle 1 square unit Square 2 square units Parallelogram 2 square units Medium Triangle 2 square unit Large Triangle 4 square unit 4. Say: You can use all seven pieces to make a figure or your can use a given number to make a figure. We are going to make a square of different sizes using a defined number of pieces. Let s try these tasks together. Select one or more based upon time constraints. Possible solutions follow. Can you make a square using one piece? (use the square piece) Can you make a square using two pieces? (two small triangles or two large triangles) Can you make a square using three pieces? (two small triangles and one medium triangle) Can you make a square using four pieces? Can you make a square using five pieces? Note: Using six pieces can t be done Can you make a square using seven pieces? (see below)
5 Math Buddies -Grade Say: Please use the seven tangram pieces to make one of the figures you select on Lesson#20: Student Activity Sheets #3A or #3B. You must use all seven pieces for each figure. I will check your answers once you inform me that you have completed a figure. 6. Answers: Activity 2.2: Transformations: Translations (Slides) 1. Say: You have been working with the Tangram pieces. While you worked to manipulate the shapes to create the different figures, often you were visualizing what they would look like once you had transformed them. You had a chance to move around the tangram pieces using a variety of transformations. 2. Say: Transformations is a word used to describe a category of movements that you can make with a shape. We will be studying three transformations: translations, rotations, and reflections. 3. Take out Lesson #13: Teacher Sheet #1. Refer to the top of the sheet as you describe translation transformations. Say: Translations are like slides, like sliding down a playground slide where you move from high to low but you are still sitting upright when you hit the bottom. A translation "slides" an object a fixed distance in a given direction. The original object and its translation have the same shape and size, and they face in the same direction. [Note: The word "translate" in Latin means "carried across".] When you are sliding down a water slide, you are experiencing a translation. Your body is moving a given distance (the length of the slide) in a given direction. You do not change your size, shape or the direction in which you are facing. Translations can be seen in wallpaper designs, textile patterns, mosaics, and artwork. 4. Say: Open your student books to Lesson #13: Student Activity Sheet #4. Look at the pentagons (five sided figures) at the top left hand side of the page. In mathematics, the translation of an object is called its image. If the original object was labeled with letters, such as ABCDE, the image may be labeled with the same letters followed by a prime symbol (like an apostrophe), A'B'C'D'E'.
6 Math Buddies -Grade Think of polygon ABCDE as sliding two inches to the right and one inch down. Its new position is labeled A'B'C'D'E'. 5. Say: A translation moves an object without changing its size or shape and without turning it or flipping it. Take out the Pattern Blocks and say: Here are some pattern blocks. Take out a blue parallelogram, a green triangle and a red trapezoid. On the pattern block grid paper, draw the translations of each shape by first placing it on the original figure and then sliding the pattern blocks the distance and the direction indicated by the arrow. To simplify this process we have only labeled one vertex of the shape with a letter. The image of the shape should have the same letter followed by the prime symbol in its new position as it had in it s original position. Check for accuracy of drawing. Ask: Did your shapes look different as a result of your translations? (no they do not change size or shape, just position) 6. Say: Now look at Part B on Activity Sheet #4. For each of the four problems, check yes or no to indicate whether one figure is the translation of the other. 7. Answers: 1. yes 2. no (change in size) 3. yes (doesn t need a slide line) 4. yes Activity 2.3: Transformations: Rotations (Turns) 1. Again take out Lesson #13: Teacher Sheet #1. Refer to the middle of the sheet as you describe the rotation transformation. Say: Rotations are turns, like when a basketball player pivots on one foot, or when a Ferris wheel turns around the center of the wheel. Look at this picture on the teacher page. To rotate a shape, you need to identify three things. First you must identify the point around which you are turning the shape, called the center of rotation. Second, you need to know the direction of the turn, clockwise or counterclockwise. Third, you need to know the angle, the number of degrees of the turn, or the fractional part of 1 whole turn (e.g. turn, or turn). Notice that the picture displays a clockwise rotation of the R around a center point, and where the angle of the turn is 90 degrees, or a one-quarter turn. 2. Say: Open your student books to Lesson #13: Student Activity Sheet #5: Discover Rotation. Notice the letter B being rotated four times around the center of the two intersecting lines. What is the direction of the rotation, clockwise or counterclockwise? (clockwise) What is the angle of the rotation for each turn? (90 degrees, or a onequarter turn) You might think of a rotation like putting an object on a plate or a Lazy
7 Math Buddies -Grade Susan, and then spinning the plate (or Lazy Susan ) around while the plate's center (or Lazy Susan s center) stays in one place. The center of the object doesn't have to be at the center of rotation (i.e. the center of your plate). Any point can be used to mark the center of rotation. 3. Say: Now look at the pattern block arrangement. Using the pattern blocks, make this same arrangement on the left side of a piece of paper. Wait until made Now, move this pattern block arrangement in a clockwise direction for an angle of 90 degrees or of a turn. Did it move off the paper? (yes) Did you arrangement stay the same distance from the center of rotation which is the bottom left hand corner of the paper as it was when you first made it? (yes) 4. Say: Now open your student books to Lesson #13: Student Activity Sheet #6: Rotation and Reflection With Pattern Blocks. In Part A, I would like Math Buddy A to make a pattern block figure on line A. Once the pattern is complete, I would like Math Buddy B to make this same pattern block figure on line B showing the pattern after a onequarter rotation in a clockwise direction. Wait until this is complete. Ask: Take a look at your work. Do you think it represents a clockwise rotation of 90 degrees and that the figures are an equal distance from the center of rotation? If not, what must be changed? If yes, you have demonstrated a rotation. 5. Say: In Summary, how can a rotation of an object be described? (There are three essential parts: 1)the object must move in a direction, clockwise or counterclockwise; 2) the object must move around a point called the center of rotation; and the object must turn some number of degrees or a fractional part of 1 whole turn.) 6. Say: Now, go back to the bottom of Student Activity Sheet #5. Decide which of the four problems represent rotations and which are not. Answers: 1. yes (1/4 turn clockwise) 2. yes (3/4 turn clockwise, or turn counterclockwise) 3. No, a translation 4. Yes (1/2 turn clockwise, or turn counterclockwise) Activity 2.4: Transformations: Reflections (Flips) 1. Take out Lesson #13: Teacher Sheet #1. Refer to the third transformation called reflection. Say: Reflection is the third transformation we will study. Reflections are like flips: like the picture of a gymnast doing a handstand. Look at the happy face and the R on this page. Each has been reflected across a line of reflection. 2. Say: In the real world, a reflection can be seen in water, in a mirror, in glass, or in a shiny surface. An object and its reflection have the same shape and size, but the figures face in opposite directions. 3. Say: When you look in the mirror what do you notice that is the same and is different about your face? (Discuss answers) In a mirror, right and left are switched. Under a reflection in a mirror, the figure does not change size. It is simply flipped over the line of reflection. In mathematics, the reflection of an object is called its image.
8 Math Buddies -Grade Say: Now open your student books to Lesson #13: Student Activity Sheet #6: Rotation and Reflection With Pattern Blocks. At the bottom of the page in Part B, I would like Math Buddy A to place a red trapezoid on the left side of the line, touching the line. Once the trapezoid is placed, say: Now, I would like Math Buddy B to place a red trapezoid on the right side of the line to show a reflection of this pattern block. Does everyone agree that this is the reflection image of the pattern block on the left. If not, make the corrections. This is one example; other placements of the trapezoid lead to other arrangements. Line of Reflection Reflection Image of Reflection 5. Say: Now let s try a more challenging task. At the bottom of the page in Part B, I would like Math Buddy A to make a pattern block design on the left side of the line so that the design touches the line. Once the pattern design is complete, I would like Math Buddy B to make the reflection of this pattern block design on the right side of the line to show the designs reflection. Once complete, say: Does everyone agree that this is the reflection of the pattern block design across the line of reflection? Do we need to make any corrections? 6. Say: Now, switch rolls, and Math Buddy B will create the design on the right side, and Math Buddy A will create it s reflection on the left side. Once complete, say: Does everyone agree that this is the reflection of the pattern block design across the line of reflection? Do we need to make any corrections? Activity 2.5: Rotation or Reflection? 1. Say: Now open your student books to Lesson #13: Student Activity Sheet #7: Rotation or Reflection? Here is a table of figures. Use your knowledge to decide whether the second figure, the image, is a rotation or a reflection of the first. Once you decide, check under the column heading of this transformation. Some images may represent a rotation and a reflection, so check both. 2. Answers: 1. Rotation (1/4 turn) 2. Reflection (across a horizontal line) or Rotation (1/2 turn) 3. Reflection (across a vertical line) 4. Rotation (3/4 turn clockwise; or turn counterclockwise) 5. Rotation (1/4 turn clockwise) 6. Reflection (across a horizontal line) 7. Reflection (across a horizontal line) 8. Reflection (across a vertical line) 9. Reflection (across a vertical line) 10. Rotation (3/4 turn clockwise; or turn counterclockwise) Activity 2.6: Is the Shape Reflection-Congruent? 1. Give each student a geo-reflector and introduce the students to its parts. Place the georeflector in front of the student so that the beveled edge is down (touching the desk) and the beveled edge is facing the student.
9 Math Buddies -Grade Point to the parts of the geo-reflector as you describe them to the students. Say: This is a geo-reflector. Feel the top edge of the geo-reflector. It has square corners for edges. Feel the Bottom edge of the geo-reflector. Is it the same as the top edge? (No) Notice that it is not as thick as any other edge on the geo-reflector. It has a beveled edge on the front face of the geo-reflector and a square corner edge on the back face of the georeflector. When you are working, always keep the beveled edge of the geo-reflector facing you so that you are looking into the front face of the geo-reflector. 3. Then review the parts of the geo-reflector by asking: a. How can you tell the top from the bottom? (The beveled edge is on the bottom.) b. How is the beveled edge different from all the other edges of the geo-reflector? (It is a different thickness.) c. How can you tell which face is the front? (By finding the beveled edge that is on the front face.) 4. Say: Now, go back to the bottom of Student Activity Sheet #6. Place the Geo-Reflector on the line of reflection and rotate your book around so that the Geo-Reflector is sitting horizontally, parallel to the table s edge. 5. Say: Now take out a yellow pattern block and place it anywhere between you and the Geo-Reflector. Using a pencil draw the perimeter of the yellow hexagon. Now, making sure the hexagon stays in this same spot, look through the Geo-Reflector and what do you see? (reflection of the hexagon) Yes, you see the reflection of the hexagon. Now I would like you to draw the perimeter of the reflection of the hexagon free hand. Once this is done, say: Remove the Geo-Reflector and pattern block leaving the drawing of the original figure, the line of reflection, and the drawing of the figure s reflection, called the image of reflection. 6. Say: Now take out a few pattern blocks and place them in front of the Geo-Reflector and look through the Geo-Reflector at their reflection. 7. Say: Now we are going to check to see whether two shapes are congruent as a result of a reflection. Take out Lesson #13: Student Activity Sheet #8: Is the Shape Reflective- Congruent. Place the Geo-Reflector between the two figures and move it around so that when you look into the Geo-Reflector you can see whether the one figure fits on top of the other. The figure between you and the Geo-Reflector, or what is in front of the Geo-Reflector is called the object. Notice that the object is outlined in black. The figure behind the Geo-Reflector is the image. What color is the image? (It is outlined in the color of the Geo-Reflector as a result of looking through the colored plastic.) 8. If the object and the image are congruent (e.g. same size and same shape), the pair of shapes are reflective-congruent. Use the Geo-Reflector to determine whether the other pairs of figures are reflective-congruent. Check Yes if they are and No if they are not. If they are congruent as a result of the reflection, draw the line of reflection by placing your pencil on the beveled edge and drawing along that edge when the object reflects onto the image.
10 Math Buddies -Grade Answers: A.) Yes B.) No C.) No D.) Yes E). No F.) Yes G.) No H.) Yes Goal 3: Identify and Draw Lines of Symmetry Activity 3.1: Lines of Symmetry 1. Ask students: What is a line of symmetry? (A line of symmetry divides a symmetrical figure, object, or arrangement of objects into two parts that are congruent if one part is reflected (flipped) over the line of symmetry.) Symmetry is everywhere in nature, art, music, mathematics, and beyond. Can you think of anything that is symmetrical? (Answers might include a butterfly, the letter H, a pair of pants, etc.) 2. In this activity, students will enhance their understanding of symmetry, particularly, reflectional symmetry, using the Geo-Reflector. Say: In our last activity, when shapes were congruent as a result of a reflection, we were able to draw a line of reflection. This line represented the line across which the objects were flipped. In this activity we will use the Geo-Reflector on individual shapes as a line of symmetry. The reflection will produce the other congruent half of the shape. Consequently, we will learn that a line of symmetry is a line that divides a figure in to congruent halves, each of which is the reflection image of the other. 3. Say: Take out Lesson #13: Student Activity Sheet #9: Line of Symmetry. The dotted line on each shape is the line of symmetry. Place your Geo-Reflector on the dotted line and draw the other side of the shape by tracing its reflection. 4. Answers: Line of Symmetry Activity 3.2: Polygons: How Many Lines of Symmetry? 1. Say: Take out Lesson #13: Student Activity #10: Polygons: How Many Lines of Symmetry? The polygons on this page are regular polygons. Regular polygons are polygons that have congruent sides and congruent angles; that is sides of the same lengths and angles of the same angle measure. 2. Say: You are going to determine how many lines of symmetry each of these polygons has using the Geo-Reflector. Move your Geo-Reflector around on the shape to find
11 Math Buddies -Grade lines of symmetry. When you find a line of symmetry, where one side can be reflected on the other, draw that line of symmetry by placing your pencil on the recessed (beveled) edge of the Geo-Reflector and drawing that line. As you complete each polygon, report the number of lines of symmetry for the identified shape in the table below. Work on this activity now and then we will summarize your findings in the table once you have finished. Lines of Symmetry: Triangle: 3 Square: 4 Pentagon: 5 Hexagon: 6 3. Say: Now, let s review the data you have collected in the table. How many lines of symmetry did you find for the equilateral triangle? (3) As you look back at these lines, notice that each line went through one vertex and through the midpoint of the side opposite the vertex. Now look at the five sided pentagon. How many lines of symmetry did you find for the pentagon? (5) How are these lines of symmetry similar to the lines of symmetry in the triangle? (Each line of symmetry went through one vertex and through the midpoint of the side opposite the vertex.) 4. Say: How many lines of symmetry did you find for the square? (4) As you look back at these lines, notice that two line went from one vertex through to the other vertex, and two line went from one midpoint through to the other midpoint on the opposite side. Now look at the six sided hexagon. How many lines of symmetry did you find for the hexagon? (6) How are these lines of symmetry similar in the hexagon similar to the lines of symmetry in the square? (Each line of symmetry went from one vertex to the opposite vertex, or from one midpoint to the opposite the midpoint.) 5. Say: Now, let s look at the numbers. Is there any relationship between the number of sides in a regular polygon and the number of lines of symmetry? (Yes, when finding lines of symmetry in regular polygons, the number of lines of symmetry equals the number of sides in the polygon.) Lesson #13: Assessment of Student Learning 1. Have students complete the thirteen multiple-choice assessment items independently by circling the correct answer. 2. Once complete, discuss the items that the students answered incorrectly, asking them to explain their thinking and reasoning about how they chose each answer. Answer Key: 1. B 2. C 3. A 4. C 5. B 6. A 7. C 8. B 9. C 10. J 11. G 12. A 13. J
12 Math Buddies -Grade Lesson #13: Student Activity Sheet #1 Congruent or Not? Look at these figures and see if you can pick congruent figures. Check yes if the figures are congruent and no if the figures are not congruent. Congruent or Not? Yes No Congruent or Not? Yes No
13 Math Buddies -Grade Lesson #13: Student Activity Sheet #2 Tantalizing Triangles Find out if the tantalizing triangles are congruent using tracing paper. Color any congruent triangles you find the same color. Hint: There are four congruent shapes for each of four different shapes! A B C D E H F G L I J K P M N O Four Congruent Triangles are: Four Congruent Triangles are: Four Congruent Triangles are: Four Congruent Triangles are:
16 Math Buddies -Grade Lesson #13: Student Activity Sheet #4 Translation With Pattern Blocks Translation "slides" an object a fixed distance in a given direction. The original object (A) and its translation (A ) have the same shape and size, and they face in the same direction. Part A: Translate the pattern blocks the distance and the direction indicated by the arrows and draw the image of the translation. Part B: Check yes or no to indicate whether one figure is a translation of the other. Translation or Not? Yes No Translation or Not? Yes No
18 Math Buddies -Grade Lesson #13: Student Activity Sheet #6 Rotation and Reflection With Pattern Blocks Part A: Math Buddy A makes a pattern block figure on line A. Math Buddy B makes the one-quarter rotation of Math Buddy A s pattern block figures on line B. Line A Center of Rotation Line B Part B: Math Buddy A makes a pattern block figure on one side of the line. Math Buddy B makes its reflection on the other side of the line. Line of Reflection
20 Math Buddies -Grade Lesson #13: Student Activity Sheet #8 Is the Shape Reflective-Congruent? Use your Geo-Reflector to check if the shapes are congruent as a result of a reflection. Check Yes if they are and draw the line of reflection; otherwise check No. A. Yes No E. Yes No B. Yes No F. Yes No C. Yes No G. Yes No D. Yes No H. Yes No
22 Math Buddies -Grade Lesson #13: Student Activity Sheet #10 Polygons: How Many Lines of Symmetry? Use the Geo-Reflector to draw as many lines of symmetry as you can find for each regular polygon. Complete the chart identifying the number of lines of symmetry. Shape Name of Shape Number of Sides Triangle 3 (Equilateral Triangle) Number of Lines of Symmetry Square 4 Pentagon (Regular Pentagon) Hexagon (Regular Hexagon) 5 6
23 Math Buddies -Grade Lesson #13: Student Assessments 1. The arrow below moved 90 degrees clockwise or turn. 4. The example below is a demonstration of what? This is an example of what? A. Translation B. Rotation C. Reflection A. Translation B. Rotation C. Reflection 2. The example below is a demonstration of a. 5. In the example below, the triangles going from left to right is an illustration of a. A. Translation B. Rotation C. Reflection A. Translation B. Rotation C. Reflection 3. The change in the position of the triangles in Set A to the position of the triangles in Set B is an illustration of a. Set A Set B 6. What is it called when the arrow in picture A is moved up to the position in picture B? Picture A Picture B A. Translation B. Rotation C. Reflection A. Translation B. Rotation C. Reflection
24 Math Buddies -Grade The arrow below in picture B is a mirror image of the arrow in picture A. This transformation is called a. 10. Picture A Picture B A. Translation B. Rotation C. Reflection 8. The example below is a demonstration of a. A. Translation B. Rotation C. Reflection 9. In which figure below is the line NOT a line of symmetry? Figure A Figure B Figure C 11. Which pair of figures does NOT show a translation? A. Figure A B. Figure B C. Figure C
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26 Math Buddies -Grade Lesson #13: Teacher Sheet #1 Transformations: Translations, Rotations and Reflections Translation: To translate an object means to move it a given distance in a given direction without rotating or reflecting it. Rotation To rotate an object means to turn it around. Every rotation has a center of rotation and an angle of rotation. 90 o angle is of a turn. Reflection To reflect an object means to flip it to produce its mirror image. Every reflection has a line of reflection along which it is flipped. Translation Slides a given distance in a given direction Rotation Turns around a center point of rotation, for a given angle or identified turn (example of turn ) reflection Flips across a line of reflection
HOME LINK Line Segments, Rays, and Lines Family Note Help your child match each name below with the correct drawing of a line, ray, or line segment. Then observe as your child uses a straightedge to draw
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Grade 8 Mathematics Geometry: Lesson 2 Read aloud to the students the material that is printed in boldface type inside the boxes. Information in regular type inside the boxes and all information outsideTessellating with Regular Polygons You ve probably seen a floor tiled with square tiles. Squares make good tiles because they can cover a surface without any gaps or overlapping. This kind of tiling is
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39 Symmetry of Plane Figures In this section, we are interested in the symmetric properties of plane figures. By a symmetry of a plane figure we mean a motion of the plane that moves the figure so thatGLOSSARY Appendix A Appendix A: Glossary Acute Angle An angle that measures less than 90. Acute Triangle Alternate Angles A triangle that has three acute angles. Angles that are between parallel lines,
HPTER 7 1 Exploring Tangrams Solve tangram puzzles. You will need scissors and a ruler. 1. Trace and cut out the 7 tans. t-home Help tangram is an ancient hinese puzzle. It has the 7 shapes, or tans, shownE XPLORING QUADRILATERALS E 1 Geometry State Goal 9: Use geometric methods to analyze, categorize and draw conclusions about points, lines, planes and space. Statement of Purpose: The activities in this
Angle - a figure formed by two rays or two line segments with a common endpoint called the vertex of the angle; angles are measured in degrees Apex in a pyramid or cone, the vertex opposite the base; in
Problem of the Month: The Problems of the Month (POM) are used in a variety of ways to promote problem solving and to foster the first standard of mathematical practice from the Common Core State Standards:Problem 1.1 Definitions: regular polygons - polygons in which all the side lengths and angles have the same measure edge - also referred to as the side of a figure tiling - covering a flat surface withGrade FCAT 2.0 Mathematics Sample Answers This booklet contains the answers to the FCAT 2.0 Mathematics sample questions, as well as explanations for the answers. It also gives the Next Generation Sunshine
Perimeter and Area Page 1 of 57 PERIMETER AND AREA Objectives: After completing this section, you should be able to do the following: Calculate the area of given geometric figures. Calculate the perimeter
Connect the dots on the isometric dot paper to represent the edges of the solid. Shade the tops of 12-1 Representations of Three-Dimensional Figures Use isometric dot paper to sketch each prism. 1. triangular
Geometry of Minerals Objectives Students will connect geometry and science Students will study 2 and 3 dimensional shapes Students will recognize numerical relationships and write algebraic expressions37 Basic Geometric Shapes and Figures In this section we discuss basic geometric shapes and figures such as points, lines, line segments, planes, angles, triangles, and quadrilaterals. The three pillars
Dear Grade 4 Families, During the next few weeks, our class will be exploring geometry. Through daily activities, we will explore the relationship between flat, two-dimensional figures and solid, three-dimensional
Illinois State Standards Alignments Grades Three through Eleven Trademark of Renaissance Learning, Inc., and its subsidiaries, registered, common law, or pending registration in the United States and other
Name: ate: 1. Parallelogram ABC was translated to parallelogram A B C. 2. A shape is shown below. Which shows this shape transformed by a flip? A. B. How many units and in which direction were the x-coordinates
Fractions In Action! Dawn Jesse Fractions In Action Dawn Jesse Fractions In Action is an interactive activity that consists of direct instruction, cooperative learning and is inquire based. As the students
Brief Overview: Warning! Construction Zone: Building Solids from Nets In this unit the students will be examining and defining attributes of solids and their nets. The students will be expected to haveHigh School - Circles Essential Questions: 1. Why are geometry and geometric figures relevant and important? 2. How can geometric ideas be communicated using a variety of representations? ******(i.e maps,
Geometry Chapter 1 Section Term 1.1 Point (pt) Definition A location. It is drawn as a dot, and named with a capital letter. It has no shape or size. undefined term 1.1 Line A line is made up of points
A Correlation of to the Minnesota Academic Standards Grades K-6 G/M-204 Introduction This document demonstrates the high degree of success students will achieve when using Scott Foresman Addison Wesley
Investigating Relationships of Area and Perimeter in Similar Polygons Lesson Summary: This lesson investigates the relationships between the area and perimeter of similar polygons using geometry software.Performance Assessment Task Which Shape? Grade 3 This task challenges a student to use knowledge of geometrical attributes (such as angle size, number of angles, number of sides, and parallel sides) to
Activity: TEKS: Exploring Transformations Basic understandings. (5) Tools for geometric thinking. Techniques for working with spatial figures and their properties are essential to understanding underlying
Problem of the Month: William s Polygons The Problems of the Month (POM) are used in a variety of ways to promote problem solving and to foster the first standard of mathematical practice from the CommonoGebra in 10 lessons Gerrit Stols Acknowledgements GeoGebra is dynamic mathematics open source (free) software for learning and teaching mathematics in schools. It was developed by Markus Hohenwarter
Geometry and Measurement of Solid Figures Activity Set 4 Trainer Guide Mid_SGe_04_TG Copyright by the McGraw-Hill Companies McGraw-Hill Professional Development GEOMETRY AND MEASUREMENT OF SOLID FIGURES
The fundamental purpose of the course is to formalize and extend students geometric experiences from the middle grades. This course includes standards from the conceptual categories of and Statistics and
how August/September Demonstrate an understanding of the place-value structure of the base-ten number system by; (a) counting with understanding and recognizing how many in sets of objects up to 50, (b)
Ohio Standards Connection Patterns, Functions and Algebra Benchmark E Solve open sentences and explain strategies. Indicator 4 Solve open sentences by representing an expression in more than one way usingNew York State Student Learning Objective: Regents Geometry All SLOs MUST include the following basic components: Population These are the students assigned to the course section(s) in this SLO all students | 677.169 | 1 |
question_answer
A variable plane is at a constant distance p from the origin and meets the axes in A, B and C. The locus of the centroid of the tetrahedron \[OABC\] is
A) \[{{x}^{-2}}+{{y}^{-2}}+{{z}^{-2}}=16{{p}^{-2}}\]
B) \[{{x}^{-2}}+{{y}^{-2}}+{{z}^{-2}}=16{{p}^{-1}}\]
C) \[{{x}^{-2}}+{{y}^{-2}}+{{z}^{-2}}=16\]
D) None of these
Correct Answer:
A
Solution :
Plane is\[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\], where \[p=\frac{1}{\sqrt{\sum\limits_{{}}^{{}}{\left( \frac{1}{{{a}^{2}}} \right)}}}\] or \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}}=\frac{1}{{{p}^{2}}}\] ?..(i) Now according to equation, \[x=\frac{a}{4},\ \ y=\frac{b}{4},\ \ z=\frac{c}{4}\] Put the values of x, y, z in (i), we get the locus of the centroid of the tetrahedron. | 677.169 | 1 |
Circular Arc Calculator
Calculations at a circular arc. A circular arc is a part of the outer line of a circle. It is enclosed by two straight lines starting at the center of the circle in a certain angle α. At 360° or 2π, the circle is complete. Enter two of the three values radius, length and angle. Choose the number of decimal places, then click Calculate. The angle can be entered in degrees, as radian or as multiples of π, it also will be calculated in those units.
Radius (r):
Length (l):
Angle α in degrees:
Angle α in radian:
Angle α in multiples of π:
Round to decimal places.
Formulas:
360° ≙ 6.283185307179586 ≙ 2π
l = 2 π r * α/360°
oder
l = 2 π r * α/(2π)
pi:
π = 3.141592653589793...
Radius and arc length have the same one-dimensional unit (e.g. meter). | 677.169 | 1 |
In this session, viewers will work with various constructions including the heptagon, the ellipse, a cyma recta profile, and a torus profile.
Physically drawing these geometric constructions will help the viewer to understand these mathematical relationships. Viewers are encouraged to download the accompanying booklet below and actually draw the constructions demonstrated in the video. Only a straightedge and a compass are required to draw the constructions. Rulers, protractors, and triangles are not necessary, though they may be useful for checking one's work. | 677.169 | 1 |
Single Idea 13907
Gist of Idea
If you pick an arbitrary triangle, things proved of it are true of all triangles | 677.169 | 1 |
CSMC 2022 Part A - Question 5, CEMC UWaterloo
Difficulty: 6
A circle has centre $O$ and diameter $AB = 2 \sqrt{19}$. Points $C$ and $D$ are on the upper half of the circle. A line is drawn through $C$ and $D$, as shown. Points $P$ and $Q$ are on the
line so that $AP$ and $BQ$ are both perpendicular to $PQ$. $QB$
intersects the circle at $R$. If $CP = DQ = 1$ and $2AP = BQ$, what is the length of $AP$?
Answer Submission Note(s)
If your answer contains a square root, type it as "sqrt(x)", and fractions should be typed as "x/y". Please view this page to learn more about formatting your answer.
Please login or sign up to submit and check if your answer is correct. | 677.169 | 1 |
Salinon
The salinon is the figure illustrated above formed from four connected semicircles. The word salinon is Greek for "salt cellar," which the figure resembles.
If the radius of the large enclosing circle is and the radius of the small central circle is , then the radii of the two small side circles are .
In his Book of Lemmas, Archimedes proved that the salinon has an area equal to the circle having the line segment joining the top
and bottom points as its diameter (Wells 1991), namely | 677.169 | 1 |
What Geometric Shape Is A Pencil
Unveiling the Geometry of Pencils: What Shape Lies Beneath?
From the utilitarian to the artistic, the humble pencil is a tool that transcends boundaries. But have you ever pondered the geometric form that underpins its design? Join us as we delve into the intricate world of pencil geometry, uncovering the shape that defines this ubiquitous writing instrument.
The Anatomy of a Pencil: A Closer Look
Before we unravel the mysteries of its shape, let's familiarize ourselves with the components of a pencil:
Core: Also known as the "lead," though typically made of graphite, not lead.
Wood Casing: Surrounds the core, providing structure and protection.
Ferrule: Metal part at the end that holds the eraser in place.
Eraser: Often attached to the ferrule, though not always.
Now, let's embark on our journey to unveil the geometric secrets of the pencil.
Exploring the Core Geometry
At the heart of every pencil lies its core, the element responsible for leaving its mark on paper. While the core itself may be cylindrical in shape, it's essential to consider the overall form of the pencil.
Is a Pencil Truly Cylindrical?
While many may assume that a pencil is a perfect cylinder, a closer inspection reveals a different reality. In fact, most pencils possess a slightly hexagonal or octagonal shape, rather than being perfectly round. This design serves both functional and ergonomic purposes:
Prevents Rolling: The flat edges of a hexagonal or octagonal pencil reduce the likelihood of it rolling off a desk or table, providing stability during use.
Enhanced Grip: The faceted surface offers a comfortable grip, allowing for greater control and precision when writing or drawing.
The Role of Geometry in Pencil Design
The decision to shape pencils in this manner is not arbitrary; it reflects a thoughtful consideration of geometry's practical implications. By incorporating facets into the design, pencil manufacturers optimize usability and functionality, aligning form with function in a harmonious balance.
FAQs: Addressing Common Queries
Q: Why are pencils not perfectly round?
A: Pencils are often crafted with a hexagonal or octagonal shape to prevent rolling and enhance grip.
Q: What material is typically used for pencil cores?
A: Pencil cores are primarily composed of graphite, though historically, they were made from a mixture of graphite and clay.
Q: Are there pencils with different core shapes?
A: While most pencils feature a cylindrical core, specialized pencils such as carpenter's pencils may have rectangular cores, suited for specific tasks.
Q: Do all pencils have erasers?
A: No, not all pencils come equipped with erasers. Some are designed solely for marking and lack an attached eraser.
Q: How are pencil cores manufactured?
A: Pencil cores are formed through a process of mixing graphite with binders, extruding the mixture into rods, and then curing and cutting them into individual cores.
Conclusion: Unveiling the Geometry of Pencils
In conclusion, the geometric shape of a pencil transcends mere aesthetics; it embodies functionality, ergonomics, and practicality. By understanding the underlying geometry, we gain a deeper appreciation for this everyday tool's design and utility. Whether sketching a masterpiece or jotting down notes, the geometry of the pencil plays a pivotal role in shaping our creative endeavors. | 677.169 | 1 |
Exploring Rotations
Directions.
Use the applet to learn how coordinates are affected when you rotate around the ORIGIN (Do NOT move Point F for this activity. Do NOT go to a different Problem other than #1)
Use the Google Form (in Google Classroom) to answer the questions. Be sure to submit when you are done.
What to Notice...
First, move the slider for the Angle of Rotation. When you do this, the shape is being rotated around Point F in a counterclockwise direction.
We could also do a clockwise rotation (but not with this particular app) .
Rotating a shape 90 degrees counterclockwise would be the SAME as rotating it 270 degrees clockwise.
Rotating a shape 180 degrees counterclockwise would be the SAME as rotating it 180 degrees clockwise.
Rotating a shape 270 degrees counterclockwise would be the SAME as rotating it 90 degrees clockwise. | 677.169 | 1 |
Copy of Investigation 3.1 In Looking for Pythagoras
The three vertices of the triangle can be dragged to change the triangles side lengths. You can also adjust the angle to create acute, obtuse, and right triangles.
Be sure to record two acute, two obtuse and two right triangles. You may not use the same measurements for side lengths as people at your table so you'll need to check in with table mates before recording side lengths and areas of squares in your book | 677.169 | 1 |
right-handed system of vectors
Three ordered non-coplanar vectors u→, v→ and w→, which have a common , are said to form a right-handed or dextral system (Latin dexter = ), if a right-threaded screw rotated through an angle less than 180o from u→ to v→ will advance in the direction of w→. For instance, the usual basis vectors i→, j→ and k→ form a dextral system. | 677.169 | 1 |
Assertion: Two vectors are said to be equal if , and only if, they have the same magnitude and the same dirction. Reason: Addition and subtraction of scalars make sense only for quantities with same units.
Video Solution
Text Solution
Verified by Experts
The correct Answer is:B
Addition and subtraction of scalars make sense only for quantities with same unit. But you can multiply and divide scalars of different units. | 677.169 | 1 |
Angles Worksheets Promote A Better Understanding Of The Various Types Of Angles And How To Differentiate Among Them.
Math angles worksheets for grade 6 students: Web now, we will learn more pairs of angles for grade 6 to grade 8 like linear, vertically opposite and adjacent angles here. Angles can be broadly classified into five.
Web Free Printable Angles Worksheets For 6Th Grade.
Web this unit test is based of the ontario grade 6 curriculum and uses the nelson 6 textbook as a key resource. These worksheets will help you learn how to. A complete angle measures __________. | 677.169 | 1 |
Relationship between constructibility of reg. polygons and cot(pi/N)
In summary, the conversation touched upon the relationship between the constructibility of regular polygons and the reducibility of trigonometric functions into expressions of square roots. The speaker discovered a general pattern for finding the area of an N-gon, and noticed that the function cot(pi/N) was reducible if and only if the polygon was constructible. They also posed four challenging questions for further exploration and discussion.
Sep 9, 2014
#1
ellipsis
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Full title: Relationship between the constructibility of regular polygons and the reducability of trigonometric functions into expressions of square roots.
I stumbled upon this after I derived the formula for the area of a triangle given it's side length x as a trigonometry exercise. ## A = \frac{1}{2}\sin{(60°)}x^2 = \frac{\sqrt{3}}{4}x^2 ##. I challenged myself to find the area formulas for every N-gon until I found a pattern. I did square, pentagon, and hexagon using basic trigonometry, and reducing the trigonometric functions into square root expressions using Wolfram.
Things started to get weird when I got to heptagon or the 7-gon, though. Wolfram wouldn't reduce it, and my area equation was made ugly by the presence of a trig function. (I spent a good few hours researching how to find exact trig values by hand using the half-angle formula et al, before I realized it was impossible)
Then I tried to find a general angle formula, but I couldn't write the later trigonometric functions in terms of sine. Eventually I figured out cot() is the only one that works (but I don't know why). This general pattern popped out at me:
$$ A = \frac{N}{4}\cot{\frac{\pi}{N}} $$
After I got to the 11-gon I started noticing another pattern: Whether or not the function ##\cot(\frac{\pi}{N})##, where N is the number of sides, is reducible has the same truth value as whether or not a given polygon is constructible. (I was amazed by this). Also, the relative complexity of each square-root expression is correlated with the relative complexity of the construction of that polygon. (As first seen with the long and convoluted 17-gon)
The wiki page for the 17-gon mentions that its area and constructibility was determined by Gauss 200 years ago, while he was going through this thought process, sqeeee!). It also claimed he had a method of determining whether cot(pi/N) was in general reducible, based on some property of "fermat" primes.
I have four challenging questions that I think this board will be interested in:
1. Does there exist a general algorithm for determining the exact value of ##\cot(\frac{\pi}{N})## in terms of square roots if such an expression exists?
2. Likewise, is there a general algorithm to derive the steps of constructing a regular polygon with a number of sides ##N##?
3. Does there exist a linear-time algorithm for converting a square-root expression into a set of polygon construction steps?
Beyond a certain degree of self-interference and complexity, a system can be used as an analogue to a Turing machine (i.e. it can be used for arbitrary computation). If a system is beyond that point, it makes certain statements about that system undecidable, such as if it will enter into an infinite-regress or not.
4. If you had infinite time, infinite paper, and an unmarked ruler and compass - could you do the same set of problems as a Turing machine?
Here's my observations on the possible answers:
* I suspect questions 1 and 2 have the same truth value.
* I suspect question 3 is true (there is a linear time algorithm for converting between a square-root expression and a set of instruction for polygon construction)
* I suspect question 4 is the negation of questions 1 and 2.
NOTE: I won't be disappointed with answers like "One could write a whole original book on this topic, nobody knows yet." I just want to encourage discussion on an interesting problem. I'm about to leave for class, so I won't be back for a while.
Related to Relationship between constructibility of reg. polygons and cot(pi/N)
What is the relationship between the constructibility of regular polygons and cot(pi/N)?
The relationship between the constructibility of regular polygons and cot(pi/N) is that a regular polygon with an odd number of sides (N) is constructible if and only if cot(pi/N) is a rational number. This means that the regular polygon can be constructed using a compass and straightedge.
How does cot(pi/N) affect the constructibility of regular polygons?
Cot(pi/N) is a measure of the slope of a line that intersects the unit circle at an angle of pi/N. The constructibility of regular polygons is dependent on the value of cot(pi/N) being a rational number. If cot(pi/N) is irrational, then the regular polygon is not constructible.
What is the significance of cot(pi/N) in the constructibility of regular polygons?
Cot(pi/N) is significant because it determines whether a regular polygon can be constructed using a compass and straightedge. If cot(pi/N) is rational, then the regular polygon is constructible. If it is irrational, then the regular polygon is not constructible.
Can a regular polygon with an even number of sides be constructed using a compass and straightedge?
No, a regular polygon with an even number of sides cannot be constructed using a compass and straightedge. This is because the value of cot(pi/N) for an even number of sides will always be irrational, making it impossible to construct the polygon using these tools.
How does the value of N affect the constructibility of regular polygons?
The value of N, or the number of sides in a regular polygon, directly affects the constructibility of the polygon. A regular polygon with an odd number of sides (N) is constructible if and only if cot(pi/N) is a rational number. A regular polygon with an even number of sides is not constructible using a compass and straightedge. | 677.169 | 1 |
86 ... inches , respectively , is equal to π times the area of a triangle of base 3 inches and altitude 3 inches . h ) The bisectors of the interior angles of any rectangle ( not a square ) will form a square . 6. A circle has a radius of 14 ...
«·'ðÕ… 89 ... inches from C and B is the mid- point of AC . Describe fully the locus of points equidistant from A and B and 5 inches from C. ( THIS EXAMINATION IS CONTINUED ON THE NEXT PAGE ) ( 14-6 ) 6. Of the following eight statements , some are ...
«·'ðÕ… ... inches in radius ; the globe rests in a supporting ring two inches above its lowest point . What is the radius of the shadow of the ring cast on the pavement 6 feet below the light ? 3. If y = x2 - 5x + 7 , is there a value of x for ... | 677.169 | 1 |
Question 1.
The ski jumper shown makes angles with her skis as shown. What is the sum of the angles?
Answer:
The sum of angles = 180°
55° + 125° = x
x = 180°
So, the sum of the angles is 180°.
Question 2.
In the roof framework shown, ∠ABD and ∠DBC have the same measure. What is the measure of ∠DBA? What is the measure of ∠ABC?
Answer:
Given that,
∠ABD and ∠DBC have the same measure.
So, ∠ABD = 60°
60° + 60° = 120°
So, the measure of ∠DBA is 60°
Question 3.
In the simple bridge structure shown, the measure of ∠RSV is 30° and ∠VST is a right angle. What is the measure of ∠RST?
Answer:
∠RSV is 30°
∠VST is 90°
30° + 90° = x°
x° = 120°
So, the measure of ∠RST is 120°
The circle at the right represents all of the students in a class. Each section represents the students in the class who chose a certain type of animal as their favorite type of pet. The angle measures for some sections are given.
Question 4.
What is the sum of the angle measures for Cat, Dog, and Horse?
Answer:
The angle measure of cat is 85°
The angle measure of dog is 150°
The angle measure of horse is 60°
85° + 150° + 60° = x°
x° = 295°
So, the angle measures for Cat, Dog, and Horse is 295°
Question 5.
What is the total angle measure for the circle? What is the angle measure for Fish?
Answer:
The total angle measure for the circle is 360°
The angle measure for Fish is 360° – 295° = f
360° – 295° = f
f = 65°
Subtract Angle Measures
Use an equation to solve.
Question 6.
In the roof framework shown, the measure of one angle is 80°. What is the unknown angle measure?
Answer:
Given, the measure of one angle is 80°.
Sum of angle = 180°
180 °- 80° =x°
x° = 100°
So, the unknown angle is 100°
Question 7.
The railing on a stairway makes a 50° angle with the upright post. What is the unknown angle measure in the diagram?
Answer:
Given,
The railing on a stairway makes a 50° angle with the upright post.
Sum of angle = 180°
180° – 50° = x
x° = 130°
So, the unknown angle is 130°.
Question 8.
When different items are poured, they form a pile in the shape of a cone. The diagram shows a pile of sand. What is the angle the sand makes with the ground?
Answer:
Sum of angle = 180°
One of the angle is 146°
180° – 146° = x°
x = 34°
So, the angle the sand makes with the ground is 34°
Question 9.
In a miniature golf game, a player hits a ball against a wall at an angle with measure 35° and it bounces off at an angle of 20°. What is the unknown angle measure in the diagram?
Answer:
Given,
In a miniature golf game, a player hits a ball against a wall at an angle with measure 35° and it bounces off at an angle of 20°.
Sum of angle = 180°
180° – (35° + 20°) = x
x = 180° – 55°
x = 125°
So, the unknown angle measure in the diagram is 125°
Question 10.
In a reclining chair, you can push back from an upright position to sit at an angle. In the chair shown, the whole angle between the back of the chair and the seat of the chair is 130°. Find the unknown angle measure to find by how much the chair is reclined from upright.
Answer:
Given,
In a reclining chair, you can push back from an upright position to sit at an angle.
In the chair shown, the whole angle between the back of the chair and the seat of the chair is 130°.
130° – 90° = x
x = 40°
So, the unknown angle measure to find by how much the chair is reclined from upright is 40°. | 677.169 | 1 |
Navigation
A Trigonometric Function (also called a Circular Function)
is a function of an angle. A Trigonometric Function will
relate the angles of a triangle to the lengths of its sides,
or the angle of a ray from the centre of the unit circle
either to the circumference, or to a tangent line. | 677.169 | 1 |
4-4 skills practice proving triangles congruent.
Students will practice the necessary skills of proving triangles are congruent to be successful in Geometry and to continue student success and growth. • Answer Sheet Included •.. ... 4.1 Classifying Triangles4.2 Angles in Triangles4.3 Congruence and Triangles4.4 Congruent Triangles: SSS and SAS4.5 Congruent Triangles: AAS, ...Practice proving congruence. Let's leave the safety of spring training and try our skills with some real major league games. Here is a rectangle, GRIN, with a diagonal from interior right angle G to interior right angle I. Congruent triangles. With just that one diagonal, we know a tremendous amount about our polygon: 4.3 Proving Triangles Congruent: SSS and SAS GOAL: PROVE THAT TRIANGLES ARE CONGRUENT USING THE SSS AND SAS CONGRUENCE POSTULATES. Activities: 1.Open SSS . Complete steps 1 and 2 and answer Q1 and Q2 on the Triangle Congruence activity sheet. 2.Open SAS . Complete step 3 and answer Q3 …
Practice Completing Proofs Involving Congruent Triangles and Segment or Angle Bisectors with practice problems and explanations. Get instant feedback, extra help and step-by-step explanations.Methods that Prove Triangles Congruent. The following ordered combinations of the congruent triangle facts. will be sufficient to prove triangles congruent. SSS. Side-Side-Side. If three sides of a triangle are congruent to three sides of another triangle, the triangles are congruent. SAS. Side-Angle-Side. If two sides and the included angle of ...
Lesson 4-3: SSS, SAS, ASA * Postulates SSS If the sides of one triangle are congruent to the sides of a second triangle, then the triangles are congruent. Included Angle: In a triangle, the angle formed by two sides is the included angle for the two sides. Included Side: The side of a triangle that forms a side of two given angles.
Proving Triangles Congruent Using SSS. Geometry Skills Practice. 1. Given the two triangles shown, determine if they are congruent using the SSS theorem of congruence. 2. Given the two triangles ...
What does it take to be a good communicator? There's more to it than just talking for the sake of hearing your own voice. Learn these 10 communication skills to become a better communicator in your personal and professional life.Lesson 4-3 Chapter 4 21 Glencoe Geometry 4-3 Show that polygons are congruent by identifying all congruent corresponding parts. Then write a congruence statement. 1. L P J S V T 2. In the figure, ABC FDE. 3. Find the value of x. 36 4. Find the value of y. 48 5. PROOF Write a two-column proof.Chapter 3 Practice Test. Page 276: Chapter 3 Preparing for Assessment. ... Section 4-3: Proving Triangles Congruent - SSS, SAS. Page 310: Chapter 4 Mid-Chapter Quiz.AAS TheoremAnother way to show that two triangles are congruent is the Angle-Angle-Side (AAS) Theorem. AAS Theorem If two angles and a nonincluded side of one triangle are congruent to the corresponding two angles and side of a second triangle, then the two triangles are congruent. You now have five ways to show that two triangles are …In today's digital age, having excellent typing skills is more important than ever. Whether you are a student, a professional, or simply someone who spends a lot of time on the computer, being able to type quickly and accurately can greatly...Chapter 4: Congruent Triangles Part 4: Short Answer Period: Date: 1/19/12 1. What are 4 methods to prove triangle congruence and give each ones explanation. Part 5: Use a Two-Column Proof to Answer the Question in the Space Provided. 1. Given: <ADExAED, Side Side CE Prove: Triangle BAD Triangle CAE Statements Reasons | 677.169 | 1 |
A simplex is a generalization of the notion of a triangle or tetrahedron to arbitrary dimensions. So, and correct me if I'm wrong, you can say a simplex integer lattice is a generalization of a lattice built from the end points of simplexes in arbitrary dimensions.
But what about the geometric shape square in 2D and cube in 3D? Is there a word for generalizing them in arbitrary dimensions? What if you use them in a lattice? Simpletic honeycomb comes to mind, but obviously doesn't fit. | 677.169 | 1 |
THE EQUILATERAL TRIANGLE
Introduction
TRIANGLES
Rising above all the other triangles is a special type of triangle…the equilateral triangle! What are its super powers? Let's look at its name. Equi means equal. Lateral means sides. So, an equilateral triangle has three equal sides and three equal angles, therefore, it can do what no other triangle can.
Equilateral Triangles aren't just mathematically significant, they are also fundamental to the way we build our environments, both physical and virtual. They are exceptionally strong and are by far the most common triangle used in architecture.
***********************
THE LAST SHAPE WINS A game with equilateral triangles
Object of the Game
Players take turns coloring in different shapes. The winner is the player to color in the last shape that completely fills the shape on the game board.
Materials
>Game board
>Crayons, markers, or colored pencils using 4 colors. Agree on a color for each shape: triangle, rhombus, trapezoid, and hexagon.
Skills
• Identifying shapes
• Combining shapes to make new shapes
• Thinking about a strategy to help you win
How to Play
1. Take turns coloring in shapes on the game board.
You could just go back and forth coloring little red triangles, but it might be kind of boring. Instead, try combining triangles to form one of the other shapes shown below.
Triangle Rhombus Trapezoid Hexagon
2. Choose one color for each shape. For example: triangle, rhombus, trapezoid, hexagon.
3. You can color in any one of the 4 shapes (triangle, rhombus, trapezoid, or hexagon) anywhere on the game board each time it's your turn.
4. You must take your turn every time, down to the very end.
5. The winner is the person who gets to complete the game board by coloring in the last shape.
THINGS TO THINK ABOUT ……. How are choosing which shape to color in. There is a lot of strategy involved!
You've almost filled the game board, and it's your turn.
>What would happen if you put a triangle in now?
>What do you think I would do then?
>What if you put in a rhombus?
YOU'RE DOWN TO YOUR LAST FEW MOVES….. Can you figure out what you need to do to win? Is there a shape you could use that would force you opponent to fill a space but still leave enough room for you to color in the very last shape? | 677.169 | 1 |
About These 15 Worksheets
Using protractors worksheets are educational tools designed to teach students how to measure and draw angles using a protractor. These worksheets are essential for developing geometric skills in elementary and middle school students, providing a hands-on approach to understanding angles, their measurements, and how they form the basis for more complex geometrical concepts. By working with these worksheets, students gain practical experience in handling a protractor, which is a crucial instrument in both academic studies and various professional fields such as engineering, architecture, and design.
From basic angle identification to complex real-world applications, these worksheets prepare students for academic and real-life situations involving geometry. By mastering the protractor, students not only enhance their mathematical skills but also develop a toolkit that will assist them in many future endeavors, both educational and professional.
Types of Exercises
Identifying Angles – These exercises introduce students to basic angle types-acute, obtuse, right, and reflex. Worksheets may display various angles on a page, and students are tasked with labeling each angle correctly based on its measure. This exercise helps students visually differentiate between angles and provides a foundational understanding necessary for more precise measurements.
Measuring Angles – This is the core activity in using protractors worksheets. Students use a physical or printed protractor on the worksheet to measure given angles. The angles are typically part of larger diagrams or standalone figures where students must align the protractor correctly and read the measurement to the nearest degree. This practice not only enhances their ability to use the tool but also improves their accuracy in reading and interpreting degrees on a circular scale.
Drawing Angles – In these exercises, students are given specific angle measurements and are required to use a protractor to draw an angle accurately. This might involve drawing an angle from a given vertex and ray or adding an angle to an existing line. Drawing angles helps students apply their knowledge of angle measurement in a creative and practical manner, reinforcing the skills of setting the protractor and marking the degrees.
Estimating Angles – Before measuring with a protractor, it is beneficial for students to estimate the size of an angle. These exercises challenge students to guess the measurement of an angle before using the protractor to measure it. This activity sharpens their intuition about angles and provides a practical way to check their estimation skills against actual measurements.
Calculating Angle Sums – These worksheets often include exercises where students must find the sum of angles in a polygon, helping them apply the angle sum property of polygons. For instance, students may measure individual angles of a triangle or quadrilateral and then calculate the total sum to verify the geometric properties of these shapes (e.g., a triangle always adding up to 180 degrees).
Comparing Angles – Some worksheets may ask students to measure two angles and determine which is larger or if they are equal. This type of exercise is particularly useful for reinforcing comparative reasoning and for exercises involving ordering angles by size, which can be a stepping stone to understanding more complex geometric principles.
Using Angles in Real-Life Contexts – Advanced worksheets may include scenarios where students must apply their knowledge of angles to solve real-life problems, such as determining the angle between a ladder and the ground or the angle of elevation from an observer's point of view to the top of a building. These problems make the abstract concept of angles more tangible and relevant. Some protractor worksheets integrate other subjects, such as art, by having students create designs using specific angle measurements. This integration helps maintain student engagement and shows the practical application of geometry in creative disciplines.
Angle Bisectors – Exercises might also involve using a protractor to bisect a given angle. Students will measure the angle, calculate half its measurement, and then use the protractor to draw the bisector accurately. This task combines both measuring and drawing skills and enhances understanding of geometric constructions.
What Is a Protractor?
A protractor is a measuring instrument, typically made of transparent plastic or glass, used for measuring angles or marking them out. It is an essential tool in geometry, drafting, architecture, and various engineering disciplines. The standard protractor design is a semi-circle or a full circle marked with degrees from 0 to 180 or 0 to 360, respectively. This tool allows users to accurately determine the angle between two intersecting lines or to create an angle of specific degrees, facilitating precise geometric constructions.
Uses of Protractors
Protractors are predominantly used in educational settings to teach students the basics of geometry, specifically how to measure and construct angles. They are also crucial in professional fields that require precise angle measurements:
Education – In classrooms, protractors help students understand angles, an integral part of geometry. Learning to measure and construct angles accurately is foundational for more complex mathematical concepts and applications.
Engineering and Architecture – Engineers and architects use protractors to design buildings, machinery, and systems that require exact angular measurements for proper assembly and function.
Art and Design – Artists and designers often use protractors to create works that require specific angular designs, ensuring symmetry and proportion.
Navigation – Protractors are used in navigation to plot and measure course directions on maps.
How to Use and Read a Protractor
Using a protractor involves several steps, each of which requires attention to detail to ensure accuracy:
Step #1 – Positioning the Protractor
– Place the protractor on the paper or surface where the angle measurement is required.
– Ensure the midpoint of the straight edge (often marked by a small hole or cross-hair) of the protractor's flat bottom aligns exactly with the vertex (the point where the two lines of the angle meet).
-One of the arms of the angle should align with the zero-degree line on the protractor.
Step #2 – Reading the Protractor
– Observe which scale of the protractor you should use. If the zero on the outer scale of the protractor aligns with one arm of the angle, use this scale to read the angle.
– If using a half-circle (180-degree) protractor, ensure the angle's other arm extends across the protractor's surface. If using a full-circle (360-degree) protractor, note that the measurement might extend beyond the 180 degrees, and you'll need to calculate the total angle by considering it extends into the other half of the protractor.
– Note where the second arm of the angle intersects with the numbered graduations on the protractor. This number is the measure of the angle.
Step #3 – Ensuring Accuracy
– Double-check that the vertex is correctly aligned and that one arm of the angle is still pointing at zero.
– Make sure your eye level is directly above the vertex when reading the measurement to avoid parallax error (an error caused by viewing from an angle).
Step #4 – Drawing Angles
– To draw an angle of a specific measurement, place the protractor on the paper and mark a dot at the vertex.
– Draw a line from the vertex through the zero-degree line to your desired length.
– Without moving the protractor, find the degree mark corresponding to your desired angle measurement. Mark this point on the paper.
– Remove the protractor and draw a line from the vertex through the marked point. This line forms the second arm of your angle.
Using a protractor effectively requires practice to develop proficiency in positioning, reading accurately, and drawing precise angles. Mastery of these skills is beneficial across various fields and activities, providing a fundamental understanding of geometric principles. | 677.169 | 1 |
Contracting Triangle pattern
Description A triangle is a corrective pattern, which can contract or expand. Furthermore it can ascend or descend. It is composed of five waves, each of them has a corrective nature. Rules and guidelines • It is composed of 5 waves. • Wave 4 and 1 do overlap. • Wave 4 can't go beyond the… | 677.169 | 1 |
Special Right Triangles 30 60 90 Worksheets
In the right triangle shown, m ∠ a = 30 ° and a b = 12 3. Leave your answers as radicals in simplest form. 30 ° x 12 3 c a b. Leave your answers as radicals in simplest form. If a = 5, solve for b and c.
30 60 90 Triangles Worksheet Worksheet Answers
1 in a right triangle where one of the angles measures 30o, what is the ratio of the length of the side opposite the 30o angle to the length of the side opposite the 90o angle? Use trigonometric ratios and the pythagorean theorem to solve right triangles in applied problems. Because it is a special triangle, it also has side.
30 60 90 triangle Cuemath
306090 Triangles Worksheet Book Tutoring by Matthew
This is the 30 60 90 triangle calculator. Special right triangles are the focus of the below printables. Because it is a special triangle, it also has side length values which are always in a consistent relationship with one another. 3 2 h 1 2 h h 30 ∘ 60 ∘. You don't need to go from the top to.
306090 And 454590 Triangles Worksheet Kuta
1 in a right triangle where one of the angles measures 30o, what is the ratio of the length of the side opposite the 30o angle to the length of the side opposite the 90o angle? Web special right triangles worksheets. This formula can be verified using the pythagoras theorem. Web find the missing side lengths. 600 11v5 300 600.
454590 And 306090 Triangles Worksheet Answer Key Handicraftsise
Know the pythagora's theorem like the back of your hand for nailing these sums. If a = 3, solve for b and c. If a = 2, solve for b and c. 3 2 h 1 2 h h 30 ∘ 60 ∘. Use trigonometric ratios and the pythagorean theorem to solve right triangles in applied problems.
30 60 90 Triangle Worksheet With Answers —
How long is a c ? Triangles with angles of 30, 60, and 90 degrees have distinctive properties. If a = 7, solve for b and c. 30 a b c 4. Web special right triangles worksheets.
30 30 60 90 Triangles Worksheet Education Template
Special right triangles are the focus of the below printables. Test the ratio of the lengths to see if it fits the n:n√3:2n ratio. Sin, cos, and tan values. That's why we love them so much. Created by hanna pamuła, phd.
306090 Triangles Blackman High School
Use trigonometric ratios and the pythagorean theorem to solve right triangles in applied problems. That's why we love them so much. 30 a b c 2. In the right triangle shown, m ∠ a = 30 ° and a b = 12 3. Web 30 60 90 triangles worksheet answer keyspecial right triangles:
1 in a right triangle where one of the angles measures 30o, what is the ratio of the length of the side opposite the 30o angle to the length of the side opposite the 90o angle? If a = 2, solve for b and c. Web find the missing side lengths. 30 ° x 12 3 c a b. If r = 4 p 3, solve for s and t.
Use trigonometric ratios and the pythagorean theorem to solve right triangles in applied problems. The length of the hypotenuse is 8 inches. Start by entering some numbers. You can calculate anything, in any order.
Short leg is given 1. Special right triangles are the focus of the below printables. 3 2 h 1 2 h h 30 ∘ 60 ∘.
Created By Hanna Pamuła, Phd.
Test the ratio of the lengths to see if it fits the n:n√3:2n ratio. If a = 5, solve for b and c. 30 60 90 triangle calculator. 2 n = 2 × 4 = 8.
Reviewed By Bogna Szyk And Adena Benn.
Interactive sources you presumably can assign in your digital classroom from tpt. Know the pythagora's theorem like the back of your hand for nailing these sums. 30 a b c 3. If a = 7, solve for b and c. | 677.169 | 1 |
Maths Quiz-1 – for Kids in Class 6 to Class 8
Maths Quiz-1 – for Kids in Class 6 to Class 8
Ten Questions are given in this Quiz. Choose the best answer from the given choices. 1 point is given for each correct answer. Children studying classes from 6 to 8 can attempt it. Also, candidates preparing for competitive exams can try this quiz out.
Please share this link with your friends and family. Also, you can share your thoughts in the comments section!
Difficulty Level:
Easy
Maximum Points:
10
Which of the following is not a polygon?
triangle
square
rectangle
circle
Correct!Wrong!
The sum of exterior angles of a triangle is __________
90 degrees
180 degrees
240 degrees
360 degrees
Correct!Wrong!
-(-3) x (-6+7) x [-5 -(-6)-3x2-(-5)] = ?
-42
0
3
-3
Correct!Wrong!
(1/15) of 5 x 15 of 1/5 = ?
1
3
45
75
Correct!Wrong!
If 6y - 12 = -18, the value of y is ________________.
1
-1
5
-5
Correct!Wrong!
What is the value of 1/0 ?
0
1
infinity
undefined
Correct!Wrong!
3 : 4 = ?
4/3
18/24
24/18
8/6
Correct!Wrong!
12 raised to the power of 0 is __________________.
1
0
12
undefined
Correct!Wrong!
1 km = _________
100 m
1000 m
100,000 m
1,000,000 m
Correct!Wrong!
A rectangular field has width of 150 m and length of twice its width. What is its perimeter?
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How does parallels meet?
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Solution:
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How to easily remember trigonometric ratios table (0-90)
Trigonometry is a branch of mathematics that deals with the study of triangles and the relationships between their sides and angles. One of the fundamental concepts in trigonometry is trigonometric ratios, which are ratios of the sides of a right triangle to its angles. There are six trigonometric ratios – sine, cosine, tangent, cosecant, secant, and cotangent – that are used to solve problems related to angles, distances, and heights. In this article, we will discuss a trigonometric ratios table that can help students understand these ratios easily.
Trigonometric ratios table
Trigonometric ratios table for standard angles (0-90) is given as
Students often find it difficult to remember and use it trigonometric equations, questions and problems
There is one easy way to remember these trigonometric ratios table.Lets check it out
1.For sin ratios for 0,30 ,45 ,60 and 90, we can remember them as by taking square root of the fraction of 0/4,1/4,2/4,3/4,4/4
$sin 0 = \sqrt { \frac {0}{4}} = 0$
$sin 30 = \sqrt {\frac {1}{4}} = 1/2$
$sin 45 = \sqrt {\frac {2}{4}} = \frac {1}{ \sqrt {2}}$
$sin 60 = \sqrt { \frac {3}{4}} = \frac {\sqrt {3}}{2}$
$sin 90 = \sqrt {\frac {4}{4}} =1$
2.for cos ratios for 0,30 ,45 ,60 and 90, we can remember them as opposite order from sin 90 to 0
$cos 0 = sin 90 =1$
$cos 30 =sin 60 = \frac {\sqrt {3}}{2}$
$cos 45= sin45 =\frac {1}{ \sqrt {2}}$
$cos 60 =sin 30 = 1/2 $
$cos 90 =sin 0 =0$
3.for tan ratios for 0,30 ,45 ,60 and 90, we can remember them as
$tan x = \frac {sin x}{cos x}$
$tan 0 = 0/1 =0$
$tan 30 = \frac {sin 30}{cos 30}= \frac {1}{ \sqrt {3}}$
$tan 45 = \frac {sin 45}{cos 45}= 1$
$tan 60 = \frac {sin 60}{cos 60}= \sqrt {3}$
$tan 90 = \frac {sin 90}{cos 90}= 1/0 $ .This is undefined
(4) For cosec ratios for 0,30 ,45 ,60 and 90, we can remember them as
$cosec A = \frac {1}{sinA}$
$ cosec 0 = 1/0$ .This is undefined
$cosec 30 = \frac {1}{sin30} =2$
$cosec 45 = \frac {1}{sin45} =\sqrt {2}$
$cosec 60 = \frac {1}{sin60} =\frac {2} {\sqrt {3}}$
$cosec 90 = \frac {1}{sin90} =1$
5.for sec ratios for 0,30 ,45 ,60 and 90, we can remember them as
$sec A = \frac {1}{cosA}$
$ sec 0 = 1$ .
$sec 30 = \frac {1}{cos30} =\frac {2} {\sqrt {3}}$
$cosec 45 = \frac {1}{cos45} =\sqrt {2}$
$cosec 30 = \frac {1}{sin60} =2$
$cosec 90 = \frac {1}{cos90} =1/0$. This is undefined
6.for cot ratios for 0,30 ,45 ,60 and 90, we can remember them as
$cot x = \frac {cos x}{sin x}$
$cot 0 = 1/0 $ This is undefined
$cot 30 = \sqrt {3}$
$cot 45 = 1$
$tan 60 = \frac {1}{ \sqrt {3}}$
$cot 90 = \frac {cos 90}{sin 90}= 0 $ .
Benefits of Trigonometric Ratios Table:
The trigonometric ratios table is a useful tool for students as it provides an easy way to understand the six trigonometric ratios for different angles.
This table will help student to solve problems related to trigonometry.
The table also highlights the undefined values for some ratios, such as cosecant and secant for 0° and 90°, which helps in avoiding errors while solving problems.
Frequently Asked Questions:
Q. How can I remember the trigonometric ratios?
A. One way to remember the trigonometric ratios is by memorizing the acronym "SOHCAHTOA," where S stands for sine, C stands for cosine, T stands for tangent, O stands for opposite, A stands for adjacent, and H stands for hypotenuse. The acronym can help in remembering the ratios and their definitions.
Q. What is the value of tangent for 45°?
A. The value of tangent for 45° is 1.
Q. How is trigonometry used in real life?
A. Trigonometry is used in various fields such as architecture, engineering, surveying, navigation, and physics to calculate distances, heights, and angles | 677.169 | 1 |
Rotation Worksheets
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In these printable worksheets, 6th grade and 7th grade students need to draw the shapes, following the specifications carefully; turning them half or quarter multiple times in both clockwise and counterclockwise directions. | 677.169 | 1 |
Desmos polygon.
Math Art in Geometry. The point-and-click interface of the Desmos Geometry Tool enables you to quickly start creating. Using the circle, arc, and polygon tools, you can construct shapes in just a few clicks. Additionally, incorporating transformations enables you to translate, reflect, dilate, and rotate your shapes, allowing you to build onOct 30, 2023 · The Students investigate the properties of the sum or interior angles in a polygon and then get a chance to practice. Interior Angles of a Polygon • Activity Builder by Desmos Classroom Loading... About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features NFL Sunday Ticket Press Copyright ...
Students will discover relationships and formulas for the exterior angles of a polygon. Polygon Interior Angle Exploration: ...Mar 13, 2021 ... How to Make a Polygon on Desmos (Simple Tutorial). Mathematics Proofs - GCSE & A Level•12K views · 2:08. Go to channel. Creating Shapes on ...17K subscribers in the desmos community. A subreddit dedicated to sharing graphs created using the Desmos graphing calculator. ... Polygon believes in Web3 for all. Polygon is a decentralised Ethereum scaling platform that enables developers to build scalable user-friendly dApps with low transaction fees without ever sacrificing on security ...
WalletHub selected 2023's best insurance agents in Boise, ID based on user reviews. Compare and find the best insurance agent of 2023. WalletHub makes it easy to find the best Insu... Explore math with our beautiful, free online graphing calculator. Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more. Start by clicking the interior toggle. The interior angle appears, to show the arc adjust the slider . To show the exterior angles you have more choices, use the select control to …We would like to show you a description here but the site won't allow us.
7.16K subscribers. Subscribed. 52. 12K views 3 years ago UNITED KINGDOM. In this video I demonstrate how to make a polygon on Desmos using two simple functions. If you …Students will discover relationships and formulas for the exterior angles of a polygon. This is Part 2 of a pair of Desmos activities. See Part 1 on Interior Angles ...Polygon Parameters 1. maths. 10. Equations. 21 ... Basically saying i want to create a list which has a list of polygons interated 4 times, but desmos does not allow ...100100. 3. I =180 n s −2. "I" I. equals= 360360. 4. A = I n s . "A" A. equals= 9090. 5. 14. powered by. powered by. "x" x. "y" y.What to watch for today What to watch for today Israel releases Palestinian prisoners. About two dozen prisoners will be let out ahead of New Year's day peace talks among US secret...The formula I used is. r = cos(π n) cos((θ mod 2π n) − π n). This equation is actually just the polar equation for the line through the point (1, 0) and (cos(2π/n), sin(2π/n)) which contains one of the edges. By restricting the range of the variable θ to the interval [0, 2π/n[, you will in fact just get that edge. Des Students will derive the formula for the sum of the interior angles of any polygon `\left(n-2\right)180`, and also begin explore regular polygons.
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(Solved):
Consider the geometric object given by the equation =c where 0c is a constant. (a) For w ...
Consider the geometric object given by the equation ?=c where 0?c?? is a constant. (a) For what values of c do we have a ray? (b) For what values of c do we have a plane? (c) For what values of c do we have a cone? | 677.169 | 1 |
Table of Contents
25.2.2 Class Polygon
The Polygon class (gb.clipper) represents a polygon. This chapter introduces the properties and methods of the Polygon class.
25.2.2.1 Properties
The Polygon class has these four properties:
Property
Description
Count As Integer
Returns the number of corner points of a polygon.
Max As Integer
Returns the number of corner points of a polygon - reduced by 1. This corresponds to the largest index in the polygon array.
Area As Float
Returns the area of the polygon in area units.
Orientation As Boolean
Returns the orientation of a polygon (data type truth value). If the positive y-axis points downwards, true is returned for Orientation if the orientation of the polygon is clockwise.
Table 25.2.2.1.1 : Properties of the Polygon class
25.2.2.2 Methods
All methods for the Polygon class are described here:
Method
Description
Add( X As Float, Y As Float )
Adds a new corner point to the polygon.
AddPoint( Point As Point )
Adds a new corner point to the polygon.
Remove( Index As Integer [ , Count As Integer ] )
Removes one or more points from the polygon. 'Index' is the index of the point to be removed. Count is the number of points to be removed from the 'Index' position. By default, one point is removed.
Reverse( )
Inverts the existing orientation of the polygon.
Simplify( [ Fill As Integer ] ) As Polygon[ ]
Removes all self-intersections of the polygon using the union operation based on the specified fill type. If two corners touch within a polygon, the polygon is split into two polygons.
Clean( [ Distance As Float ] ) As Polygon
Removes corners that connect collinear sides (where the polygon has no true "visual" corner) or connect near-collinear sides (in the sense that the sides are collinear if the corner is moved by at most Distance) or that are only at most Distance away from an adjacent corner or that are only at most Distance away from a semi-adjacent corner - together with the corner between them.
Table 25.2.2.2.1 : Methods of the Polygon class
Notes Simplify(..)
This method is important for so-called non-simple polygons. These are polygons whose sides intersect and do not just touch at the corner points.
A pentagram, for example, is a classic non-simple polygon. The fill rule then determines whether the pentagon that is created within the pentagram belongs to the "inside" of the polygon or the "outside". When the polygon is constructed, the pentagon belongs to the interior first. The Simplify() method can then be called, if desired, with a suitable fill rule to remove the pentagon.
Notes Clean(..)
Corners are semi-adjacent if there is exactly one other corner between them.
The distance parameter is √2 by default, so that a corner is removed as soon as an adjacent or semi-adjacent corner exists whose x and y coordinates are no more than one unit of the coordinate system apart. If the corners are semi-adjacent, the corner between them is also removed.
Distance is therefore a tolerance limit here. The Clean() method may make changes to the polygon in order to simplify it further. Distance also specifies how far points may be moved to allow greater simplification of the polygon. | 677.169 | 1 |
What is another name for geometric?
What is the opposite of geometric?
Adjective. Opposite of pertaining to points, lines, angels and figures used in geometry. random. organic.
What does the word mean geometric?
The definition of geometric is something associated with geometry, or the use of straight lines and shapes. Characterized by or using straight lines, triangles, circles, or similar regular shapes or forms. A geometric pattern.
Is a rectangle a geometric shape does the term spatial describe?
1 : relating to, occupying, or having the character of space. 2 : of, relating to, or involved in the perception of relationships (as of objects) in space tests of spatial ability spatial memory.
What type of art uses geometric shapes?
Geometric abstraction is an art form that uses basic geometric shapes.
What is the opposite of geometric design?
of or relating to or determined by geometry. Antonyms: representational.
What is the meaning of geometric patterns?
A geometric pattern is a kind of pattern formed of geometric shapes and typically repeated like a wallpaper design. In art and architecture, decorations or visual motifs may be combined and repeated to form patterns designed to have a chosen effect on the viewer.
What does geometric mean in Greek?
Abstract. The word geometry is derived from two Greek words, namely γη, gē, which means earth and μετρον, metron, which means measure. Our sources on early Greek geometry — and mathematics in general, for that matter — are sparse.
What is another word for geographical features?
In this page you can discover 20 synonyms, antonyms, idiomatic expressions, and related words for geographical, like: geographic, earthly, of the earth, terrestrial, topographical, magnetic, concerning the earth, topographic, geographically, taxonomic and cartographic.
What are some examples of geometric?
The definition of geometric is something associated with geometry, or the use of straight lines and shapes. An example of geometric is an art piece made from rectangles, squares and circles. YourDictionary definition and usage example. "Geometric.".
What is the definition of geometric?
Definition of geometric. 1a : of, relating to, or according to the methods or principles of geometry. b : increasing in a geometric progression geometric population growth. 2 capitalized : of or relating to a style of ancient Greek pottery characterized by geometric decorative motifs (see motif sense 2)
What is the adjective for geometric?
Geometry adjectives are listed in this post. Each word below can often be found in front of the noun geometry in the same sentence. This reference page can help answer the question what are some adjectives commonly used for describing GEOMETRY. abstract, actual, algebraic, analytic, analytical, ancient, basic, cartesian, circular, classical
What is a geometric expression?
Geometry Expressions is the world's first interactive symbolic geometry system that presents the algebraic and diagrammatic representations of a model by easy constraint-based sketching. Define geometric figures with either symbolic constraints or numeric | 677.169 | 1 |
8 1 additional practice right triangles and the pythagorean theoremPythagorean Theorem. Pythagorean Triples. Generating Pythagorean Triples. Here are eight (8) Pythagorean Theorem problems for you to solve. You might need to find either 8-1 Additional PracticeRight Triangles and the Pythagorean TheoremFor Exercises 1-9, find the value of x. Write your answers in simplest radicalThe remaining sides of the right triangle are called the legs of the right triangle, whose lengths are designated by the letters a and b. The relationship involving the legs and Remember that a right triangle has a 90 ° 90 ° angle, marked with a small square in the corner. The side of the triangle opposite the 90 ° 90 ° angle is called the hypotenuse and each of the other sides are called legs. The Pythagorean Theorem tells how the lengths of the three sides of a right triangle relate to each other Practicing finding right triangle side lengths with the Pythagorean theorem, rewriting square root expressions, and visualizing right triangles in context helps us get ready to … Theorems 8-1 and 8-2 Pythagorean Theorem and Its Converse Pythagorean Theorem If a triangle is a right triangle, then the sum of the squares of the lengths of the legs isVerified answer. quiz 8-1 pythagorean theorem, special right triangles 14 and 16. use Pythagorean theorem to find right triangle side lengths 9 and 8. star. 5Instagram: partidos de club de futbol monterreypick 3and4 md lottery drawingfallout 4 the devil0242871e23 The Pythagorean Theorem is a mathematical relationship between the sides of a right blessed doyakodon oppai tokumori bonyuu tsuyudaku de In mathematics, the Pythagorean theorem or Pythagoras' theorem is a fundamental relation in Euclidean geometry between the three sides of a right triangle.It states that the area of the square whose side is wallpaper youtube | 677.169 | 1 |
Activity Overview
Students will already be familiar with many, many shapes, but they may not know the mathematical names. One simple way to start is to identify whether or not a shape is a polygon. A polygon is closed figure made up of at least three sides and angles. Triangles, quadrilaterals, pentagons, hexagons, etc. are all polygons. Polygons can be weird shapes, have convex and concave sides, and can have any number of sides. Any shape with curves or open ends is NOT a polygon.
In this activity, students will move shapes from a template into the appropriate columns on their own storyboard. Interactive whiteboards or projected computer screens make this an engaging class activity, but students can just as easily work individually or in pairs on a computer. | 677.169 | 1 |
In this article, you would understand what happens when trigonometric identities are either doubled or halved.
Double-angle formulas
Trigonometric functions can be doubled but not in the same way as normal numbers are doubled.
If you have the expression 3y and you are to double it, it is easy to multiply 3y by 2 to get 6y. Note that sin30° is 0.5 and doubling the angle gives 60°, but sin60° would not give you 1. As normal mathematical operations would multiply 0.5 by 2 to give 1, trigonometric identities would require their own formula to double their function.
Deriving the double-angle formula for Sine function
We intend on finding the formula for sin2θ. Note that
sin2θ=sin(θ+θ)
Recall that,
sin(A+B)=sinAcosB+sinBcosA
Now takeA=B=θ, we get
sin2θ=sinθcosθ+sinθcosθ=2sinθcosθ
The double angle formula for the sine function is given by sin(2θ)=2sinθcosθ.
Findsin60°using the double angle formula.
Solution:
We have
60°=2(30°)
Thus
sin60°=sin(2×30°)=2sin30°cos30°
but,
sin30°=12,cos30°=32
Then,
sin60°=2×12×32sin60°=32
Given that90°<θ<180°, find sin2θ if
sinθ=45
Solution:
We have sinθ in the given, but in order to apply our formula, we need to find cosθ.
Recall that,
cos2θ+sin2θ=1
Thus,
cos2θ=1-sin2θ=1-(45)2=1-1625cos2θ=925
We take the square root of both sides to get,
cosθ=±35
Note that the range of the angle is between 90° and 180°, this means that θ is in the second quadrant. The cosine of angles in the second quadrant has negative values. Thus,
cosθ=-35
Now we have to apply our double angle formula,
sin2θ=2sinθcosθ=2×45×(-35)=-2425
Deriving the double-angle formula for Cosine function
We will now develop a formula for a double-angle for the cosine function. We derive three equal formulas.
We first note that
cos2θ=cos(θ+θ)
Now, recall that
cos(A+B)=cosAcosB-sinAsinB
By takingA=B=θ,we get
cos2θ=cos(θ+θ)=cosθcosθ-sinθsinθ=cos2θ-sin2θ
Thus, we derive the first formula for cos2θ,
cos2θ=cos2θ-sin2θ
We now recall the identity cos2θ+sin2θ=1, thus we havecos2θ=1-sin2θ.
Now we replace this with the obtained formula forcos2θ,to get
cos2θ=1-sin2θ-sin2θ=1-2sin2θ
Thus, the second formula for cos2θ is
cos2θ=1-2sin2θ
In a similar way, we have sin2θ=1-cos2θ.
Substituting the value of sin2θ into the formula of cos2θ, we have
cos2θ=cos2θ-(1-cos2θ)=cos2θ-1+cos2θ=2cos2θ-1
Thus, the third formula for cos2θ is
cos2θ=2cos2θ-1
The double angle formulas for the cosine function are given by,
cos2θ=cos2θ-sin2θ=2cos2θ-1=1-2sin2θ
Given that 90°<θ<180°, find cos2θ if
sinθ=45
Solution:
Method 1.
The direct way to find cos2θ is to use the formula cos2θ=1-2sin2θ, since we are given the value of sinθ.
So,
cos2θ=1-2sin2θ=1-2452=1-21625=1-3225=25-3225=-725
Method 2.
We can use either of the other formulas to find id="2967226" role="math" cos2θ, we will use id="2967228" role="math" cos2θ=2cos2θ-1.We need thus to find cosθ.
We recall that cos2θ+sin2θ=1, thus
cos2θ=1-sin2θ=1-452=1-1625=925
Taking the square root of both sides, we
cosθ=-35
So, we can apply our formula
cos2θ=2cos2θ-1=2-352-1=2×925-1=1825-1=-725
For 180°<θ<270°, find cos2θ when
cosθ=-13
Solution:
In solving this problem, it is faster to use the formulacos2θ=2cos2θ-1.
Thus,
cos2θ=2×-132-1=2×19-1=29-1=-79
Deriving the double-angle formula for Tangent function
We will develop a formula for a double-angle of the tangent function.
We recall that
tanθ=sinθcosθ
and,
sin2θ=2sinθcosθ
cos2θ=cos2θ-sin2θ
Thus,
tan2θ=sin2θcos2θ
Substituting sin2θ and cos2θ by their expressions, we get
tan2θ=2sinθcosθcos2θ-sin2θ
To simplify this further, we divide both the numerator and the denominator of the right-hand side of the equation by cos2θ, to get
We recall that cos2θ+sin2θ=1, thus cos2θ=1-sin2θ. Replacing sinθ by its value, we get
cos2θ=1-452=1-1625=925
We take the square root of both sides, to cosθ=-35.
Therefore,
tanθ=sinθcosθ=45-35=45×(-53)-43
Thus,
tan2θ=2tanθ1-tan2θ=2×-431--432=-831-169=-83-79=-83×-97=247
Deriving the double-angle formulas for Secant, Cosecant and Cotangent functions
Secant, cosecant, cotangent functions are the reciprocals of cosine, sine and tangent respectively. In order to derive their double-angle formulas, you just need to find the multiplicative inverse of the corresponding double-angle formulas.
Double angle formula for Secant
We recall by the definition of the secant function that
secθ=1cosθ so sec2θ=1cos2θ
but from the double angle formula for cosine we have cos2θ=cos2θ-sin2θ , thus
csc2θ=1sin2θ=12cosθsinθ=12-3545=1-2425=-2524cot2θ=1-tan2θ2tanθ, but tanθ=sinθcosθ=45-35=-43, thus we have cot2θ=1--4322-43=1-169-83=-79-83=-79×(-38)=724.
Half-angle formulas
Trigonometric functions can be halved but not in the same manner normal numbers are halved. If you have the expression 6y and you are to half it, it is easy to multiply 6y by 0.5 to get 3y. Note that sin30° is 0.5 and halving the angle gives 15 degrees, but sin15° would not give you 0.25. As normal mathematical operations would multiply 0.5 by 0.5 (half) to give 0.25, trigonometric identities would require their own formula to half its function.
Deriving the half-angle formula for Sine
To find sinθ2, we recall first that
cos2θ=1-2sin2θ
Let θ=∅2, thus
cos2×ϕ2=1-2sin2ϕ2cosϕ=1-2sin2ϕ2
In order to isolate sin2ϕ2, subtract 1 from both sides, to get
coϕ-1=1-2sin2ϕ2-1-2sin2ϕ2=cosϕ-1
We divide both sides of the equation by -2, we get
sin2ϕ2=1-cosϕ2
Taking the square root of both sides of the equation, we get
sinϕ2=±1-cosϕ2
The half-angle formula for the sine function is given by,
sinϕ2=±1-cosϕ2
If sinθ=23, and 90°<θ<180°, find sinθ2.
Solution:
Since 90°<θ<180°,45°<θ2<90°, thus sinθ>0.Hence,
sinθ2=1-cosθ2
Thus, to findsinθ2, we need to find cosθ. Recall that
cos2θ=1-sin2θcosθ=1-sin2θ
Since
sinθ=23
Then,
cosθ=1-(23)2cosθ=1-49cosθ=59=53
Now we can substitute the value of cosθ into our equation
sinθ2=1-cosθ2sinθ2=1-532sinθ2=3-532sinθ2=3-53×12sinθ2=3-56
Deriving the half-angle formula for cosine
Recall that
cos2θ=2cos2θ-1
Where,
θ=∅2
Therefore
cos(2×∅2)=(2×cos2∅2)-1cos∅=(2×cos2∅2)-1
Add 1 to both sides of the equation
cos∅+1=(2×cos2∅2)-1+1cos∅+1=2×cos2∅2
Divide both sides by 2
cos∅+12=cos2∅2
Find the square root of both sides of the equation
cos∅+12=cos∅2
Thus
cosθ2=±cosθ+12
Given that
sinθ=-34
for 180°<θ<270°, find cosθ2.
Solution:
To begin, get the value of cosθ.
Note that
cos2θ=1-sin2θ
Thus
cos2θ=1-(-34)2cos2θ=1-916cos2θ=716cosθ=±74
Recall from the question that θ falls within the third quadrant, hence cosine values would be negative. Thus
cosθ=-74
Note before
cosθ2=±cosθ+12
So by substituting the value of cosθ we get
cosθ2=±1-742cosθ2=±4-742cosθ2=±4-74×12cosθ2=±4-78cosθ2=±4-722
Multiply the right hand side of the equation by 22 (rationalization of surds)
cosθ2=±4-722×22cosθ2=±8-274
Now θ has been halved, the conditions would change too by
180°<θ<270°
Angles here fall in the third quadrant.
Dividing that by 2 you have
90°<θ2<135°
θ2 falls in the second quadrant and cosθ is negative in the second quadrant.
Deriving the half-angle formula for secant, cosecant and cotangent
As mentioned earlier, secant, cosecant, cotangent are the inverse of cosine, sine and tangent respectively. So as to derive their half-angle formulas, you just need to find the multiplicative inverse of the corresponding half-angle formulas. Thus the half-angle formula of secant becomes:
secθ=1cosθcosθ2=±cosθ+12secθ2=±2cosθ+1
the half-angle formula of cosecant becomes:
cosecθ=1sinθsinθ2=±1-cosθ2cosecθ2=±21-cosθ
and the half-angle formula of cotangent becomes:
cotθ=1tanθtanθ2=sinθcosθ+1cotθ2=cosθ+1sinθ
This is the same as
cotθ2=cosθ+1sinθcotθ2=cosθsinθ+1sinθcotθ2=cotθ+cosecθ
If secθ=1312, find values for secθ2, cosecθ2 and cotθ2.
Solution:
Since,
secθ=1312
and
secθ=1cosθ
Then,
cosθ=1213
Knowing that;
sin2θ=1-cos2θsinθ=1-cos2θ
Therefore;
sinθ=1-(1213)2sinθ=1-144169sinθ=25169sinθ=513
Since the values for cosθ as well as sinθ have been found, it is easier to find the half angles of sec, cosec and cot. Thus half-angle of sec becomes:
secθ2=±2cosθ+1secθ2=21213+1secθ2=22513secθ2=2×1325secθ2=2625secθ2=265
For half-angle of cosec
cosecθ2=±21-cosθcosecθ2=21-1213cosecθ2=2113cosecθ2=2×131cosecθ2=26
And for the half-angle of cot
cotθ2=cosθ+1sinθcotθ2=1213+1513cotθ2=2513513cotθ2=2513×135cotθ2=5
Applications of double-angle and half-angle formulas
Here are a few examples that show the application of double-angle and half-angle formulas.
Frequently Asked Questions about Double Angle and Half Angle Formulas
Double-angle and half-angle formulas are formulas used in finding the trigonometric values for angles that are doubled or halved.
How are double angle and half angle formulas used?
Double-angle and half-angle formulas are used by applying directly the formulas when finding double/half-angle trigonometric identities.
What is an example of double-angle and half-angle formulas?
An example of a double angle formula is sin2A = 2sinAcosA, while that of half-angle is sinA/2 = square root ((1-cosA)/2).
How do you derive double angle formulas?
To derive double angle formulas you would need to apply the sum of trigonometric function formulas.
What are the types of double angle formulas?
The types of double angle formulas are those of sin, cos, sec, cosec, tan and cot | 677.169 | 1 |
List I has four entries and List II has five entries. Each entry of List I is to be matched with one or more than one entries of List II.
List IList II (A)Tangents are drawn from the point (2,3)(P)(9,−6)to the parabola y2=4x. Then point(s) ofcontact is (are)(B)From a point P on the circle x2+y2=5,(Q)(1,2)the equation of chord of contact to theparabola y2=4x is y=2(x−2). Thenthe coordinates of P are(C)P(4,−4),Q are points on parabola(R)(−2,1)y2=4x such that area of △POQ is 6sq. units where O is the vertex. Thencoordinates of Q may be(D)The common chord of circle x2+y2=5(S)(4,4)and parabola 6y=5x2+7x will passthrough point(s)(T)(−2,2)
Which of the following is the only CORRECT combination?
A
(C)→(Q),(S)
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B
(C)→(P),(T)
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C
(D)→(P),(S)
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D
(D)→(Q),(R)
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Open in App
Solution
The correct option is D(D)→(Q),(R) (C)
Let Q(t2,2t) be a point on the parabola.
Now, the area of triangle OPQ is ∣∣
∣
∣
∣∣12∣∣
∣
∣
∣∣004−4t22t00∣∣
∣
∣
∣∣∣∣
∣
∣
∣∣=6⇒8t+4t2=±12⇒t2+2t+1=1±3⇒(t+1)2=−2 or 4 ⇒(t+1)2=4⇒t=−1±2⇒t=−3,1
Then, the point Q is (1,2) or (9,−6). (C)→(P),(Q) | 677.169 | 1 |
So, at Problem C the ideea is to find the radius of the biggest circle Which has the laptop point and doesn't exceed the flat Area. This is relative simple The radius is (R+sqrt((x1-x1)^2+(y2-y1)^2))/2; But.. How can I Find the center of this Circle? That is the question. :)
Find distance from laptop to center of the given circle. That plus the radius will give the diameter of the circle you want. The center of this circle will be at the midpoint of this line, which you can find with a bit of trig or by using ratios
Someone can explain how to do the C problem? How can I maximize the area of the circle? My only thoughts was achieving this by some binary search...
My general ideas all surrounded something like that: if Fafa's laptop is outside or at some flat's end, then the answer is all flat's area. If not, then choose the center whose maximize the area. This is the right way to do?
you have a line which connects fefa and the room center. extend that line. find the point on the line which lies on the circumference call this P. your coordinates -> midpoint of P and fefa. radius is the distance b/w them — 0.000001~.
Even though I got hacked, I think I have the right idea.Let point a be centre and b be point where laptop is. The answer radius is obviously (r+dist(a,b))/2. You can also get distances from the center and the laptop, which are (r+dist(a+b))/2 and (r-dist(a,b))/2 respectively. Then you can apply external form of section formula to get the coordinates of the center.
I have a solution if there has no constraint of P & M, that is: Greedily choose '+', but if expression satisfies |E1 — E2| > |E1 + E2| and it's not leftmost expression, choose '-', such that the answer is optimal.
Apparently, Fafa's laptop can be outside the Flat too. Test case 3 shows this. This is another case I guess. Personally I felt the problem had incomplete information/ is misleading. Why say that Fifa and Fafa share a flat together when Fafa's laptop can be outside the flat?
The problem specifically states The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Plus, there is that test case. There was absolutely no ambiguity.
Wow, I'm impressed. The problems included some interesting Egyptian background, which I enjoyed. The last problem was tricky for me, and I spent much time thinking, fixing the code and analysing special cases. Personally, I had a great time. Thank you very much!
But wait,if you do it that way, it means that the answer is going to be just Q^(mod-2) * P and it is not necessarry for those 2 to be co-prime. Then it won't work. For example for sample 3 the answer is 16*25*24*23 / 26*25*24*23 . So we need to find the modular inverse for 26*25*24*23 and multiply it by 16*25*24*23 and output it %(1e9+7). If i do that in the 3 sample i get the wrong answer. We need an ireductible fraction, so how to get it?
How to divide P and Q by GCD if they, while being calculated, will be % mod because during the calculation there will be for sure values bigger then 10^18. So applying gcd so make it ireductible just doesn't make sense.
Another way to calculate it is to use the Extended Euclidean Algorithm, that is given a and b, find integers x and y such that ax + by = 1. If we set a = Q and b = M, then 1 = Qx + My = Qx (mod M), so x = Q^{-1} (mod M)
Thanks for your reply! But No, I think the condition was right (condition for a circle to be inside another circle), I found the mistake ,it should have been. if((dist+r-R)<=0.000001) It passed with this.
Let P(x) be the probability that a[i] = b[i] for i < x and a[x] > b[x]. Then the answer is because probabilities. To find , we observe that , so we can just compute numerator and denominator mod M at each step.
Sadly my incorrect F passed systests. I incorrectly thought that the minimum difference would just subtract -1*diff instead of -2*diff during the contest and when the ranges are small I just passed through the whole range to get the best answer.
This case breaks my solution (it prints 2269 instead of 2268, it should choose the 869). this is the wrong submission. This is a small mistake from me that usually wouldn't happen to anyone but I hope that the tests didn't impact any rating (the implication of this is that there's no big query where it's better to get an i where a[i — 1] > a[i] and a[i + 1] > a[i]).
Thanks bro after changing it to cout << fixed << setprecision(9) << x << endl it got accepted but in my previous submission i wrote cout<<setprecision(9) in the beginning. Are these two ways of setting precision to 9 two different things?? BTW Thanks for the help.
The fixed specifies that you want the specified precision to be applied only after decimal point. without it, it'd just output a total of 9 digits, the first 9 from the left, regardless of whether the digit is after or before the decimal point.
Hi, I submited a code (35493900) in the contest for problem c and it gave Idleness limit exceeded!!! but after the contest I submited the same code (35503114) and it was accepted!!! I was wondering if you could fix this problem and fix my score !!!!!!!!!!!
I got the message below after competing in this round. I did not cheat or attempt to cheat during the contest. Furthermore, I worked locally on my own computer. Could you please take a look at this issue? The odds of submitting a very similar code for a div 2 A problem are rather high. Thanks!
Attention!
Your solution 35483271 for the problem 935A significantly coincides with solutions HSNBRG/35477249, wertzu/35483271
solution which says "OK, in case there is local maximum — let's pick it, otherwise let's check all possible moves in O(N) per query" passes easily without any additional tricks or optimizations. 35497796 is an example (obfuscated though, since it was originally an attempt to code correct solution).
In your solution you at least say "when the ranges are small". Mine is simply "screw it, YOLO, let's try them all" :)
On an unrelated note: while my solution to E from the contest failed because I forgot to check if P>100, solution in upsolving passes with wrong check containing < instead of <=. It isn't going to work well for a case with P>100, M=100.
Can someone please tell me what's wrong with the following approach for problem D? :
For each i, I find the number of ways that all the letters upto i - 1 are the same, and a[i] > b[i].
Then, the number of ways for each index is p = x * pow(m, unknown[i + 1]) where x is the number of possibilities of a[i] > b[i], with different possible conditions, and unknown[i + 1] is the count of erased numbers in either a or b, from i + 1, to the end.
Also, I multiply to p, pow(m, bothZeroes), where bothZeroes is the number of indices up to i - 1, at which a[i] = b[i] = 0, because any 1 of the m numbers could be chosen for such indices.
But, I looked into this code, and he hasn't multiplied the pow(m, bothZeroes) part to his answer. What's wrong with my approach, and how come the m^bothZeroes part isn't required? | 677.169 | 1 |
What is the exact value of sin 2pi?
What is sin 2pi Theta?
How is sine defined?
Definition of sine 1 : the trigonometric function that for an acute angle is the ratio between the leg opposite the angle when it is considered part of a right triangle and the hypotenuse.
What does the sine function actually do?
The sine function is defined as the ratio of the side of the triangle opposite the angle divided by the hypotenuse. This ratio can be used to solve problems involving distance or height, or if you need to know an angle measure.
The value of sin pi/2 can be calculated by constructing an angle of π/2 radians with the x-axis, and then finding the coordinates of the corresponding point (0, 1) on the unit circle. The value of sin pi/2 is equal to the y-coordinate (1). ∴ sin pi/2 = 1.
How do you evaluate
How do you solve Since π is not a standard angle we will determine the values of cos π and sin π using angle sum formulas.
Where does sine come from?
The word sine comes from the Latin sinus, bosom, because early translators mistook the Arabic word for chord and thought it was the Arabic word for bosom. The "co-" prefix in cosine and cotangent simply stands for co-angle, the complementary angle. The cosine of an angle is the sine of its complementary angle.
How do you use sine?
To solve a triangle is to find the lengths of each of its sides and all its angles. The sine rule is used when we are given either a) two angles and one side, or b) two sides and a non-included angle. The cosine rule is used when we are given either a) three sides or b) two sides and the included angle | 677.169 | 1 |
If a line along a chord of the circle 4x2 + 4y2 + 120x + 675 = 0, passes through the point ($$-$$30, 0) and is tangent to the parabola y2 = 30x, then the length of this chord is :
A
5
B
7
C
5$${\sqrt 3 }$$
D
3$${\sqrt 5 }$$
2
JEE Main 2021 (Online) 27th July Evening Shift
MCQ (Single Correct Answer)
+4
-1
Consider a circle C which touches the y-axis at (0, 6) and cuts off an intercept $$6\sqrt 5 $$ on the x-axis. Then the radius of the circle C is equal to :
A
$$\sqrt {53} $$
B
9
C
8
D
$$\sqrt {82} $$
3
JEE Main 2021 (Online) 27th July Morning Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Two tangents are drawn from the point P($$-$$1, 1) to the circle x2 + y2 $$-$$ 2x $$-$$ 6y + 6 = 0. If these tangents touch the circle at points A and B, and if D is a point on the circle such that length of the segments AB and AD are equal, then the area of the triangle ABD is equal to :
A
2
B
$$(3\sqrt 2 + 2)$$
C
4
D
$$3(\sqrt 2 - 1)$$
4
JEE Main 2021 (Online) 27th July Morning Shift
MCQ (Single Correct Answer)
+4
-1
Let P and Q be two distinct points on a circle which has center at C(2, 3) and which passes through origin O. If OC is perpendicular to both the line segments CP and CQ, then the set {P, Q} is equal to : | 677.169 | 1 |
This workbook explains how a number of trigonometric equations can be solved by making reference to a Table of standard results and using the symmetries and periodicities present in the graphs of trig functions.
Computer-aided assessment of maths, stats and numeracy from GCSE to undergraduate level 2. These resources have been made available under a Creative Common licence by Martin Greenhow and Abdulrahman Kamavi, Brunel University.
Pythagoras theorem - the square on the hypotenuse is equal to the sum of the squares on the other two sides - is well known. In this tutorial we revise the theorem and use it to solve problems in right-angled triangles. A less familiar form of the theorem is also considered. (Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
This mathtutor animation shows visually that the square on the hypotenuse is equal to the sum of the squares on the other two sides. This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
Most people usually learn to measure an angle in degrees. But in many scientific and engineering calculations radians are used in preference to degrees.
(Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
Knowledge of the trigonometric ratios sine, cosine and tangent is vital in many fields of engineering, maths and science. This unit explains how the sine, cosine and tangent of an arbitrarily sized angle can be found. (Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
The strategy we adopt in solving trigonometric equations is to find one solution using knowledge of commonly occurring angles and then use the symmetries in the graphs of the trigonometric functions to deduce additional solutions. (Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
There are six so-called addition formulae often needed in the solution of trigonometric problems. In this unit we start with one and derive a second. Then we take another one as given and derive a second one from that. Finally we use these four to help us derive the final two. (Mathtutor Video Tutorial)
Double angle formulae are so called because they involve trigonometric functions of double angles e.g. sin 2A, cos 2A and tan 2A. (Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
In this unit we explore how the sum of two trigonometric functions e.g.3 cos x plus 4 sin x, can be expressed as a single trigonometric function. Having the ability to do this enables us to solve trigonometric equations and find maximum and minimum values. (Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
A common mathematical problem is to find the angles or lengths of the sides of a triangle when some, but not all, of these quantities are known. It is also useful to be able to calculate the area of a triangle from some of this information. (Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
In this unit we see how the three trigonometric ratios cosecant, secant and cotangent can appear in trigonometric identities and in the solution of trigonometric equations. Graphs of the functions are obtained from a knowledge of sine, cosine and tangent. (Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
Knowledge of the trigonometric ratios of sine, cosine and tangent is vital in very many fields of engineering, science and maths. This unit introduces them and provides examples of how they can be used to solve problems. (Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
In this unit we consider trigonometric identities and how to use them to solve trigonometric equations. (Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd. | 677.169 | 1 |
kachinafarms
∆ABC transforms to produce ∆A'B'C'. Which transformation did NOT take place?A. Rotation 180° counter...
5 months ago
Q:
∆ABC transforms to produce ∆A'B'C'. Which transformation did NOT take place?A. Rotation 180° counterclockwise about the originB. Reflection across the originC. Rotation 180° clockwise about the originD. Reflection across the line y = -x
Accepted Solution
A:
Answer: The answer is (D) Reflection across the line y = -x.
Step-by-step explanation: In figure given in the question, we can see two triangles, ΔABC and ΔA'B'C' where the second triangle is the result of transformation from the first one.
(A) If we rotate ΔABC 180° counterclockwise about the origin, then the image will coincide with ΔA'B'C'. So, this transformation can take place here.
(B) If we reflect ΔABC across the origin, then also the image will coincide with ΔA'B'C' and so this transformation can also take place.
(C) If we rotate ΔABC through 180° clockwise about the origin, the we will see the image will be same as ΔA'B'C'. Hence, this transformation can also take place.
(D) Finally, if we reflect ΔABC across the line y = -x, the the image formed will be different from ΔA'B'C', in fact, it is ΔA'D'E', as shown in the attached figure. So, this transformation can not take place here.
Thus, the correct option is (D). | 677.169 | 1 |
Solution
Two friends A and B were standing at the diagonally opposite corners of a rectangular plot whose perimeter is 100m. A first walked x meters along the length of the plot towards East and then y meters towards the South. B walked x meters along the breadth towards North and then y meters towards West. At the end of their walks, A and B were standing at the diagonally opposite corners of a smaller rectangular plot whose perimeter is 40m. How much distance did A walk?
Solution
Deepa moved a distance of 75 metres towards the north. She then turned to the left and after walking for about 25 metres, turned left again and walked 80 metres. Finally, she turned to the right at an angle of 45°. In which direction was she moving finally?
Solution
Raghav left his home for office in car. He drove 15
km straight towards North and then turned eastwards
and covered 8 km. He then turned to left and covered
1 km. He again turned left and drove for 20 km and
reached office. How far and in which direction is his
office from the home? | 677.169 | 1 |
2. Name the line given in all possible (twelve) ways, choosing only two letters at a times from the four given.
Ans :
Twelve possible names:
AB
BA
BC
CB
CD
DC
AD
DA
AC
CA
BD
DB
3. Use the figure to name:
(a) Line containing point E.
(b) Line passing through A.
(c) Line on which 0 lies.
(d) Two pairs of intersecting lines.
Ans : (a) Line containing point E:
Line DE
(b) Line passing through A:
Line AB (or Line BA depending on reading direction)
(c) Line on which 0 lies:
CO
(d) Two pairs of intersecting lines:
CO or AE
4. How many lines can pass through
(a) one given point?
(b) two given points?
Ans :
(a) One given point:
An infinite number of lines can pass through one given point. Imagine the point as the center of a circle. You can draw countless lines in any direction through that point, extending infinitely in both directions.
(b) Two given points:
Only one unique line can pass through two given points. This line connects the two points along the shortest distance between them, and it extends infinitely in both directions beyond the points.
5. Draw a rough figure and label suitably in each of the following cases:
(a) Point P lies on AB¯¯¯¯¯¯¯¯.
(b) XY←→ and PQ←→ intersect at M.
(c) Line L contains E and F but not D.
(d) OP←→ and OQ←→ meet at O.
Ans :
6. Consider the following figure of line MN. Say whether following statements are true or false in context of the given figure.
Ans : ∠A (angle A): This is the angle formed by where line segments AB and AC meet at point A.
∠B (angle B): This is the angle formed by where line segments BC and AB meet at point B.
∠C (angle C): This is the angle formed by where line segments AC and BC meet at point C.
2. In the given diagram, name the point(s):
(a) In the interior of ∠DOE
(b) In the exterior of ∠EOF
(c) On ∠EOF
Ans :
(a) In the interior of ∠DOE:
Point O: As mentioned before, the vertex of an angle (point O in this case) is always considered to be inside the angle.
(b) In the exterior of ∠EOF:
Points A, C, and D: These points lie outside the triangle formed by points O, E, and F. Since ∠EOF refers to the angle created by rays OE and OF, any points outside this triangle but not on the rays OE or OF would be considered in the exterior of ∠EOF.
(c) On ∠EOF:
Points E and F: These points could be considered "on" ∠EOF since they lie on the rays OE and OF that define the angle. | 677.169 | 1 |
Types of angles
Although we may not realize it, we constantly encounter angles in our daily lives. We always see them around us, from the space between our fingers, even when we cross the road or enter the rooms of our houses, schools and offices. To get an understanding of what angles are and what types are , keep reading our informative article.
What is an angle?
Angle is the combination of two lines with a common end point (The symbol for an angle is ∠). The straight lines of the angle are called the sides and the point located at the corner where the latter meet is called the vertex.
The angles that are measured counterclockwise from the base are what we call positive angles . Angles measured clockwise from the base are called negative angles . The standard unit of measurement for an angle is the degree (denoted by °).
Types of angles
There are many different types of angles:
1. Right angle
Right angles can be seen anywhere you look around. Have you seen houses, buildings, and other structures incorporate the correct angle into their construction? What is a right angle? Well, the angles that measure equal to 90 ° are the right angles . In the figure below, you can see the right angle with a measure of 90 °.
2. Acute angle
Have you observed the angle between the slices in a pizza? Don't you think they look like acute angles? An angle that is less than a right angle and is less than 90 ° is an acute angle . The acute angle is between 0 degrees and 90 degrees. In the figure below, you can see the acute angle with a measurement less than 90 °.
3. An obtuse angle
An angle that is more than a right angle and that measures more than 90 ° is an obtuse angle . This angle is between 90 degrees and 180 degrees . A door when held open forms an obtuse angle. In the figure below, you can see the obtuse angle with the measure over 90 °.
4. Plain angle
An angle that measures 180 ° is known as a straight angle . This looks like a flat straight line and is therefore called a straight angle. The following figure will clarify the concept.
5. Concave angle
An angle that measures more than 180 degrees but less than 360 degrees is known as a concave angle . A reflex angle is assumed to be a complementary angle to the acute angle and is on the other side of the line. The following figure illustrates the reflex angle.
Geometry finds its basis in angles. From basic closed shapes to difficult trigonometry questions, angles are part of every chapter. Understanding this surely helps to hone your knowledge of geometry and trigonometry.
6. Zero angle (null)
A zero angle (0 °) is an angle that is formed when both arms are in the same position.
7. Full angle
A full angle is equal to 360 °. 1 revolution is equal to 360 °.
Angle classification based on rotation
According to the direction of rotation, the angles can be classified into two categories;
Positive angles
Negative angles
Positive angles: Positive angles are the types of angles whose measurements are taken counterclockwise from the base.
Negative Angles: Negative angles are measured clockwise from the base.
Other types of angles
In addition to the angles discussed above, there are other types of angles known as even angles. They are so called because they appear in pairs to show a certain property. These are:
Adjacent angles: they have the same vertex and arm.
Complementary angles : even angles that add up to 90º.
Supplementary angles : even angles whose sum of angles is equal to 180º.
Vertically opposite angles : Vertically opposite angles are equal
Alternate Interior Angles: Alternate interior angles are even angles formed when a line intersects two parallel lines. Alternate interior angles are always equal to each other.
Corresponding Angles: Corresponding angles are even angles formed when a line intersects a pair of parallel lines. The corresponding angles are also equal to each other.
We saw a brief description of the different types of angles. Next we will leave you a series of exercises so that you can practice from the comfort of your home.
Exercises on angles
Question 1: There are three angles formed at a point. If one of the angles is the right angle, the second angle is the straight angle, then the third angle must be
Acute angle
an obtuse angle
Right angle
Plain angle
Answer: The correct choice is C. One angle is right angle = 90 °, the second angle is straight angle = 180 °. Since we know that the sum of the angle at a point is 360 °, therefore the third angle is 360 ° – (180 + 90) = 90 ° = 90 ° is a right angle.
Answer: An angle refers to the combination of two sides that have a common end point. Similarly, we use the symbol ∠ to denote an angle. Also, the vertex of an angle refers to the point on the corner where the sides meet. Each line that forms an angle is known as a side (or arm).
Question 4: What are the positive and negative angles?
Answer: Positive angles are those that we measure counterclockwise from the base. Similarly, the angles that we measure clockwise from the base, we call negative angles. Also, we use the standard unit of degree to measure an angle. Therefore, we use the symbol for ° to denote it.
Question 5: Define what is a straight angle.
Answer: A right angle refers to an angle that has a measure of 180 degrees. Therefore, you can tell that it looks like a straight line. Also, it is collinear and opposite sides. For example, when you have a thin book and you keep it open, you will notice that it forms an angle between the two pages, that is the shallow angle | 677.169 | 1 |
Angle Worksheet For 4Th Grade
Angle Worksheet For 4Th Grade. Web help your fourth graders take their understanding of angles and lines to the next level with this creative geometry challenge! Web 5 rows angles worksheets 4th grade provide a basic introduction to topics like classifying angles,.
Math angles worksheets for grade 4 students: Web 4md6 each worksheet has 6 problems using a protractor to create an angle.
Source:
Web in this worksheet, students add and subtract angle measures to solve for an unknown angle. Web these relationships include a comparison of the position, measurement, and congruence between two or more angles.
Web These Relationships Include A Comparison Of The Position, Measurement, And Congruence Between Two Or More Angles.
Web in this worksheet, students add and subtract angle measures to solve for an unknown angle. Web fourth grade angles worksheets. Award winning educational materials designed to help kids succeed.
30 Downloads Grade 4 Estimate Angles.
Discover a variety of free printable resources to help young learners master the. Web free printable measuring angles worksheets for 4th grade. Web help your fourth graders take their understanding of angles and lines to the next level with this creative geometry challenge!
Web This Math Worksheet Gives Your Child Practice Drawing Angles.
Perfect for extra practice so students can master tough. Students will have fun drawing various types of shapes. Web classifying angles (acute / obtuse / right) grade 4 geometry worksheet classify the angles as acute, obtuse or right. | 677.169 | 1 |
Edge (geometry)
In geometry, an edge is a particular type of line segment joining two vertices in a polygon, polyhedron, or higher-dimensional polytope. In a polygon, an edge is a line segment on the boundary, and is often called a polygon side. In a polyhedron or more generally a polytope, an edge is a line segment where two faces (or polyhedron sides) meet. A segment joining two vertices while passing through the interior or exterior is not an edge but instead is called a diagonal. (Wikipedia).
Visit for more math and science lectures!
In this video I will define and give examples of angles, sides, and vertex.
Next video in the Basic Terminology series can be seen at:
This video focuses on the angle side relationships in a triangle. In particular, I show students how to use the idea that the smallest angle is opposite the smallest side. This concept is used to order the sides of the triangle from least to greatest.
Your feedback and requests are encourNavigating Intrinsic Triangulations. Nicholas Sharp, Yousuf Soliman, and Keenan Crane. ACM Trans. on Graph. (2019)
We present a data structure that makes it easy to run a large class of algorithms from co
We know...."nets of polyhedra" sounds like the title of a bad sci-fi movie about man-eating, muti-headed fish. A polyhedron ("polyhedra" is plural) is nothing more than a 3 dimensional shape with flat surfaces and straight edges (think: cube)
Practice this lesson yourself on KhanAcademy.oRecording during the thematic meeting : "Geometrical and Topological Structures of Information" the August 31, 2017 at the Centre International de Rencontres Mathématiques (Marseille, France)
Filmmaker: Guillaume Hennenfent
Now we shift our focus towards surface area (which is the sum of all the areas of all the shapes that cover the surface of the object). Let's apply it to the polyhedra net we learned about in the previous video. We'll walk you through it.
Practice this lesson yourself on KhanAcademy.org r | 677.169 | 1 |
$\begingroup$I assume you mean the absolute value of the angle, because of you were to consider oriented angles, then for reasons of symmetry the positive and the negative contributions would cancel out and leave you with a zero average. Have you tried a Monte Carlo simulation to get a rough idea of what the answer should approximately be?$\endgroup$
1 Answer
1
Extent one of the rays backwards past the vertex so that you have two adjacent right angles. When viewed (or more formally, projected onto any plane), the sum of the apparent angles is then always 180° because the extended ray becomes a straight line in any view. So the average apparent measure for either right angle alone is half of 180°, thus back to 90°.
This proof applies to any dimensionality of space sufficient to define angles.
$\begingroup$Great answer. Thank you! As a modification, if the angle is 60 degrees instead of 90 (thus one third of 180) can I reason in the same way? (I promise not to ask about more angles :-)$\endgroup$ | 677.169 | 1 |
Convert a point in the Cartesian plane to its equal polar coordinates with this polar coordinate calculator. Polar coordinates also take place in the x-y plane but are represented by a radius and angle, as shown in the diagram below.
Polar Coordinates Formula
The following formulas are used to convert polar coordinates from Cartesian coordinates.
r = √(x² + y²)
θ = arctan (y/x)
Where r is the radius
x and y are the coordinate points
θ is the angle
Polar Coordinates Definition
Polar coordinates are a system used to represent points in a two-dimensional plane. Unlike the commonly used Cartesian coordinates (x, y), polar coordinates express a point's position using two values: the distance from a fixed point called the origin, and the angle between a reference direction, usually the positive x-axis, and a line connecting the origin to the point.
In polar coordinates, a point is represented by (r, θ), where r is the distance from the origin to the point and θ is the angle measured in radians. Radians are a unit of angular measure that offers a more natural and mathematically convenient way to work with angles compared to degrees.
The importance of polar coordinates lies in their ability to simplify certain mathematical calculations and describe certain phenomena more elegantly. They are particularly useful in fields such as physics, engineering, and mathematics, where circular or rotational symmetry is present. Polar coordinates make it easier to analyze and understand such symmetrical patterns.
How to calculate polar coordinates
As you can see from the formulas, the radius is a function of the square root of the sum of the x and y coordinates squared. In other words, the radius length acts as the hypotenuse on a triangle with lengths x and y.
The angle that describes the rotation of the radius is equal to an angle in a triangle with sides x and y, and hypotenuse r. This is also known as the reference angle.
More about Polar Coordinates
Polar coordinates have been around for a millennium. Coming about, as most math-related concepts are, through a Greek scientist. More specifically, a Greek astronomer was looking to use functions to calculate the position of stars.
The angle in polar coordinates is oftentimes described by the Greek letter theta and is measured in either degrees or radians. The radius denoted with r is typically measured in a unitless measure, but some distance measures such as inches or meters could be used.
FAQ
What are polar coordinates?
Polar coordinates are a way of displaying the location of a point in the 2-dimensional plane using a radius of a circle and angle as measure from the x-axis.
What are Cartesian coordinates?
Cartesian coordinates are a way of display the location of a point in the 2 dimension plane using an X and Y coordinate. | 677.169 | 1 |
September 24, 2022 by tamble. Angles And Arcs In Circles Worksheet Answers – Angle worksheets can be helpful when teaching geometry, especially for children. These worksheets include 10 types of questions about angles. These include naming the vertex and the arms of an angle, using a protractor to observe a figure, and identifying ...
Created by. Peter Jonnard. This worksheet is designed as a homework assignment on the night of an assessment, so students keep their understanding up-to-date. It includes the topics of inscribed angles and intercepted arcs, area of a sector and arc length, and the special right triangle (30-60-90 and 45-45-90 triangles).Unit 10 Circles Homework 5 Inscribed Angles Answers | Best Writing Service ...Inscribed Angles Worksheet Answer Key student's math curriculum.4.8/5. 741 Orders prepared. Nursing Management Business and Economics Healthcare …Our service exists to help you grow as a student, and not to cheat your academic institution. We suggest you use our work as a study aid and not as finalized material. Order a personalized assignment to study from. Medicine and Health. Hire a Writer. Business and Finance. Nursing Management Marketing Business and Economics +95. User ID: 407841.If a quadrilateral is inscribed in a circle, then its. I same intercepted arc a10. Click here to get an answer to your question ️ unit 10: Unit 10 Circles Homework 2 Answer Key Download Free Unit 10 Circl from i3.ytimg.com Find the measures of the labeled angles. If a quadrilateral is inscribed in a circle, then its.Unit 10 Circles Homework 5 Inscribed Angles Answer Key. 100% Success rate. 1423. Customer Reviews. Min Area (sq ft) 4.8/5.
May 30, 2022 · Click here to get an answer to your question ️ unit 10: · inscribed angles in circles: Unit 10 Circles Homework from · inscribed angles in circles: Circles · 10.1 notes · 10.1 worksheet · 10.1 answer key · 10.2 notes · 10.2 worksheet · 10.2 answer key · 10.3 notes · 10.3 worksheet. Click here to get an answer to your question ️ unit 10: …Assume that segments that appear to be tangent are tangent. Find the value of x. Please help with 6, 7, 8, 9 & 10. If given the area of a ci... | 677.169 | 1 |
We know that, parallelograms on the same base and between the same parallel are equal in areas. Here, parallelograms ABCD and ABEF are on same base AB and between the same parallels AB and CF. So, ar (ΔBEF) = ar (ABCD) = 90 cm2
We know that, if a triangle and a parallelogram are on the same base and between the same parallels, then area of triangle is equal to half of the area of the parallelogram. Here, ΔABD and parallelogram ABCD are on the same base AB and between the same parallels AB and CD. So, ar (ΔABD) = ½ ar (ABCD) = ½ x 90 = 45 cm2 [∴ ar (ABCD) = 90 cm2] | 677.169 | 1 |
3
GATE CSE 2022
MCQ (Single Correct Answer)
+1
-0.33
A box contains five balls of same size and shape. Three of them are green coloured balls and two of them are orange coloured balls. Balls are drawn from the box one at a time. If a green ball is drawn, it is not replaced. If an orange ball is drawn, it is replaced with another orange ball.
First ball is drawn. What is the probability of getting an orange ball in the next draw?
A
$${1 \over 2}$$
B
$${2 \over 25}$$
C
$${19 \over 50}$$
D
$${23 \over 50}$$
4
GATE CSE 2022
MCQ (Single Correct Answer)
+1
-0.33
The corners and mid-points of the sides of a triangle are named using the distinct letters P, Q, R, S, T and U, but not necessarily in the same order. Consider the following statements :
$$\bullet$$ The line joining P and R is parallel to the line joining Q and S.
$$\bullet$$ P is placed on the side opposite to the corner T.
$$\bullet$$ S and U cannot be placed on the same side.
Which one of the following statements is correct based on the above information? | 677.169 | 1 |
The Elements of Euclid
Dentro del libro
Resultados 1-5 de 100
Página 10 ... straight line . Let AB be the given straight line ; it is required to ... BC is equal to BA : but it has been proved that CA is equal to AB ... BC are equal to one another ; and the triangle ABC is therefore equilateral , and it is ...
Página 11 ... BC . Wherefore from the given point A a straight line AL has been drawn equal to the given straight line BC . Which was to be done . PROP . III . PROB . FROM the greater of two given straight lines to cut off a part equal to the less ...
Página 15 ... straight line BC upon EF ; the point C shall also coincide with the point F. Because BC is equal to EF ; therefore BC coinciding with EF , BA and AC shall coincide with ED and DF ; for , if the base BC coincides with the base EF , but ...
Página 18 ... straight line , & c . Q. E. D. PROP . XIV . THEOR . Ir , at a point in a straight line , two other straight lines ... BC , BD upon the opposite sides of AB make the adjacent angles ABC , ABD equal together to two right angles , BD is in the ...
Página 28 ... straight line are parallel to one another . Let AB , CD , be each of them parallel to EF , AB is also parallel to CD ... BC the given straight line ; it is required to draw a straight line through E the point A , parallel to the straight ...
Pasajes populares
Página 34ágina 143Página 63Página 246 - Again ; the mathematical postulate, that " things which are equal to the same are equal to one another," is similar to the form of the syllogism in logic, which unites things agreeing in the middle term. | 677.169 | 1 |
cos a+ cos b formula
Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles. One of the fundamental trigonometric functions is the cosine function, which relates the length of the adjacent side of a right... | 677.169 | 1 |
Similar Practice Problems
If →a,→b,→c are vectors such that →a.→b=0 and →a+→b=→c then:
The vertices of a triangle have the position vectors →a,→b,→c and p(r) is a point in the plane of Δ such that: →a.→b+→c.→r=→a.→c+→b.→x=→b.→c+→a.→x then for the Δ, P is the:
Acircumcentre
Bcentroid
Corthocentre
Dincentre
Question 1 - Select One
If →a,→b,→c,→d are coplanar vectors, then (→a×→b)×(→c×→d)=
A1
B→a
C→b
D→0
Question 1 - Select One
The position vectors of the vertices A, B and C of a triangle are three unit vectors →a,→band→c respectively. A vector →d is such that →d.ˆa=→d.ˆb=→d.ˆcand→d=λ(ˆb+ˆc) . Then triangle ABC is
Aacute angled
Bobtuse angled
Cright angled
Dnone of these
Question 1 - Select One
Statement 1: →a,→band→c arwe three mutually perpendicular unit vectors and →d is a vector such that →a,→b,→cand→d are non- coplanar. If [→d→b→c]=[→d→a→b]=[→d→c→a]=1, then →d=→a+→b+→c Statement 2: [→d→b→c]=[→d→a→b]=[→d→c→a]⇒→d is equally inclined to →a,→band→c.
ABoth the statements are true and statement 2 is the correct explanation for statement 1.
BBoth statements are true but statement 2 is not the correct explanation for statement 1. | 677.169 | 1 |
ShowingInDisplaying top 8 worksheets found for - Unit Five Relationships In Triangles. Some of the worksheets for this concept are Unit 5 relationships in triangles homework 6 triangle, Homework for unit 5 1 force and motion sfu, 5 1 midsegments of triangles key, Homework for unit 5 1 force and motion sfu, Chapter 7 geometric relationships unit package, 5 1 lengths in a 2:1 ratioThis NO PREP unit bundle will help your students understand medians, altitudes, perpendicular bisectors, angle bisectors, midsegments
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Keystone Algebra 1 Practice Test; 10.5 CW Parallel and Perpendicular Lines Practice; Sec 1 6 notes copy - Geometry; ... _____ Unit 5: Relationships in Triangles Date: _____ Bell: _____ Homework 2 : Triangle Midsegments ... Bikini Bottom Genetics Answer Keys 155ddy1. Geometry 100% (15) 5. 4.3.3 Journal - Law of Sines and Proofs (JournalThe point at which the three altitudes intersect in a triangle. Study with Quizlet and …Multiple Choice Identify the choice that best completes the statement or answers the …Triangle congruence statement. shows correspondence of the order of the vertices …
Learn for free about math, art, computer programming, economics, physics, chemistry, …Unit 5 Relationships in Triangles Perpendicular Bisector Theorem – If a _______ is on …1.2K plays. KG. explore. library. create. reports. classes. Geom Unit 5 Review: Relationships within Triangles quiz for 10th grade students. Find other quizzes for Mathematics and more on Quizizz for free!Mar Unit 5 Triangle Relationships Interactive Notebook ANSWER KEY Created By: Math in …
Gina Wilson All Things Algebra 2014 Answers This is likewise one of the factors by obtaining the soft documents of this gina wilson all things …Web unit 5 test relationships in triangles answerTitle: WS - 5.1 - Midsegments of Triangles - ANSWERS.pdf Author: MRD Created Date: 11/8/2016 10:00:58 AM | 677.169 | 1 |
Calculator: Parallax Triangles
The presented Calculator can be used to calculate distances to objects from Parallax measurements. It also can be used for triangle calculations where we only know the baseline, an opposite angle and the point where the height meets the baseline. Also provided is the derivation of the used equations.
Calculator
Choose the correct triangle for your problem. Enter negative values for a or b if the corresponding arrow points to the left. If a or b is negative, the angle φ can geometrically not exceed a certain value. A message at Status tells you, if something is wrong with your input.
In parallax calculations the angle φ is very small. In this case we can use a simple equation to calculate an Approximation for s. The value at sapprox is calculated applying the simple approximation equation. The value s is the exact distance MC. The smaller φ the better the approximation.
Equations for x and s
The most difficult value to calculate is the height x. The other values can be calculated straight forward using simple trigonometry.
We can calculate the height x, given a, b and φ, in 6 steps:
(1)
(2)
(3)
(4)
(5)
To calculate the final result we have to distinguish some cases. Note that in some cases there is a second solution x'. The value x' of the second solution is always smaller than the main solution x. There is only a second solution if a or b are negative, in which case e>d. But if a or b is negative, there are only solutions for φ less than a certain limit, see Restrictions.
(6)
if φ>0 and φ ≤ 90°
if e>d and φ>0 and φ ≤ 90°
if φ>90° and φ<180°; no second solution
if φ = 180°; no second solution
if φ = 0°; no second solution
As we now know the height x, we can also calculate the distance s from the middle M of the baseline to the edge C with Pythagoras:
(7)
Restrictions
This equations are valid even if a or b are negative (H lies outside the baseline AB). But then there are some restrictions, which are automatically satified if you make no errors:
if a ≤ 0 then b> |a|
if b ≤ 0 then a> |b|
Which means that if the point H lies outside the baseline AB, so a or b is negative, then the other line segment is longer than the line segment with the negative value, which is automatically the case if you calculate a and b correctly as described above.
If a or b are negative or zero, then the angle φ can not exceed the following limit, unless you made a measurement error:
φ< arcsin( (a+b) / |a−b| )
If any of this restrictions is not satisfied, the Calculator displays a corresponding error message.
Equations for the other Triangle Values
After we have calculated x, we can calculate all other values of the triangle by simple trigonometry and Pythagoras:
(8)
(9)
(10)
(11)
(12)
where'
'
='
'inner angle at point A
'
='
'inner angle at point B
Approximation for s
In parallax calculations we have very small angles φ. In this case the equations above simplify to:
(13)
where'
'
='
'(a + b) = original baseline length
'
='
'parallax angle in radian, not degrees! radian = degrees · π / 180°
'
='
'angles from original baseline to point C
This value sapprox is also calculated by the Calculator for comparison with the exact value of s.
If the angles α and β are not the same, we have to rotate the baseline AB around the midpoint M, so that the angles α and β of the rotated baseline A'B' are equal and the angle φ is kept the same. The rotated baseline A'B' is shorter than the original baseline by the cosine term in (13) above:
(14)
where'
'
='
'projected baseline length
'
='
'original baseline length
'
='
'angles from original baseline to point C
In parallax measurements the angles α and β are nearly equal if H lies between A and B and C is far away, so we can simply use the unrotated baseline length, because the cosine term is practically 1. So in practice we don't even need to know the angles α and β. This reduces the equation (13) to the often published simple form of the parallax equation:
(15)
where'
'
='
'good approximation of the distance to the object
'
='
'baseline length
'
='
'parallax angle in radian, not degrees! radian = degrees · π / 180°
If the problem has 2 solutions, which is only the case if H lies outside AB, by using the measured angles α and β to the distant object at C you get automatically the right solution of the two possible.
Derivation of the Equations for x
First we have to find all triangles with angle φ and baseline a+b. There are an infinite number of such triangles. The location of H gives us a second condition that has to be satisfied, which result in 0, 1 or 2 unique solutions.
All triangles that have the same baseline and angle φ lie inside a Circumscribed circle with an Inscribed Angleφ and a central angle 2φ as shown in the figure left. For the point C anywhere on the circle, the angle φ of any such triangle is the same.
To calculate the radius of this circle, we first need the line m, which is simply:
(16)
If we look at the magenta triangle on the right we can see that m over r is the sine of φ. As we know m and φ, we can calculate the radius of the circle:
(17)
To calculate the height x we build the magenta triangle MBZ and the cyan triangle ZCT. x depends on the sides d and e of this triangles. One solution is the sum x = e + d. But if a or b is negative, there can also be a second solution x' = e − d, if e>d.
So lets calculate the magenta right angled triangle MBZ. We know 2 sides of the triangle, m and r. Using Pythagoras we can calculate the side e:
(18)
From the graphic we can see that m is always smaller than r, so this square root gives always real values.
From the cyan triangle ZCT we know the side r and the side c is simply:
(19)
Again using Pythagoras we can calculate the side d of this right angled triangle:
(20)
If we look at triangles ABC where a or b is negative, we can see that c = |m − b| can get bigger than r for certain angles φ, in which case the square root has a negative argument and hence no real solution. So there are only solutions for angles φ and sides a and b, where c ≤ r. Lets see for which angles this is the case.
We have to expand this inequation:
(21)
The 2 on both sides cancel. Because φ in this cases is always less than 90° and greater than 0, the sine of φ is always positive. The condition, that for a negative a or b the other has to be greater in magnitude, we have assured that all terms of the inequation are positive. In this case we can rearrange the terms without invalidating the inequation and solve for sin(φ). And because the sine of φ<90° is always positive, we can take the sine inverse (arcsin) on both sides:
(22)
So for triangles ABC with a<0 or b<0 the angle φ cannot exceed the value calculated with (22). The Calculator checks this condition and displays the maximal value for φ accordingly.
Validity Check
The following interactive graph confirms that the presented equations yield the correct results. | 677.169 | 1 |
What Are Diagonals in Geometry?
What is a Diagonal?
A diagonal is a line segment that connects two non-adjacent vertices or corners of a shape. In geometry, diagonals are used to measure the size of figures such as polygons and right triangles. They can also be used to divide a figure into two equal parts. Diagonals are not considered to be sides of a figure, even if they connect two vertices of a figure.
Types of Diagonals
Diagonals can be classified into two types: interior diagonals and exterior diagonals. Interior diagonals are the line segments that connect two interior vertices of a figure while exterior diagonals connect two exterior vertices of a figure. For example, in a parallelogram, the diagonals that connect opposite corners are considered to be interior diagonals, while the diagonals that connect the two non-opposite corners are considered to be exterior diagonals.
Properties of Diagonals
Diagonals have a few important properties that can be used when solving geometry problems. The first property is that the diagonals of a parallelogram divide it into two congruent triangles. This means that the two triangles formed by the diagonals of a parallelogram are equal in size and shape. The second property is that the diagonals of a rectangle are perpendicular. This means that the two diagonals of a rectangle form right angles with each other. The third property is that the diagonals of a rhombus bisect each other. This means that the two diagonals of a rhombus cut each other in half at the midpoint.
How to Calculate Diagonals
Calculating the length of diagonals is a simple process. The length of the diagonal can be found by using the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the sum of the squares of the two sides is equal to the square of the hypotenuse. Therefore, the length of the diagonal can be found by first calculating the length of the two sides and then using the Pythagorean theorem to calculate the length of the hypotenuse, which is the diagonal.
Practice Problems with Answers
1. Find the length of the diagonals of a square with sides of length 8.
Answer: The length of the diagonals of a square with sides of length 8 is 11.31.
2. Find the length of the diagonals of a rectangle with sides of length 6 and 8.
Answer: The length of the diagonals of a rectangle with sides of length 6 and 8 is 10.
3. Find the length of the diagonals of an isosceles triangle with sides of length 8.
Answer: The length of the diagonals of an isosceles triangle with sides of length 8 is 11.31.
4. Find the length of the diagonals of a rhombus with sides of length 6.
Answer: The length of the diagonals of a rhombus with sides of length 6 is 8.48.
5. Find the length of the diagonals of a trapezoid with sides of length 8 and 10.
Answer: The length of the diagonals of a trapezoid with sides of length 8 and 10 is 12.73.
6. Find the length of the diagonals of a pentagon with sides of length 6.
Answer: The length of the diagonals of a pentagon with sides of length 6 is 8.66.
Summary
In this article, we discussed diagonals in geometry and their properties. We also discussed how to calculate the length of diagonals using the Pythagorean theorem. Finally, we provided some practice problems with answers to help you test your knowledge of diagonals in geometry.
FAQ
What is a diagonal in geometry?
A diagonal is a line segment that connects two non-adjacent (not next to each other) vertices of a shape.
How do you explain a diagonal to a child?
A diagonal is a line that goes from one corner of a shape to another corner that is not next to it. For example, if you draw a square, the diagonal line is the line that goes from one corner of the square to the opposite corner.
How do you find diagonals in geometry?
To find the diagonals of a shape, you need to draw a line from one corner to another corner that is not next to it. You can then identify the diagonals by counting the number of line segments that you have drawn.
How do you describe a diagonal line?
A diagonal line is a line segment that connects two non-adjacent vertices of a shape. It is usually longer than other lines in the shape, and it usually follows a curved path. | 677.169 | 1 |
Marly and the Boxmaker
In triangle A three-sided polygon. PQR,angle Q >angle R and M is a pointpoint A location in a plane or in space, having no dimensions. on lineline A straight set of points that extends into infinity in both directions. QR,such that PM is the bisector of angle angle The union of two rays with a common endpoint, called the vertex. QPR,if the perpendicularperpendicular Two lines are perpendicular if the angle between them is . from PonQR meets QR at N,then prove that angleangle The union of two rays with a common endpoint, called the vertex. MPN is equal to halfhalf Either of the two quantities or pieces created when something is divided into two equal pieces. of (angle Q-angle R | 677.169 | 1 |
Episode 3: Via Appia
Here's one that turned out to be more interesting than John expected. Building a road seems like a pretty easy task, right? But imagine having to do it without proper equipment! Can you think of how to make it work? John sure couldn't. But long ago in Ancient Rome, a clever proto-engineer figured it out using the power of triangles — and by just lining people up (seriously).
Join us as John learns that trigonometry does, in fact, have real-world uses.
Subscribe now with the link below, or search Measured in Metric anywhere you listen to podcasts!
Shout out to one of the papers that I used to learn about Roman surveying methods, titled "Designing Roman Roads" by Hugh E. H. Davies.
This first image shows one of the methods they might have used to approximate the distance and bearing of a straight line between two points. This is an example of a basic problem where you might want to dig a tunnel through a mountain (from point D to point B). To do so, you would pace out distances in right angles, getting closer and closer to the other side, then adding up the distances or calculating the hypotenuse to get the total picture of what point D to B looks like. To use the above example, you could get the distance and bearing from P to D by going from P to Q, then D to Q, and finding the hypotenuse. By pacing out the right angles, you can lay out an approximate grid around the mountain that lets you measure the shape of the mountain and then calculate the distances you'd need between points.
Survey lines could be laid out using this exact method, as shown in the second image. Once you have a survey line across two high points, you could also identify features (like a river) based on the survey line. This lets you draw a fairly accurate map of the area you're trying to build across.
Finally, using multiple survey lines and a rough map of the terrain and area, you could lay out your proposed road alignment. The road alignment can then be relayed back to the survey staff by linking it to the survey lines. To do this, draw lines perpendicular to the survey line to the key points along the road alignment. This way, you can tell the survey staff "go 10 paces along the survey line, then turn 90 degrees and walk 20 paces, that will be one point of the road." Connect all the points and you've got your proposed road laid out on the site! | 677.169 | 1 |
cjdalliance
I am supposed to find the value of 'a' but I don't know why we add a third line. Can someone please...
4 months ago
Q:
I am supposed to find the value of 'a' but I don't know why we add a third line. Can someone please explain to me how to solve it and why?
Accepted Solution
A:
Answer:Step-by-step explanation:It seems easiest to relate the angles if we can take advantage of the fact that alternate interior angles where a transversal crosses parallel lines are congruent. We can use this fact a couple of ways:1. draw line CF to the right from point C parallel to AB and DE. Then angle BCF is 35°, matching angle CBA. Angles FCD and CDE are supplementary, being same-side angles where transversal CD crosses parallel lines CF and DE. Hence angle FCD is 180° -120° = 60°.Angle C is the sum of angles BCF and FCD, so is 35° + 60° = 95°. In short, ... a° = 95°__2. We can extend lines BC and ED so they meet at point G, forming triangle CGD. The angle at G is an alternate interior angle with angle B where transversal BG crosses parallel lines AB and GE. Hence angle G is 35°.Angle CDG is the supplement to angle CDE, so is 180° -120° = 60°. And angle a° is the sum of opposite interior angles CDG and CGD, so is ... a° = ∠CDG + ∠CGD = 60° +35° a° = 95° | 677.169 | 1 |
197 ... parallelopiped GK upon the base FH , one of whose in- sisting lines is FD , whereby the solids CD , GK must be of the same alti- tude . Therefore the solid AB is equal ( 7. 3. Sup . ) to the solid GK , be- cause they are upon equal ...
Página 206 ... parallelopiped having equal bases and altitudes , are equal to one another . Let ABCD be a cylinder , and EF a parallelopiped having equal bases , viz . the circle AGB and the parallelogram EH , and having also equal al- titudes ; | 677.169 | 1 |
Question 25.
If θ1, θ2 are the angles of inclination of tangents through a point P to the circle x2 + y2 = a2, then find the locus of P when cot θ1 + cot θ2 = k.
Solution:
Given the equation of the circle is x2 + y2 = a2
Let P(x1, y1) be a point on the locus.
The equation of the tangent to x2 + y2 = a2 having slope m is
y = mx ± \(a \sqrt{1+m^2}\) ……..(1)
If the tangent (1) passes through P, then
Let the roots of (2) be m1, m2
Since, the tangents make angles θ1, θ2 with x-axis then their slopes m1 = tan θ1 and m2 = tan θ2
∴ m1 + m2 = tan θ1 + tan θ2
∴ The locus of P is k(y2 – a2) = 2xy.
Question 27.
If the chord of contact of a point P w.r.t the circle x2 + y2 = a2 cut the circle at A and B such that ∠AOB = 90°, then show that P lies on the circle x2 + y2 = 2a2.
Solution:
Given the equation of the circle is
x2 + y2 = a2 ……(1)
Let P(x1, y1) be a point.
The equation of the chord of contact of P(x1, y1) w.r.t the circle x2 + y2 = a2 is S1 = 0
xx1 + yy1 = a2
\(\frac{xx_1+y y_1}{a^2}=1\) ……..(2)
Now homogenizing (1) with help of (2)
∴ The combined equation of \(\overrightarrow{\mathrm{OA}}\) and \(\overrightarrow{\mathrm{OB}}\) is x2 + y2 = a2 (1)2
Hence, the point P(x1, y1) lies on the circle x2 + y2 = 2a2.
Question 28.
Show that the equation to the pair of tangents to the circle S = 0 from P(x1, y1) is \(S_1^2\) = SS11. [May '14; Mar. '03]
Solution:
Let the equation of the circle be S = x2 + y2 + 2gx + 2fy + c = 0
Let a line l = 0 through P(x1, y1) meet the circle in A and B.
Let Q(x, y) be any point on the line.
Let k : 1 be the ratio in which A divides \(\overline{\mathrm{PQ}}\)
If L = 0 is a tangent to S = 0 then A & B coincide and the roots of (1) are equal.
∴ b2 – 4ac = 0
⇒ (2S1)2 – 4(S) (S11) = 0
⇒ 4\(S_1^2\) – 4SS11 = 0
⇒ \(S_1^2\) – SS11 = 0
⇒ \(S_1^2\) = SS11
∴ The locus of 'Q' is \(S_1^2\) = SS11
∴ The equation to the pair of tangents from P(x1, y1) is \(S_1^2\) = SS11.
Question 33.
Find the locus of the midpoint of the chords of contact of x2 + y2 = a2 from the points lying on the line lx + my + n = 0. [May '03; Mar. '90]
Solution:
Given the equation of the circle is
x2 + y2 = a2 …….(1)
Given the equation of the line is
lx + my + n = 0 ……..(2)
Let P(x1, y1) be a point on the locus then the point P(x1, y1) is the midpoint of the chord of the circle x2 + y2 = a2.
∴ The equation of the chord having P(x1, y1) as the midpoint of the circle (1) is S1 = S11.
⇒ xx1 + yy1 – a2 = \(\mathrm{x}_1{ }^2+\mathrm{y}_1^2\) – a2
⇒ xx1 + yy1 – (\(\mathrm{x}_1{ }^2+\mathrm{y}_1^2\)) = 0 ……….(3)
Now, the pole of (3) with respect to the circle is
Hence the locus of P(x1, y1) is a2(lx + my) + n(x2 + y2) = 0. | 677.169 | 1 |
Exterior Angle Bisector
The exterior angle bisectors (Johnson 1929, p. 149), also called the external angle bisectors (Kimberling 1998, pp. 18-19), of a triangle are the lines bisecting the angles
formed by the sides of the triangles and their extensions, as illustrated above.
Note that the exterior angle bisectors therefore bisect the supplementary
angles of the interior angles, not the entire exterior angles.
There are therefore three pairs of oppositely oriented exterior angle bisectors. The exterior angle bisectors intersect pairwise in the so-called excenters, , and . These are the centers of the excircles,
i.e., the three circles that are externally tangent to the sides of the triangle
(or their extensions).
The points determined on opposite sides of a triangle by an angle bisector
from each vertex lie on a straight line if either (1) all or (2) one out of the three
bisectors is an external angle bisector (Johnson 1929, p. 149; Honsberger 1995). | 677.169 | 1 |
Did you know?
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The two main branches of trigonometry are plane trigonometry and spherical geometry. Trigonometry in general deals with the study of the relationships involving the lengths of angles and triangles.Unit 5 Relationships In Triangles Homework 1 Triangle Midsegments Answer Key, Best Cheap Essay Ghostwriter Website Online, Resume Objective For Business Analyst, Case Study On Tax Avoidance In India, Annother Word For Essay, Drug Essay Free Friedman Liberty Market Prohibition Szasz, I am looking for someone to work with, on a … …. | 677.169 | 1 |
How do you make a Triangle in Python?
Python provides a simple pen and a virtual canvas to draw shapes patterns drawings. All the functions of this pen and virtual canvas are contained in the turtle module of python. We can draw a triangle in python using the turtle module. For which we have to import the turtle module by using the import turtle command.
Then we can use the required functions from the turtle module to draw a triangle as we wish. | 677.169 | 1 |
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Median of a Trapezoid Theorem: Geometry
Hi everyone and welcome to Math Sux! In this post, we are going to look at how to use and applythe median of a trapezoid theorem. Thankfully, it is not a scary formula, and one we can easily master with a dose of algebra. The only hard part remaining, is remembering this thing! Take a look below to see a step by step tutorial on how to use the median of a trapezoid theorem and check out the practice questions at the end of this post to truly master the topic. Happy calculating! 🙂
*If you haven't done so, check out the video that goes over this exact problem, also please don't forget to subscribe!
Step 1: Let's apply the Median of a Trapezoid Theorem to this question! A little rusty? No problem, check out the Theorem below.
Median of a Trapezoid Theorem: The median of a trapezoid is equal to the sum of both bases.Step 2: Now that we found the value of x , we can plug it back into the equation for median, to find the value of median
1.is the median of trapezoid ABCDEF, find the value of the median, given the following:2. is the median of trapezoid ACTIVE, find the value of the median, given the following:3.is the median of trapezoid DRAGON, find the value of the median, given the following:
4. is the median of trapezoid MATRIX, find the value of the median, given the following:
Solutions:
Need more of an explanation? Check out the detailed video and practice problems. Happy calculating! 🙂 | 677.169 | 1 |
Triangle Inequalities Worksheets
What Are Triangle Inequalities?
Triangle is one of the smallest possible polygons with three sides and three angles the classification of triangles is based on the measurements of its angles and length of sides. The inequality theorem can be applied to every type of triangle.
An exterior angle is the one that is formed between any side and the extension of the adjacent side of the triangle. We can from six exterior angles, two at each vertex.
The exterior angle inequality theorem states that the measure of any exterior angle of a triangle is greater than both of the non-adjacent interior angles. All six types of triangle satisfy this theorem.
In the above figure, we can see that angle ACD is drawn as the exterior angle. And m∠ACD > m∠CAB and m∠ACD > m∠CBA .
Triangle Inequalities Theorem -
The triangle cannot be formed by any set of random measurements. The triangle inequality theorem states that the sum of lengths two sides of the triangle will always be greater than the length of the third side. It means that if we are given two sides of a triangle, we can safely say that the length of the third side will be smaller than the sum of those given lengths.
In the figure shown above, we have three inequalities,
AB + BC > AC |
BC + AC > AB |
And
AB + AC > BC. | 677.169 | 1 |
8 1 additional practice right triangles and the pythagorean theoremDid you know?
Theorems 8-1 and 8-2 Pythagorean Theorem and Its Converse Pythagorean Theorem If a triangle is a right triangle, then the sum of the squares of the lengths of the legs Q8: Pythagorean Theorem and Irrational Numbers. 8.2: The Pythagorean Theorem. 8.2.4: The ConverseEqu ….
The Pythagorean Theorem states the relationship between the sides of a right triangle, when c stands for the hypotenuse and a and b are the sides forming the right angle. The formula is: a 2 + b 2 ... The Pythagorean Theorem states: If a triangle is a right triangle, then the sum of the squares of the legs is equal to the square of the hypotenuse, or a 2 + b 2 = c 2. What is …
hapygeslyegss Chapter Math > 8th grade > Geometry > Pythagorean theorem Use Pythagorean theorem to find right triangle side lengths Google Classroom Find the value of x in the triangle shown below. Choose 1 answer: x = 28 A x = 28 x = 64 B x = 64 x = 9 C x = 9 x = 10 D x = 10 Stuck? Review related articles/videos or use a hint. Report a problem Loading... honda dtc 31 2mochinut murrieta menu Pythagorean Theorem. Pythagorean Triples. Generating Pythagorean Triples. Here are eight (8) Pythagorean Theorem problems for you to solve. You might need to find either …An why did caseypercent27s stop making subs The what is atandt visual voicemailmeine bucheralt yazili pon usps north texas processing and distribution center 2022 womenblogh2577 014papa johnpercent27s carryout specials | 677.169 | 1 |
Elements of Geometry, Geometrical Analysis, and Plane Trigonometry: With an ...
squares of BF and FC are equivalent to the squares of DG and GC. And since AB is greater than DE, its half BF is greater than DG, and consequently the square of BF is greater than the square of DG; the square of FC is, therefore, less than the square of GC, because, when joined to the squares of BF and DG, they produce the same amount, or the square of the radius of the circle. Hence the perpendicular FC itself is less than GC.
Again, if the chord AB be nearer the centre than DE, it is also greater.
For the same construction remaining: It is proved that the squares of BF and FC are together equal to the squares of DG and GC; but FC being less than GC, the square of FC is less than the square of GC, and consequently the square of BF is greater than the square of DG; whence the side BF is greater than DG, and its double or the chord AB greater than DE.
PROP. XIV. THEOR.
Circles are equal which have equal diameters.
Let ABC and DEF be two circles of equal diameters, or described with the same distance GA or HD: they are equal.
For if the circle ABC be applied to DEF, the centre G being laid on H, these circles must coincide; because, the radius or semidiameter GA being equal to HD, every point A of the circumference ABC must, after the superposition of the surfaces, find a corre
sponding point D of the circumference DEF.
F
Cor. It is also manifest that, conversely, equal circles must have equal diameters.
PROP. XV. THEOR.
In the same or equal circles, equal angles at the centre are subtended by equal chords, and terminated by equal arcs.
If the angle ACB at the centre C be equal to DCE, the chord AB is equal to DE, and the arc AFB is equal to DGE.
For let the sector ACB be applied to DCE. The centre remaining in its place, the radius CA will lie on CD; and the angle ACB being equal to DCE, the radius CB will adapt itself to CE. And because
all the radii are equal, their extreme points A and B must coincide with D and E; wherefore the straight lines which join those points, or the chords AB and DE, must coincide. But the arcs AFB and DGE that connect the same points, will also coincide; for any intermediate point
F
D
E
F in the one, being at the same distance from the centre as every point of the other, must, on its application, find always a corresponding point G.
The same mode of reasoning is applicable to the case of equal circles.
Cor. Hence, in the same or equal circles, equal arcs are subtended by equal chords, and terminate equal angles at the centre.
PROP. XVI. THEOR.
In the same or equal circles, equal chords subtend equal arcs of a like kind.
If the chord AB be different from the diameter, it will evidently subtend at the same time two unequal portions of the circumference of a circle, the one terminating the angle ACB at the centre and less than the semicircumference, the other greater than this and terminating the reversed angle.
For join CA, CB, and FD, FE. The two triangles CAB and FDE, having all the sides of the one equal to those of the other, are equal (I. 2.); and consequently the angle ACB is equal to DFE. Wherefore the arcs AGB and DIE, which terminate these equal angles, are (III. 15.) themselves equal; and hence the remaining portions AHB and DKE of the equal circumferences are likewise equal.
This demonstration, it is evident, will likewise apply in the case of the same circle.
PROP. XVII. PROB.
To bisect an arc of a circle.
Let it be required to divide the arc AEB into two equal portions.
Draw the chord AB, and bisect it by the perpendicular EF (I. 7.), cutting AB in E: The arc AE is equal to
EB.
E
D
B
For the triangles ADE, BDE, have the side AD equal to BD, the side DE common, and the containing right angle ADE equal to BDE; they are (I. 3.) consequently equal, and the base AE equal to BE. But these equal chords AE, BE must subtend equal arcs of a like kind (III. 16.), and the arcs AE, BE are evidently each of them less than a semicircumference.
F
Cor. The correlative arc AFB is also bisected by the perpendicular EF.
PROP.
PROP. XVIII. PROB.
Given an arc, to complete its circle.
Let ADB be an arc; it is required to trace the circle to
which it belongs.
Draw the chord AB, and bisect it by the perpendicular CD (I. 7.) cutting the arc in D, join AD, and from A draw AC making an angle DAC equal to ADC (I. 4.): The intersection C of this straight line with the perpendicular, is the centre of the circle required.
For join CB. The triangles ACE and BCE, having the side EA equal to EB, the side EC common, and the contained angle AEC equal to BEC, are equal (I. 3.), and consequently AC is equal to BC. But AC is also equal
to CD (I. 9.) because the angle DAC was made equal to
ADC. Wherefore (III. 9. cor.) the three straight lines CA, CD, and CB being all equal, the point C is the centre of the circle.
PROP. XIX. THEOR.
The angle at the centre of a circle is double of the angle which, standing on the same arc, has its vertex in the circumference.
Let AB be an arc of a circle; the angle which it terminates at the centre, is double of ADB the corresponding angle at the circumference.
For join DC and produce it to the opposite circumference. This diameter DCE, if it lie not on one of the sides of the angle ADB, must either fall within that angle or without it.
First, let DC coincide with DB. And because AC is equal to DC, the angle ADC is equal to DAC (I. 8.); but the exterior angle ACB is equal to both of these (I. 34.) and therefore equal to double of either, or the angle ACB at the centre is double of the angle ADB at the circumference.
Next, let the straight line DCE lie within the angle ADB. From what has been demonstrated, it is apparent, that the angle ACE is double of ADE, and the angle BCE double of BDE; wherefore the angles ACE, BCE taken together, or the whole angle ACB, . are double of the collected angles | 677.169 | 1 |
DISTANCE BETWEEN 2 POINTS
The distance formula PQ=√(x2−x1)2+(y2−y1)2. Set of axes with the points P(x1, y1) and B(x2. It is clear from the distance formula that. Solution Thus, d = (5 − (− 2)) 2 + (1 − 3) 2 = 53 d = \sqrt{\left(5 - \left(-2\right)\right)^{2} + \left(1 - 3\right)^{2}} = \sqrt{53} d=(5−(−2))2+. In mathematics, the Euclidean distance between two points in Euclidean space is the length of the line segment between them. It can be calculated from the. The formula for the distance between the two points P1=(x1,y1) and P2=(x2,y2) can be derived using a combination of the Pythagorean Theorem and the. Derived from the Pythagorean Theorem, the distance formula is used to find the distance between two points in the plane. The Pythagorean Theorem, a2+b2=c2 a.
Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of distance between two points. Natural Language; Math Input. Distance between two points and the midpoint The distance formula is an algebraic expression used to determine the distance between two points with the. Enter latitude and longitude of two points, select the desired units: nautical miles (n mi), statute miles (sm), or kilometers (km) and click Compute. The Distance Formula takes two points and implicitly assigns them the role of the hypotenuse. Content Continues Below. gilno.ru Distance Formula on. Distance Between Two Points To find the distance of AB using the toolbar, click on the icon with an angle and click on distance. After that, click on the two. If the two points lie on the same vertical or horizontal line, the distance formula is simplified and is calculated by subtracting the coordinates. To calculate. Measure distance between points · On your computer, open Google Maps. · Right-click on your starting point. · Select Measure distance. · To create a path to measure. Distance between two points and the midpoint The distance formula is an algebraic expression used to determine the distance between two points with the. How it works: Just type numbers into the boxes below and the calculator will automatically calculate the distance between those 2 points. How to enter numbers. Think of the distance between any two points as a line. The length of this line can be found by using the distance formula: \sqrt((x_2 - x_1)^2 + (y_2. Use distance formula calculator to find the distance between 2 points. Calculate the distance using (x2-x1)^2+(y2-y1)^2 formula.
Steps. Start by setting up the formula. Substitute the values for the two points into the formula. Evaluate the subtraction. Evaluate the absolute value. The distance between them will appear just above the map in both miles and kilometers. The tool is useful for estimating the mileage of a flight, drive, or walk. This uses the 'haversine' formula to calculate the great-circle distance between two points – that is, the shortest distance over the earth's surface – giving. In Euclidean geometry, the distance from a point to a line is the shortest distance from a given point to any point on an infinite straight line. Measure distance between points · On your Android phone or tablet, open the Google Maps app · Touch and hold anywhere on the map that isn't a place's name or icon. Calculate the distance between two points or one point and a number of points, sorted by closest. More Help This tool will help you calculate the distance. Quick Explanation. Imagine you know the location of two points (A and B) like here. What is the distance between them? We can run lines down from A, and along. Distance Between 2 Points Formula · To find the distance between two points, take the coordinates of two points such as (x1, y1) and (x2, y2) · Use the distance. A point (x,y) is at a distance r from the origin if and only if √x2+y2=r, or, if we square both sides: x2+y2=r2. This is the equation of the circle of radius r.
Since a real line is a one-dimensional object, the distance between two points uses a variation of the formula for the distance between two points shown above. Distance between two points is the length of the line segment that connects the two given points. Learn to calculate the distance between two points formula. d = (x 2 − x 1) 2 + (y 2 − y 1) 2 We can think of x2 - x1 as the horizontal distance between the two points on the coordinate plane. Similarly, y2 - y1 is. The Distance Formula is a useful tool for calculating the distance between two points that can be arbitrarily represented as points [latex]A[/latex]. If you want to calculate the distance between two locations and you use the city blocks as your units of measurement, then two points on a map can be used to.
Practice using the distance formula to find the distance between two points. Ace your Math Exam! | 677.169 | 1 |
Library Function Math h Tan()
February 1, 2023
Tan() library function in Math
The tan() function is a part of the C math library and is used to calculate the tangent of a given angle. The angle is specified in radians. The tan() function takes a single argument, which is the angle in radians for which the tangent is to be calculated. The tangent of an angle is the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle in a right triangle. Let us understand more about Library Function Math h Tan().
Library Function Math h Tan()
The tan() function takes a single argument of type double, float, or long double and returns the tangent of the angle as a value of the same type. Here is the syntax for the tan() function in C: | 677.169 | 1 |
What is Rotation matrix: Definition and 81 Discussions
In linear algebra, a rotation matrix is a transformation matrix that is used to perform a rotation in Euclidean space. For example, using the convention below, the matrix
R
=
[
cos
θ
−
sin
θ
sin
θ
cos
θ
]
{\displaystyle R={\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \\\end{bmatrix}}}
rotates points in the xy-plane counterclockwise through an angle θ with respect to the x axis about the origin of a two-dimensional Cartesian coordinate system. To perform the rotation on a plane point with standard coordinates v = (x, y), it should be written as a column vector, and multiplied by the matrix R:
R
v
=
[
cos
θ
−
sin
θ
sin
θ
cos
θ
]
[
x
y
]
=
[
x
cos
θ
−
y
sin
θ
x
sin
θ
+
y
cos
θ
]
.
{\displaystyle R\mathbf {v} \ =\ {\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \end{bmatrix}}{\begin{bmatrix}x\\y\end{bmatrix}}\ =\ {\begin{bmatrix}x\cos \theta -y\sin \theta \\x\sin \theta +y\cos \theta \end{bmatrix}}.}
If x and y are the endpoint coordinates of a vector, where x is cosine and y is sine, then the above equations become the trigonometric summation angle formulae. Indeed, a rotation matrix can be seen as the trigonometric summation angle formulae in matrix form. One way to understand this is say we have a vector at an angle 30° from the x axis, and we wish to rotate that angle by a further 45°. We simply need to compute the vector endpoint coordinates at 75°.
The examples in this article apply to active rotations of vectors counterclockwise in a right-handed coordinate system (y counterclockwise from x) by pre-multiplication (R on the left). If any one of these is changed (such as rotating axes instead of vectors, a passive transformation), then the inverse of the example matrix should be used, which coincides with its transpose.
Since matrix multiplication has no effect on the zero vector (the coordinates of the origin), rotation matrices describe rotations about the origin. Rotation matrices provide an algebraic description of such rotations, and are used extensively for computations in geometry, physics, and computer graphics. In some literature, the term rotation is generalized to include improper rotations, characterized by orthogonal matrices with a determinant of −1 (instead of +1). These combine proper rotations with reflections (which invert orientation). In other cases, where reflections are not being considered, the label proper may be dropped. The latter convention is followed in this article.
Rotation matrices are square matrices, with real entries. More specifically, they can be characterized as orthogonal matrices with determinant 1; that is, a square matrix R is a rotation matrix if and only if RT = R−1 and det R = 1. The set of all orthogonal matrices of size n with determinant +1 forms a group known as the special orthogonal group SO(n), one example of which is the rotation group SO(3). The set of all orthogonal matrices of size n with determinant +1 or −1 forms the (general) orthogonal group O(n).
Hello.
I know that a 3×3 orthogonal matrix with determinant = 1 (so a 3×3 special orthogonal matrix) is a rotation in 3D.
I was wondering if there is a 3×3 orthogonal matrix with determinant = –1 could be visualised in some way.
Thank you!
I have three frames. The first is the fixed global frame. the second rotates an angle PHIZ with respect to the first. And the third first rotates a PHIX angle with respect to the x axis of the second frame, and then rotates a PHIY angle with respect to the last y axis. That is, there are a total...
Goldstein 3rd Ed pg 161.
Im not able to understand this paragraph about the ambiguity in the sense of rotation axis given the rotation matrix A, and how we ameliorate it.
Any help please.
"The prescriptions for the direction of the rotation axis and for the rotation angle are not unambiguous...
If we change the orientation of a coordinate system as shown above, (the standard eluer angles , ##x_1y_1z_1## the initial configuration and ##x_by _b z_b## the final one), then the formula for the coordinates of a vector in the new system is given by
##x'=Ax##
where...
In Rigid body rotation, we need only 3 parameters to make a body rotate in any orientation. So to define a rotation matrix in 3d space we only need 3 parameters and we must have 6 constraint equation (6+3=9 no of elements in rotation matrix)
My doubt is if orthogonality conditions...
Below is the attempted solution of a tutor. However, I do question his solution method. Therefore, I would sincerely appreciate it if anyone could tell me what is going on with the below solution.
First off, the rotation of the matrix could be expressed as below:
$$G = \begin{pmatrix} AB & -||A...
I have asked this question twice and each time, while the answers are OK, I am left dissatisfied.
However, now I can state my question properly (due to the last few responses).
Go to this page and scroll down to the matrix for sixth row of the proper Euler angles...
Hi,
if I have a equation like (just a random eq.) p_dot = S(omega)*p. where p = [x, y, z] is the original states, omega = [p, q, r] and S - skew symmetric.
How does the equation appear if i only want a system to have the state z? do I get z_dot = -q*x + p*y. Or is the symmetric not valid so I...
Hello
This could very well be an idiotic question, but here goes...
Consider a general matrix M
Consider a rotation matrix R (member of SO(2) or SO(3))
Is it possible to split M into the product of a rotation matrix R and "something else," say, S?
Such that: M = RS or the sum M = R + S...
Good Day
I have been having a hellish time connection Lie Algebra, Lie Groups, Differential Geometry, etc.
But I am making a lot of progress.
There is, however, one issue that continues to elude me.
I often read how Lie developed Lie Groups to study symmetries of PDE's
May I ask if someone...
Homework Statement
Show that every matrix A ∈ O(2, R) is of the form R(α) = cos α − sin α sin α cos α (this is the 2d rotation matrix -- I can't make it in matrix format) or JR(α). Interpret the maps x → R(α)x and x → JR(α)x for x ∈ R 2
Homework EquationsThe Attempt at a Solution
So I know...
Hello! I need to find the rotation matrix around a given vector v=(a,b,c), by and angle ##\theta##. I can find an orthonormal basis of the plane perpendicular to v but how can I compute the matrix from this? I think I can do it by brute force, rewriting the orthonormal basis rotated by...
Say I have {S_{x}=\frac{1}{\sqrt{2}}\left(\begin{array}{ccc}
0 & 1 & 0\\
1 & 0 & 1\\
0 & 1 & 0\\
\end{array}\right)}
Right now, this spin operator is in the Cartesian basis. I want to transform it into the spherical basis. Since, {\vec{S}} acts like a vector I think that I only need to...
First, I'd like to say I apologize if my formatting is off! I am trying to figure out how to do all of this on here, so please bear with me!
So I was watching this video on spherical coordinates via a rotation matrix:
and in the end, he gets:
x = \rho * sin(\theta) * sin(\phi)
y = \rho*...
The Lie Algebra is equipped with a bracket notation, and this bracket produces skew symmetric matrices.
I know that there exists Lie Groups, one of which is SO(3).
And I know that by exponentiating a skew symmetric matrix, I obtain a rotation matrix.
-----------------
First, can someone edit...
I want to rotate an inclined plane to achieve a flat surface.
I think I can use the Euler angles to perform this operation.
Using following data:
and following rotation matrix
I think you can make the plane flat by following rotations:
1: rotation around x-axis by 45°
2: rotation around...
I have a velocity vector as a function of a latitude and longitude on the surface of a sphere. Let us assume I have a point V(lambda, phi) where V is the velocity. The north pole of this sphere is rotated and I have a new north pole and I have a point V'(lambda, phi) in the new system. I have...
The question mentions an orthogonal matrix describing a rotation in 3D ... where $\phi$ is the net angle of rotation about a fixed single axis. I know of the 3 Euler rotations, is this one of them, arbitrary, or is there a general 3-D rotation matrix in one angle?
If I build one, I would start...
Homework Statement
Let A∈M2x2(ℝ) such that ATA = I and det(A) = -1. Prove that for ANY such matrix there exists an angle θ such that
A = ##
\left( \begin{array}{cc}
cos(\theta) & sin(\theta)\\
sin(\theta) & -cos(\theta)\\
\end{array} \right) ##
It is not sufficient to show that this matrix...
I'd like to prove the fact that - since a rotation of axes is a length-preserving transformation, the rotation matrix must be orthogonal.
By the way, the converse of the statement is true also. Meaning, if a transformation is orthogonal, it must be length preserving, and I have been able to...
Apologies up front for the long question … I have tried to be brief.
I want to define camera angles for Google Earth (GE) when rotated about an aircraft yaw axis. The input is Latitude, Longitude, Altitude plus Heading, Pitch and Bank angles, actually coming from Flight Simulator. These drive...
I am reading a paper and am stuck on the following snippet.
Given two orthonormal frames of vectors ##(\bf n1,n2,n3)## and ##(\bf n'1,n'2,n'3)## we can form two matrices ##N= (\bf n1,n2,n3)## and ##N' =(\bf n'1,n'2,n'3)##. In the case of a rigid body, where the two frames are related via...
I'm working through Meisner Thorne and Wheeler (MTW), but have been temporarily sidetracked by a problem with rotation matrices.
I've worked through the maths and produced the matrices by multiplying the three individual rotation matrices, (no problem there) but I have been trying to work out...
Suppose a position vector v is rotated anticlockwise at an angle ##\theta## about an arbitrary axis pointing in the direction of a position vector p, what is the rotation matrix R such that Rv gives the position vector after the rotation?
Suppose p = ##\begin{pmatrix}1\\1\\1\end{pmatrix}## and...
Hey, let's say that in 2D space we have a 2x2 rotation matrix R. Normally you could multiply this rotation matrix by a 2x1 column matrix / vector X. In that case it would be XR to get the vector rotated in the way described by R. Now what I'm wondering is, what if I had 3 column vectors that I...
I was trying to deduce the 2D Rotation Matrix and I got frustrated. So, I found this article: Ampliación del Sólido Rígido/ (in Spanish).
I don't understand the second line. How does he separate the matrix in two different parts?
Thanks for your time.
Homework Statement
The probelm is to show, that a rotation matrix R, in a odd-dimensional vector space, leaves unchanged the vectors of at least an one-dimensional subspace.
Homework Equations
This reduces to proving that 1 is an eigenvalue of Rnxn if n is odd. I know that a rotational...
Hi,
I have two questions related to angular velocity:
1. According to rotational damper, Torque = Viscous Damping Coefficient * Angular Velocity. This equation gives the unit of Angular Velocity as meter square per second. How is it equivalent to rad per second?
2. If I have an angular...
Homework Statement
Which matrix represents a rotation?
Homework Equations
The Attempt at a Solution
It seems odd that this matrix has somewhat the form for rotation about z-axis, just that you need to swap the cos θ for the sin θ.
Homework Statement
There are two coordinate systems which have different euler angles. Approximately find the euler angles of the second coordinate system with respect to the first coordinate system. Do this by taking the fact that you are able to plot points and know the position of the...
Homework Statement
Prove that a rotation matrix in R3 preserves distance.
Such that if A is a 3*3 orthogonal rotation matrix then |A.v|=|v|.
I know one can prove this is in R2 by using a trig representation of a rotation matrix and then simplifying. Is there an analogue method in R3 or some...
So I am given B=\begin{array}{cc} 3 & 5 \\ 5 & 3 \end{array}. I find the eigenvalues and eigenvectors: 8, -2, and (1, 1), (1, -1), respectively. I am then told to form the matrix of normalised eigenvectors, S, and I do, then to find S^{-1}BS, which, with S = \frac{1}{\sqrt{2}}\begin{array}{cc} 1...
Hello everyone.
I'm having some trouble with rotation matrices. I'm given three matrices
J_1, J_2, J_3
which form a basis for the set of skew-symmetric matrices (\mathfrak{so}_3).
Further, the matrix exponent function is such that
exp[\alpha J_i]=R_i(\alpha),
so taking the...
Homework Statement
In a particular coordinate frame, the moment of inertia tensor of a rigid body is given by
I = {{3,40},{4,9,0},{0,0,12}}
in some units. The instantaneous angular velocity is given by ω = (2,3,4) in some units. Find a rotation matrix a that transforms to a new coordinate...
I'm just learning this Latex(sic) formatting, so it's not ideal.
I was trying to explore the geometrical significance of the cross product when I happened upon an interesting observation. I've seen things like this before, but never had time to really examine them.
I define two vectors...
Hey guys! I'm new here, so forgive me if I'm posting in the wrong section.
I recently picked up a book on robotics and it had a section about rotation matrices. I'm having a difficult time with the decomposition of rotation matrices. Everywhere I look, I can find the the equations for the roll...
Hi, I want to calculate the coordinates of an object after a particular translation.
I have the 3D coordinates at the origin: x0,y0,z0
and i have the 3x3 rotation matrix: (r11, r12, r13; r21, r22, r23; r31, r32, r33)
If I want to move 3 units forward, in the direction i am facing and two...
I am trying to use FEA with space frame element. I know that for rotating an angle a around the z-axis, the translational displacements of the local and global coordinates are related through the rotation matrix:
\begin{bmatrix}cos(a) & sin(a) & 0 \\ -sin(a) & cos(a) & 0 \\ 0 & 0 &...
Homework Statement
Find the eigenvalues and normalized eigenvectors of the rotation matrix
cosθ -sinθ
sinθ cosθ
Homework Equations
The Attempt at a Solution
c is short for cosθ, s is short for sinθ
I tried to solve the characteristic polynomial (c-λ)(c-λ)+s^2=0, and... | 677.169 | 1 |
chapter outline
Hypotenuse of a Triangle
What is the Hypotenuse of a Triangle
A hypotenuse is the longest side of a right triangle. It is the side opposite the right angle (90°). The word 'hypotenuse' came from the Greek word 'hypoteinousa', meaning 'stretching under', where 'hypo' means 'under', and 'teinein' means 'to stretch'.
Hypotenuse of a Triangle
Formulas
How to Find the Hypotenuse of a Right Triangle
a) When Base and Height are Given
To calculate the hypotenuse of a right or right-angled triangle when its corresponding base and height are known, we use the given formula.
Hypotenuse of a Right Triangle Formula
Derivation
By Pythagorean Theorem,
(Hypotenuse)2 = (Base)2 + (Height)2
Hypotenuse = √(Base)2 + (Height)2
Thus, mathematically, hypotenuse is the sum of the square of base and height of a right triangle.
The above formula is also written as,
c = √a2 + b2, here c = hypotenuse, a = height, b = base
Let us solve some problems to understand the concept better.
Problem: Finding the hypotenuse of a right triangle, when the BASE and the HEIGHT are known.
What is the length of the hypotenuse of a right triangle with base 8m and height 6m. | 677.169 | 1 |
Ex 6.4, 1 (ii) - Chapter 6 Class 7 Triangle and its Properties
Last updated at April 16, 2024 by Teachoo
Transcript
Ex 6.4, 1 Is it possible to have a triangle with the following sides? (ii) 3 cm, 6 cm, 7 cmGiven 3 sides
3 cm, 6 cm, 7 cm
If these sides form a triangle, them
Sum of two sides > Third side
Here,
Sum of two sides is greater than third side in all three cases
So, possible | 677.169 | 1 |
One way to triangulate a simple polygon is based on the two ears theorem, as the fact that any simple polygon with at least 4 vertices without holes has at least two "ears", which are triangles with two sides being the edges of the polygon and the third one completely inside it.[5] The algorithm then consists of finding such an ear, removing it from the polygon (which results in a new polygon that still meets the conditions) and repeating until there is only one triangle left.
This algorithm is easy to implement, but slower than some other algorithms, and it only works on polygons without holes. An implementation that keeps separate lists of convex and concave vertices will run in O(n2) time. This method is known as ear clipping and sometimes ear trimming. An efficient algorithm for cutting off ears was discovered by Hossam ElGindy, Hazel Everett, and Godfried Toussaint.[6]
A simple polygon is monotone with respect to a line L, if any line orthogonal to L intersects the polygon at most twice. A monotone polygon can be split into two monotone chains. A polygon that is monotone with respect to the y-axis is called y-monotone. A monotone polygon with n vertices can be triangulated in O(n) time. Assuming a given polygon is y-monotone, the greedy algorithm begins by walking on one chain of the polygon from top to bottom while adding diagonals whenever it is possible.[1] It is easy to see that the algorithm can be applied to any monotone polygon.
If a polygon is not monotone, it can be partitioned into monotone subpolygons in O(n log n) time using a sweep-line approach. The algorithm does not require the polygon to be simple, thus it can be applied to polygons with holes.
Generally, this algorithm can triangulate a planar subdivision with n vertices in O(n log n) time using O(n) space.[1]
A useful graph that is often associated with a triangulation of a polygon P is the dual graph. Given a triangulation TP of P, one defines the graph G(TP) as the graph whose vertex set are the triangles of TP, two vertices (triangles) being adjacent if and only if they share a diagonal. It is easy to observe that G(TP) is a tree with maximum degree 3.
Bernard Chazelle showed in 1991 that any simple polygon can be triangulated in linear time, though the proposed algorithm is very complex.[12] A simpler randomized algorithm with linear expected time is also known.[13]
The time complexity of triangulation of an n-vertex polygon with holes has an Ω(n log n)lower bound, in algebraic computation tree models of computation.[1] It is possible to compute the number of distinct triangulations of a simple polygon in polynomial time using dynamic programming, and (based on this counting algorithm) to generate uniformly random triangulations in polynomial time.[15] However, counting the triangulations of a polygon with holes is #P-complete, making it unlikely that it can be done in polynomial time.[16] | 677.169 | 1 |
How will light react in this situation....
In summary, the individual is seeking help with calculating shadows on a flat surface when the light source is at a 0 or 180 degree angle. They mention their knowledge of light traveling in straight lines and the difficulty in finding the intersection of the shadow volume with the ground. They also mention the possibility of using GPUs for these types of calculations. The conversation touches on the concept of a point source versus an extended source and the potential difficulty in achieving a 0 or 180 degree angle in nature.
Jul 23, 2020
#1
xrayspecs
2
2
Dear Physics Forum,
I need help with this problem. In the diagrams above I try to show my difficulty. My main problem is working out how shadow will fall on a completely flat surface with the light source at more or less a 0 or 180 degree angle (depending on how you want to look at it) . Through internet searches I have seen how it is possible to calculate the shadow in a natural setting at around 45 degrees.
From knowledge acquired at school and light research... (pun intended) I know light travels in straight lines. So if it met a completely flat surface how would it be possible to work out where the shadow of the shape would appear on the ground? And how long it would measure on said ground?
Please let me know if you need any more information or drawings to try and make this situation easier to understand.
Any help will be most appreciated!
Yes, but the problem is that different straight lines don't necessarily have the same direction.
xrayspecs said:
So if it met a completely flat surface how would it be possible to work out where the shadow of the shape would appear on the ground?
In general you need to find the intersection of the shadow volume with the ground. In case of a sphere and a point source the shadow volume is a cone and the intersection with the ground is an ellipse, parabola or hyperbola. In case of real bodies and light sources it will me much more complex. This kind of calculations is what GPUs are build for.
xrayspecs said:
And how long it would measure on said ground?
As mentioned above this can be quite difficult to answer, especially if the shape of the object is not clearly defined. But you can at least estimate the maximum length of the umbra. It is the distance where the object has the same angular diameter as the light source. In sunlight that would be 106 to 109 times the size of the object.
You have to understand that the picture you drew is misleading. The Sun is so far away that its rays are parallel at any given location. A modified picture is shown below. What do you think the shadow would be like on the flat plane? Remember that shadow is the absence of light. In the picture below, what light, which would normally fall on the plane, is blocked by the object?
If we're treating the sun as a point source (which it isn't), then the shadow would lie between the two thin lines in the OP -- the ones that eminate from the lower-left and lower-right "corners" of the object. And as A.T. said, it would extend to infinity.
If we're treating the sun as an extended source, then each of those lines becomes a pair of lines, at roughly 0.5 degrees to each other. In between the two lines, there is a transition from fully bright to fully dark. But the shadow still extends to infinity.
The answer to the question "How would I show this curve?", is that you would not show it at all.
... how shadow will fall on a completely flat surface with the light source at more or less a 0 or 180 degree angle...
The problem I see is the difficulty to achieve that angle in nature.
Even before the Sun rises, some light hits our eyes and objects.
That light comes from reflection on dust particles, bottom of clouds, etc.
If we are treating the sun as a point source then there is no shadow in the first place. The surface is not illuminated.
Aug 2, 2020
#8
xrayspecs
2
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Thank you for your comments!
Related to How will light react in this situation....
1. How will light react when passing through a prism?
When light passes through a prism, it will refract or bend, separating into its component colors. This is due to the different wavelengths of light being bent at different angles as they pass through the prism.
2. How will light behave when reflecting off a smooth surface?
When light reflects off a smooth surface, it will follow the law of reflection, which states that the angle of incidence (incoming light) is equal to the angle of reflection (outgoing light). This results in a clear and undistorted reflection.
3. How will light react when traveling through a vacuum?
In a vacuum, light will travel at its maximum speed of approximately 299,792,458 meters per second. This is because there are no particles in a vacuum to slow down or obstruct the path of light.
4. How will light behave when passing through a medium with a higher refractive index?
When light passes through a medium with a higher refractive index, it will slow down and bend towards the normal (a line perpendicular to the surface of the medium). This is known as refraction and is responsible for phenomena such as the bending of light in a glass of water.
5. How will light react when encountering an object with a rough surface?
When light encounters an object with a rough surface, it will scatter in all directions due to the uneven surface. This results in a diffuse reflection, where the light is not reflected in a clear and organized manner, but rather in multiple directions. | 677.169 | 1 |
Triangle is a basic shape in geometry with three sides and three vertices. Calculating the vertices and angles can be done manually using formulas. Still, to assist students in calculating easily, allcalculator.net has created a simple tool known as the Triangle calculator that helps to calculate all the required values of a triangle within seconds of entering the data.
How to use the Triangle calculator?
Using the Triangle calculator is very simple as all you need to do is enter any three data as per your requirement and click calculate to get
All the sides of the triangle
all the angles of the triangle
area, perimeter, and semi perimeter
height
median
inradius
circumradius
Based on all desired values, the final type of triangle will be displayed for your easy understanding.
What is the purpose of this triangle calculator?
The triangle calculator will help you learn about triangles and the techniques used in CGI to create characters that resemble the constructed work commonly used in movies and television series, with which characters can be easily modified, stored, and used digitally.
This can be used to create complex structures that are used in architectural design, especially in making shapes and curves and mathematical shapes.
This can also be used to compute a right-angled triangle, an important shape in trigonometry that uses the Pythagorean theorem.
Who can use this triangle calculator?
This allcalculator.net triangle calculator was designed keeping in mind the school students who need help understanding the basic concepts of triangles in the chapter geometry. They can use this calculator without guidance.
Mathematics teachers can also use these triangle calculators to get to the result quickly and cross-verify during evaluation after a test.
What are the formulas used in the triangle calculator?
This triangle calculator uses four main theorems, which are
Heron's formula
Cosine theorem
Sine theorem
Pythagorean theorem
Can this calculator be used in solving geometric word problems?
Definitely, yes, this triangle calculator can be used to solve word problems, provided the kids understand the questions well right before entering the appropriate values in the given spaces. One wrong value can result in a different outcome, so one must be extra careful while assigning the values to the required field. | 677.169 | 1 |
problem solving for angles
Appendix B: Geometry
Using properties of angles to solve problems, learning outcomes.
Find the supplement of an angle
Find the complement of an angle
Are you familiar with the phrase 'do a [latex]180[/latex]?' It means to make a full turn so that you face the opposite direction. It comes from the fact that the measure of an angle that makes a straight line is [latex]180[/latex] degrees. See the image below.
[latex]\angle A[/latex] is the angle with vertex at [latex]\text{point }A[/latex].
We measure angles in degrees, and use the symbol [latex]^ \circ[/latex] to represent degrees. We use the abbreviation [latex]m[/latex] to for the measure of an angle. So if [latex]\angle A[/latex] is [latex]\text{27}^ \circ [/latex], we would write [latex]m\angle A=27[/latex].
If the sum of the measures of two angles is [latex]\text{180}^ \circ[/latex], then they are called supplementary angles. In the images below, each pair of angles is supplementary because their measures add to [latex]\text{180}^ \circ [/latex]. Each angle is the supplement of the other.
The sum of the measures of supplementary angles is [latex]\text{180}^ \circ [/latex].
The sum of the measures of complementary angles is [latex]\text{90}^ \circ[/latex].
Supplementary and Complementary Angles
If the sum of the measures of two angles is [latex]\text{180}^\circ [/latex], then the angles are supplementary .
If angle [latex]A[/latex] and angle [latex]B[/latex] are supplementary, then [latex]m\angle{A}+m\angle{B}=180^\circ[/latex].
If the sum of the measures of two angles is [latex]\text{90}^\circ[/latex], then the angles are complementary .
If angle [latex]A[/latex] and angle [latex]B[/latex] are complementary, then [latex]m\angle{A}+m\angle{B}=90^\circ[/latex].
In this section and the next, you will be introduced to some common geometry formulas. We will adapt our Problem Solving Strategy for Geometry Applications. The geometry formula will name the variables and give us the equation to solve.
In addition, since these applications will all involve geometric shapes, it will be helpful to draw a figure and then label it with the information from the problem. We will include this step in the Problem Solving Strategy for Geometry Applications.
Use a Problem Solving Strategy for Geometry Applications.
Read the problem and make sure you understand all the words and ideas. Draw a figure and label it with the given information.
Identify what you are looking for.
Name what you are looking for and choose a variable to represent it.
Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
Solve the equation using good algebra techniques.
Check the answer in the problem and make sure it makes sense.
Answer the question with a complete sentence.
The next example will show how you can use the Problem Solving Strategy for Geometry Applications to answer questions about supplementary and complementary angles.
An angle measures [latex]\text{40}^ \circ[/latex].
1. Find its supplement
2. Find its complement
Write the appropriate formula for the situation and substitute in the given information. [latex]m\angle A+m\angle B=90[/latex] Step 5. Solve the equation. [latex]c+40=90[/latex]
[latex]c=50[/latex] Step 6. Check:
[latex]50+40\stackrel{?}{=}90[/latex]
In the following video we show more examples of how to find the supplement and complement of an angle.
Did you notice that the words complementary and supplementary are in alphabetical order just like [latex]90[/latex] and [latex]180[/latex] are in numerical order?
Two angles are supplementary. The larger angle is [latex]\text{30}^ \circ[/latex] more than the smaller angle. Find the measure of both angles.
Tyler thinks that this figure has enough information to figure out the values of \(a\) and \(b\).
Do you agree? Explain your reasoning.
Exercise \(\PageIndex{2}\): What Does It Look Like?
Elena and Diego each wrote equations to represent these diagrams. For each diagram, decide which equation you agree with, and solve it. You can assume that angles that look like right angles are indeed right angles.
1. Elena: \(x=35\)
Diego: \(x+35=180\)
2. Elena: \(35+w+41=180\)
Diego: \(w+35=180\)
3. Elena: \(w+35=90\)
Diego: \(2w+35=90\)
4. Elena: \(2w+35=90\)
Diego: \(w+35=90\)
5. Elena: \(w+148=180\)
Diego: \(x+90=148\)
Exercise \(\PageIndex{3}\): Calculate the Measure
Find the unknown angle measures. Show your thinking. Organize it so it can be followed by others.
Lines \(l\) and \(m\) are perpendicular.
Are you ready for more?
The diagram contains three squares. Three additional segments have been drawn that connect corners of the squares. We want to find the exact value of \(a+b+c\).
Use a protractor to measure the three angles. Use your measurements to conjecture about the value of \(a+b+c\).
Find the exact value of \(a+b+c\) by reasoning about the diagram.
To find an unknown angle measure, sometimes it is helpful to write and solve an equation that represents the situation. For example, suppose we want to know the value of \(x\) in this diagram.
Using what we know about vertical angles, we can write the equation \(3x+90=144\) to represent this situation. Then we can solve the equation.
Line \(l\) is perpendicular to line \(m\). Find the value of \(x\) and \(w\).
Exercise \(\PageIndex{6}\)
If you knew that two angles were complementary and were given the measure of one of those angles, would you be able to find the measure of the other angle? Explain your reasoning.
Exercise \(\PageIndex{7}\)
For each inequality, decide whether the solution is represented by \(x<4.5\) or \(x>4.5\).
\(-24>-6(x-0.5)\)
\(-8x+6>-30\)
\(-2(x+3.2)<-15.4\)
(From Unit 6.3.3)
Exercise \(\PageIndex{8}\)
A runner ran \(\frac{2}{3}\) of a 5 kilometer race in 21 minutes. They ran the entire race at a constant speed.
How long did it take to run the entire race?
How many minutes did it take to run 1 kilometer?
(From Unit 4.1.2)
Exercise \(\PageIndex{9}\)
Jada, Elena, and Lin walked a total of 37 miles last week. Jada walked 4 more miles than Elena, and Lin walked 2 more miles than Jada. The diagram represents this situation:
Find the number of miles that they each walked. Explain or show your reasoning.
(From Unit 6.2.6)
Exercise \(\PageIndex{10}\)
Select all the expressions that are equivalent to \(-36x+54y-90\).
\(-9(4x-6y-10)\)
\(-18(2x-3y+5)\)
\(-6(6x+9y-15)\)
\(18(-2x+3y-5)\)
\(-2(18x-27y+45)\)
\(2(-18x+54y-90)\)
(From Unit 6.4.2)
Maths Questions
Angles Questions
Angles questions and answers are available here to help students understand how to solve basic angles problems. These questions cover the different types of angles and their measures, and finding the missing angles when a pair or more than two angles are given in a specific relation. You will also get some extra practice questions at the end of the page. These will help you to improve your geometry skills and get a clear understanding of angles.
What are angles?
In geometry, angles are the figures formed by two rays that are connected by a common point called the vertex. We can measure the angles between two lines, rays or line segments using one of the geometric tools called a protractor. Based on the measure of these angles, we can classify them.
The different types of angles are listed below:
Acute angle (< 90°)
Obtuse angle (> 90° and < 180°)
Right angle (= 90°)
Straight angle (= 180°)
Reflex angle (> 180° and < 360°)
Full rotation angle (= 360°)
Also, check: Angles
Angles Questions and Answers
1. Classify the following angles:
55° < 90°
Thus, 55° is an acute angle.
90° < 146° < 180°
So, 146° is an obtuse angle.
90° is a right angle.
180° < 250° < 360°
Thus, 250° is a reflex angle.
2. Write two examples of obtuse angles and reflex angles.
As we know, obtuse angles are the angles that measure less than 180° and greater than 90°.
Examples: 112°, 177°
Reflex angles measure less than 360° and greater than 180°.
Examples: 210°, 300°
3. Find the measure of an angle which is complementary to 33°.
If the sum of two angles is 90°, they are called complementary angles.
Let x be the angle which is complementary to 33°.
So, x + 33° = 90°
x = 90° – 33° = 57°
Therefore, the required angle is 57°.
4. What is the measure of an angle that is supplementary to 137°?
If the sum of two angles is 180°, they are called supplementary angles.
Let x be the angle which is supplementary to 137°.
So, x + 137° = 180°
x = 180° – 137° = 43°
Hence, the required angle is 43°.
5. If three angles 2x, 3x, and x together form a straight angle, find the angles. Solution:
We know that straight angle = 180°
Given that the angles 2x, 3x, and x form a straight angle.
That means 2x + 3x + x = 180°
2x = 2 × 30° = 60°
3x = 3 × 30° = 90°
Therefore, the angles are 60°, 90° and 30°.
6. Are 125° and 65° supplementary angles?
As we know, the condition for supplementary angles is that they add up to 180°.
8. Find the value of y if (4y + 22)° and (8y – 10)° form a linear pair.
According to the given,
(4y + 22)° + (8y – 10)° = 180°
4y + 22° + 8y – 10° = 180°
12y + 12° = 180°
12y = 180° – 12°
y = 168°/12
9. Three angles at a point are 135°, 75° and x. Find the value of x.
Given angles at a point are: 135°, 75°, x
As we know, the sum of angles at a point = 360°
So, 135° + 75° + x = 360°
210° + x = 360°
x = 360° – 210° = 150°
Therefore, the value of x is 150°.
10. If 3x + 24° and 5x – 16° are congruent, then find the value of x.
Congruent angles mean equal angles.
So, 3x + 24° = 5x – 16°
⇒ 5x – 3x = 24° + 16°
⇒ x = 40°/2
Thus, the value of x is 20°.
Practice Problems on Angles
Are 42° and 58° complementary angles?
How do you find the measure of an angle which is supplementary to 92°?
What is the condition for a reflex angle?
If 38° and 2x + 26° form a right angle, find the value of x.
Classify the following angles: (i) 72° .
Here we will learn about angles, including angle rules, angles in polygons and angles in parallel lines.
There are also angles worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you're still stuck.
What are angles?
Angles measure the amount of turn required to change direction. At GCSE we can measure angles using a protractor using degrees . If the diagram is not drawn to scale we can determine missing angles by using angle facts (also referred to as angle properties or angle rules ).
There can be multiple different approaches to find a missing angle.
Angle types
There are different types of angles.
Step-by-step guide: Types of angles
Angle rules
We can use angle rules to work out missing angles.
Angle rules are facts that we can apply to calculate missing angles in a diagram.
The five key angle facts that are used widely within the topic are:
Angles on a straight line
The sum of angles on a straight line is always equal to \bf{180^{o}.}
A straight line would be considered to be half of a full turn; if you were standing on the line facing towards one end, you would have to turn 180 degrees to face the other end of the line.
A straight line can be called a straight angle if there is a vertex on the line and the turn around that vertex is 180^{o}.
Angles at a point
The sum of angles at a point is always equal to \bf{360^{o}} .
A point would be considered to be a full turn; if you were standing at the point facing in one direction, you would have to turn 360 degrees to return back to your original position.
Complementary angles
The sum of complementary angles is always equal to \bf{90^{o}} .
Complementary angles therefore make up a right angle.
These angles do not need to be together and form a right angle. If any two angles sum to 90^o they are complementary.
Supplementary angles
The sum of supplementary angles is always equal to \bf{180^{o}} .
Supplementary angles therefore make up a straight line.
These angles do not need to be together on a straight line. If any two angles sum to 180^o they are supplementary.
Vertically opposite angles
Vertically opposite angles are equal .
This occurs when two straight lines meet (intersect) at a point known as a vertex , forming an x shape where the opposite pairs of angles are the same size. Also, two adjacent angles are supplementary (they add to equal 180^o ).
Step-by-step guide: Angle rules
Angles in polygons
We can calculate the interior and exterior angles of any polygon.
Interior angles
The sum of interior angles in any n -sided shape is determined using the formula,
One interior angle of a regular polygon with n -sides is determined using the formula,
For an irregular polygon, the missing angle is calculated by subtracting all of the known angles from the total sum of the interior angles of the polygon.
Exterior angles
The sum of exterior angles for any polygon is \bf{360^{o}} .
Whereas the interior angle sum is different for each n -sided shape, the exterior angle sum is always 360^{o}, regardless of how many sides the polygon has.
This is because, as you walk around the perimeter of the shape, the exterior angle is the turn from the direction of one edge to the next edge of the polygon.
For a regular polygon, each exterior angle is equal to 360 divided by the number of sides, n, and so
For an irregular polygon, the unknown exterior angle is calculated by subtracting the known exterior angles from 360^{o}.
The sum of an exterior angle and its adjacent interior angle is 180^{circ} , because they both lie on a straight line.
Angles in a triangle
The sum of angles in a triangle is \bf{180^{o}} .
Remember that there are four different types of triangles, each with a specific angle property.
Angles in a quadrilateral
The sum of angles in a quadrilateral is \bf{360^{o}} .
Any quadrilateral can be constructed from two adjacent triangles. This means that as the angle sum of a triangle is equal to 180^{o}, two triangles would have an angle sum of 360^{o}.
There are several different types of quadrilaterals, each with a specific angle property.
Step-by-step guide: Angle in polygons
Angles in parallel lines
Angles in parallel lines are facts that can be applied to calculate missing angles within a pair of parallel lines. The three key angle facts that are used when looking at angles in parallel lines are,
Corresponding angles
Alternate angles
Co-interior angles
Corresponding angles are equal .
When we intersect a pair of parallel lines with a transversal (another straight line), corresponding angles are the angles that occur on the same side of the transversal line. They are either both obtuse or both acute.
Alternate angles are equal .
When we intersect a pair of parallel lines with a transversal, alternate angles occur on opposite sides of the transversal line. They are either both obtuse or both acute.
The sum of two co-interior angles is \bf{180^{o}} .
Co-interior angles occur in between two parallel lines when they are intersected by a transversal . The two angles that occur on the same side of the transversal always add up to 180^{o}.
Step-by-step guide: Angles in parallel lines
How to use angles
We can use angles in lots of different contexts.
We will learn about,
Types of angles
Explain how to use angles
Angles worksheet
Get your free angles worksheet of 20+ questions and answers. Includes reasoning and applied questions.
Angle rules examples
Example 1: angles on a straight line.
Calculate the value of x.
Add all known angles.
The angle highlighted as a square is a right angle. A right angle measures 90^{o}. Adding the angles together, we can form the expression
2 Subtract the angle sum from \bf{180^{o}} .
As the sum of angles on a straight line total 180^{o},
3 Form and solve the equation.
Here we have no equation to solve as the missing angle is 20^{o}.
Example 2: angles at a point
Subtract the angle sum from \bf{360^{o}} .
Form and solve the equation.
Here there is no equation to solve as we know the angle x=60^{o}.
Example 3: complementary angles
Angle AOB is complementary to BOC. Determine the size of the angle AOB.
Identify which angles are complementary .
Here, the angles AOB and BOC are complementary.
Clearly identify which of the unknown angles the question is asking you to find the value of.
The question would like us to calculate the angle AOB.
Solve the problem and give reasons where applicable .
As the sum of the two angles is 90 degrees, forming an equation, we have
x+10+3x=90,
Solving this for x, we have
\begin{aligned} 4x&=90-10\\\\ 4x&=80\\\\ x&=20. \end{aligned}
Clearly state the answer using angle terminology.
We need to determine the size of angle AOB for the solution and so we need to substitute x=20 into x+10.
Angle AOB = x+10=20+10=30^{o}.
Example 4: supplementary angles
Given that AC is a straight line, determine the size of angle AOB.
Identify which angles are supplementary .
As the line AC is a straight line, the two angles AOB and BOC are supplementary. This means that we can use the angle rule, "the sum of supplementary angles is 180^{o} ".
We need to determine the size of angle AOB.
Solve the problem and give reasons where applicable.
Adding the two angles 5x+35 and x+25 is equal to 180^{o}, which gives us the equation
5x+35+x+25=180.
Simplifying the left side of the equation, we have
Subtracting 60 from both sides of the equation, we have
6x=180-60=120.
Dividing both sides by 6, we have
x=120\div{6}=20.
We need to determine the size of angle AOB for the solution and so we need to substitute x=20 into angle AOB \ (5x+35).
Angle AOB = 5x+35=(5\times{20})+35=135^{o}.
Example 5: vertically opposite angles
Given that AC and BD are straight intersecting lines at the point O, determine the size of angle COD.
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Solving Triangles
Six Different Types
If you need to solve a triangle right now choose one of the six options below:
Which Sides or Angles do you know already? (Click on the image or link)
... or read on to find out how you can become an expert triangle solver :
Your Solving Toolbox
Want to learn to solve triangles?
Imagine you are " The Solver " ... ... the one they ask for when a triangle needs solving!
In your solving toolbox (along with your pen, paper and calculator) you have these 3 equations:
1. Angles Add to 180° :
A + B + C = 180°
When you know two angles you can find the third.
2. Law of Sines (the Sine Rule):
When there is an angle opposite a side, this equation comes to the rescue.
Note: angle A is opposite side a, B is opposite b, and C is opposite c.
3. Law of Cosines (the Cosine Rule):
This is the hardest to use (and remember) but it is sometimes needed to get you out of difficult situations.
It is an enhanced version of the Pythagoras Theorem that works on any triangle.
Six Different Types (More Detail)
There are SIX different types of puzzles you may need to solve. Get familiar with them:
This means we are given all three angles of a triangle, but no sides.
AAA triangles are impossible to solve further since there are is nothing to show us size ... we know the shape but not how big it is.
We need to know at least one side to go further. See Solving "AAA" Triangles .
This mean we are given two angles of a triangle and one side, which is not the side adjacent to the two given angles.
Such a triangle can be solved by using Angles of a Triangle to find the other angle, and The Law of Sines to find each of the other two sides. See Solving "AAS" Triangles .
This means we are given two angles of a triangle and one side, which is the side adjacent to the two given angles.
In this case we find the third angle by using Angles of a Triangle , then use The Law of Sines to find each of the other two sides. See Solving "ASA" Triangles .
This means we are given two sides and the included angle.
For this type of triangle, we must use The Law of Cosines first to calculate the third side of the triangle; then we can use The Law of Sines to find one of the other two angles, and finally use Angles of a Triangle to find the last angle. See Solving "SAS" Triangles .
This means we are given two sides and one angle that is not the included angle.
In this case, use The Law of Sines first to find either one of the other two angles, then use Angles of a Triangle to find the third angle, then The Law of Sines again to find the final side. See Solving "SSA" Triangles .
This means we are given all three sides of a triangle, but no angles.
In this case, we have no choice. We must use The Law of Cosines first to find any one of the three angles, then we can use The Law of Sines (or use The Law of Cosines again) to find a second angle, and finally Angles of a Triangle to find the third angle. See Solving "SSS" Triangles .
Tips to Solving
Here is some simple advice:
When the triangle has a right angle, then use it, that is usually much simpler.
When two angles are known, work out the third using Angles of a Triangle Add to 180° .
Try The Law of Sines before the The Law of Cosines as it is easier to use.
Missing Angles
Related worksheets.
Find the missing angles in a triangle, around a point, in a quadrilateral, find opposite and supplementary angles, or find angles which require multi-step problem solving skills.
For more shape and space resources click here.
Game Objectives
New Maths Curriculum:
Year 3: Identify right angles, recognise that two right angles make a half-turn, three make three quarters of a turn and four a complete turn; identify whether angles are greater than or less than a right angle
Year 4: Identify acute and obtuse angles and compare and order angles up to two right angles by size
Year 5: Identify: multiples of 90°, angles at a point on a straight line and ½ a turn (total 180°); angles at a point and one whole turn (total 360°); reflex angles; and compare different angles
Year 6: Find unknown angles where they meet at a point, are on a straight line, and are vertically opposite
Angle Word Problems
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Find and Identify Angles in the Real World
How are Angles Used in the Real World?
Angles are everywhere! It's great to have an angle hunt or to search for angles in the real world because students and teachers start to see angles that they may not have noticed before. When you look for angles in objects, you can find patterns and see how angles work together.
I love helping students find and identify angles in real life objects because they start to see how one angle might have another angle that goes with it. One time I had a drawing of a house and asked students to find one right angle. I was thinking that they would point out the right angle inside the house, but I had some students find the right angle of the outside, where the house meets the ground, or a right angle inside the door or window. Just this simple perspective of finding the same angle in different places helps reveal more views of angles and of the house. This helps students see how angles are made and even introduces concepts like supplementary and complementary angles.
As students discover angles in real life objects, they also learn about different types of angles and how angles help to make up shapes. I've had students search for and find types of angles in everyday objects and share what they found. If I can get my hands on a camera, then I have students take pictures of objects, print them out and outline and identify angles inside their photos.
Set up for the Find and Identify Angles Lesson
Set up themed stations around the room with photographs of items from those places (photographs are provided on pages 2-11 of the Find and Identify Angles in the Real World worksheet , but you can add your own, especially if you have photographs of local spots that students would recognize): Aquarium, At the Park, Amusement Park, In the City, At the Beach.If possible, print the photographs in color for each station. You can also put the photographs into plastic sleeves. If you use page protectors, have thin tipped dry erase markers available.
Provide each student with the Find and Identify Angles in the Real World worksheet (these can be printed in black and white).
Provide students with colored pencils or thin tipped markers to record the angles onto the pictures on their worksheet.
If you have a projector, set it up to show the image of a playground (shown below or a similar one) to project the image and then be able to draw images over the projection on the board, remove the image and see the angles on their own.
Provide students with blank paper and rulers, and, if available, protractors. (Students do not need to use the protractors to measure the angles, but this is a good time to introduce the protractor and how it helps students identify acute, obtuse, and right angles.)
Launch the Activity: Angles in the Real World
Project a photograph of a playground on a whiteboard or large piece of paper. The students have the same photograph on page 1 of their Find and Identify Angles worksheet and should follow along with as you trace angles onto the photograph, they can do so on their worksheet. If you want students to be able to practice before drawing onto the worksheets, you can provide plastic sleeves and students can use dry erase markers.
Here is an image that provides examples of each type of angle:
With the image projected, you can start to name and identify types of angles. StartHere is an example of an angle in this image:
Ask students if they can identify or name this type of angle. It is an acute angle .
Then show an acute angle outside of the picture. You can do this by turning off the projector and showing the angle by itself. Like this:
Have the students sketch an angle similar to this on their worksheet and identify it as an acute angle .
Now help students to identify an obtuse angle , a right angle , and a straight angle in this image, reviewing the definition of each as you find it. You can identify and show these angles yourself, or ask for volunteers to come up and find an angle in the photo. Students can find these angles in their own image on the worksheet too.
Make sure every student fills out an example and definition for each type of angle:
Here is an example of the image with each type of angle.
You can color code if you use plastic sleeves, but students should sketch angles with pencil first on their worksheet and then color code.
After finding angles on the playground, and naming angles on the worksheet, divide the students into groups and have them visit each themed station. Students should use the color photographs at each station to find and identify angles. They should find the corresponding photograph on their worksheet, which will give them instructions for which types of angles to find. Then they will draw those angles (preferably in color) onto the image on their worksheet.
Each station should have multiple photographs so students have the option to work independently, even if they are grouped at a station.
Photographs for each station are located on pages 2-11 of the Find and Identify Angles worksheet , but you can also include more of your own, especially if you have photographs of local places that students would recognize and that correspond to the themes below. The instructions for the types of angles students should search for in each photograph are located beneath each photograph on the student worksheets.
Below are the station themes and lists of objects that would likely have angles to identify, in case you want to supplement with local pictures:
Set a timer for every 8-10 minutes, depending on the length of your class period, and have students rotate to a new station. Students do not need to do every photograph and do not need to work in the order the photographs appear on the worksheet, but they should keep working and tracing angles until the time is up at that station.
While students are working at the stations, circulate through the room and ask questions, encouraging students to find new angles they have not yet noticed.
Angles in the Real World Activity Reflection
Lead students in small or whole group discussions around reflective questions, such as the following:
Did you find more acute or more obtuse angles?
Are there more right angles than other kinds of angles?
Did you find one type of angle and then see another one that goes alongside it?
Did each station get easier or harder?
Which station did you find the most challenging?
There are also reflection questions on the worksheet that students can complete on their own or in groups.
Angles in the Real World Extensions
Students can go around school, at home, or in the community to take pictures with a tablet or camera and draw and sketch angles on the photos. Students could also add their own photographs (with angles sketched on) to the existing themed stations.
Students can create a poster or slides presentation that shows images that have angles. Maybe one image highlights only acute angles, another shows only obtuse and another shows only right angles.
Create a "Seek and Find" book or poster where there are "hidden" angles and include an answer key.
FREE Find and Identify Angles in Real Life Objects Worksheets and Resources
These are all PDF Files. They will open and print easily. The Student Edition Files are labeled SE and the Teacher Editions Files are labeled TE. Click the links below to download the different resources.
7-2 Assignment SE – Angles
7-2 Assignment TE – Angles ( Member Only )
7-2 Bell Work SE – Angles
7-2 Bell Work TE – Angles ( Member Only )
7-2 Exit Quiz SE – Angles
7-2 Exit Quiz TE – Angles ( Member Only )
7-2 Guided Notes SE – Angles
7-2 Guided Notes TE – Angles ( Member Only )
7-2 Interactive Notebook SE – Angles
7-2 Lesson Plan – Angles
7-2 Online Activities – Angles
7-2 Slide Show – Angles
Find and Identify Angles in Real Life Objects Worksheets and Resources
To get the Editable versions of these files Join us inside the Math Teacher Coach Community! This is where we keep our full curriculum of 4th Grade Math Lessons and Activities.
Don't Forget to Pin this lesson on Find and Identify Angles in Real Life Objects…
Want to see the rest of the activities for Unit 7– Geometry ?
7-1 The Undefined Terms in Geometry
7-3 Parallel and Perpendicular Lines
7-4 Measuring and Sketching Angles
7-5 Addition of Angle Measures
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Master Geometry Formulas: A Comprehensive Guide for Academic Success
Geometry Formulas Unraveled: Your Go-To Resource for Assignments
Geometry, a captivating branch of mathematics, ventures into the nuanced exploration of shapes, sizes, and dimensions within the space we occupy. The intricate world of geometry presents students with a labyrinth of concepts, demanding a profound comprehension of formulas. These formulas are not mere mathematical abstractions; they are the key to unlocking the secrets hidden within angles, polygons, circles, and three-dimensional spaces. Whether engrossed in challenging assignments, gearing up for examinations, or simply striving to grasp the fundamental principles, students stand to gain significantly from a comprehensive resource. This blog serves as a beacon, illuminating the path toward mastery in this mathematical domain by providing practical insights, real-world applications, and a holistic approach that transcends rote memorization. For those seeking assistance with your Geometry assignment , this guide aims to be an invaluable tool in achieving success and mastering the fascinating world of geometry.
As we conclude this exploration, it becomes evident that geometry is not just a theoretical construct but a dynamic tool. Armed with a profound understanding of these formulas, students can navigate the intricate terrain of mathematical problem-solving with confidence. Whether unraveling the mysteries of angles, exploring the symmetry of shapes, or delving into the realms of trigonometry, this blog encapsulates the essence of geometry, offering a comprehensive resource that transcends the boundaries of a traditional academic guide. In the tapestry of mathematical knowledge, geometry unfolds as a vibrant thread, weaving together the abstract and the concrete, and in this journey, we hope to be your steadfast companion, illuminating the path to geometric proficiency in your academic endeavors.
The Basics of Geometry
In laying the groundwork for understanding the intricate realm of geometry, it's imperative to revisit the fundamental concepts that underpin this mathematical discipline. At its core, geometry explores the properties and relationships of various elements in the space we inhabit. Beginning with the basic building blocks such as points, lines, and angles, students gain a foundational understanding that serves as a prerequisite for more advanced geometric principles. Euclidean geometry, with its axioms and postulates, forms the bedrock upon which many geometric theorems rest. Delving into the terminology and classifications of shapes, including triangles, quadrilaterals, and polygons, provides the scaffolding for the application of formulas governing perimeter, area, and other characteristics of these fundamental geometric entities. This section serves as an essential refresher, emphasizing the importance of mastering these basic concepts before venturing into the more complex and nuanced realms of geometric reasoning. As students grasp the basics, they not only develop the necessary skills for problem-solving but also cultivate a solid foundation upon which the edifice of geometric knowledge can be constructed, paving the way for a deeper exploration of the geometrical intricacies that lie ahead.
Properties of Shapes
We delve into the multifaceted realm of shape properties, unraveling the fundamental formulas that underpin geometric understanding. From the simplicity of triangles to the intricacies of polygons, each shape brings a unique set of properties and challenges. Exploring the concept of perimeter, we uncover the formulaic essentials for calculating the total length of a shape's boundary, shedding light on its significance in practical applications. Moving on to the realm of area, we navigate the intricacies of surface measurement, illustrating how these formulas are not just mathematical abstractions but crucial tools for quantifying space within shapes. Through a comprehensive examination of shapes like rectangles, circles, and irregular polygons, we decode the formulas for area, shedding light on the interplay between shape dimensions and space occupancy. Additionally, we unveil the importance of understanding the Pythagorean theorem in the context of right-angled triangles, highlighting its role in calculating sides and angles. As we unravel the properties of shapes, we emphasize the practical relevance of these formulas, showcasing how they extend beyond the classroom into real-world scenarios, from calculating material requirements in construction to determining land areas in urban planning. This section serves as a foundational guide, empowering students and enthusiasts alike with the tools needed to navigate the dynamic landscape of shape properties and apply them adeptly in problem-solving and analysis.
Circle Formulas and Applications
In the realm of geometry, circles stand out as ubiquitous and intriguing entities, playing a pivotal role in numerous facets of our daily lives. Circle formulas, encapsulating essential parameters like circumference, area, and sector calculations, serve as the mathematical backbone for understanding and manipulating these geometric wonders. The circumference, representing the boundary of a circle, is determined by the formula C = 2πr, where r is the radius. This seemingly simple formula finds applications in a myriad of fields, from calculating the length of a bicycle tire to determining the orbits of celestial bodies. The area of a circle, expressed as A = πr², elucidates the extent of space enclosed by the circular boundary, proving indispensable in fields such as land surveying and urban planning. Sector calculations, offering insights into portions of a circle, involve formulas like the area of a sector (A_sector = 0.5r²θ) and the arc length (L = rθ), where θ represents the central angle. These formulas find resonance in diverse areas, ranging from pie-chart constructions to the assessment of angles of elevation in trigonometry. As we unravel the applications of circle formulas, their ubiquity becomes apparent, resonating not only in mathematical problem-solving but also in the intricate tapestry of our tangible, real-world experiences. Whether in the precision of scientific measurements or the aesthetics of design, the understanding and application of circle formulas serve as a cornerstone in comprehending the geometrical symphony that surrounds us.
Angle Relationships and Trigonometry
Angle relationships and trigonometry form a crucial aspect of geometry, offering a deeper understanding of the relationships between angles and their applications in real-world scenarios. In the realm of angle relationships, the study involves exploring the fundamental theorems governing angles, such as the vertical angle theorem, complementary and supplementary angles, and the transversal angle theorems. These principles lay the groundwork for solving intricate geometric problems, providing a systematic approach to angle-related inquiries. Trigonometry, on the other hand, introduces the study of the ratios and functions of angles within right-angled triangles, including sine, cosine, and tangent. These trigonometric functions serve as powerful tools in measuring distances, heights, and angles, extending their applicability to fields ranging from physics and engineering to astronomy. Understanding the relationships between angles and the application of trigonometric functions empowers individuals to solve complex problems involving triangles and circular motion. Moreover, trigonometry serves as a bridge between geometry and algebra, showcasing the interconnectedness of mathematical concepts. As students delve into angle relationships and trigonometry, they not only gain proficiency in handling geometric challenges but also acquire valuable problem-solving skills with broader implications across various disciplines.
Three-Dimensional Geometry
Three-dimensional geometry adds a layer of complexity to our understanding of space, introducing a dynamic realm of shapes and structures. In this multidimensional space, volumes and surfaces take center stage, challenging us to explore formulas that go beyond the confines of traditional two-dimensional geometry. From calculating the volume of prisms and cylinders to determining the surface areas of spheres and cones, three-dimensional geometry provides a rich tapestry of mathematical concepts. The formulas associated with these shapes offer practical insights into real-world applications, influencing fields ranging from architecture to physics. Understanding spatial relationships becomes crucial as we navigate the intricacies of three-dimensional space, requiring a keen awareness of how shapes interact and transform. As we delve into this dimension, the significance of these formulas becomes evident in their role as problem-solving tools for architects designing structures, engineers conceptualizing complex systems, and scientists modeling the behavior of physical objects. Three-dimensional geometry not only expands our mathematical toolkit but also deepens our appreciation for the spatial dimensions that shape the world around us, fostering a connection between abstract mathematical concepts and their tangible manifestations in our physical reality. In this realm of mathematical exploration, students and professionals alike find themselves unraveling the complexities of volumes, surfaces, and spatial relationships, opening doors to new dimensions of understanding and problem-solving.
Transformational Geometry
Transformational geometry is a captivating branch that takes geometry beyond static shapes and explores the dynamic world of spatial transformations. In this realm, geometric figures undergo changes in position, size, and orientation, opening up a rich array of mathematical possibilities. Transformations include translations, where figures move parallel to themselves; rotations, where figures pivot around a fixed point; reflections, which involve flipping figures across a line; and dilations, where figures scale up or down. Each transformation has its set of formulas and rules, serving as tools to manipulate and analyze shapes in various contexts. Understanding transformational geometry is not only vital for solving intricate mathematical problems but also has practical applications in fields like computer graphics, art, and design. In computer-aided design (CAD), for instance, transformations play a crucial role in modeling and rendering three-dimensional objects. Furthermore, in art and animations, transformations bring life to characters and scenes by seamlessly altering their appearance. Transformational geometry adds a dynamic layer to the study of shapes, providing a bridge between mathematical abstraction and real-world applications. Mastery of these transformations empowers individuals to explore the fluidity and adaptability inherent in geometric figures, enhancing problem-solving skills and fostering a deeper appreciation for the dynamic nature of our geometric universe.
Real-World Problem Solving
In the realm of geometry, the application of formulas extends far beyond the confines of a classroom. Section Engineers rely on geometric principles to calculate stress distributions, optimize designs, and construct efficient structures. The application extends to physics, where geometric formulas come into play when determining distances, trajectories, and spatial relationships between objects in motion. From the intricate calculations involved in satellite orbit trajectories to the design of everyday objects like bridges and vehicles, geometry underpins the fabric of our built environment. Moreover, advancements in technology leverage geometric concepts, enabling innovations in computer graphics, 3D modeling, and virtual simulations. Whether it's the precise mapping of geographical landscapes or the development of cutting-edge virtual reality applications, geometry serves as the silent architect shaping the digital and physical worlds. As students immerse themselves in real-world problem-solving scenarios, they not only reinforce their understanding of geometric principles but also recognize the profound impact these formulas have on shaping the world we inhabit. Thus, Section 7 illuminates the transformative role of geometry, bridging the gap between theoretical knowledge and its tangible manifestations in the complex, multidimensional landscapes of architecture, engineering, physics, and technology.
Conclusion:
In conclusion, the journey through the intricacies of geometry formulas has revealed the underlying structure and beauty inherent in the mathematical world. From foundational concepts to advanced three-dimensional applications, this comprehensive guide serves as a valuable resource for students grappling with assignments and seeking a deeper understanding of geometric principles. As we've unraveled the mysteries of shapes, angles, circles, and transformations, it's evident that geometry is not merely an abstract discipline but a dynamic tool with practical applications across various fields. The real-world problem-solving section highlights the versatility of geometric formulas in architecture, engineering, and physics, showcasing their relevance beyond the classroom. Armed with this knowledge, students can approach assignments with confidence, recognizing the connections between theoretical concepts and their tangible implications. Moreover, the exploration of transformational geometry underscores the dynamic nature of shapes, paving the way for applications in computer graphics and animation. As we navigate the vast landscape of geometric principles, one can appreciate how these formulas are not just academic exercises but essential tools for understanding and shaping the world around us. With this go-to resource in hand, students are equipped to conquer the challenges of geometry, unveiling the elegance and precision that underlie our geometric reality.
Mathematics > Metric Geometry
Abstract: After having investigated the geodesic triangles and their angle sums in Nil and $Sl\times\mathbb{R}$ geometries we consider the analogous problem in Sol space that is one of the eight 3-dimensional Thurston geometries. We We also discuss the behavior of this surface. In our work we will use the projective model of Sol described by E. Molnár in \cite{M97}. new virtual interpolation technology with range as object
Original Paper
Published: 17 May 2024
Cite this article
Yunxiu Yang 2 ,
Wendong Chen 3 &
Qin Shu 1 interpolating step according to the mapping between the angle and the range, a high-performance interpolating method with less computation is realized. For large region virtual interpolation, on the other hand, based on multi-region interpolation technology, the region is divided into multiple nested regions to achieve high-performance transformation results and can be used to improve the interpolation accuracy in the process of range interpolation technology. The simulation results show that the range interpolation transform method has obvious advantages over the classical transform method in both transformation error and DOA estimation performance, and the angle-sensitive problem is effectively alleviated. The nested virtual interpolation can be used to achieve high-performance virtual interpolation and improve the accuracy of interpolation in large regions.
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Contributions
TL and QS contributed to the conception of the study; TL performed the data analyses and wrote the manuscript; YY and WC contributed significantly to analysis and manuscript preparation.
Corresponding author
Correspondence to Qin Shu .
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The authors declare no conflict of interest.
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Moms are flocking to use desks that have cribs attached. They say it allows them to balance motherhood with their careers.
The desks have a play area attached to keep babies and tots entertained.
Parents say that the desks allow them to work near their children.
They're best for short spurts of work, parents say.
When Maegan Moore returned to work about two months ago after the birth of her first child , she found a unique solution for balancing her career with mothering: a work-play desk that lets her care for her baby, Eleanor, while also having a dedicated workspace.
"For her age, it's been awesome," Moore told Business Insider. "I'll try to time it so I can feed her, put her down in the play desk area, and do some work that doesn't involve calls."
The work and play desk is a sort of cubicle designed for parents and children. The desk has a flat work area for parents and a play-pen-like attachment for babies.
Moore uses the desk at a coworking space in New York, but the idea originally started at a library in Virginia that largely serves a disadvantaged population that has trouble accessing both childcare and reliable internet.
Moore doesn't use the desk for long stretches of time — about an hour or two is her current limit. Although she has childcare during the middle of the day, she uses the desk most mornings and afternoons. She says that being able to have Eleanor nearby and nurse her, rather than pumping, has eased her transition back to work.
"That's been a real gift," she said.
A mom uses the desk to cope with days off school
Bethany Crystal , who contracts for multiple companies in tech and education, uses the work-and-play desk for her 21-month-old, Sydney. Because of the nature of her work, she doesn't have an office to report to. Before finding the work and play desk, she had instances of trying to nurse her baby in a WeWork coworking space or struggling to find a place to put her while she interviewed with firms. The work and play desk has solved that.
"It's really useful to be able to go places for an hour or two and have a place to put a baby," Crystal told BI.
Now that Sydney is a toddler, she's enrolled in a local Montessori school , but the frequent days off mean that Crystal still utilizes the work and play desk regularly.
"There's a lot of inservice days and holidays, which wreaks havoc for professional, entrepreneurial parents like myself," Crystal said.
Sydney is "so happy" to be next to her mother, and Crystal is able to get solid two-hour blocks of work done, which add up over the course of her 60-hour workweek .
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"Even being able to put her in the crib for little stints made such a big difference," she said.
To Crystal, the desk is representative of a modern work-life balance.
"I believe we are in a new era of work, where it's no longer about what job you want, it's about what kind of lifestyle you want," she said. "For me, it's critically important to have spaces to do work that lean into the messy complications of an imperfect life with a lot of demands."
Solving the problem of the generation
Both Moore and Crystal use the work and play desk at Workplayce , a family-friendly coworking facility on the Upper West Side of Manhattan. In addition to the desks, the building offers quiet work space, a communal area for parents to work while kids play, and dedicated on-site childcare that can be booked by the hour.
For Crystal, who also has a 4-year-old, the community aspect has been as important as the physical space.
"Not only do I have a place where kids can be kids and I can be at work, but I have fellow parents who are in work-hybrid mode," she said. "It's the first time I've felt like I've been part of a peer-parent community."
"It's refreshing to me," she added, "and helpful for us to see each other and learn how we are all making it work."
When Moore heard about Workplayce, she thought, "This is incredible. What a cool tool to equip working parents."
Prior to this, Moore tried working from home with a babysitter for Eleanor. But living in a small apartment, it was difficult "to not look around and see all the tasks and to-dos," which could distract from work, she said.
Now, she typically goes to the coworking space about four times a week, enrolling Eleanor in childcare when she has a meeting, a call, or work that requires deeper concentration.
"It's great having her around and getting to pop in and see her, but when I need full separate space, I have the ability to do that as well," Moore said.
That setup more accurately matches the lives of many modern working parents, Crystal said.
"There's a unique opportunity for parents to make work and kids fit with their lives. Decoupling work and kids from these arbitrary 9-to-5 work days or 9-to-3 day care days is the first step to making a life that works for you," she said. "This is a really important thing for our generation to solve for."
Streamlining and improving the billing process in law firms is often perceived as a daunting task, fraught with complexities that can challenge even the most seasoned legal professionals. Billing is a time consuming, labor-intensive necessary evil that is often fraught with inaccuracies due to inefficiencies in process and extensive operational silos which can cost a law firm potentially millions of dollars each year.
According to a recent Thomson Reuters survey , the average law firm experiences an 18% realization loss due to billing challenges. This means that nearly one-fifth of billable work is not converted into revenue, negatively impacting the firm's profitability. 81% of law firms report issues with a significant portion of invoices remaining unpaid or delayed, creating cash flow challenges. This is a huge revenue loss that is creating major problems for law firms, but there are solutions.
Why is legal billing so complex?
Most unique to attorneys is that their work is often time-consuming difficult to tangibly track. Tracking billable hours accurately requires meticulous attention to detail and can be particularly challenging when dealing with tasks that don't neatly fit into predefined categories. This creates problems because clients often have unique billing preferences and expectations ranging from traditional hourly billing to flat fees, contingency fees and/or blended rates. Accommodating these distinctive preferences for each client and integrating them into a billing system adds a layer of complexity to billing that makes it extremely labor intensive. Additionally, silos that exist within law firms in the form of practice groups or other operational units can make regularity in billing guidelines either difficult to understand or completely nonexistent.
Another complexity in legal billing is the regulatory compliance and ethical guidelines that must be adhered to. Some of these issues include fee agreements, billing transparency, client confidentiality, and conflicts of interest. Staying compliant with these regulations while meeting client expectations and maintaining profitability requires a thorough understanding of legal and professional standards.
While most law firms rely on billing software to streamline the process, integrating these systems with other practice management tools and ensuring data accuracy and security can present technical challenges that require ongoing maintenance and support.
Reasons to outsource billing
By streamlining the billing processes , law firms can reduce the risk of errors, and focus on core competencies. Technical expertise and best practices ensure process optimization and service quality. Outsourcing can be a valuable solution for law firms aiming to stabilize their cash flow, reduce costs, and improve overall efficiency.
Access to Advanced Technology: Outsourcing billing often provides access to advanced billing technologies and software platforms that may be cost-prohibitive for smaller law firms to implement independently. These technologies can automate billing processes, provide real-time reporting and analytics, and offer insights into billing trends and patterns that allow firms to optimize their billing practices for greater efficiency and profitability Additionally, predictive AI can provide billing staff with powerful tools to wade through complex billing guidelines in a timely manner, increasing productivity within billing departments.
Expertise and Accuracy: Billing processes can be complex, especially in law firms where billing requirements may vary from client to client and case to case. Outsourcing billing to specialized professionals ensures that billing is handled by experts who are well-versed in legal billing practices. This expertise leads to greater accuracy in invoicing, reducing the likelihood of errors or discrepancies that could lead to disputes or delays in payment. This also leads to more standardized output.
Centralization: Often, billing departments in law firms are segregated either geographically or by practice area, leading to inefficiencies and lack of scalability. Centralizing billing can assist by allowing billing to be looked at holistically as a firm rather than piecemeal, leading to a more effective business model for financial processes.
Efficiency and Timeliness: Outsourcing billing allows law firms to streamline their billing processes and improve efficiency. Billing professionals have the necessary tools and systems in place to generate invoices promptly and ensure timely submission to clients. This helps expedite the payment cycle, leading to improved cash flow for the firm.
Cost Savings: While some may initially view outsourcing as an additional expense, it often proves to be a cost-effective solution in the long run. By outsourcing billing, law firms eliminate the need to hire and train in-house billing staff, invest in billing software, and bear the overhead costs associated with maintaining billing infrastructure, lowering both soft and hard operational costs and simplifying a billing budget. Additionally, outsourcing billing allows firms to scale their billing operations up or down as needed without incurring significant fixed costs.
Compliance and Regulation: Legal billing is subject to various regulations and compliance requirements, including those related to client confidentiality, billing transparency, and fee agreements. Outsourcing billing to professionals who are experts in these regulations helps ensure compliance and mitigates the risk of potential legal or ethical issues arising from billing practices.
Focus on Core Competencies: Attorneys and legal staff excel in providing legal counsel and representation, but are not usually trained in managing billing and administrative tasks, and even if they are, legal staff turning their attention to these tasks from an administrative perspective harms the firm's bottom line by taking time away from client-facing, billable projects. By outsourcing billing, law firms can free up valuable time and resources to concentrate on what they do best, which is serving clients legal needs.
Legal billing is a challenge that costs firms millions of dollars every year. It has become such an issue that many of law firms have said this is their main priority to fix by 2025. Centralized teams face challenges due to the varied requirements of different attorneys and clients. The benefits to outsourcing billing not only will help realization rates but will foster a better relationship with clients and will help alleviate headaches of both lawyers and legal administrators as well.
The contents of this article are intended to convey general information only and not to provide legal advice or opinions.
Easy mistakes to make. Mixing up the sum of interior angles and exterior angles Students may use 360^{\circ} instead of 180^{\circ} for the sum of the interior angles of the triangle and vice versa.; Equal angles in an isosceles triangle Selecting the wrong angles when identifying the equal angles in an isosceles triangle (particularly a problem when the equal angles are not at the bottom).
Using Properties of Angles to Solve Problems
If the sum of the measures of two angles is 90∘ 90 ∘, then the angles are complementary. If angle A A and angle B B are complementary, then m∠A+m∠B =90∘ m ∠ A + m ∠ B = 90 ∘. In this section and the next, you will be introduced to some common geometry formulas. We will adapt our Problem Solving Strategy for Geometry Applications.
Elena and Diego each wrote equations to represent these diagrams. For each diagram, decide which equation you agree with, and solve it. You can assume that angles that look like right angles are indeed right angles. 1. Elena: \(x=35\) Diego: \(x+35=180\) Figure \(\PageIndex{2}\) 2. Elena: \(35+w+41=180\) Diego: \(w+35=180\) Figure \(\PageIndex ...
Angle Properties
The sum of angles on a straight line is 180˚. Interior angles are supplementary. Supplementary angles are angles that add up to 180˚. The sum of angles in a triangle is 180˚. An exterior angle of a triangle is equal to the sum of the two opposite interior angles. The sum of interior angles of a quadrilateral is 360˚.
Angles Questions (Angles Questions with Solutions)
Linear pair of angles: Sum of angles = 180°. Sum of angles at a point = 360°. 3. Find the measure of an angle which is complementary to 33°. Solution: If the sum of two angles is 90°, they are called complementary angles. Let x be the angle which is complementary to 33°. So, x + 33° = 90°. x = 90° - 33° = 57°.
Angles, polygons, and geometrical proof Angles
Hand Swap. My train left London between 6 a.m. and 7 a.m. and arrived in Paris between 9 a.m. and 10 a.m. At the start and end of the journey the hands on my watch were in exactly the same positions but the minute hand and hour hand had swopped places.
Angles
Here we have no equation to solve as the missing angle is 20^{o}. Example 2: angles at a point. Calculate the value of x. Add all known angles. 90+90+120=300. Subtract the angle sum from \bf{360^{o}} . ... Solve the problem using the information you have already gathered.
This means we are given two angles of a triangle and one side, which is the side adjacent to the two given angles. In this case we find the third angle by using Angles of a Triangle, then use The Law of Sines to find each of the other two sides. See Solving "ASA" Triangles. 4. SAS. This means we are given two sides and the included angle.
Art of Problem Solving
A straight angle is an angle formed by a pair of opposite rays, or a line. A straight angle has a measure of . A right angle is an angle that is supplementary to itself. A right angle has a measure of . An acute angle has a measure greater than zero but less than that of a right angle, i.e. is acute if and only if .
Missing Angles
Find the missing angles in a triangle, around a point, in a quadrilateral, find opposite and supplementary angles, or find angles which require multi-step problem solving skills. For more shape and space resources click here. Scan to open this game on a mobile device. Right-click to copy and paste it onto a homework sheet. Play game ...
Angle Word Problems
Students also learn the exterior angle theorem, which states that the exterior angle of a triangle is equal to the sum of the measures of the remote interior angles. Students are then asked to solve problems related to the exterior angle theorem using Algebra. Example: The measure of the angles of a triangle are in the ratio 2:5:8.
Angle notation and problem solving
5 Questions. Q1. The exterior angles of a hexagon sum to 540 degrees. Q2. A triangle ALWAYS has each exterior angle as 60 degrees. Q3. The general formula for working out the mean exterior angle of an n-sided polygon is... Q4. The calculation to work out the sum of the interior angles for an octagon would be...
PDF Lesson 10: Angle Problems and Solving Equations
NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 10•3. Lesson 10: Angle Problems and Solving Equations. Student Outcomes. Students use vertical and adjacent angles and angles on a line and angles at a point in a multi-step problem to write and solve simple equations for an unknown angle in a figure.
Find and Identify Angles in the Real World
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A new virtual interpolation technology with range as object ...
The Viral Desk Comes With a Crib; Parents Can Get Work Done
Solving the problem of the generation. Both Moore and Crystal use the work and play desk at Workplayce, a family-friendly coworking facility on the Upper West Side of Manhattan. In addition to the ...
Solving Law Firm's Biggest Problem: Outsourcing Legal Billing to
Angle. Solving Law Firm's Biggest Problem: Outsourcing Legal Billing to Improve Collection and Profitability. Legal Operations; 2 Mins; Streamlining and improving the billing process in law firms is often perceived as a daunting task, fraught with complexities that can challenge even the most seasoned legal professionals. Billing is a time ... | 677.169 | 1 |
The Elements of Plane Geometry ...
From inside the book
Results 1-5 of 28
Page 10 ... sides and contained angle ; ( 2 ) two angles and side adjacent ; ( ) two angles and side opposite . 7. The drawing of tangents to circles , under various condi- tions . 8. The inscription and circumscription of figures in and about ...
Page 24 ... sides , a pentagon one of five sides , a hexagon one of six sides , and so on . DEF . 27. A triangle is a figure contained by three straight lines . DEF . 28. Any side of a triangle may be called the base , and the opposite angular ...
Page 25 ... sides equal , then the triangles are identically equal , and of the angles those are equal which are opposite to the equal sides . Let ABC , DEF be two triangles having the side AB equal to the side DE , the side AC to the side DF , and ...
Page 27 ... sides between the vertices of these angles equal , then the triangles are identically equal , and of the sides those are equal which are opposite ... side BC to the side EF : A E then shall the triangles be identically equal , having the ...
Page 28 ... side AC to the side DF , and the side AB to the side DE . Q.E.D. Ex . 10. If the bisector of an angle of a triangle is also perpendi- cular to the opposite side , the triangle is isosceles . THEOR . 7. If two sides of a triangle are | 677.169 | 1 |
3D Geometry Basics with MCQs
Three-dimensional geometry is a branch of mathematics that deals with the properties and relationships of objects in three-dimensional space. It is an essential part of many fields such as physics, engineering, and computer graphics. Understanding the concepts of three-dimensional geometry is crucial for solving problems related to distance, volume, and surface area.
One of the fundamental concepts in three-dimensional geometry is the distance formula. This formula is used to find the distance between any two points in three-dimensional space.
The formula is given by:
Distance = √((x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2)
For example, if two points in three-dimensional space are given as (3, 5, 8) and (6, 8, 12), the distance between the two points can be found by substituting these values into the
In conclusion, three-dimensional geometry is an essential branch of mathematics that deals with the properties and relationships of objects in three-dimensional space. Understanding the concepts of distance, volume, and surface area is crucial for solving problems related to three-dimensional geometry. With practice and persistence, anyone can master the concepts and solve problems with ease.
Practice Questions on 3D Geometry:
1. What is the formula for the distance between two points in three-dimensional space? | 677.169 | 1 |
Understanding the Concept of Perpendicular Bisectors
A perpendicular bisector is a line that divides another line into two equal parts at a 90° angle. By breaking down the term, we can see that "perpendicular" means intersecting at a right angle, and "bisector" means dividing into two equal parts.
Take a look at the image below to see a visual representation of a perpendicular bisector. | 677.169 | 1 |
There are also special cases of right triangles, such as the 30° 60° 90, 45° 45° 90°, and 3 4 5 right triangles that facilitate calculations. Where a and b are two sides of a triangle, and c is the hypotenuse, the Pythagorean …
The online 30-60-90 triangle calculator does the following computations: Length of second side Hypotenuse Area Height Perimeter Inradius Circumradius But what to do if you are …
30 60 90 Triangle | 677.169 | 1 |
Q3. Find the area of a triangle ABC whose vertices are A(2,7),B(3,-1) and C(-56).
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