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Unit 1 geometry basics quiz 1 1 answer key.
Name: _____ Unit 1: Geometry Basics Date: _____ Per: _____ Homework 1: Points, Lines, and Planes. Use the diagram to answer the following questions. Use the diagram to answer the following questions. z3. Use the diagram to answer the following questions. a) How many points appear in the figure?
Oct 1, 2020 · Name I Unit 1 Test Study Guide (Geometry Basics) Dole: 2 October 1020 re Topic is Points, Lines & Planes Use the diagram to the right to envever questions 1-4. 1. Name two point collhear to poli K. J 2. Give another name for Ine b. J. Name the Intersection of ne e and plane R. _E point 4. Esther Bensadon Unit 1: Geometry Basics Date: Per: Homework 2: Segment Addition Postulate ** This is a 2-page document! * * Use the diagram below to 1. If LM = 22 and MN = 15, find LN. answer questions 1 and 2. 22+15 = 37 2. If LN = 54 and LM = 31, find MN. M N 54 +31 : 85 3. If RT = 36, find the value of x. 4.Ge –AE –OY by EB 3. Name the intersection of line and plane X. ... Device 1 Test Review Answers ... geometry unit 1. Study. Flashcards ...
plane and line. Undefined Geometric Term described as a location on a coordinate plane that is designated by an ordered pair (x,y) point. Undefined Geometric Term described as a two-dimensional set of points that has no beginning or end. plane. Dimensions of a Line. A line has one dimension because it is made up of all points that extend ... Unit 1. 6.1 Area and Surface Area. Reasoning to Find Area. Lesson 1 ... Open Up Resources 6-8 Math is published as an Open Educational Resource.To cut something into more than five pieces. It is a plane with two sets of wings. A shape that has three sides. To cut something into two congruent pieces or in half. 14. Multiple-choice. 30 seconds. 1 pt.Unit 1. 6.1 Area and Surface Area. Reasoning to Find Area. Lesson 1 ... Open Up Resources 6-8 Math is published as an Open Educational Resource. Quick HereFor example, 4x2 and −9y2 don't have a common factor, but the whole polynomial is still factorable: 4x2−9y2 = (2x + 3y)(2x−3y). 3. Divide the x term into the sum of two terms, factor each portion of the expression separately, and then factor out the GCF of the entire expression. 5. 7m. 7.idellemaldonado42. Answer: ur mom doesn't like you. Step-by-step explanation: 1: u found out u were. arrow right. Explore similar answers. messages. Get this answer verified by an Expert.Unit 1 Test Geometry Basics Answers. Gina Wilson All Things Algebra 2015 Unit 1. Houghton Mifflin provides answer keys online as well as for printable resources. Test 1 units Rate lesson poor Rate Geometry Unit 2 Test Quizlet Chapter 2 standardized test practice answers geometry 0. Substitution PoE or transitive 3.Geometry CC RHS Unit 1 Points, Planes, & Lines 5 Regents Test Basic Concepts and Definitions Questions 1 through 9 refer to the following: 1) PR and TS determine a plane. TRUE FALSE 2) Points T, Q, and S TRUEare collinear. FALSE 3) S is between P and R. TRUE FALSE 4) Points P, Q, and R TRUEare collinear. FALSEWorksheets are Lets practice, Unit 1 points lines and planes homework, Lines and angles, Gina wilson all things algebra 2014 answers pdf, Gina wilson all things algebra 2013 answers, Geometry unit 10 notes circles, Geometry unit answer key, Identify points lines and planes. *Click on Open button to open and print to worksheet. 1.
Mathleaks provides student friendly solutions, answers, and hints to all exercises for commonly used textbooks in Integrated Mathematics. Our solutions consist of clear and concise explanations, always explained simply with figures and graphs, with step-by-step calculations and the included theory. Using Mathleaks gives more benefits than just | 677.169 | 1 |
The Definition Of A Circle Uses The Undefined Term
The term "circle" is used when you want to define a shape, but you don't know how to write it. You can use the undefined term, or just say "circle," which is pretty confusing. You have to have a point and a plane on which to draw the circle. You can also use the undefined term to talk about triangles. A triangle is a sphere whose perimeter is the same as its diameter.
The definition of a circle, for instance, uses the undefined term to refer to the slopes of a circle. This is a common practice in mathematics. For example, a circle's surface area is not exactly flat. A more correct definition would use a radial distance. A tangential length, on the other hand, is a line whose length is one half of its radius.
If you want to find out how to define a circle, first consider the definition of points. These terms are important because they refer to positions, such as a plane. A point can also be a small or large point. In addition to the dimensions, a point can represent a plane. This is why the definition of a circle uses the undetermined term "plane." A line is a set of parallel points.
Another important fact to remember is that the definition of a circle should be precise and accurate. It must include the units and their edges. Then, we can use the undefined term "unit circle" to describe the right-angled triangle and other shapes. It is a necessary prerequisite for theology and other fields based on geometry. So, it's crucial to have a precise and clear definition of a circle to be able to make your own interpretations and definitions of geometric shapes.
In addition to this, you should also consider the different types of circles in order to create the right concept. The term "unit circle" refers to a circle with two convex edges. A second unit circle, in contrast, is a cylinder with no point. The term 'unit circle' is a polygonal unit with a fixed axis. This means that the definition of a circle is convex.
The definition of a circle is an undefined term. It is important to understand how a circle is defined and how it is constructed. For example, a square is a sphere and a hexagon is a polygon. A rectangular sphere is a rectangle. A triangle is a square whose circumcenter is the centre. The unit of a triangle is a tangent.
The undefined term is also used in dynamic geometry. The undefined term in a triangle is not used when describing a circle. In fact, the undefined term in a circle refers to the undefined term in a triangle. It is not necessary to define the term "circle" in your textbook. The definition of a circle is the same as the triangle's.
The definition of a circle is not the same as the definition of a triangle. The undefined term of a triangle refers to an angle. It is a set of points that intersect in the same plane. This type of triangle is also known as a dot. However, the undefined term of a circle is not the same as a point. The word "point" is defined as a dot.
A circle is a closed, two-dimensional curved shape. It has no corners and is divided into three parts by a plane. It is a two-dimensional figure that separates triangles. The definition of a triangle, on the other hand, is a plane that divides a square into three parts by the undefined term. Likewise, a line has no defined slope. | 677.169 | 1 |
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Consecutive Interior Puzzle
In this activity students will solve 10 different consecutive angles questions. They will be given a set of parallel lines and have to solve for x. If they solve the question correctly, the box will turn from red to green and part of the puzzle will reveal. The questions have to be done in order for the puzzle to reveal. The picture is large so students will need to scroll to see all of the questions and the picture | 677.169 | 1 |
The Elements of Euclid; viz. the first six books,together with the eleventh and twelfth, with an appendix
Dentro del libro
Resultados 1-5 de 67
Página 10 ... base BC shall coincide with the base EF ; because the point B coinciding with E , and C with F , if the base BC did not coincide with the base EF , two straight 10 Ax . lines would enclose a space , which is impossible . * Therefore the ...
Página 11 ... bases shall likewise be equal , and the triangles shall be equal , and their ... base shall be equal . Let ABC be an isosceles triangle , of which the side ... B C * 4. 1 . F G D E ACF to the angle ABG , and the angle AFC BOOK I. PROP ...
Página 12 ... base of the triangle ABC : and it has also been proved that the angle FBC is equal to the angle GCB , which are the ... BC common to both , the two D A sides DB , BC are equal to the 12 EUCLID'S ELEMENTS .
Página 13 Euclides William Rutherford. D A sides DB , BC are equal to the two AC , CB , each to each ; and the angle DBC is equal to the angle ACB ; therefore the base DC is equal to the base AB , and the triangle DBC is equal to the triangle ACB ...
Página 14 ... base CD are equal to one another ; but the * angle ECD is greater than the angle BCD ; therefore the angle FDC is ... BC equal to the base A EF ; the angle BAC shall be equal to the angle EDF . For , if the triangle ABC be applied to DEF | 677.169 | 1 |
What will be the face correction made on bright line, if the radius of the signal is given as 24m, angle with which the station points make with the sun is given as 24?12? and the distance between the station points is given as 85m? | 677.169 | 1 |
Monthly Archives: June 2015 downIt's been about 3 years since I taught Geometry and there are people doing a lot more creative and deep-thinking stuff than myself (see: Jim, Julie, Lisa,Mimi, Shireen, and many others), but ugh, having to draw a geometry diagram in Word?
(Hint: draw it in Geogebra, then use the snipping tool to grab it and insert, but even then it's a pain). So just to save you some time and hassle, I'll also be trying to share my Geometry files this summer; stay tuned to this page! Since it has been three years, much of it will be presented without comment. 🙂
Ah, the dreaded first chapter full of definitions. I guess I hated it so much I didn't keep many files, but here's what I do have:
Then finally (finally!) we can start on some real from-scratch proofs:
File here. Also, look below at me being clever with using "G.I.F.T." for proofs! And also Color With A Purpose. I also tell them that you should use every line in a proof, which is what the arrows are for. It seems to help if they're stuck, "Well, is there anything you haven't used yet?" | 677.169 | 1 |
C program to check whether triangle is valid or not if sides are given
Write a C program to input side of a triangle and check whether triangle is valid or not using if else. How to check whether a triangle can be formed or not if sides of triangle is given using if else in C programming. Logic to check triangle validity if sides are given in C program.
Example Input
Input first side: 7
Input second side: 10
Input third side: 5
Output
Triangle is valid
Required knowledge
Property of triangle
A triangle is valid if sum of its two sides is greater than the third side. Means if a, b, c are three sides of a triangle. Then the triangle is valid if all three conditions are satisfied a + b > c a + c > b and b + c > a
Logic to check triangle validity using nested if – best approach
Despite of easiness, the above code is messy and less readable. In above code printf("Triangle is not valid."); statement is unnecessarily repeated for various conditions. You can cut the extra printf("Triangle is not valid."); statement using a flag variable.
Let us suppose a temporary variable valid initialized with 0 indicating triangle is not valid. The main idea is to check triangle validity conditions and if triangle is valid then set valid variable to 1 indicating triangle is valid. Finally, check if(valid == 1) then triangle is valid otherwise not valid. | 677.169 | 1 |
What is Circumcentre of a circle?
Introduction to Circles
The circumcenter of a circle is an important concept in geometry, closely related to triangles and circles. It is a point at the intersection of the perpendicular bisectors of the sides of a triangle. This point is equidistant from the three vertices of the triangle, and it also lies on the circumference of the circle that passes through the three vertices. The circumcenter plays an important role in several geometric properties and constructions involving triangles and circles.
To better understand the concept of the circumcenter, it is important to have a solid understanding of triangles, their properties, and the relationship between triangles and circles. Let's take a closer look at the circumcenter.
Properties of the Circumcenter
The circumcenter has several important properties that make it a valuable point of interest in geometry. Here are some important properties of the circumcenter:
1. Equidistance: The circumcenter is equidistant from the three vertices of the triangle. This means that the distance from the circumcenter to each vertex is the same. This property is crucial in many geometric constructions and proofs.
2. Circle intersection: The circumcenter lies on the circumference of the circle that passes through the three vertices of the triangle. This circle is called the circumcircle. The circumscribed circle is unique to each triangle and its center is the circumcenter.
3. Perpendicular bisectors: The circumcenter is the intersection of the perpendicular bisectors of the sides of the triangle. A perpendicular bisector is a line that divides a side into two equal parts and is perpendicular to that side. The circumcenter is the only point that satisfies this condition for all three sides of the triangle.
4. Type of triangle: The position of the circumcenter relative to the triangle can tell you the type of triangle. For example, if the circumcenter is inside the triangle, the triangle is acute. If it is outside, the triangle is obtuse. If the circumcenter coincides with the center of the hypotenuse, the triangle is a right triangle.
Construction of the Circumcenter
The circumcenter can be constructed using a compass and a straightedge. Here's a step-by-step procedure for constructing the circumcenter:
Step 1: Draw the given triangle on a piece of paper or in a geometric construction software.
Step 2: Construct the perpendicular bisectors of the sides of the triangle. To do this, take each side of the triangle and draw a line perpendicular to it through its center. Repeat this step for each of the three sides.
Step 3: Find the intersection of the three perpendicular bisectors. This point is the circumcenter of the triangle.
Step 4: Check the properties of the circumcenter by measuring the distances from the circumcenter to each vertex of the triangle. The distances should be equal, confirming the equidistance property.
Applications of the Circumcenter
The concept of the circumcenter finds applications in various areas of mathematics, engineering, and computer science. Here are some notable applications:
1. Triangulation: In computational geometry, the circumcenter is used in triangulation algorithms. Triangulation is the process of dividing a complex shape into a series of triangles, which can aid in various tasks such as mesh generation, terrain modeling, and computer graphics.
2. Geometric Analysis: The circumcenter is a fundamental tool in geometric analysis. It helps to understand and prove properties of triangles and circles, such as the relationship between angles, side lengths, and circumradii.
3. Engineering and architecture: The concept of circumcenter is essential in fields such as civil engineering and architecture. It plays a role in structural analysis, determining centers of gravity, and designing geometrically balanced structures.
4. Navigation and Surveying: The circumcenter assists in navigation and surveying applications. It can be used to calculate the radius and position of the circumcenter, which can be helpful in determining distances, angles, and positions in the field.
Conclusion
The circumcenter is an important point in geometry, located at the intersection of the perpendicular bisectors of the sides of a triangle. It has unique properties that make it a valuable concept in various geometric applications. The circumcenter is equidistant from the three vertices of the triangle and lies on the circumference of the circumcircle. Its construction involves drawing the perpendicular bisectors of the triangle's sides and locating their intersection. The circumcenter has applications in fields such as computational geometry, geometric analysis, engineering, architecture, navigation, and surveying. Understanding the circumcenter and its properties enhances our understanding of triangles, circles, and their relationships in science.
FAQs
What is the Circumcentre of a circle?
The circumcentre of a circle is the point where the perpendicular bisectors of the sides of a triangle intersect. It is the center of the circle that passes through all the vertices of the triangle.
How is the circumcentre of a circle determined?
The circumcentre of a circle can be determined by finding the intersection point of the perpendicular bisectors of any two sides of a triangle. The perpendicular bisector is a line that cuts a side of a triangle into two equal halves at a 90-degree angle. The circumcentre is the point where these bisectors intersect.
Does every triangle have a circumcentre?
Not every triangle has a circumcentre. A triangle must be non-degenerate, which means it cannot be a straight line or a point. If a triangle is non-degenerate, it will have a unique circumcentre.
Where is the circumcentre located in different types of triangles?
In an acute triangle, the circumcentre lies inside the triangle. In an obtuse triangle, the circumcentre lies outside the triangle. In a right triangle, the circumcentre is located at the midpoint of the hypotenuse.
What are the properties of the circumcentre?
The circumcentre of a triangle has several interesting properties:
It is equidistant from the three vertices of the triangle.
It is the center of the circle that passes through all the vertices of the triangle.
The circumradius, which is the radius of the circle passing through the vertices, is the distance between the circumcentre and any of the vertices. | 677.169 | 1 |
Let b and a be end-to-end on a side of an arbitrary angle. Call the vertex of the angle A, the point between b and a B, and the end of a C. Describe how to construct a segment of length x that satisfies the given proportion using this angle. | 677.169 | 1 |
White Right-Pointing Small Triangle ▹
Symbol Meaning
In chatting and posting this shape is used at the start of a bulleted list item, and it can also be used to create symbol combinations that decorate web pages.
IQ tests present different types of shapes, including complex ones made up of two figures. The small white triangle can be inserted into a larger shape to create a complex figure that can be used to establish various patterns.
Finally, this symbol is used in creating pictograms, rebuses, and riddles.
The symbol "White Right-Pointing Small Triangle" is included in the "Geometric shapes" subblock of the "Geometric Shapes" block and was approved as part of Unicode version 1.1 in 1993. | 677.169 | 1 |
PARALLEL
Parallel
In geometry, parallel lines are lines in a plane which do not meet. The assumed existence and properties of parallel lines are the basis of Euclid's parallel postulate. Two lines in a plane that do not intersect or touch at any point are said to be parallel. By extension, a line and a plane, or two planes, in three-dimensional Euclidean space that do not share a point are said to be parallel.
One of a series of long trenches constructed before a besieged fortress, by the besieging force, as a cover for troops supporting the attacking batteries. They are roughly parallel to the line of outer defenses of the fortress.
A character consisting of two parallel vertical lines, used in the text to direct attention to a similarly marked note in the margin or at the foot of a page | 677.169 | 1 |
Q&A: What are some Common Errors and Misconceptions about the Pythagorean Theorem?
Applying the Pythagorean Theorem on non-right triangles. (They may also think that the word "hypotenuse" means the longest side of *any* triangle.)
Assuming that the missing side is always the square root of the sum of squares of the other sides -- even when that missing side is a leg, not the hypotenuse. (They think the unknown side is always $c$ in the formula $a^2 + b^2 = c^2.$)
Writing their final answer with $\pm$ even though the side length cannot be negative. (They use $c = \pm \sqrt{a^2 + b^2}$ because they're used to solving quadratic equations with no domain constraints, e.g., $x^2 = 9 \Rightarrow x = \pm 3.$)
Mindless Mistakes in Algebra
Forgetting to take the square root at the end (i.e., $c = a^2 + b^2$).
"Canceling out" exponents of all terms in an equation (i.e., $a^2 + b^2 = c^2$ $\Rightarrow a + b = c$). Note that while this is technically the same error as distributing the square root, most students who make this error won't understand that and will continue to make it even if they understand that they cannot distribute a square root. | 677.169 | 1 |
A point P moves inside a triangle formed by A(0,0) , B(1,√3.) , C(2,0) satisfying PA<=1 ,PB<=1 and >PC<=1 .If the area bounded by the curve traced by P is equal to aπ/b , then find the minimum value of (a+b).
I have tried to consider P as (h,k) and then related the distances of AP , BP , CP According to the condition given in the question.
Since only one option amongst AP<= 1,BP<= 1,CP<= 1 has to be satisfied , therefore I tried considering only one option at a time.
That is:
1) if AP<=1:
But then again , I am overwhelmed by the number of variables in the question since I need to find the locus of the point P and find the area of the curve traced by P.
Can Anyone give me a hint on how to solve this problem? | 677.169 | 1 |
Answer:x = 63z = 117x + z = 180Step-by-step explanation:x is a "corresponding" angle for the one marked 63°. Here, corresponding angles are congruent, so x = 63°.__x and z are "same-side interior angles," so are supplementary. Their sum is 180°. x + z = 180°.__Because x and z are supplementary and z and 63° are supplementary, you know that ... z = 180° -63° z = 117°_____Comments on the other answer choicesThe value of y can be computed using the fact that the sum of angles interior to the triangle is 180°. The unmarked triangle interior angle is a vertical angle to the one labeled 63°, so it is 63°. Then the value of y is ... y° = 180° -47° -63° = 70° . . . . . . . not 47°__We already know that 63° + y + 47° = 180° and x = 63°. That means ... x + y + 47° = 180°so it cannot be true that x+y = 180. | 677.169 | 1 |
...to the angle CHG; and they are adjacent angles ; but when a ftraight line ftanding cet on a ftraight line makes the adjacent angles equal to one another, each of them is a right angle, and the ftraight line which flands upon the other is called a perpendicular to it; therefore from the given...
...two, thus : ABC, or СВЛ. 11. When one ftraight line falls upon another ftraight line, fo as to make the adjacent angles equal to one another, each of them is a right angle and the ftraight Une which falls upon the other is perpendicular to it. See plate i.fg- 4. I г. An angle which...
...the angle AFE is equal b to the angle BFE : But when a flraight line ftanding upon another nv.ik.es the adjacent angles equal to one another, each of them is a right c angle: Therefore each of the angles AFE, Bl' E is a right angle ; wherefore the ftraight line CD,...
...to the angle , CHG ; and they are adjacent angles ; but when a ftraight line ftanding on a ftraight line makes the adjacent angles equal to one another, each of them is a right angle, and the ftraight line which ftands upon the other, is called a perpendicular to it ; therefore from the given...
...angle CHG; and they are adjacent e. 8. i. :angles. but when a ftraight line flanding on a ftraight line makes the adjacent angles equal to one another, each of them is a right angle, and the ftraight line which ftands upon the other is called af perpen- f. to.Def.t. dicular to it. thereforefrom...
...therefore the angle AFE is equal b to the angle BFE. And when a ftraight line flanding upon another makes the adjacent angles equal to one another, each of them is a right c angle : Therefore each of the angles AFE, BFE is a right angle; wherefore the ftraight line CD, drawn...
...by a letter placed at that point ; as the angle at E.' VII. When a straight line standing on another straight line makes the adjacent angles equal to .one another, each of the angles is called a right angle ; and the straight line which stands on the other is called a perpendicular...
...meet together, but are not in the same straight line. 10. When a straight line standing on another straight line makes the adjacent angles equal to one another, each of the angles is called a right angle; and the straight line which stands on the other is called a perpendicular...
...Therefore the angle ADG is equal0 to the angle GDB : But when a straight line standing upon another straight line makes the adjacent angles equal to one another, each of the angles is a right angle d: Therefore the angle GDB is a right angle: MoDef. ' But FDB is likewise...
...equal (8. 1.) to the angle CHG ; and thry are adjacent angles ; now when a straight lina standing on a straight line makes the adjacent angles equal to one...perpendicular to it ; therefore from the given point C a pendicular CH has been drawn to the given straight line AB. Which ivas to be done. PROP. XIII. THEOR.... | 677.169 | 1 |
...in the line bisecting the angle between them, are equal to each other. 4. The straight lines which join the extremities of two equal and parallel straight...same parts, are also themselves equal and parallel. 5. To describe a square upon a given straight line. 6. If a straight line be divided into any two parts,...
...circles made use of in the construction must necessarily cut each other. 4. The straight lines which join the extremities of two equal and parallel straight lines towards the same parts, are themselves equal and parallel. What parallelograms have their diagonals equal ? 6. If a straight line...
...of the other to which the equal sides are opposite. XXXV.— EUCLID I. 33. The straight lines which join the extremities of two equal and parallel straight lines towards the same parts are themselves also equal and parallel. Let AB, CD (Fig. 28) be equal and parallel straight lines joined...
...are either two right angles, or are together equal to two right angles. 2. The straight lines which join the extremities of two equal and parallel straight...same parts, are also themselves equal and parallel. VOLUNTARY PORTION. 1. Describe an isosceles triangle having each of the angles at the base double of...
...3, and which is equal to ^ when 2 is added to the denominator. EUCLID. 1. The straight lines which join the extremities of two equal and parallel straight...same parts are also themselves equal and parallel. 2. If a straight line be bisected and produced to any point, the rectangle contained by the whole line...
...figure are equal to four right angles, (ax. 3.) PROPOSITION XXXIII. THEOREM. The straight lines which join the extremities of two equal and parallel straight...towards the same parts by the straight lines AC, BD. Then AC, BD shall be equal and parallel. Join BC. Then because AB is parallel to CD, and BC meets them,...
...triangle are together equal to a right angle. PROPOSITION XXXIII. THEOREM. The straight lines which join the extremities of two equal and parallel straight...same parts, are also themselves equal and parallel. Given AB and CD, two equal and parallel straight lines, joined towards the eame parts by the straight...
...From a given point to draw a straight line equal to a given straight line. 2. Straight lines which join the extremities of two equal and parallel straight lines towards the same parts are themselves equal and parallel. 3. Every parallelogram having one angle a right angle has all its angles...
...angles are equal to four right angles, (ax. 3.) PROP. XXXIII.— THEOREM. The straight lines which join the extremities of two equal and parallel straight...same parts, are also themselves equal and parallel. (References— Prop. i. 4, 27, 29.) Let A I"., CD, be two equal and parallel straight lines, and joined...
...perpendicular let fall on it from the opposite angle, and the acute angle. 10. The straight lines which join the extremities of two equal and parallel straight...same parts, are also themselves equal and parallel. V. Algebra. 1. If x = 4, y = 3, z = i, n = 2, what is the value of (x1 -y')-(xy)' + n(xyz) ? Simplify... | 677.169 | 1 |
Wings of most butterflies are identical on the left and right sides. HI- if you fold them they will have a horizontal line of symmetry. Another symmetry that children sometimes use in their Pattern Block designs is Rotational Symmetry. A figure has rotational symmetry if some rotation (other than a full 360 turn) produces the same figure. This center line or axis can be located either vertically or horizontally. This line may also be called a sleeping straight line that parts an image or shape into identical halves. A, H, M, O, U, V, W, T, Y are some of the alphabets that can be divided vertically in symmetry. The word OHIO is symmetrical in all three senses, because each letter displays horizontal, vertical, and rotational symmetry. They all have vertical symmetry. Email this page to a friend.Also: Sign up for our free web site updates here. A line of symmetry is one that divides a figure into two halves which are mirror images of each other. H-If you fold this on top of eachother I the H's line in the middle will go on top of I. The word symmetry originates from the ancient Greek word, 'symmetria' and approximately means 'equal proportions' or an 'agreement in dimensions'. Which is more common horizontal symmetry or vertical symmetry? Radial Symmetry. Answers is the place to go to get the answers you need and to ask the questions you want answered by Sara February 6, 2010 Here are letters that have symettry about a vertical fold O, I, M, W, T, A, H,U,V,X,Y Does that help? Vertical Line of Symmetry :The axis of the shape which divides the shape into two identical . Which alphabet has horizontal symmetry line? Alternatively, you can use the Ctrl+E keyboard shortcut. 3. Some horizontally symmetrical words are: CHEEK, BIKE, BOX, BOOK, HIDE, X-BOX. 3) Three mutually perpendicular planes of reflection symmetry. The symmetry line or horizontal axis of a shape which divides the shape into two identical halves is known as horizontal line of symmetry. 4) Five planes of reflection symmetry. Types of Line of Symmetry Vertical Line of Symmetry. Functional cookies help to perform certain functionalities like sharing the content of the website on social media platforms, collect feedbacks, and other third-party features. 1455 Quebec Street B, C, D, E and K have horizontal lines of symmetry. Each symmetry operation has a corresponding symmetry element, which . Letters that have vertical lines of symmetry include A, M, T, U, V, W and Y. Since we launched in 2006, our articles have been read billions of times. 2 has no symmetry. Wordplay: Words with vertical symmetry AUTOMATA AUTOTOMY HIMATIA HOITY-TOITY HOMOTAXIA MAHATMA MAHIMAHI MAMMATI MAMMOTH MATAMATA MOTIVITY MOUTH-TO-MOUTH MYOMATA y=-2 y= 2. Theres also an easy way to make this happen. Press ESC to cancel. And some, like P, R, and N, have no lines of symmetry. boolean horizontal = true; boolean vertical = true; // Checking for Horizontal Symmetry. Which English letter has horizontal and vertical symmetry? They all have vertical symmetry. A common misconception found even in many glossaries and texts: Not all lines that divide a figure into two congruent parts are lines of symmetry. 5 Which alphabet has horizontal symmetry line? Letters with horizontal symmetry are B, C, D and E. The letters K, F, G, J, K, L, N, P, Q, R, S and Z all do not have lines of symmetry. Letters with tails or embellishments may lose their lines of symmetry, while fonts that produce uniform letters retain them. This new shape the combination of the triangular half of the original rectangle and its image in the mirror is called a kite. If you drew a horizontal line across the word and folded it over, it would overlap on itself. We also use third-party cookies that help us analyze and understand how you use this website. Analytical cookies are used to understand how visitors interact with the website. 5 Tips to Making Your Writing More Exciting. For example, in general, human faces are identical on the left and right sides. The Diagonal Line of Symmetry His sister has wallet-sized photos with a width of 2.5 inches made of the same photograph. The Horizontal Line of Symmetry When a horizontal line divides an object into two identical halves, it is called a horizontal line of symmetry. 6 has no symmetry. A symmetry operation is an action that leaves an object looking the same after it has been carried out. 1) A single plane of reflection symmetry. Vertical symmetry can be observed in the majority of simple shapes, like squares, some triangles, circles, hearts, hexagons and octagons. 2) Two perpendicular planes of reflection symmetry. Needless to say, it is relevant whether the word is written in upper or lower case. The cookie is used to store the user consent for the cookies in the category "Performance". figure (1 point) yes no 3. Word records: consecutive letters. (There are no vowels in the bottom row.) So, there are only 5 capital English alphabets having a single and horizontal line of symmetry. Vancouver, BC Definitions of symmetric. RELATED: How to Change the Alignment of the Numbers in a Numbered List in Microsoft Word. This photograph shows a simple picture with a mirror placed along the line of symmetry. True or false: The given shape has a vertical line of symmetry. Angry, Hungry, and GRY words What is the third word ending in GRY. Using only the fingers of the right hand, one can type the 12-letter word JOHNNY-JUMP-UP (common name for a flower) including the hyphens, and also PHYLLOPHYLLIN (13 letters). Closely related to palindromes, symmetrical words are words that have an axis of symmetry, or point of rotational symmetry. We gratefully acknowledge that Science World is located on the traditional, unceded territory of the xmkym (Musqueam), Swxw7mesh (Squamish) and slilwta (Tsleil-Waututh) peoples. A vertical line of symmetry is that line that runs down vertically, divides an image into two identical halves. The alphabets T, U, V, W, X and Y also possess vertical symmetry as the line divides then equally in both the parts. Figures with one line of symmetry are symmetrical only about one axis. The longest words with vertical symmetry are OTTO, MAAM, and TOOT. Which alphabet has horizontal line of symmetry? 1 What words have vertical lines of symmetry? For example, the diagonal of a (non-square) rectangle is not a line of symmetry. There are four types of symmetry that can be observed in various cases. The cookie is set by the GDPR Cookie Consent plugin and is used to store whether or not user has consented to the use of cookies. F,G,J,L,N,P,Q,R,S, and Z letters dont have line of symmetry. Choose active voice and vivid verbs. What words have vertical lines of symmetry? Others include MOM, WOW, AHA, AHA, AIA, AMA, AVA, AWA, HAH, HOH, HUH, MAM, MIM, MUM, OHO, OXO, TAT, TIT, TOT, TUT, UTU, VAV, and WAW! This tool also allows you to be selective with the text you align horizontally, giving you more control over what you can do to your document. If hyphens are allowed, then TEETER-TOTTER (12 letters) is possible, as is TEETER-TOTTERER (14 letters), although the latter is not found in dictionaries. In 2001, the ten most frequently looked up words in the Cambridge Dictionaries Online were SERENDIPITY, IDIOM, PARADIGM, UBIQUITOUS, DICTIONARY, PRAGMATIC, EFFECT, GRY, JINGOISM, and FOIBLE. This is a straight line, but it's neither horizontal or vertical. That means the horizontal line of symmetry goes from left to right (or vice versa) in an object. Given below is a list of statements explaining the symmetry of letters from A-Z. For example, the following shape can be split into two identical halves by a standing straight line. If you turn or rotate the starfish about point P, it will still look the same from all directions. But opting out of some of these cookies may affect your browsing experience. By clicking Accept All, you consent to the use of ALL the cookies. Remember vertical means from top to. And some, like P, R, and N, have no lines of symmetry. Needless to say, it is relevant whether the word is written in upper or lower case. Which is a word with a point of symmetry? The same study showed the words most frequently misspelled (by gross count) to be: RECEIVE, A LOT, AMATEUR, SEPARATE, REALIZE, THEIR, DEFINITE, INDEPENDENT, WEIRD, EMBARRASS, ARGUMENT, NO ONE, ACQUIRE, ACCIDENTALLY, OCCURRENCE, COLLECTIBLE, RIDICULOUS, MANEUVER, LIAISON, GAUGE, ATHEIST, GRAMMAR, SUPERSEDE, KERNEL, and CONSENSUS. How much longer should the Sun remain in its stable phase? This opens the Page Setup dialog box. CODEBOOK, COOKBOOK, EXCEEDED and ICEBOX are among the others, while HOITY-TOITY, provided it is written in a column, would be the longest word with a line of vertical symmetry. Axis of Symmetry : THe line that divides the figure into two identical parts is called Axis of Symmetry. 6 Which of the following has a horizontal line of symmetry? However, you may visit "Cookie Settings" to provide a controlled consent. A line of symmetry is one that divides a figure into two halves which are mirror images of each other. How does violence against the family pet affect the family? Vertical Line of Symmetry. For example, a square has a symmetrical shape. There are multiple ways to identify the line of symmetry in an item, shape or object. The two concepts have nothing to do with one other. The longest is NONSUPPORTS (11 letters), although if place names are allowed then the 13-letter words TUTTOQQORTOOQ (in Greenland) and ROSSOUWSPOORT (in South Africa) are valid. The longest words with vertical symmetry are OTTO, MAAM, and TOOT. You can use words or make a diagram. 1 Which words have both vertical and horizontal symmetry? Horizontal symmetry follows the same rules as vertical symmetry, only along a horizontal axis. About a half inch below the lines,draw a half inch horizontal line that is centered between the lines 3. Rotational symmetry exists when a shape is turned, and the shape is identical to the origin. We say that the original figure is symmetric with respect to the mirror; it has reflective symmetry. His illustration work has been published in the Walrus, The National Post, Readers Digest and Chickadee Magazine. Some, like P, R, and N, have no symmetry lines. The selected text will now reflect the selected vertical alignment option. Nature has plenty of objects having symmetry. These cookies help provide information on metrics the number of visitors, bounce rate, traffic source, etc. Angry, Hungry, and GRY words What is the third word ending in GRY? Out of these, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. However, i loses only its horizontal symmetry, as it can still be split into symmetrical halves along its vertical axis. 9 Which letters have horizontal lines of symmetry? So, the slope of a vertical line is m = (y 2 - y 1) / (x - x) = (y 2 - y 1) / 0 which is not defined as . Other uncategorized cookies are those that are being analyzed and have not been classified into a category as yet. A shape is said to be symmetric if it can be divided into two more identical pieces which are placed in an organized way. Letters like B and D have a horizontal line of symmetry: their top and bottom parts match. You wont notice a difference in the text alignment if you use it on a page thats already full of text. Most of these leaves depict symmetric patterns if we take the middle vein as the line of vertical symmetry. Advertisement cookies are used to provide visitors with relevant ads and marketing campaigns. V6A 3Z7 Map . In other words, it is a straight standing line that divides an image or shape into two identical halves. By submitting your email, you agree to the Terms of Use and Privacy Policy. This cookie is set by GDPR Cookie Consent plugin. For example, the image below shows a horizontal line of symmetry. H-If you fold this on top of eachother Rotational symmetry is the quality a design has if it maintains all characteristics when it is rotated about a point. The lines are always written in the form y=a y = a where a a represents a real number. Symmetry Line in Numbers From 0 To 9 0 has 2 lines of symmetry 1 horizontal and 1 vertical. We select and review products independently. Symmetrical Words Symmetry is when one side of a shape is a mirror image of the other side. In such a case, the line of symmetry is vertical. Remember vertical means from top to bottom. The lines may vertical, horizontal, or diagonal lines. That means the second half or the missing part of the figure will be exactly the same as given on the other side. Email this page to a friend.Also: Sign up for our free web site updates here. the question mark broke the symmetry) Now, for right-left symmetry, the letters are (also capital letters): W, T, Y, U, I, O, A, H, X, V and M Examples: HA <-> AH YAM <-> MAY The cookies is used to store the user consent for the cookies in the category "Necessary". This also means that it has perfect rotational symmetry, a concept well explore more in a following section. What if you only want to align specific text in the document vertically? Some letters, for example, X, H, and O, have both vertical and horizontal lines of symmetry. Horizontal, Vertical, Perpendicular, Parallel, Diagonal - 1 Week Unit of WorkThis download includes a Math unit on different types of lines, which is made up of 3 tasks. This figure has two lines of symmetry: the horizontal line of symmetry cuts the figure into a top and bottom that are mirror images of each other the vertical | 677.169 | 1 |
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2013 AIME I Problems/Problem 9
Contents
Problem
A paper equilateral triangle has side length . The paper triangle is folded so that vertex touches a point on side a distance from point . The length of the line segment along which the triangle is folded can be written as , where , , and are positive integers, and are relatively prime, and is not divisible by the square of any prime. Find .
Solution 1
Let and be the points on and , respectively, where the paper is folded. Let be the point on where the folded touches it.
We have and , so . Denote ; we get .
In triangle , , and .
In triangle , we get and then use sine-law to get ; similarly, from triangle we get . Thus Since , we get
Then
The answer is .
Solution 2
Let and be the points on and , respectively, where the paper is folded.
Let be the point on where the folded touches it.
Let , , and be the lengths , , and , respectively.
We have , , , , , and .
Using the Law of Cosines on :
Using the Law of Cosines on :
Using the Law of Cosines on :
The solution is .
Solution 3
Proceed with the same labeling as in Solution 1.
Therefore, .
Similarly, .
Now, and are similar triangles, so
.
Solving this system of equations yields and .
Using the Law of Cosines on :
The solution is .
Note
Once you find and , you can scale down the triangle by a factor of so that all sides are integers. Applying Law of cosines becomes easier, you just need to remember to scale back up.
Solution 4 (Coordinate Bash)
We let the original position of be , and the position of after folding be . Also, we put the triangle on the coordinate plane such that , , , and .
Note that since is reflected over the fold line to , the fold line is the perpendicular bisector of . We know and . The midpoint of (which is a point on the fold line) is . Also, the slope of is , so the slope of the fold line (which is perpendicular), is the negative of the reciprocal of the slope of , or . Then, using point slope form, the equation of the fold line is
Note that the equations of lines and are and , respectively. We will first find the intersection of and the fold line by substituting for :
Therefore, the point of intersection is . Now, lets find the intersection with . Substituting for yields
Therefore, the point of intersection is . Now, we just need to use the distance formula to find the distance between and .
The number 39 is in all of the terms, so let's factor it out:
Therefore, our answer is , and we are done.
Solution by nosaj.
Solution 5
Note: this requires lots of calculations that increase your chance of errors, but it only requires simple understanding of areas, similar triangles, and Heron's formula. I'll just put the strategy here because I am too lazy to calculate it myself right now.
1. notice that the two triangles on the sides of the folded corner are similar. using this, we can find that the side lengths of them are 9,7.8,4.2 and 3, 45/7, 39/7
2. use heron's formula to find the areas of those two triangles. remember that it is sqrt[s(s-a)(s-b)(s-c)]
3. using the area of these triangles, we can find the area of the triangle with the length we need.
4. use heron's formula again, with the unknown length as x, and since we know the area and the other two side lengths, we can just solve for x with this equation. | 677.169 | 1 |
Calculating COGO curve parameters
Available with Standard or Advanced license.
When you enter curves for a line or a polygon, you need to provide at least two curve parameters to create the curve segment. Occasionally, you might want to know what all the parameters are or calculate missing parameters.
The Curve Calculator lets you determine the complete characteristics of a curve from two known parameters. This illustration shows the parts of a curve:
Chord—Also referred to as the chord distance, the straight line distance between the endpoints of the curve.
Angle—The angle formed between the endpoints of the curve and the center point.
Arc—The length along the curve.
Radius—The length from the center point to the curve.
Tangent—The distance between the endpoint and the point of intersection. The point of intersection is determined by intersecting a perpendicular line from each of the endpoints of the curve.
Chord height—Also referred to as the arc height, this is the distance between the curve and the chord segment.
There are several cases where the curve calculated is based on an angle greater than 180°:
The angle is greater than 180°.
The chord distance is greater than the arc length.
In these cases, the tangent distance is negative and represents the point of intersection on the opposite side of the curve.
Click Curve Calculator on the COGO toolbar.
Click the drop-down arrows and choose two parameters to determine the curve.
Type the appropriate values for the parameters. Angles are reported in the current direction type and units. Distances are always reported in the units of the map's spatial reference.
Click Calculate.
The curve is calculated and shown in the list below the Calculate button. | 677.169 | 1 |
Why is trig hard?
Trigonometry is hard because you have to memorize various values of different functions in both degrees and radians. If you don't remember them, or if you mix them up, then your calculations will be incorrect.
Why there are only 6 trigonometric functions?
There are only 6 trigonometric ratios because only 6 ratios can define ratios of all sides. For example, If you want ratio between perpendicular and hypotenuse there is sin. If you want ratio between perpendicular and base there is tan.
Who is father of trigonometry?
Who invented sine?
Sine was introduced by Abu'l Wafa in 8th century, as a more convenient function, and gradually spread first in the Muslim world, and then to the West. (But apparently it was used in India centuries before him), as a more convenient function. However this new notation was adopted very slowly, it took centuries.
Who inventedIs trig useless?
Granted that trigonometry is complex and difficult to understand, but it's certainly not useless. In fact, it has practical applications in almost every industry. If someone tells you that learning trigonometry is not going to help you in the real world, do not listen.
Is trigonometry used in physics?
Physics lays heavy demands on trigonometry. Optics and statics are two early fields of physics that use trigonometry, but all branches of physics use trigonometry since trigonometry aids in understanding space. Related fields such as physical chemistry naturally use trig.
Is calculus a trigonometry?
It seems as though calculus applies trigonometric functions in the same way algebra does, i.e. it's a system of operations which doesn't depend on things like trig' in any way, but rather serves as a foundation/context for using trig'.
What is Versin Theta?
The versine or versed sine is a trigonometric function already appearing in some of the earliest trigonometric tables. It is written as versin(θ), sinver(θ), vers(θ), ver(θ) or siv(θ). In Latin, it is known as the sinus versus (flipped sine), versinus, versus or the sagitta (arrow).
What is cosec in math?
Cosecant Formula. Cosecant is one of the six trigonometric ratios which is also denoted as cosec or csc. The cosecant formula is given by the length of the hypotenuse divided by the length of the opposite side in a right triangle. | 677.169 | 1 |
Unlike the projective case, ideal points form a boundary, not a submanifold. So, these lines do not intersect at an ideal point and such points, although well-defined, do not belong to the hyperbolic space itself.
if all vertices of a quadrilateral are ideal points, the quadrilateral is an ideal quadrilateral.
While all ideal triangles are congruent, not all convex ideal quadrilaterals are. They can vary from each other, for instance, in the angle at which their two diagonals cross each other. Nevertheless all convex ideal quadrilaterals have certain properties in common:
The interior angles of a convex ideal quadrilateral are all zero.
Any convex ideal quadrilateral has an infinite perimeter.
Any convex ideal quadrilateral has area −2π/K{\displaystyle -2\pi /K} where K is the (always negative) curvature of the plane.
Given two distinct points p and q in the open unit disk the unique straight lineGiven two distinct points p and q in the open unit disk then the unique circle arc orthogonal to the boundaryIn the Poincaré half-plane model the ideal points are the points on the boundary axis. There is also another ideal point that is not represented in the half-plane model (but rays parallel to the positive y-axis approach it). | 677.169 | 1 |
Recognizing Angles
$1.95
With our Recognizing Angles lesson plan, students learn how to recognize and identify right, obtuse, and acute angles. Students practice identifying each type by sight and answer questions about them add an art project to this lesson by having students use a certain number of angles to create a piece of art, like a painting or picture.
Description
What our Recognizing Angles lesson plan includes
Lesson Objectives and Overview: Recognizing Angles teaches students what angles are, and how to recognize different types of angles, such as obtuse and acute, by sight. At the end of the lesson, students will be able to define an angle, explain how they are formed, and identify the basic concepts of angle measurement The first suggested lesson adjustment is for the activity worksheet. You can make the activity longer by creating, or having your students create, additional activity pages and holding several different rounds of competition. An optional addition to this lesson is to give your students index cards showing the different types of angles and have them use the cards to identify various right, obtuse, and acute angles in the classroom. You can also have them match their cards to angles outside, on a playground, or anywhere else you have access to. Finally, you can add an art project to this lesson by having students use a certain number of angles to create a piece of art, like a painting or picture.
Teacher Notes
The teacher notes page includes a paragraph with additional guidelines and things to think about as you begin to plan your lesson. It notes that this lesson revisits some things that your students may have already learned about, such as rays and endpoints. You can pair this lesson with other geometry lessons that relate to angles and angle types. This page also includes lines that you can use to add your own notes as you're preparing for this lesson.
RECOGNIZING ANGLES LESSON PLAN CONTENT PAGES
What is an Angle?
The Recognizing Angles lesson plan includes two pages on content. The lesson begins by listing some things that students see every day that have angles, like corners of rooms and edges of boxes. Students will learn that an angle is the space between two lines that share a common endpoint. Angles can be large or small. For example, the angle that forms between two fingers is smaller than the one between your arm at your elbow! Angles are a geometric shape that forms when two rays have a common endpoint. A ray, as students may already know, is a line with a start point but no end point. Rays can be short or long, but no matter what, they form an angle when joined at their endpoints. The lesson includes a few examples of rays that have formed angles by joining at their endpoints.
Next, students learn that we determine the different sizes of angles by the amount of space between two rays. We measure this space in degrees, represented by the degree symbol (°). Students will learn that these degrees are not the same things as the degrees we use to measure temperatures. We measure angles in math using the amount of space in a circle, which has a total of 360°.
The lesson shows a green circle broken into four right angles. Each of these right angles are 90°, so the total of all four is 360°. All right angles measure 90°. Acute angles measure less than 90° and obtuse angles measure greater than 90°. The lesson shows examples of each type of angle.
Angles can have any measurement larger than 0° and less than 360°. You can find angles everywhere once you know to look for them! We measure angles using a helpful tool called a protractor. We use these to find the exact number of degrees in an angle. Knowing how to identify angles and understanding how to measure them is an important skill. People who work in construction, architecture, art, design, and many other fields use them every day!
Key Terms
Here is a list of the vocabulary words students will learn in this lesson plan:
Angle: the space between two lines that share a common endpoint
Ray: a line with a start point but no end point
Right angles: angles with measurements of 90°
Acute angles: angles measuring less than 90°
Obtuse angles: angles measuring greater than 90°
Protractor: a tool used to find the exact number of degrees for an angle
RECOGNIZING ANGLES LESSON PLAN WORKSHEETS
The Recognizing Angles lesson plan includes three worksheets: an activity worksheet, a practice worksheet, and a homework assignment. You can refer to the guide on the classroom procedure page to determine when to hand out each worksheet.
LABELLING COMPETITION ACTIVITY WORKSHEET
Students work in pairs to complete the activity worksheet. The student pairs will compete with each other to see who can complete the activity the fastest. They will look at the activity worksheet and identify all of the right, acute, and obtuse angles. They will label right angles with an "R," acute angles with an "A," and obtuse angles with an "O."
Students may also work either alone or in larger groups for this activity if you'd prefer.
ANGLES PRACTICE WORKSHEET
The practice worksheet includes two exercises. For the first exercise, students will tell whether various statements about angles are true or false. For the second exercise, students will look at different letters and determine whether the shape of the letters include right, acute, obtuse angles, or no angle at all.
RECOGNIZING ANGLES HOMEWORK ASSIGNMENT
For the homework assignment, students will answer five questions about recognizing and measuring angles. They will also look at six different angles and determine if they're acute, right, or obtuseMath
State Educational Standards
LB.MATH.CONTENT.4.MD.C.5
Lessons are aligned to meet the education objectives and goals of most states. For more information on your state objectives, contact your local Board of Education or Department of Education in your state. | 677.169 | 1 |
I have tried searching for a similar question but couldn't find any which helps me with my problem. I've got a point P on (1,2,3), A plane with a normal towards point (1,1,0) and a point Q on the plane which goes through (1,0,3). I have to calculate the distance between the point and the plane, and even though I even tried drawing in a 3d renderer, I can't seem to solve it.
This question was taken right from my exam. However, I just solved the answer myself. For people also looking for this problem:
The idea here is to project the line PQ onto the normal N. The formula for projection is: " Projection of v onto W = |v dotproduct w| / || v ||. So in this particular question; Projecting the line PQ onto the normal which is given by; | 677.169 | 1 |
In the regular quadrangular pyramid SABCD, point O is the center of the base
In the regular quadrangular pyramid SABCD, point O is the center of the base, S is the vertex, SO = 12, SB = 15. Find the length of the line segment AC.
Let us construct the AC and ВD diagonals at the base of the pyramid.
In a right-angled triangle SOB, according to the Pythagorean theorem, we determine the length of the leg OB.
OB ^ 2 = SB ^ 2 – SO ^ 2 = 15 ^ 2 – 12 ^ 2 = 225 – 144 = 81.
OB = 9 cm.
Since the base of a regular quadrangular pyramid is the square ABCD, point O divides the diagonals AC and BD in half, then BD = 2 * OB = 2 * 9 = 18 cm.
Since BD = AC, then AC = 18 cm.
Answer: The length of the AC segment is 18 | 677.169 | 1 |
Number of Altitudes in a Triangle Explained
Triangles are basic geometric shapes that we encounter frequently in mathematics, engineering, and various other fields. One of the fundamental concepts related to triangles is the altitude. In this article, we will delve into the concept of altitudes in a triangle, their properties, and how they can be used to solve geometric problems.
Understanding Altitudes in a Triangle
In geometry, an altitude of a triangle is a line segment drawn from one of the vertices of the triangle perpendicular to the opposite side. Each triangle has three altitudes, one from each vertex. These altitudes may or may not lie within the interior of the triangle, depending on the type of triangle.
Properties of Altitudes in a Triangle
Concurrency: The three altitudes of a triangle are always concurrent, meaning they intersect at a single point called the orthocenter.
Right Angles: Each altitude of a triangle forms a right angle with the side it intersects.
Bisecting the Base: The altitude drawn from the vertex of a triangle bisects the side it intersects at 90 degrees.
Relationship with Orthocenter: The orthocenter of a triangle may lie inside, outside, or on the triangle itself, depending on the type of triangle.
Types of Triangles based on Altitudes
Acute Triangle: In an acute triangle, all three altitudes lie within the triangle's interior, and the orthocenter is inside the triangle.
Right Triangle: In a right triangle, one of the altitudes coincides with a side of the triangle, and the orthocenter is located at the vertex forming the right angle.
Obtuse Triangle: In an obtuse triangle, one of the altitudes lies outside the triangle, and the orthocenter is outside the triangle as well.
Applications of Altitudes in Problem Solving
Altitudes play a crucial role in various geometric problem-solving techniques. Here are some common applications:
1. Area Calculation:
By using the altitude of a triangle, you can easily calculate the area of the triangle using the formula: $Area = 0.5 * base * height$.
2. Orthocenter determination:
Altitudes help in locating the orthocenter of a triangle, which in turn aids in further geometrical derivations and constructions.
3. Angle Calculations:
Altitudes can be used to calculate various angles within a triangle, especially right angles formed with the sides.
Frequently Asked Questions (FAQs) About Altitudes in a Triangle
How many altitudes does a triangle have?
A triangle has three altitudes, each drawn from a vertex to the opposite side.
Do all triangles have altitudes?
Yes, all triangles have altitudes, though in some cases, they may lie outside the triangle.
What is the significance of the orthocenter in a triangle?
The orthocenter of a triangle is the point where all three altitudes intersect. It has important geometric implications and properties.
How do altitudes help in calculating the area of a triangle?
Altitudes provide the height of the triangle, which, when multiplied by the base and halved, gives the area of the triangle.
Can an acute triangle have its orthocenter outside the triangle?
No, in an acute triangle, the orthocenter always lies inside the triangle as all altitudes are contained within the triangle.
How does the length of the altitude affect the area of a triangle?
The longer the altitude of a triangle, the greater the area of the triangle, as area is directly proportional to the height of the triangle.
Can altitudes of a triangle be equal in length?
Yes, in an isosceles triangle, where two sides are equal, the altitudes drawn from the base to the apex are equal in length.
What happens if two altitudes of a triangle are perpendicular to each other?
If two altitudes of a triangle are perpendicular to each other, it indicates that the triangle is a right-angled triangle.
Are altitudes and medians of a triangle the same thing?
No, altitudes and medians of a triangle are different. Altitudes are perpendicular segments from a vertex to the opposite side, while medians are line segments from a vertex to the midpoint of the opposite side.
How do altitudes help in proving the congruence of triangles?
Altitudes contribute to proving the congruence of triangles by ensuring that the corresponding sides and angles are equal due to the perpendicular nature of altitudes.
In conclusion, altitudes in a triangle are essential elements that help determine geometric properties, calculate areas, and establish important points like the orthocenter. Understanding the concept of altitudes enhances our grasp of triangle geometry and aids in solving various mathematical problems efficiently. | 677.169 | 1 |
Introduction
A starting point for the introduction of 3D graphics would be to delve into its historical origins, the foundation of geometry was laid during ancient Greece with significant contributions from figures such as Thales, Pythagoras, and Euclid, who is often regarded as the "father of geometry".
"Geometry" has its roots in Ancient Greek, γεωμετρία (geometria) composed of two elements:
Γῆ (Ge): This element means "Earth's land"
Μέτρον (Metron): This element means "measure" or "measurement"
When combined, the term γεωμετρία means the study of spatial relationships defined by practical measure, analyze and describe features of the Earth's land, for real-world applications such as agriculture, construction, and navigation.
Euclid wrote a book called Element where he provided specific definitions for several fundamental geometric terms:
«Σημεῖόν ἐστιν, οὗ μέρος οὐθέν» could be translated by «A point is, of which part is none», this definition emphasizes the geometric indivisibility of a point, hence its only measurable attribute is the distance between two points.
«Γραμμὴ δὲ μῆκος ἀπλατές» «A line, however, has length without breadth» emphasizing the idea that a line has only one measurable attribute, its length.
«Γραμμῆς δὲ πέρατα σημεῖα» «The extremities of a line are points», underlying that a line is composed of points, since a line can have different length.
«Σῶμά ἐστι τὸ μῆκος, πλάτος, βάθος ἔχον» «A solid is that which has length, breadth, and depth», meaning that a solid has 3 measurable attributes that do not overlap between them
«Τριγώνου δεξιοῦ τὸ τετράγωνον τῆς ὑποτείνουσης τοὺς τετραγώνους τῶν περὶ τὰς ἄλλας πλευρὰς ἰσότητα πεποιήκαμεν» «In a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides», we will keep this proposition to explain rotation spatial transformation later on.
With these in mind, we can build any geometric elements with just points.
Hence I do not rely on modern 3D graphics techniques to render geometric elements but the definition of points as Euclid set in his "Element" book.
Introduction
Euclid did not introduce the concepts of axes, coordinates and dimensional space as known as Cartesian coordinate system, but a comprehensive analysis of points and lines can lay down its foundation.
Euclid's statement "Γραμμῆς δὲ πέρατα σημεῖα" also highlights the spatial relationship between the endpoints of a line, the relative spatial position between them where the length tell that position.
Hence, if points do have a spatial relationship between them and if any geometric elements can be shaped with points, each points inside a geometric element possesses a unique spatial position, relative to that element.
To precisely determine the relative spatial position of a point inside a more complex geometric element other than a line, we may trace the shortest line from this point to the other point within each of the existing or imaginary measurement lines, establishing a network of interconnected points that defines their relative spatial position.
And the best candidate to build any geometric elements with a set of points is a στερεὰ (sterea), a geometrical solid such as παραλληλεπίπεδον (parallelepiped), since its lines draw the length, breadth, and depth without overlapping each others, hence each point inside a παραλληλεπίπεδον will have a set of 3 measurements/distances between them and the corresponding points within the length's line, breadth's line and depth's line of the παραλληλεπίπεδον.
Later on, those measurement components will be called coordinates, coordinates relative to the measurement lines, which will be called axes, where each axis open a dimensional space.
In ancient Greece, measurements were often based on geometric and proportional relationships rather than standardized units as we have today, hence πῆχυς (Cubit) and στάδιον (Stadion) were used as counting units to set the distance in construction.
Euclid had a Κανών (ruler) to draw his geometrical element that had no numerical value markings, a πῆχυς could not fit the size of a ruler to draw geometrical elements, hence the metric system that came in the late of 19th century was the best choice as measurement units.
Now that we have lay down the foundation of Cartesian coordinate system, we have one more study to achieve, while we logically can move a point on a straight line by adding a variable number to one of its 3 measurement components, we have one spatial transformation left, moving a point on curved line, known as rotation.
And the rotation do have the circle as perfect geometric candidate since it has a single reference point located in its center for all points inside, visually rotating around the said reference point.
As Euclid stated above «In a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides», emphasizing that a right-angled triangle had a spatial relationship between its three line's length, such as, known as Pythagorean theorem:
Hence if the ὑποτείνουσα (hypotenuse) has always the same length (known as circle's size), a circle can be drawn if we vary the length of the first side between 0 to 1 and find the corresponding length of the second side, such as:
SecondSideLength = sqrt(1 - FirstSideLength × FirstSideLength)
In programming, it could be translated with the c++ code below that I have added in the demo:
While we could use that way to rotate points, the lack of a rotation measurement known as angle, will rotate the points without accuracy
Unfortunately, I haven't found a connecting bridge between the Pythagorean theorem and trigonometric functions that is using angles instead of using a right-angled triangle properties to draw a circle, thus the comprehensive analysis of Euclid's book stop here and we must leap toward modern geometry era.
Trigonometric functions such as sin(θ) and cos(θ) where θ represent the angle of a point on the circle's line relative to the measurement X line of the circle and return the corresponding relative position to its measurement lines.
Those measurement lines are commonly called X and Y
Hence, we will start with a visual study of the trigonometric circle to achieve rotation of points.
First we set a point at 0° (x = 1, y = 0) and we apply a rotation on this point by+90° (+ is anticlockwise and - is clockwise) around the center, the result to find is x = 0, y = 1.
And the general formulas to find this result is:
X′ = X × cos(θ) - Y × sin(θ)
Y′ = X × sin(θ) + Y × cos(θ)
It's relatively simple to retrieve this equation from scratch and later, we will see that this formulas is just the linear form of the Z rotation matrix.
Here's the methodology:
We have a circle with two measurement lines (known as 2D) with a radius of 4.
We set a point, rotate it and try to retrieve the equation that lead to its new position, thus once we have operate on a P(X, 0) and P(0, Y) combinations, we will find the equation to rotate any P(X, Y)
Let's begin with P(4, 0), visually a red dot:
We rotate this point of +90° and try to find the equation of its new position, visually the result to find is P(0, 4):
Finally, in order to achieve a rotation on Euclid's points in a 3D coordinate system, we do need to multiply the 2D rotation matrices above, between them and where the order of the multiplication matters, but we will take a closer look on that in the next tutorial heading this one.
Comments and Discussions
To get the same resolution for objects in this engine like for rasterization, we need at least more than 10 millions of voxel for a single object, but it's impossible to work with it actually because my CPU display a 2 fps for such resolution | 677.169 | 1 |
Andhra Pradesh State Board STD 7th Maths Chapter 12 Symmetry Unit Exercise Books Solutions with Answers are prepared and published by the Andhra Pradesh Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of AP Board Class 7th Maths Chapter 12 Symmetry Unit Exercise Books Answers Solutions, then you are in the right place. Here is a complete hub of Andhra Pradesh State Board Class 7th Maths Chapter 12 Symmetry Unit Exercise solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Andhra Pradesh Board STD 7th Maths Chapter 12 Symmetry Unit Exercise Textbooks. These Andhra Pradesh State Board Class 7th Maths Chapter 12 Symmetry Unit Exercise Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Andhra Pradesh Board.
Find below the list of all AP Board Class 7th Maths Chapter 12 Symmetry Unit Exercise Textbook Solutions for PDF's for you to download and prepare for the upcoming exams:
Question 1. Fill in the blanks. (i) Afigure has _______ symmetry if there is a line about which the figure may be folded so that the two parts of the figure coincide. Answer: Line
(ii) A regular pentagon has _______ line of symmetry. Answer: Five
(iii) A figure has _______ symmetry if after a rotation through a certain angle. The figure looks exactly the same. Answer: Rotational
(iv) A _______ triangle has no lines of symmetry. Answer: Scalene
(v) Each regular polygon has as many lines of symmetry as it has _______ Answer: Number of sides
(vi) The concept of line symmetry is closely related to _______ reflection. Answer: Mirror
(vii) The quadrilateral that has four lines of symmetry and order-four rotational symmetry is a _______ Answer: Square
(viii) The angle of rotational symmetry for letter S is _______ Answer: 180°
(ix) A line segment is symmetrical about its _______ Answer: Perpendicular bisector
(x) Station turns an object about a fixed point. The fixed point is called _______ Answer: Point of rotation
(xi) Each of the letters H, N, S and Z has a rotational symmetry of order _______ Answer: 2
(xii) The line of symmetry of an isosceles triangle is it's _______ from the vertex having the equal sides. Answer: Median/altitude/angular bisector
Question 2. Cut the capital letters of English and paste them in your note book. Draw possible number of lines of symmetry for each of the letter. (i) How many letters have no line of symmetry? What are thpy? Answer: 10. They are F, G, J, L, N, P, Q, R, S, Z.
(ii) How many letters have one line of symmetry? What are they? Answer: 12. They are A, B, C, D, E, K, M, T, U, V, W, Y.
(iii) How many letters have two lines of symmetry? What are they? Answer: 3. They are H, I, X.
(iv) How many letters have more than two lines of symmetry? What are they? Answer: 1. That is 0.
(v) Which of them have rotational symmetry? What are they? Answer: 4. They are H, I, 0, X.
(vi) Which of them have point symmetry? What are they? Answer: 7. They are H, I, N, 0, S, X, Z.
Question 3. Draw some natural objects which have at least one line of symmetry. Answer:
Question 4. Draw three tessellations and expose the basic shapes used on your tessellation. Answer: | 677.169 | 1 |
Chapter 5.6 Proving Triangle Congruence by ASA and AAS
5.6 Proving Triangle Congruence by ASA and AAS Exercise 1 (a) The AAS Congruence Theorem and ASA Congruence Theorem are similar in the sense that they are used to prove that two triangles are congruent. Both AAS and ASA require two pairs of corresponding angles to be congruent. (b) When applying AAS Congruence Theorem, we consider a non-included side to prove that triangles are congruent. But when applying ASA Congruence Theorem, we consider an included side to prove that triangles are congruent. Exercise 2 To show that triangles are congruent using AAS Congruence Theorem or ASA Congruence, we need to know the pair of corresponding congruent side in addition to the pairs of congruent corresponding angles. Exercise 3 To prove that ∆ ABC ≅ ∆ QRS, we need to know the corresponding congruent parts. From the diagram, we can see that they have equal corresponding sides and angles: ( 𝐴𝐴𝐴𝐴 ���� ) ≅ ( 𝑄𝑄𝑄𝑄 ���� ) ∠ B ≅∠ R ∠ A ≅∠ Q Hence, enough information is given to prove that the triangles are congruent by AAS Congruence Theorem. Exercise 4 To prove that ∆ ABC ≅ ∆ DBC, we need to know the corresponding congruent parts. From the diagram, we can see that they have equal corresponding sides and angles: ( 𝐴𝐴𝐶𝐶 ���� ) ≅ ( 𝐴𝐴𝐶𝐶 ���� ) (common sides) ∠ ACB ≅∠ DCB (right angles) ∠ A ≅∠ D Hence, enough information is given to prove that the triangles are congruent by AAS Congruence Theorem. Exercise 5 To prove that ∆ XYZ ≅ ∆ JKL, we need to know the corresponding congruent parts. From the diagram, we can see that they have equal corresponding sides and angles: ( 𝑋𝑋𝑋𝑋 ���� ) ≅ ( 𝐽𝐽𝐽𝐽 ��� ) ( 𝑋𝑋𝑌𝑌 ���� ) ≅ ( 𝐽𝐽𝐾𝐾 ���� ) ∠ Z ≅∠ L The triangle is congruent based on SSA Congruent Theorem Hence, there is not enough information is given to prove that the triangles are congruent by AAS Congruence Theorem. Exercise 6 To prove that ∆ RSV ≅ ∆ UTV, we need to know the corresponding congruent parts. From the diagram, we can see that they have equal corresponding sides and angles: ( 𝑄𝑄𝑆𝑆 ���� ) ≅ ( 𝑇𝑇𝑆𝑆 ���� ) ∠ S ≅∠ T ∠ RVS ≅∠ UVT(cross angles) The triangle is congruent based on ASA Congruence Theorem. Hence, enough information is given to prove that the triangles are congruent by ASA
So, the image of the construction, Exercise 14 1) Construct AB so that it is congruent to DF. 2) Construct ∠ A with vertex A and side AB so that it is congruent to ∠ D. 3) Construct ∠ B with vertex B and side BD so that it is congruent to ∠ F. 4) Label the intersection of the sides of ∠ A and ∠ B that you constructed in Step 2 and Step 3 as C. 5) Join A and C, B and C. \triangle ABC is the required triangle. So, the image of the construction, Exercise 15 Using ASA Congruence Theorem, we can prove that ∆ JKL and ∆ FGH are congruent if two angles and the included sides given are congruent. From the diagram, we can see that the corresponding congruent parts given are: ( 𝐽𝐽𝐽𝐽 ��� ) ≅ ( 𝐺𝐺𝐺𝐺 ���� ) ∠ K ≅∠ G ∠ J ≅∠ F Based on ASA Congruence Theorem, ∆ JKL and ∆ FGH are congruent. But, the error is in that the vertices of ∆ FGH is written in a wrong order. It was written as
∆ FHG instead of ∆ FGH. Exercise 16 Using AAS Congruence Theorem, we can prove that ∆ QRS and ∆ VWX are congruent if two angles and the non-included sides given are congruent. From the diagram, we can see that the corresponding congruent parts given are: ( 𝑄𝑄𝑄𝑄 ���� ) ≅ ( 𝑆𝑆𝑉𝑉 ����� ) ∠ Q ≅∠ V ∠ R ≅∠ W The error is that ( 𝑄𝑄𝑄𝑄 ���� ) ≅ ( 𝑆𝑆𝑉𝑉 ����� ) is not the non-included sides based on AAS Congruence Theorem. Hence, ∆ QRS and ∆ VWX are not congruent using AAS Congruence Theorem. But, the triangles are congruent based on ASA Congruence Theorem. Exercise 17 Using ASA Congruence Theorem, we can prove that ∆ NQM and ∆ MPL are congruent if two angles and the included sides given are congruent We are given that: M is the midpoint of ( 𝑀𝑀𝐾𝐾 ���� ) ( 𝑀𝑀𝐾𝐾 ���� ) ⊥ ( 𝑀𝑀𝑄𝑄 ���� ) ( 𝑀𝑀𝐾𝐾 ���� ) ⊥ ( 𝑀𝑀𝑀𝑀 ����� ) (QM) ⊥ (PL) From the diagram, we can see that ∠ QNM and ∠ PML are right angles. Hence, ∠ QNM ≅ ∠ PML because all right angles are congruent Recall that if two parallel lines are cut by a transversal, then the pairs of corresponding angles are congruent. ⟹∠ QMN ≅ ∠ PLM Therefore, ∆ NQM and ∆ MPL are congruent based on ASA Congruence Theorem. Exercise 18 Using ASA Congruence Theorem, we can prove that ∆ ABK and ∆ CBJ are congruent if two angles and the included sides given are congruent We are given that: M is the midpoint of ( 𝑀𝑀𝐾𝐾 ���� ) ( 𝐴𝐴𝐽𝐽 ��� ) ≅ ( 𝐽𝐽𝐴𝐴 ���� ) ∠ BJK ≅ ∠ BKJ ∠ A ≅ ∠ C From the diagram, we can see that: ( 𝐽𝐽𝐽𝐽 ��� ) ≅ ( 𝐽𝐽𝐽𝐽 ��� ) (By Reflexive property of equality) ( 𝐴𝐴𝐽𝐽 ���� ) ≅ ( 𝐴𝐴𝐽𝐽 ��� ) (By definition of congruent segments) Therefore, ∆ ABK and ∆ CBJ are congruent based on ASA Congruence Theorem. Exercise 19 Using AAS Congruence Theorem, we can prove that ∆ XWV and ∆ ZWU are congruent if two angles and the non-included sides given are congruent We are given that: ( 𝑆𝑆𝑉𝑉 ����� ) ≅ ( 𝑈𝑈𝑉𝑉 ����� ) ∠ X ≅ ∠ Z
From the diagram, we can see that by the reflexive property of congruence: ∠ W ≅ ∠ W Hence, based on AAS Congruence Theorem ∆ XWV ≅ ∆ ZWU. Exercise 20 Using AAS Congruence Theorem, we can prove that ∆ NMK and ∆ LKM are congruent if two angles and the non-included sides given are congruent We are given that: ∠ L ≅ ∠ N ∠ NKM ≅ ∠ LMK From the diagram, we can see that by the reflexive property of congruence: ( 𝐽𝐽𝑀𝑀 ����� ) ≅ ( 𝑀𝑀𝐽𝐽 ����� ) Hence, based on AAS Congruence Theorem ∆ NMK ≅ ∆ LKM . Exercise 21 Hypotenuse Angle Theorem is true for the right triangles only because, the hypotenuse is one of the right-angled triangle legs. It is a criterion used to prove whether a given set of right triangles are congruent. Given that a triangle is a right triangle, only 1 angle out of the 3 angles is a right angle. Recall that the sum of interior angle s of a triangle =180° Since a right triangle has 1 right angle, the other 2 angles will be acute. To find the 3rd angle in a right triangle =180°-90°-m ∠ 2 Hence all the corresponding angles will be congruent. Given that the hypotenuse is congruent, then the 2 other legs of the triangle will also be congruent. Exercise 23 Given that ∆ ABC and ∆ XYZ are right triangles, we will know that: ( 𝐴𝐴𝐶𝐶 ���� ) ≅ ( 𝑋𝑋𝑋𝑋 ���� ) (leg),( 𝐶𝐶𝐴𝐴 ���� ) ≅ ( 𝑋𝑋𝑌𝑌 ���� ) (leg) and ∠ B ≅ ∠ Y(right angle) Since we have 2 legs, it will help us prove that the triangles are congruent based on leg- leg theorem. We can conclude that leg-leg congruence theorem is equivalent to SAS Congruence Theorem. Exercise 23 Assuming that ∆ ABC and ∆ DEF are right angles, we will know that: ∠ B ≅ ∠ E(right angle),( 𝐴𝐴𝐶𝐶 ���� ) ≅ ( 𝐷𝐷𝐷𝐷 ���� ) (leg) and ∠ A ≅ ∠ D(acute angle) Since the sum of the interior angles of a triangle is 180, we know that C and F must also be congruent to each other. Hence, angle-leg congruence theorem is equivalent to ASA Congruence Theorem Exercise 24 Given that ∆ ABC ≅ ∆ DEF using ASA Congruence Theorem, the corresponding congruent parts from the diagram are: ∠ L ≅∠ L(common angle) ( 𝐾𝐾𝐽𝐽 ���� ) ≅ ( 𝐾𝐾𝑀𝑀 ���� ) The additional information we need to prove that he triangles are congruent using ASA Congruence Theorem is: ∠ LKJ ≅∠ LNM
∠ STV ≅∠ UVW(right angle) ∠ TSV ≅∠ VUW Since we are given an included sided and 2 angles , we can conclude that ∆ TVS and ∆ VWU are congruent using ASA Congruence Theorem. The correct option is C. Exercise 27 Converse of the base angles theorem: If two angles in a triangle are congruent, then the sides opposite those angles are also congruent. Statement & Reasons ∠ 𝐶𝐶 ≅ ∠ 𝐴𝐴 & Given Draw a bisector AD of ∠ A & Construction ∠ 𝐴𝐴𝐴𝐴𝐷𝐷 ≅ ∠ 𝐶𝐶𝐴𝐴𝐷𝐷 & 𝐴𝐴𝐷𝐷 is the bisector of ∠ A 𝐴𝐴𝐷𝐷 ≅ 𝐴𝐴𝐷𝐷 & 𝑄𝑄𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 property of Congruence ∆ ABD ≅ ∆ ACD & AAS congruence theorem AB ≅ AC & Corresponding sides of two congruent triangles Exercise 28 It is not possible to interchange AAS Congruence Theorem with ASA Congruence Theorem. This is because the congruent side in AAS is not between the two congruent angles and ASA has the congruent side between the two angles. Hence, the claim is wrong Exercise 29
The required sketch, a) Statement & Reasons ∠ 𝐴𝐴𝐷𝐷𝐶𝐶 ≅ ∠ 𝐴𝐴𝐷𝐷𝐶𝐶 , 𝐷𝐷𝐶𝐶 ⊥ 𝐴𝐴𝐴𝐴 & Given ∠ 𝐴𝐴𝐶𝐶𝐷𝐷 ≅ ∠ 𝐴𝐴𝐶𝐶𝐷𝐷 & Both are right angles 𝐶𝐶𝐷𝐷 ≅ 𝐶𝐶𝐷𝐷 & Same side △ 𝐴𝐴𝐶𝐶𝐷𝐷 ≅△ 𝐴𝐴𝐶𝐶𝐷𝐷 & AAS congruence theorem b) As ∠ 𝐴𝐴𝐶𝐶𝐷𝐷 ≅ ∠ 𝐴𝐴𝐶𝐶𝐷𝐷 , So AD = AC, corresponding sides of △ 𝐴𝐴𝐶𝐶𝐷𝐷 and \triangle CBD. So, △ 𝐴𝐴𝐷𝐷𝐴𝐴 is an isosceles triangle. c) No, because the person is already seeing the complete reflected image of his or her body. But the size of the image will get smaller. Exercise 30 The pairs of congruent triangles that can be formed from the triangle are: ∆ PQS ≅ ∆ RSQ (Based on ASA Congruence Theorem) ∆ PSR ≅ ∆ RQP (Based on ASA Congruence Theorem) ∆ PTS ≅ ∆ RTQ(Based on SAS Congruence Theorem) ∆ PQT ≅ ∆ RST (Based on SAS Congruence Theorem) Exercise 31 The required image, From the picture, we see that all the angles in both triangles have the same measure, but they are not congruent. Exercise 32 | 677.169 | 1 |
How do you label the circumference of a circle?
There are two formulas that can be used to calculate the circumference of a circle: C = 2πr or C = πd, where π is the mathematical constant approximately equal to 3.
What are the terms of circles?
This section of Revision Maths defines many terms in relation to circles, including: Circumference, Diameter, Radius, Chord, Segment, Tangent, Point of contact, Arc, Angles on major and minor arcs, Angle of Centre and Sectors. Circumference: The circumference of a circle is the distance around it.
What does it mean to name a circle?
A circle is named by the point in the center. A radius is a line segment from the center of the circle to the edge. A diameter is a line segment that passes through the center of a circle. It has two points on the outside edge of the circle.
How circles are used in everyday life?
Answer. Some examples of circles in real life are camera lenses, pizzas, tires, Ferris wheels, rings, steering wheels, cakes, pies, buttons and a satellite's orbit around the Earth. Circles are simply closed curves equidistant from a fixed center.
Why do we learn about Circles?
To the Greeks the circle was a symbol of the divine symmetry and balance in nature. Greek mathematicians were fascinated by the geometry of circles and explored their properties for centuries. ... Circles are still symbolically important today -they are often used to symbolize harmony and unity.
Why are circles so important?
Because of their symmetry, circles were seen as representations of the "divine" and "natural balance" in ancient Greece. Later on, the shape would become a vital foundation for the wheel and other simple machines. A focus on circles is evident among structures built throughout history
What will be our world if there are no circles?
Life without circles would be as a square. All the planets including earth would not exist in a circular shape. There would be no movement of wheels of cars and bicycles on the road. Also scientific terms like rolling friction would not exist.
Why are circles the strongest shape?
The arc (think: circle) is the strongest structural shape, and in nature, the sphere is the strongest 3-d shape. The reason being is that stress is distributed equally along the arc instead of concentrating at any one point.
What is the strongest shape in the universe?
triangles
What is the weakest 3d shape?
Triangle
Whats stronger circle or triangle?
The answer is the triangle, because of the way it distributes pressure. I'm assuming you're referring to an equilateral triangle? The circle ie the strongest shape because it maintains constantly pressure , There is no inherent leverage as all points are at a constant distance from each other.
What is the strongest point of a triangle?
An equilateral triangle is structurally the most sound for obvious reasons. Equal angles and vectors naturally resist gravitational and lateral forces equally in all directions. This is true in theory and practice if one doesn't consider any other variables.
Do pyramids fit all other shapes?
Pyramid Jim - "A shape that fits all other shapes inside of it."
What do you call a triangle with a curved side?
Reuleaux triangles have also been called spherical triangles, but that term more properly refers to triangles on the curved surface of a sphere. ... Because of this property of rotating within a square, the Reuleaux triangle is also sometimes known as the Reuleaux rotor.
Can a triangle have a curved side?
A "side" should be defined as a straight line, or a segment that falls between 2 corners. Any curved side of a shape represents an infinite number of sides, so it would not be a triangle, but in fact an infinity-gon.
Can a pyramid have 3 sides?
The A triangle-based pyramid with equal sides is called a tetrahedron.
What do you call a four sided pyramid?
Can a pyramid have more than 4 sides?
One very unusual feature of the Great Pyramid is a concavity of the core that makes the monument an eight-sided figure, rather than four-sided like every other Egyptian pyramid. That is to say, that its four sides are hollowed in or indented along their central lines, from base to peak.
What is a three sided polygon called?
triangle
What is a 1000000000000000 sided shape called?
In geometry, a chiliagon (/ˈkɪliəɡɒn/) or 1000-gon is a polygon with 1,000 sides. Philosophers commonly refer to chiliagons to illustrate ideas about the nature and workings of thought, meaning, and mental representation.
What is 12 sided shape called?
dodecagon
What is a shape with 13 sides?
tridecagon
What is a one sided shape called?
henagon monogon
What is a 1 million sided shape called?
megagon
What is the sum of a 13 sided polygon?
For a pentagon, it's (5-2)180=(3)180=540 degrees. So plug in n=13 to see the sum of the angle measures for a 13-sided polygon. | 677.169 | 1 |
Showing top 8 worksheets in the category - Grade 10 Trigonometry. Some of the worksheets displayed are Work 7 trigonometry grade 10 mathematics, Trigonometry ... Showing top 8 worksheets in the category - Grade 10 Trigonometry. Some of the worksheets displayed are Work 7 trigonometry grade 10 mathematics, Trigonometry ...
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Scalar quantities are added according to the ordinary rules of arithmetic. For example , a mark of 50 added to a mark of 40 produces a mark of 90 –no directional property .But a force of 50N combined with a force of 40N may produce 90N if they are acting in same direction. But they are acting in opposite direction it would produce a different result. These vectors are combined or added by a special law the parallelogram law of addition of vectors.
VECTOR REPRESENTATION
A vector quantity can be graphically represented by a line drawn so that the length of the line denotes the magnitude of the quantity . The direction of the line indicates the direction in which the vector quantity act and it is shown by an arrow head . E.g a distance of 5km west represented by 5cm length of line where 1km = 1cm
N
5cmw
ADDITION AND SUBTRACTION OF VECTORS
Two or more vectors acting on a body in a specified direction can be combined to produce a single vector having the same effect .The single vector is called the resultant.
For example:
(a) Two forces Y and X with magnitude of 3N and 4N respectively acting along the same direction will produce a resultant of 7N (algebraic sum of the two vectors).
3N+4N = 7N
(b) If Y and X act in opposite direction, the resultant will be 1N.
4N_3N1N
-
(c) If the two vectors are inclined at an angle less than 900 or more than 900 , the resultant cannot be obtained by Pythagoras theorem but by vector addition,. Parallelogram law of vector, trigonometric or scale drawings can be used to calculate the magnitude and direction of the resultant 4N
Φ3Nφ < 900
VECTORS AT RIGHT ANGLES
Parallelogram law of vectors states that if two vectors are represented in magnitude and direction by adjacent sides of a parallelogram , the resultant is represented in magnitude and direction by the diagonal of the parallelogram drawn from the common point
Y
3N 3N R
4N X 4N
R2 = X2 + Y2= 42 + 32 = 16 + 9 = 25
R = √ 25 = 5N
Tan θ = Y/X
θ = tan-1 (Y /X) = tan-1 (3/4)
θ = tan-1 (0.75)
θ = 36.90
If the two vectors are inclined at an angle less than 900 , the scale drawing or trigonometric method can used . In using scale drawing (graphical ) methods, a convenient scale is chosen ( if the magnitude of the forces given is large ) and then draw the lengths corresponding to the magnitude of the forces . A Protractor is used to draw the angle in between the forces. The parallelogram is completed and the resultant and its fraction obtained
R R
RESOLUTION OF VECTORS
A single vector can be resolved into two vectors called components. A vector F represented as the diagonalof the parallelogram can be resolved into its component later taken as the adjacent sides of the parallelogram.
F Y
X
sin θ = Y /F
Y = F sin θ (vertical component)
cos θ = X /F
X = F cos θ (horizontal component)
The direction of F is given by
Tan θ = Y/X
Θ = tan-1 (Y/X)
THE RESULATNT OF MORE THAN TWO VECTORS
To find the resultant of more than two vectors, we resolve each vector in two perpendicular direction s add all the horizontal components X, and all the vertical components, Y.
For example, consider four forces acting on a body as shown below
F2 F1 Y
Θ2θ1
Θ3θ4
X
Add all the resolved horizontal components
X = F1 cos θ1 + (-F2 cosθ2 ) + (-F3 cos θ3 ) + F4 cos θ4
Y= F1 sin θ1 + F2 sinθ2 + (-F3 sinθ3) + (-F4 sinθ4)
R = √X2+ Y2
And the direction ∞ is given by
Tan ∞ = y/x
EVALUATION
1 Calculate the resultant of five coplanar forces of values10N, 12N , 16N , 20N , 15N on an object as shown below
20N 12N
40 O 500
30O10N
16N 15N
F(N) inclination Hor.comp. Vert. comp.
10 0 10cos θ=10.00 10 sin θ= 0
12 50 12 cos 50 =7.71 12 sin50= 9.19
20 40 -20 cos 40 =-15.32 20sin40= 12.85
16 90 16 cos 90 = 0 -16 sin 90= -16.00
15 60 15cos60 =7.50 -15 sin60 =-12.99
9.89 -6.95
R = √(19.892 + (6.952
R = 12.09
Tan ∞ = 6.95/9.89
∞ = -35.1054.9
90 – 35.1
=54.9
The direction of the resultant is S 54.90
GENERAL EVALAUTION
A body of mass 3.0Kg is acted upon by a force of 24N, if the frictional force on the body is 13N.Calculate the acceleration of the body.
For the body in question 1 above, what distance would it move if the force was applied for a period of 7s?
WEEKEND ASSIGNMENT
Which of the following is not a vector quantity (a) speed (b) velocity (c) force
(d) acceleration (e) Electric field
Which of the following is not a scalar quantity (a) density (b) weight (c) speed (d) mass
(e) temperature
Two forces , whose resultant is 100N are perpendicular to each other.If one of the | 677.169 | 1 |
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TheSiggi
The Pythagorean theorem describes the relation among the three sides of a right angle rectangle; namely that the sum of the two squares of the rectangle sides² is equal to the sum of the square of the hypotenuse. | 677.169 | 1 |
Law of cosines on a sphere
The previous post looked at the analog of the Pythagorean theorem on a sphere. This post looks at the law of cosines on a sphere.
Yesterday we looked at triangles on a sphere with sides a and b meeting at a right angle and hypotenuse c. Denote the angle opposite a side with the capital version of the same letter, so C is the angle opposite c. We assumed C is a right angle, but now we will remove that assumption.
The spherical analog of the law of cosines says
cos(c) = cos(a) cos(b) + sin(a) sin(b) cos(C).
Note that we have two kinds of angles: the arcs that make up the sides, and the angle formed by the intersection of arcs [1]. So the cos(C) term at the end is different animal from the other terms in the equation.
If C is a right angle, cos(C) = 0 and the second term drops out, leaving us with the spherical counterpart of the Pythagorean theorem. But we do not require that C be a right angle.
Application to air distance
Suppose we want to find how far a plane would travel between Los Angeles (LAX) and Houston (IAH), assuming it takes a great circle path. The lat/long coordinates of the two airports are (33.94°, −118.41°) and (29.98°, −95.34°).
Los Angeles and Houston are approximately at the same latitude, but even if they were at exactly the same latitude we couldn't just find the flight distance by finding the length of the arc of constant latitude between them because that arc would not be part of a great circle.
To find the distance between LAX and IAH we form a triangle with vertices at LAX, IAH, and the north pole. Call the arc from LAX to the north pole a and the arc from IAH to the north pole b. Since latitude is the angle up from the equator, the angle down from the pole is 90° minus latitude. So
a = 90° − 33.94° = 56.06°
and
b = 90° − 29.98° = 60.02°
The arcs a and b meet at the angle C equal to the differences in the two longitudes. That is, | 677.169 | 1 |
we draw lines from centre of the circle to the two corner of given triangle it will form an isoscales triangle with radius both sides and 6 other side. Then we need to draw perpendicular line from base to the centre of the circle to create, 30:60:90 triangle. With that we can calculate the radius of the circle, r = 6/ \sqrt{3} | 677.169 | 1 |
Question Paper from: IBPS PO Prelims 2016
S, T, U, V, W, X, Y and Z are sitting in a straight line equidistant from each other (but not necessarily in the same order). Some of them are facing south while some are facing north.Facing the same direction means, if one is facing north then the other also faces north and vice-versa. Facing the opposite directions means, if one is facing north then the other faces south and vice-versa) S faces north. Only two people sit to the right of S. T sits third to the left of S. Only one person sits between T and X. X sits to the immediate right of W. Only one person sits between W and Z. Both the immediate neighbors of T face the same direction. U sits third to the left of X. T faces the opposite direction as S. Y does not sit at any of the extremes ends of the line. V faces the same direction as W. Both Y and U face the opposite direction of Z. | 677.169 | 1 |
The solution of the cases of triangles by logarithmic sines, &c. is generally more expeditious than that by natural sines, &c., and therefore is preferable in practice. Besides, some of the best tables do not contain natural sines, but all contain logarithmic sines, which are indispensable in trigonometrical calculations.
Instrumentally.
In the first proportion, extend the compasses from 90° to 48° 17′ on the line of sines. That extent will reach from 324 to 242 on the line of numbers, and will be the length of BC.
In the second proportion, extend the compasses from 90° to 41° 43′ on the line of sines. That extent will reach from 324 to 215.6 on the line of numbers, and will be the length of AB.
Note. The radius is 90° of sines (27), and 45' of tangents (28), and 0 of cosines (30) and secants (29).
Ex. 2. Given the hypothenuse AC 121 yards, and the angle at the base A 55° 30′, to find the rest.
Answer. BC=99-719 yards, AB=68.535, angle C= 34° 30'.
Case 2. Given the side AB 125, and the angle A 51° 19′, to find the rest.
By Construction.
Draw an indefinite line AB, and from A to B set off 125, from a scale of equal parts.
Make the angle A=51 19', as in case 1. At the point B erect a perpendicular, and from A through the extremity of the arc which measures the angle A draw AC meeting the perpendicular in C. Then AC, BC, measured on the same scale of equal parts, will be 199.5 and 156. The angle C-90-51 19-38 41.
In this operation log. R, which is always 10·0000000, is subtracted, though not expressed. In other operations log. R will be seldom expressed, but will be added or subtracted tacitly.
Log. AC = log. R+ log. AB-log. cos. A orlog. sine C.
Log. R+log. AB 125
12.0969100
Log. cos. 51° 19′ or log. sine 38° 41′ 9.7958909
Log. AC 199.5603
2.3010191
Log. AC = log. sec. A+ log. AB-log. R.
Log. sec. 51° 19'
Log. AB 125
Log. AC 199-5603
10-2031641
2.0969100
2.3000741
Instrumentally.
In the first proportion, extend the compasses from 45° to 51° on the line of tangents. That extent will
* Some tables do not contain secants, but all contain sines, cosines, and tangents.
E
reach from 125 to 156 on the line of numbers, and will be the length of BC.
In the second proportion, extend the compasses from 382° to 90° on the sines. That extent will reach from 125 to 199.5 on the line of numbers, and will be the length of AC.
In the third analogy, extend the compasses from 0 to 511 on the secants. That extent will reach from 125 to 199.5 on the numbers, and will be the length of AC.
Ex. 2. Given the base AB 50, and the angle at the base A 25° 17', to find the rest.
Answer. Angle C = 64° 43', BC= 23.617, AC = 55.297.
Case 3. Given the hypothenuse AC 415, and the base AB 249, to find the rest.
By Construction.
Draw AB 249. At the point B erect a perpendicular. From the centre A, with the radius AC = 415, describe an arc intersecting the perpendicular in C. Then BC measured on the line of numbers is 332, and the angles A and C measured on the scale of chords, or with a protractor, are 53° 8' and 36° 53′ nearly. | 677.169 | 1 |
Properties and typesedit
The pitch of a helix is the height of one complete helix turn, measured parallel to the axis of the helix.
A double helix consists of two (typically congruent) helices with the same axis, differing by a translation along the axis.[3]
A circular helix (i.e. one with constant radius) has constant band curvature and constant torsion.
A conic helix, also known as a conic spiral, may be defined as a spiral on a conic surface, with the distance to the apex an exponential function of the angle indicating direction from the axis.
A curve is called a general helix or cylindrical helix[4] if its tangent makes a constant angle with a fixed line in space. A curve is a general helix if and only if the ratio of curvature to torsion is constant.[5]
A curve is called a slant helix if its principal normal makes a constant angle with a fixed line in space.[6] It can be constructed by applying a transformation to the moving frame of a general helix.[7]
Handednessedit
Helices can be either right-handed or left-handed. With the line of sight along the helix's axis, if a clockwise screwing motion moves the helix away from the observer, then it is called a right-handed helix; if towards the observer, then it is a left-handed helix. Handedness (or chirality) is a property of the helix, not of the perspective: a right-handed helix cannot be turned to look like a left-handed one unless it is viewed in a mirror, and vice versa.
Two types of helix shown in comparison. This shows the two chiralities of helices. One is left-handed and the other is right-handed. Each row compares the two helices from a different perspective. The chirality is a property of the object, not of the perspective (view-angle)
Another way of mathematically constructing a helix is to plot the complex-valued function exi as a function of the real number x (see Euler's formula).
The value of x and the real and imaginary parts of the function value give this plot three real dimensions.
Except for rotations, translations, and changes of scale, all right-handed helices are equivalent to the helix defined above. The equivalent left-handed helix can be constructed in a number of ways, the simplest being to negate any one of the x, y or z components.
Arc length, curvature and torsionedit
A circular helix of radius a and slope a/b (or pitch 2πb) expressed in Cartesian coordinates as
In aviation, geometric pitch is the distance an element of an airplane propeller would advance in one revolution if it were moving along a helix having an angle equal to that between the chord of the element and a plane perpendicular to the propeller axis; see also: pitch angle (aviation). | 677.169 | 1 |
Vector field calculations- dumb question
Since \vec{u} = \vec{\hat{x}} and \vec{v} = \vec{\hat{y}}, the magnitudes are 1 and the dot product is 0, so the resulting equation is:\cos(\theta) = \frac{0}{1 \cdot 1} = 0Therefore, the angle between the two vectors is 90 degrees. In summary, if the two vectors have a relationship of barXx-barYy and their magnitudes are calculated using the equation sqrt(x2+y2), then the angle between them
Jul 12, 2011
#1
bookdad
2
0Since the two vectors are obviously perpendicular, their dot product is zero, and hence the angle between them is 90 degrees.
One way that the dot product of two vectors is defined is:
[tex]\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}| \cos(\theta)[/tex]
You can solve this equation for cos(theta) like so:
[tex]\cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}| }[/tex]
Related to Vector field calculations- dumb question
1. What is a vector field calculation?
A vector field calculation is a mathematical process used to determine the magnitude and direction of a vector at a given point in space. This is useful in many scientific fields, including physics, engineering, and meteorology.
2. How is a vector field calculated?
A vector field is calculated by first defining the vector at each point in space and then using mathematical operations to determine the magnitude and direction of the vector. This can be done using equations, computer algorithms, or physical measurements.
3. What are some real-world applications of vector field calculations?
Vector field calculations have many practical applications, such as predicting the movement of fluids in engineering, modeling weather patterns in meteorology, and analyzing the force and direction of magnetic fields in physics.
4. Are there different types of vector fields?
Yes, there are several types of vector fields, including conservative and non-conservative fields, gradient fields, and curl fields. Each type has different properties and is used for different purposes in vector field calculations.
5. What are some common tools and techniques used in vector field calculations?
In addition to mathematical equations and computer algorithms, scientists may use specialized software, physical models, and experimental data to perform vector field calculations. Visualization tools, such as vector field plots, are also commonly used to analyze and interpret the results. | 677.169 | 1 |
4. Which of the following is not a procedure for describing a circle round a polygon? A. Bisect two external
angles. B. Bisect two sides. C. Draw a line from the determined centre, parallel to one side.
D. Draw a line from the determined centre, perpendicular to one side.
5.AB is the diameter of the circle shown below. What type of triangle is ABC? A. Scalene. B. Right
angled. C. Oblique. D. Isosceles.
Theory
1 . A regular hexagon has its sides 50mm long. Draw the hexagon and describe circles about it.
2. Draw a line AB of length 80mm and locate a point Q on it which is 20mm from end B. Draw a circle to
touch points P which is at a suitably chosen distance and Q.
WEEK THREE DATE:………………
Topic:Polygons and irregular figures
Content:
(i) Meaning and types of polygon.
(ii) Construction of polygons.
Meaning and types of polygon
Polygons could be defined as plane figures whose interior and exterior angles add up to
[(n – 2) x 180]0 and [4 right angles ]0 respectively. ( n is the number of sides of the polygon). Polygons could either be regular or irregular. Regular polygons have equal sides and angles. Examples include equilateral triangle, square, etc. Irregular polygons do not have all the sides and angles equal. An example of irregular polygon is re-entrant polygon which has one of its interior angles greater than 1800. See diagrams below for further illustration.
Polygons are named based on the number of sides they have. A polygon with three sides is called triangle, four sides is quadrilateral, pentagon 5, hexagon 6, heptagon 7, octagon 8, nonagon 9 and decagon 10 sides. These could either be regular or irregular.
There are two methods of constructing polygons and these include (i) Exterior angle method. (ii) Circumscribing circle method. In the external angle method, the number of sides, N of the required polygon is used to divide 3600 in order to determine the exterior angle to which that particular polygon is to be drawn. For instance, a regular pentagon with 5 sides will have its sides drawn to an exterior angle of 720
Similarly, a regular hexagon with six sides will have its exterior angle drawn at 600, heptagon 520, octagon 450, nonagon 400 etc.
The circumscribing circle method has the polygon enclosed in a circle of known diameter.
To construct a regular pentagon using the exterior angle method.
Method:
(i) Divide 3600 by the number of sides N which in this case is 5 to obtain an exterior angle of 720.
(ii) Draw a line AB equal to the length of one side of the regular pentagon.
(iii)With a protractor at point A, measure in a clockwise direction angle 720. Draw the line AQ and mark
off the given length on it to obtain point D.
(iv) Repeat same for point B in an anti-clockwise direction to obtain point C.
(v) Repeat the same for points C and D to obtain point E. ABCDE is the required pentagon.
To construct a regular pentagon using the circumscribing circle method.
Method:
(i) Draw the given circle that will
circumscribe the pentagon.
(ii) Bisect the radius to locate point A.
(iii)With A as centre and radius AB, swingan
arc to cut the horizontal diameter line at point C.
(iv)With B as centre and radius BC, swing an arc
to cut the circle at point D.
(v) Join DB which is the length of one side of the pentagon.
(vi)Step- off the length DB round the circle to obtain the remaining sides of the pentagon.
To construct a regular hexagon using the exterior angle method when given the length of one side.
Method:
(i) Determine the size of the exterior angle by dividing 3600 by 6 (hexagon) which gives 600.
To construct a regular heptagon when given the distance across corners.
Method:
(i) Draw the circle using the given distance across corners, (d/c). Divide the horizontal diameter AB intoequal number of sides of the polygon you are required to construct. In this case, divide it into 7 equalparts.
(ii) With points A and B in turn as centre and radius AB, draw arcs to intersect at point C.
(iii) Draw a line from point C through the second division to touch the circle at point D.
(iv) Join AD which represents one side of the heptagon.
(v) Step off the length AD round the circumference of the circle to obtain the remaining sides.
To construct any polygon when given the length of side.
Method:
(i) Draw one side AB and extend it to the left.
(ii) With A as centre and radius AB, draw a
semi-circle and divide it by trial into as many
equal parts astherequired polygon has.
(iii) Join A-2 which is the second division on the semi-circle.
(iv) Draw the bisectors of AB and A-2 and these bisectors intersect at point O.
(v) With O as centre and radius equal to OA or OB, draw a circle.
(vi) Mark-off the length AB or AD round the circumference of the circle to get points C and D.
Join 2C andCD to obtain the required polygon.
To construct a number of polygons on a given base.(two– triangle method)
A scale is used to measure or lay out a line on a drawing in full size, or larger or smaller than full size. The term scale also refers to the ratio of the size of a drawing to the actual object. This ratio or fraction is called the Representative Fraction of a scale.
Representative fraction (RF) = distance or size drawn
distance or size represented
The representative fractions of the most commonly used scales are as follows:
Plain scale: A plain scale has two units of measurement. They are metres and decimetres. Scales such as 1:2, 1:5, and 1:100 can be used direct from a standard scale rule. Other scales, for example, 1:21/2, would need to be constructed
Evaluation Questions
1. Define representative fraction.
2. What is a plain scale?
Construction of plain scale.
To construct a plain scale of 1: 21/2 to show a maximum of 400mm and a minimum of 10mm.
Method:
(i) Workout the total or maximum length of the scale. 1: 21/2 x 400mm = 160mm
(ii) Draw a line 160mm long and divide it into four (4) equal parts of 40mm units each. (iii) The height of the scale can be any convenient height.
4. The rulers in your mathematical set were constructed to measure to a scale ratio of A. 1:2
B. 1: 1 C. 2:3 D. 1:5
5. Which of the following is not an enlargement scale? A. 5: 3 B 3: 4 C. 2:1 D. 3:2.
Theory
1. Draw the diagram below to a scale of 1:2.
2. The only measurements available on the drawing of an ornamental gate shown below
are 2100mm and 300mm sizes. Find the scale of the drawing and measure the other
sizes; A, B, C, D and E. (Hints: R.F = drawn/ actual size)
WEEK FIVE DATE:………………
Topic: Diagonal and chordal scale and their uses.
Content:
(i) Meaning and construction of diagonal scales.
(ii) Scale of chords and its construction.
Meaning and construction of diagonal scales.
Diagonal scale: Diagonal scales unlike plain scale can measure to a fine degree of accuracy with three units of measurement; centimeters, millimeters and tenths of a millimeter. This is the singular advantage of diagonal scale over plain scale.
Construction of diagonal scale
To construct a diagonal scale of centimeters to read up to 7cm in millimeters and tenths of a millimeter.
Method:
(i) Draw a line 7cm long and divide it into 7 equal parts.
(ii) Divide the first part into 10 equal units
(iii) Draw the height AC of the scale rule to any convenient length and divide it into 10 equal parts.
Transfer same10 division to the first part AD
(iv) Project horizontal lines from each of these 10 divisions on the height AC.
(v) Project vertical lines from each of the division on line AB. Draw diagonal lines in between the 10
vertical lines of the first part as shown above.
Note: Diagonal scale is read from the highest unit to the smallest. The value of S is 4.66cm or 46.6mm
(4cm + 6mm + 6/10mm => 40mm + 6mm + 0.6mm)
To construct a diagonal scale of twice full size to read up to 6centimetre in millimeters and tenth of a millimeter.
Method:
(i) Draw a line 12cm long i.e. (2×6)cm and divide it into 6 equal parts.
(ii)Follow the same procedure as above. The value of X is 2.49cm or 24.9mm.
To construct a diagonal scale of 3centimetres equal to 1metre and to read up to 4metres in decimeters and centimeters.
Method:
(i) Draw a line 12cm long i.e. (3×4) cm and divide it into 4 equal parts.
(ii) Same procedure as above. The value of S is 1.93m or 19.3dm or 193cm.
To construct a diagonal scale of 1/4 full size to read up to 5dm in centimeters and millimeters
Method:
(i) Draw a line of length equal to 125mm i.e. (1/4x 5)dm => 1.25dm or 12.5cm or 125mm.
(ii) Same procedure as above. The value of Z is 3.93dm or 39.3cm or 393mm.
Evaluation Questions
1. What is diagonal scale?
2. Construct a diagonal scale of 2/5 full size to read up to 3 dm in centimeters and millimeters.
3. What is the difference between the values of k and n in the figure shown below?
The chordal scale: The chordal scale is used to construct angles that cannot easily be constructed using a ruler and a pair of compasses .For instance, to construct angles which are multiples of 50, a scale of chords showing a distance of 50 is constructed.
Construction of the scale of chords
To draw a scale of chords to show distances of 50
Method:
(i) Draw a quadrant ABC of any radius and divide the arc AC into 18 equal parts i.e. 50x18 = 900 which
is the angle in a quadrant.
(ii) Extend the base line AB of the quadrant to the right.
(iii) With 000 as centre and radius equal to each of the 18 divisions in turn, draw
arcs to touch the base line AB of the quadrant and its extension.
(iv) Construct the scale which could be of any convenient height below the quadrant.
This scale is used to construct the required angle.
Example 1
Construct angle 250 using scale of chords.
(i) Draw a line PM of any convenient length.
(ii) With P as centre and radius R of length equal to the radius of the quadrant (00-600), draw an arc to cut
line PM at Q.
(iii) With Q as centre and radius r of length equal to the required angle (00-250) measured from the
constructed scale, draw an arc to cut the previous one at S.
(iv) Draw a line from P through S. Scale of chord is a type of scale that
can be used to construct angles other than the ones done with a pair of compasses eg 240, 370
etc.
General evaluation/ revision questions
1. Construct a scale of 4cm equals 2dm to read up to 5dm in centimeters. | 677.169 | 1 |
7. It Has Infinite Set Of Points That Extends In All Direction. A. Point…
7. It Has Infinite Set Of Points That Extends In All Direction. A. Point…
7. It has infinite set of points that extends in all direction. A. point B. line C. plane D. all of the above 8. A subset of a line which consists of two endpoints. A point C. segment D. angle 9. It is a subset of a line which consists of one endpoint and extends in one direction. A. point В. ray C. segment D. angle 10. Which of the following is true? A. ray can extend in both directions C. AB is the same as BA B. line segment has two endpoints D. line CD can be written as CD For numbers 11 -12, refer to Figure 1. 11. What three points are collinear? A. Points X, Y, and Z C. Points X, V, and Z Х B. Points V, Y, and Z D. Points X, Y, and V 12. Which of the following statements is TRUE? A. Points Y and Z are coplanar. Figure 1 z B. Points X, Y, and Z are coplanar. C. Points Vand Z are non-collinear. D. Points X, Y, and Z non-collinear 13. These are points that lie on the same plane. A Collinear points C. Coplanar points B. Non-collinear points D. Non-coplanar points 14. Points R, O, and Mlie on a line with the coordinates 5, 10, and -3, respectively. Which point lies between the other two? A. Paint R B. Point O C. Point M D. cannot be determined 15. What is AB if the coordinate of Ais-9 and that of Bis -12? A.-3 B. 3 C. 21 D. 21 16.Points A, B, and C are collinear. If AB= 8 cm and BC = 12 om, what is AC? A. 4 cm B. 8 cm C. 12 cm D. 20 cm 17. Points A, B, and C are collinear. If AB = 2x + 3 and BC = 3x – 7, what is AC? A.X + 10 B. X-4 C. 5×4 D. 5x + 10 18. On a number line, point Y is between points X and Z. If XZ w 18, the coordinate of Y is 3 and the coordinate of Z is 7, what is the coordinate of X? A 15 B.B C. – 8 D.-11 19.On a number line, the length of AB is 26. If the coordinate of the midpoint Mof AB is -4, what are the coordinates of points A and B? A.-17 and 9 B. 17 and -9 C.-30 and 22 D. 30 and -22 20. Which among the given equation is the Pythagorean Equation from Pythagorean Theorem? B. a + b2=0 c. (a + b) = 0 D. a2 + b2 = 2 11 A. a + b = ca
Lines points plane line postulate intersect intersection two if point geometry planes distinct basic also ck. Plane point through lay line perpendicular. Determining whether four points lie in a plane – geogebra | 677.169 | 1 |
Questions tagged [quadrilateral]
For questions about general quadrilaterals (including parallelograms, trapezoids, rhombi) and their properties.
In Euclidean plane geometry, a quadrilateral is a polygon with four sides (or edges) and four vertices or corners. Squares, rectangles, rhombi, parallelograms and trapezoids are special kinds of quadrilaterals.
An exam for high school students had the following problem:
…
Square $ABCD$ has area $1cm^2$ and sides of $1cm$ each.
$H, F, E, G$ are the midpoints of sides $AD, DC, CB, BA$ respectively.
What will the area of the square formed in the middle be?
I know that this problem can be solved by trigonometry by using…
Below the parallelogram is obtained from square by stretching the top side while fixing the bottom.
Since area of parallelogram is base times height, both square and parallelogram have the same area.
This is true no matter how far I stretch the topAs already introduced in this post, given the series of prime numbers greater than $9$, let organize them in four rows, according to their last digit ($1,3,7$ or $9$). The column in which they are displayed is the ten to which they belong, as…
If all sides: $a, b, c, d$ are known, is there a formula that can calculate the area of a trapezoid?
I know this formula for calculating the area of a trapezoid from its two bases and its height:
$$S=\frac {a+b}{2}×h$$
And I know a well-knownTwo weeks ago our professor taught us the Japanese theorem for cyclic quadrilaterals. It states that the inscribed centres of the four triangles formed by two sides and a diagonal of a cyclic quadrilateral always form a rectangle.
After the proof…
Let $ABCD$ be a fixed convex quadrilateral and $P$ be an arbitrary point. Let $S,T,U,V,K,L$ be the projections of $P$ on $AB,CD,AD,BC,AC,BD$ respectively. Let $X,Y,Z$ be the midpoints of $ST,UV,KL$. Is it true that the angles of triangle $\triangle…
My niece is in the 10th grade, and they have to do lot of theorems related to quadrilaterals. And, I was surprised to know that they have to learn by rote some theorems. This has made her feel that maths is the worst subject (though she likes…
Does anyone know how to calculate the volume of an irregular octahedron from the lengths of the edges?
The octahedron has triangular faces, but the only information are the edge lengths.
Alternatively, how might I calculate the length of a line…
I have such an interesting observation: if I take a square grid and rotate it over itself by atan(3/4) , it forms a structure which has four axes of reflection symmetry:
The resulting structure is really mesmerizing, I see a lot of symmetric shapes…
Can you prove the following claim? The claim is inspired by Harcourt's theorem.
In any rhombus $ABCD$ construct an arbitrary tangent to the incircle of rhombus . Let $n_1,n_2,n_3,n_4$ be a signed distances from vertices $A,B,C,D$ to tangent line…
I have recently been reading about a very interesting geometry problem and have tried to solve it. I'm now in a point, in which I don't know how to move forward and would appreciate if someone could help.
The problem is the following
Consider an… | 677.169 | 1 |
Splitting polygon into triangles
The polygon coordinates divided by gaps. Can have a+ib appearance
You entered the following coordinates of a polygon
The area of the set polygon (in conventional units)
The method of partitioning an arbitrary, disjoint polygon into triangles is considered.In addition to experience in implementing such an algorithm, we still get the opportunity to further calculate the coordinate of the center of gravity such polygons.
The calculation is carried out online, and you, as the user of this site, just enter the coordinates of the vertices of the polygon.
The only condition is that the coordinates are entered only (!!!) clockwise.
If this condition is not met, the result will be incorrect if the bot answers you at all :)
In addition to the list with the coordinates of the triangle, as a result of calculations, the area of each of the triangles will be calculated. | 677.169 | 1 |
Teens Have Proven the Pythagorean Theorem With Trigonometry. That Should Be Impossible.
The Pythagorean Theorem (a2 + b2 = c2) is fundamental to mathematics, especially to the field of trigonometry.
Some mathematicians have stated that proving the theorem using trigonometry is impossible without circular reasoning, because trigonometry relies so much on the theorem itself.
Two New Orleans high school students say they've proven the theorem using trigonometry without relying on circular reasoning.
The Pythagorean Theorem—discovered by the Greek mathematician Pythagoras in the 6th century BCE—is a cornerstone of mathematics. Simply stated as a2 + b2 = c2, the theorem posits that the sum of the two shortest sides of a right triangle (a2 and b2) is equal to that triangle's longest side (c2). For centuries, this idea has been proven by some of history's greatest minds, such as Albert Einstein, U.S. President James Garfield (of all people), and even Pythagoras himself.
In fact, there have been hundreds of proofs of the Pythagoras' groundbreaking theorem, but almost none of them—if not none at all—have independently proved it using trigonometry. That's because the fundamentals of the theorem are what the entire field of trigonometry is built on, and so, the thinking goes that to use trigonometry to prove the theorem is to employ what's called "circular reasoning." It's essentially using the Pythagorean Theorem to prove the Pythagorean Theorem.
Some mathematicians argue that using trigonometry to independently prove the theorem is actually impossible, including Elisha Loomis, whose book on the topic (originally published almost a century ago) states that "there are no trigonometric proofs because all the fundamental formulae of trigonometry are themselves based upon the truth of the Pythagorean theorem."
Well, two New Orleans high schoolers say they've proven the impossible.
"It's really an unparalleled feeling, honestly, because there's just nothing like being able to do something that people don't think young people can do," Johnson told WWL in an interview. "A lot of times you see this stuff, you don't see kids like us doing it."
The American Mathematical Society said in a Facebook post that the two young mathematicians are being encouraged to submit their work to a peer-reviewed journal.
Although the full details of the proof haven't been published, YouTuber MathTrain believes he worked out an approximation of Johnson's and Jackson's Law of Sines proof by analyzing stills from a local news report.
Although the proof is an impressive bit of mathematics, other mathematicians have employed similar approaches before, using sine and cosine to independently prove the Pythagorean Theorem without relying on sin²α + cos²α = 1. But regardless of these previous attempts, Johnson and Jackson's achievement is no less impressive. Much like how Albert Einstein proved the theorem at the age of 12, an amazing mathematical proof has once again come from the ingenuity of young | 677.169 | 1 |
Evaluate rydr rdy; where C Is the rectangle with vertices (0,0), (2,0), (2,41, and (0,4)
Question:
Answers
Answers #1
Rectangle Show that the points $(2,2,2),(2,0,1),(4,1,-1),$ and (4,3,0) are the vertices of a rectangle.
.
Answers #2
So we are given a 0.345345 If we wanted to put that on our coordinate grid every three in the X direction for in the Y direction and then five in the Z direction So it would be some place like here. You notice that this kind of forms a nice rectangular prism, right? If we, um if we drew all of these straight line paths to get to this point, we could dio you know, in the X direction Z direction What? Why? Direction or Z direction? Why direction X direction Or, you know, any combination of those directions in any order we frame this rectangular prison. So now we want to know what would be Some of these, uh, were to seize all in this rectangular prison eso in for point A We see that, uh, it's nearly the same as our point p Look, all this p uh, but the why component has been pushed to zero. So we're projecting our point p onto the x Z plane eso we right away know that are why component is zero. But also we see that when we project something onto the XY plane.
We don't change the X or Z coordinates. So our Exon's in coordinate are going to stay the same. We get 305 for point B. We're doing the same thing. Except for now.
We're projecting it onto the X Z plane. Were kind of, uh, squishing it down to the xz plane. And in doing so, we're making our Z coordinate zero eso again. I can go ahead and fill that in right away. I see that it's on the X Y plane, Zorzi Cornet zero.
But I see that by projecting this point onto our x y playing, we don't change the X y x value or the Y value. Right. So our x and Y, you're going to see the same, which are three and four, respectively. And now lastly point see, doing the same thing. We projected it onto the y Z plane.
Um, so we projected onto the y Z plane. We see that we've pushed it back into this plane on X now becomes zero, so I can fill that in right away. Still, I'm left with my wine z coordinate. But when I when I make a projection onto the y Z plane. I don't change my why or z coordinates so I can fill this in just as it was originally.
So these are our three different points that are kind of defined by the vergis ease of the of this imaginary rectangular prison that you get when you're discussing point in three dimensional space..
answer from John Mcalister
Answers #3
So if we have, um, our coordinates based like this and were given two points, say 580 So five in our X direction A and R Y direction and zero in Jersey direction. We call that point p he is 580 and then another point Q, which is just 00 10 um, 00 10. So just 10 in our Z direction like that. We can use this to define a, um, rectangular prism. If all of our edges are parallel to either the X y or Z uh, access.
We use this, uh, these two points as opposite points in our rectangular prison, and we can define a rectangular prison like that. Um, so now that we have that, we can use this to use this rectangular prison in the information we know about these two points to find on the other virtus ease of this rectangular prison. So first, let's bust you this point point A. We can see that point a has the same X value as our point p, right. A path to point a would move along the X axis and then up in the Z direction, the seeming that as a path to point P would move along the X axis until it, uh, moves in the Y direction.
Eso we see that both point A and P have the same x value so we can fill that in. P has a expert, you five. So, you know hasn't expect your five. We also know right from the path that I just drew. We don't have any r y component zero.
This is in the xz plane. Let me label my axes. That may be helpful. Uh, this is point A's and xz plane, so it doesn't have any why I could put it all so I'll fill in a zero for that. And then lastly, we see that a has the same Z component as point Q right? It's on the top surface of our wrecking your prison s so we can see that it has the same see value as a point queues.
We can fill that in as 10. So that's our point. A. Now we can move on to our point B, which is now in here. Our point b, we can see is lies on the Y axis, and this is important because it tells us that both X and Z are zero.
It's something lies on the Y axis. Well, we know that right off the bat, x zero and z zero. Um, the only motion that we're interested in is the distance along the Y axis on. For this, we can see that it has the same distance along the y axis as our point p. It has the same Y value is our point p just like they have the same X values point B b has the same y value this point peace we can draw in.
You can fill in our why Value like this has the same y value as point P. So be a 080 and then lastly Ah, see, which is this one that's not in line with any of our planes x z x y or y z um, so we actually know that it's going to have a non zero component for all X, y and Z. Um, so to figure this out, we noticed that it I can't see is directly above point P. It's it is directly above point P. So if we go to point P and add only in the Z direction, we will get to point C.
So that means the X and the Y value for point. See, you're gonna be the same as they are from point P. So we have five and eight as our X and Y value. And then we also see that the Z value for Point C is the same as the Z value for point Q. Right.
It's on the top surface of our rectangular prison so we can fill in our the Z component as 10 for point C. And these are all of the points that we need to label on this rectangular prison s so we can see that we have two points like this in free space, we can define a rectangular prism in this way. Where are edges are parallel to the X, the why or the Z axis? Um, And in doing so, we can fill in where the rest of our verses would be just by knowing two opposite points like that.
answer from John Mcalister
Answers #4
So let's draw our coordinates space Ah, with our x y and Z axes. Um, just like this. So this is my xxx This is my y axis, and this is my easy access. And I have extended my, uh, Y axis in the negative direction because that's where that's where we'll be working. Um, and so we have two points.
We have the 20.3 negative 40 three negative for zero, which would be three in the X direction, negative for in the Y direction and zero in the Z direction. So let's leave at that point. P I think that and we also have the 0.5 which would simply be five in the Z direction. Just like that looks. Label that point Q zero 05 Uh, now we can use these two points, um called in the opposite points in a rectangular prison and say that that rectangular prism has edges that are parallel to you, the x y or Z axes s so we can draw these edges exists lines parallel to the X Y or Z axis.
I mean, we draw a rectangular prism just like this, and now we can use the information just from these two points to find the rest toe, find the locations of more of these vergis ease. So let's start with a point labeled a Our point A, we see is directly above the point p in the Z direction. So, uh, if we were to travel the point p, we would dio certain amount in the X direction a certain amount in or why direction, right? And then all that would be left to do to get the point a would be to move up. So from that we know that point a has the same X and why, um, component as our point peace. We can fill that in right away.
And then we moved up from Point P to point a Now point is that the same sea level the same height as the point Q. So it's gonna have the same Zeke unquote component as the point. Q. Um, and that Z component is five. So we can fill that in.
Just like that. Is three negative for five cool. So we can remind now to our point b, which is down here. This is a point B and something important to notice is that it's on the Y axis so that moons when it's on the y axis, we know that X zero and Z is zero. Let me draw that little bit nicer.
You know, the X zero nz is zero, right? It's gonna the only thing that is non zero is our why component. And we knew that our why component is in line with the Y component of PPI, right? If we take a path from the origin to our point B, it's gonna travel in the Y direction only and it's going to travel in the Y direction, the same amount that it would to to take it straight. My path from the point Pete, from the origin to the point p. Um, so we know that the you want this thing, Why component as P, which was negative force. We can fill that in zero native 40 and finally our point c, which is, uh, this for text right here.
See, we see that it's gonna have the same height as points A and Q so we could fill that in right away. Five. It's gonna be directly above be so just like we're talking about how a is a direct directly above P C is directly above the so we know it. It'll have the same X and y coordinates. The only difference between B and C is the Z coordinate eso you can just fill these in.
He sat like this and everything that is important to notice is that, uh, see is on the Z y playing so we know from that that are X component is going to be zero. Um so stepping said there's lots of ways Teoh, you know, figure out lots of logical paths to figure out the positions of a B and C Um, but anyway, that you do it, you should arrive at these positions for these were vergis ease of this rectangular prison..
answer from John Mcalister
Answers #5
For this question were given a cube in our coordinates space X, y z, um, and were given that this point is zero negative three zero. And furthermore, they tell us that all of the side of things are the same, right? They tell us to cube. So we could imagine this cube, Um, like this where this is a square base sightly sort of the same, uh, and directly above it is another square base with similar siblings. And then all of the edges between those two squares have the same selling. She right, calm.
So now we want to find where some other points in this cube are, right? So let's start with this point. We'll call this point a to this point A, we know is, um not it's non zero in the X in the Y direction, but it's in the X Y plane, so we know that it's Z component is zero, so I can fill that in right away. It's in the X y plane, so it's the component is zero. Um, and because we knew all of the cycling through the same, it's clear to see that, uh, it's moved in the Y direction. Negative three, right? This is the same.
Why component as this point that we have labeled here? We call that p has the same. Why component is point P, um, and it's removed some edge length away from point P in the X direction. Well, we know that that that length is going to be three because all of the side wings are the same and we noticed that it's moving in the negative X direction, right? This is negative. X. This is pasta vex, so it can feel in negative three there for a point.
A, um, our point b is this went right here, and what we notice about point B is that it's on the Z axis, so ran way. We actually know that X and Y are zero. We know that because it's on the Z axis. Uh, the only it's only removed from the origin in the Z direction, right? There's no change in X or changing wife relative to our origin. So we know that X and Y both could be zero in the Z component.
Well, again, we know all of this island's here are three, so it's going to be only three removed from the origin, and we notice we're going in a positive direction. So we can say this is 003 Lastly, a point C which is right here. Well, this is non zero in the extraction, None zero in the Y direction and not zero in the same direction. Um, we got to think of this as being directly above the point A. Right.
So, um, maybe already have the intuition that because it's non zero, all these are gonna be threes. We can see that it's in the negative direction in why? And Exxon deposit directions. E. So maybe we can really fill this out. But if not, we noticed that it's directly above the point A in the Z direction.
So it has the same X and y components as a so we can copy those in negative three negative three. But the Z component is different. It z component is increased by one side length. Right? And of course, we know the side length is three. We're going in the positive direction.
So we have negative three negative. 33 So these are our three points that are defined by this cube Silent three And once we know what optimize it is right octane being, um, one of these eight spaces where in this case, X's negative wise, negative and Z is positive. That's That's this Arctic that this cube is in. Once we know the doctor, never a Cuban, we know one side like we can fill in the rest of the information about the Vergis ease of the Cube..
answer from John Mcalister
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Daily Solutions for the popular LA Times Crossword Puzzle! Primary Menu. Homepage; Printable Puzzles; Posted in: Crossword Clues Greek letter for an angle crossword clue. Written by krist October 10, 2022. On this page you …Greek god of war. Massage haven. Long-eared hopper. Get out of bed Tanning lotionFraternity Greek letters are traditionally two or three letters that represent the motto of a fraternity. In some fraternity names, the Greek letters are the initials of the words Greek letter. We will try to find the right answer to this particular crossword clue. Here are the possible solutions for "Greek letter" clue. It was last seen in British general knowledge crossword. We have 20 possible answers in our database.Less than right, as an angle. The answer to this question: A C U T E. Salt Lake City's State. Go back Tri5 days ago · Greek letter This clue has appeared on Daily Themed Crossword puzzle. 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Triangle Inequality Theorem Worksheet Doc
Describe the possible lengths of the third side. State if the three numbers can be the measures of the sides of a triangle.
Math 8 Lesson Plan 69d Triangle Inequality Theorem Class Outline Doc
At the same time with the use of lego robot they learn of motor speed through the use of distance and time.
Triangle inequality theorem worksheet doc. Using the figure and the inequality theorem which angle 5 or 8 has the smallest measure. Triangle inequalities 5 5 key ideas the longer the side of a triangle the larger the angle opposite of it. The triangle inequality theorem date period state if the three numbers can be the measures of the sides of a triangle.
1 if two sides of a triangle are 1 and 3 the third side may be. The bigger the angle in a triangle the longer the opposite side. List the angles of the triangle in order from smallest to largest.
Worksheet by kuta software llc assignment 31a triangle inequality theorem name id. A triangle has one side of 23 meters and another of 17 meters. 5 5 and 5 6 notes.
Can these numbers be the length of the sides of a triangle. The triangle inequality theorem states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side. Namely the triangle inequality theorem.
Hinge theorem if two sides of one triangle are congruent to two sides of another triangle and the included angle of the first is larger than the included angle of the second. Then circle yes or no. 2 if the lengths of two sides of a triangle are 5 and 7.
Chp 7 practice test triangle inequalities determine whether the given coordinates are the vertices of a triangle. Using this theorem answer the following questions. Triangle inequality theorem the sum of the lengths of any two sides of a triangle is greater than the length of the.
1 date period k z2h0m1u5k tkzudtta bsho fzthwsakr el qlklnct p c paklylu drmi gfhyt sm yrde sjelrdvxevd. Triangle 7 5 an inequality involving an exterior angle of a triangle 7 6 inequalities involving sides and angles of a triangle chapter summary vocabulary review exercises cumulative review geometric inequalities euclid s proposition 20 of book 1 of the elements states in any triangle two sides taken together in any. Practice triangle inequality theorem triangle inequality theorem the sum of the lengths of any two sides of a triangle is than the length of the third side.
Show math to prove your answer using the triangle inequality theorem. Engineering connection a triangle is simply defined as a shape that is made up of 3 angles and 3 line segments known as its sides. A 5 b 2 c 3 d 4.
F Makes a triangle a valid triangle. Create your own worksheets like this one with infinite geometry. | 677.169 | 1 |
Angle
Our online angle converter allows you to easily convert between different units of angle, including degrees, radians, and more. Simply input the value and select the unit you wish to convert to get an accurate result.
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Select the current unit in the left column, the desired unit in the right column, and enter a value in the left column to generate the resulting conversion.
Formuladivide the Angle value by 57.29577951322445
Result:
1 radian = 57.2957795131 Degree
Divide the number by 57.29577951322445 to convert Radian to Degree.
Angle Unit Conversion chart
Check out this Angle conversion table to find out what the conversion factors are between the various Angle units:
Complete list of Angle units for conversion to see conversion of 1 of below units of Angle to Radian.
Unit
Symbol
Value in Radian
Radian
rad
1
Degree
°
0.0174532925199
Minute
"
0.000290888208666
Second
'
0.0000048481368111
Arcminute
"
0.000290888208666
Arcsecond
'
0.0000048481368111
Gradian
^g, gon or grad
0.0157079632679
Gon
0.0157079632679
sign
0.523598775598
Mil
0.000981747704247
Revolution
r
6.28318530718
Circle
6.28318530718
Turn
6.28318530718
Quadrant
1.57079632679
Right angle
1.57079632679
Sextant
1.0471975512
Radian to Degree conversion table
Radian
Degree
1
57.2957795131
2
114.591559026
3
171.887338539
4
229.183118052
5
286.478897565
6
343.774677078
7
401.070456592
8
458.366236105
9
515.662015618
10
572.957795131
20
1145.91559026
30
1718.87338539
40
2291.83118052
50
2864.78897565
60
3437.74677078
70
4010.70456592
80
4583.66236105
90
5156.62015618
100
5729.57795131
Angle Details
What are the different units of Angle?
The SI unit of Angle is degree. Alternatively, the Angle can also be expressed in. | 677.169 | 1 |
In addition, the contour point p_{m} with maximum distance to the
center of gravity c is calculated. If the angle between the
vector p_{m}c and the vector given by
Phi is
greater than Pi, the value of Pi is added
to the angle. If XLDXLDXLDXLDXLDXLD consists of only two points,
PhiPhiPhiPhiPhiphi is given by the direction from the first point towards
the second point. | 677.169 | 1 |
Trigonometry calculator sin cos tan inverse english español
This is an online trigonometry calculator to find out the equivalent values of radians and degrees for the given number. Any number can have different values of radians and degrees with respect to the trigonometric functions such as Sine(Sin), Cosine(Cos), Tangent(Tan), Cotangent(Cot), Secant(Sec), Cosecant(Cosec), etc.
RadiansDegreesTrigonometry calculator sin cos tan inverse english español
This is an online trigonometry calculator to find out the equivalent values of radians and degrees for the given number. Any number can have different values o | 677.169 | 1 |
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Description. This is a set of four trigonometry mazes to practice finding missing side and angles measures in right triangles using the sine, cosine, and tangent ratios. Students use their solutions to navigate through the maze. Find Exercise PrintFor the point ( x, y) on a circle of radius r at an angle of θ, we can define the six trigonometric functions as the ratios of the sides of the corresponding triangle: The sine function: sin(θ) = y r. The cosine function: cos(θ) = x r. The tangent function: tan(θ) = y x. The cosecant function: csc(θ) = r y.Report Teach Yourself Trigonometry is suitable for beginners, but it also goes beyond the basics to offer comprehensive coverage of more advanced topics. Each chapter features numerous worked examples and many carefully graded exercises, and full demonstrations of trigonometric proofs are given in the answer key. Product Engineering The Experiment Apr 4, 2022 · Recording sheet and teacher answer key ♢ link to an optional, . 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Language: English (en) ID: 1053CK-12 Trigonometry Concepts Answer Key Concept: Domain, Free printable worksheets (pdf) with answer keys on algebra i, geom When it comes to purchasing a new pillow, one of the key factors to consider is the warranty that comes with it. A good warranty can provide peace of mind and assurance that your i... Find step-by-step solutions and answers to Trigonometry - 97... | 677.169 | 1 |
.
Bicone
An bicone or dicone is the three-dimensional geometric shape swept by revolving an isosceles triangle around its edge of non-equal length. Alternatively, one can view a bicone as the surface created by joining two identical right circular cones base-to-base. | 677.169 | 1 |
Basic Terms And Definitions Of Lines And Angles
Quizizz is a platform that provides engaging and effective lessons by enabling educators to create custom quizzes and track student progress in real-time. It also provides resources to help teachers tackle the problems their students face in learning the concepts of math. | 677.169 | 1 |
How To Evaluate Trigonometric Functions Using Periodic Properties - Trigonometry
TLDRThe video script discusses the periodic properties of trigonometric functions, focusing on sine, cosine, tangent, and cotangent. It explains that adding 360 degrees or 2π to the angle of these functions results in the same value, a characteristic known as periodicity. The script uses examples to illustrate this concept, such as sine of 30 degrees being equal to sine of 390 degrees due to their coterminal relationship. It also demonstrates how to evaluate trigonometric functions at specific angles by finding coterminal angles less than 360 degrees, using sine and cosine examples at 420 degrees, -2π/3, and 750 degrees. The unit circle is referenced to determine the values of these functions, showing that the sine of 60 degrees and its coterminal angles like 420 and 750 are all equal to √3/2, and cosine of 4π/3 and -2π/3 is -1/2.
Takeaways
🔁 The periodic property of trigonometric functions states that the sine of an angle is equal to the sine of the same angle plus 360 degrees or 2π.
🔄 Coterminal angles, which differ by a multiple of 360 degrees, share the same trigonometric function values.
📚 This periodic nature applies to all trigonometric functions, including sine, cosine, tangent, and cotangent.
📉 To find a coterminal angle, you add or subtract multiples of 360 degrees from the original angle.
📈 The graph of sine and cosine functions demonstrates their periodicity, repeating their values as the angle increases by multiples of 360 degrees.
🧩 Evaluating trigonometric functions at specific angles can be simplified by finding a coterminal angle within the first 360 degrees.
📌 Example given: Sine of 420 degrees is equal to sine of 60 degrees, which is \( \frac{\sqrt{3}}{2} \) based on the unit circle.
➡️ Reference angles are used to find the values of cosine for angles outside the first 360 degrees, such as cosine of -2π/3 being equal to cosine of 4π/3.
📐 At a unit circle, cosine is equal to the x-value of the point corresponding to the angle, which helps in determining the sign of the cosine value.
🔢 Another example: Sine of 750 degrees simplifies to sine of 30 degrees, which is 1/2, due to the periodic nature of the sine function.
Q & A
What does the periodic property of trigonometric functions mean?
-The periodic property of trigonometric functions means that the functions repeat their values in regular intervals. For sine, cosine, tangent, cotangent, etc., adding or subtracting multiples of 360 degrees (or 2π radians) to the angle results in the same function value.
How does the periodic property apply to the sine function?
-The periodic property applies to the sine function such that sine of theta is equal to sine of theta plus any multiple of 360 degrees. This means sine of an angle and sine of its coterminal angle will have the same value.
What is a coterminal angle?
-A coterminal angle is an angle that differs from the original angle by a full rotation, which is 360 degrees. Coterminal angles have the same trigonometric function values as their corresponding angles.
How can you find a coterminal angle for a given angle?
-To find a coterminal angle for a given angle, you add or subtract multiples of 360 degrees to the original angle.
What is the value of sine 30 degrees and how does it relate to sine 390 degrees?
-The value of sine 30 degrees is 1/2. Since sine 390 degrees is a coterminal angle of sine 30 degrees (390 - 360 = 30), sine 390 degrees also has the same value of 1/2.
How does the periodic property apply to the cosine function?
-The periodic property applies to the cosine function such that cosine of an angle is equal to cosine of that angle plus any multiple of 360 degrees. This ensures that cosine of an angle and its coterminal angle will have the same value.
What is the value of cosine 50 degrees and how does it relate to cosine 410 degrees?
-Cosine 50 degrees is equal to cosine 50 degrees plus 360 degrees, which is cosine 410 degrees. Therefore, both have the same value.
How can you use the periodic property to evaluate sine of 420 degrees?
-To evaluate sine of 420 degrees, you find a coterminal angle that is less than 420 degrees by subtracting 360 degrees from 420 degrees, resulting in 60 degrees. Therefore, sine 420 degrees is equal to sine 60 degrees, which is root 3/2.
What is the value of cosine negative 2 PI/3 and how can you find it using the periodic property?
-Cosine negative 2 PI/3 can be found by adding 2 PI to the angle, which gives cosine negative 2 PI/3 plus 2 PI, or cosine 4 PI/3. Since 4 PI/3 has the same reference angle as PI/3 (60 degrees), cosine 4 PI/3 is negative 1/2, which is the same as cosine negative 2 PI/3.
How can you evaluate sine of 750 degrees using the periodic property?
-To evaluate sine of 750 degrees, you reduce the angle by subtracting multiples of 360 degrees until you get an angle less than 360. Subtracting 360 twice from 750 gives 30 degrees. Therefore, sine 750 degrees is equivalent to sine 30 degrees, which is 1/2.
What is the significance of the unit circle in evaluating trigonometric function values?
-The unit circle is a circle with a radius of 1, where the angles are measured from the positive x-axis. It is significant in evaluating trigonometric function values because it provides a visual representation of the sine and cosine values for standard angles, making it easier to determine the values for coterminal angles.
Outlines
00:00
📚 Introduction to Periodic Properties of Trigonometric Functions
This paragraph introduces the concept of periodicity in trigonometric functions, specifically focusing on the sine function. It explains that sine of an angle (theta) is equal to sine of the same angle plus any multiple of 360 degrees or 2π. This property implies that sine values remain constant for coterminal angles, which are angles that differ by full rotations (360 degrees). The paragraph also mentions that this periodic property applies to other trigonometric functions such as cosine, tangent, and cotangent. The concept is illustrated with examples, showing how to find the sine value for angles like 30 degrees and 390 degrees, which are coterminal and thus share the same sine value. The explanation transitions into a discussion of how these functions are graphed, with sine and cosine functions repeating their patterns as the angle increases by multiples of 360 degrees used to describe periodic properties and are fundamental to understanding the behavior of sine, cosine, and tangent functions. The script discusses how these functions repeat their values for different angles, emphasizing their periodic nature.
💡Periodic properties
Periodic properties refer to the characteristic of a function to repeat its values at regular intervals. In the video, this concept is central as it explains how trigonometric functions like sine, cosine, and tangent repeat their values when the angle is increased or decreased by a full cycle, such as 360 degrees or 2π radians. The script uses this property to simplify the evaluation of trigonometric functions at different angles.
💡Coterminal angles
Coterminal angles are angles that share the same terminal side in standard position. The video explains that sine of an angle is equal to sine of that angle plus any multiple of 360 degrees, which means that angles differing by full rotations (360 degrees) are coterminal. This concept is used to find equivalent angles that are easier to work with, as their trigonometric function values are the same.
💡Unit circle
A unit circle is a circle with a radius of one, centered at the origin of a coordinate system. It is used in the context of the video to determine the values of trigonometric functions for specific angles. The script mentions the unit circle when explaining how to find the sine of 60 degrees, which corresponds to a point on the unit circle with specific x and y coordinates.
💡Reference angle
The reference angle is the acute angle formed by the terminal side of a given angle and the x-axis. In the video, the reference angle is used to find the cosine of an angle greater than 90 degrees, such as 4π/3 radians. The script explains that the reference angle for 4π/3 is π/3, and the cosine value is determined based on the x-coordinate of the point on the unit circle corresponding to the reference angle.
💡Sine function
The sine function is one of the primary trigonometric functions that relates the ratio of the opposite side to the hypotenuse in a right-angled triangle. The video script uses the sine function to illustrate periodic properties, showing that sine of an angle is equal to sine of that angle plus or minus multiples of 360 degrees.
💡Cosine function
The cosine function is another fundamental trigonometric function that relates the ratio of the adjacent side to the hypotenuse in a right-angled triangle. The video demonstrates the periodic nature of the cosine function, showing that cosine of an angle is equal to cosine of that angle plus or minus multiples of 360 degrees.
💡Tangent function
The tangent function is the ratio of the sine to the cosine of an angle. Although not explicitly detailed in the script, it is implied that the tangent function also exhibits periodic properties, as it is derived from the sine and cosine functions, which are both periodic.
💡Cotangent function
The cotangent function is the reciprocal of the tangent function, or the ratio of the adjacent side to the opposite side in a right-angled triangle. Similar to the tangent function, the cotangent function is periodic, as indicated in the video script, although it is not the primary focus of the explanation.
💡Evaluating trigonometric functions
Evaluating trigonometric functions involves finding the value of a trigonometric function for a given angle. The video script provides examples of how to evaluate sine and cosine functions at specific angles by using periodic properties and the unit circle. For instance, sine of 420 degrees is evaluated by finding a coterminal angle, 60 degrees, and using the known value from the unit circle.
Highlights
Sine of theta is equal to sine of theta plus 360 degrees or 2 pi.
Adding 360 degrees to theta results in the same sine value.
Coterminal angles share the same sine value.
To find a coterminal angle, add or subtract by 360 degrees.
Sine of 30 degrees and sine of 390 degrees have the same value.
This periodic property is true for all trigonometric functions: sine, cosine, tangent, cotangent.
Cosine of 50 degrees is equal to cosine of 50 degrees plus 360 degrees or cosine of 410 degrees.
The word 'periodic' means things that repeat.
Adding 2 pi to an angle results in the same cosine value.
When graphing sine and cosine functions, the pattern repeats over and over.
Evaluating sine of 420 degrees by finding a coterminal angle less than 420 degrees.
Sine of 420 degrees equals sine of 60 degrees.
Using the unit circle, sine of 60 degrees is root 3 over 2, so sine of 420 degrees is also root 3 over 2.
Cosine of negative 2 pi over 3 equals cosine of 4 pi over 3.
At 4 pi over 3, the x and y values are negative, so cosine 4 pi over 3 is negative 1/2.
Cosine of negative 2 pi over 3 is also negative 1/2.
Sine of 750 degrees can be evaluated by repeatedly subtracting 360 degrees. | 677.169 | 1 |
A course of practical geometry for mechanics
From inside the book
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Page 11 ... degrees ; thus , the com- plement of an arc or angle of 30 degrees , is an arc or angle which contains 60 degrees ; 20 degrees are the complement of 70 degrees , and so on , 22. The supplement of an arc or angle is the difference ...
Page 12 ... 60 degrees . The three angles of any triangle , are together equal to two right angles or 180 degrees . XXV . An isosceles triangle is that which has only two sides equal . A Two of its angles are equal , namely , those opposite to the ...
Page 17 ... 60. To Produce or Prolong a straight line is to lengthen it in the same straight line . 61. A Generatrix is that by ... degree is divided into sixty equal parts , called minutes ; these into sixty equal parts , called seconds ; and these ...
Page 21 ... degrees . 1. Draw a line as A B in Problem II . 2. Take 60 degrees in the compasses , from any line or scale of chords ( marked CHO or C ) on an ivory scale , and from A as a centre , describe the arc G F. 3. Take the number of degrees ...
Page 22 ... degrees added . For example , if the angle is to contain 170 degrees , 90 degrees must first be applied on the arc , and then 80 degrees more ; each quantity to be taken from the same line of chords as the 60 degrees were , with which | 677.169 | 1 |
2D Printable Shapes
2D Printable Shapes - Web kids in 1st grade and 2nd grade are expected to identify the 2d shapes and write their names in this pdf. Here you will find our range of free printable shape sheets. ]the following printable shape sheets contain. Whether just learning to name shapes in. Web our 2d shapes worksheets will ensure they are prepared every step of the way. That is, shapes are represented on the x and y axis or on a horizontal and vertical plane. Web this complete guide to geometric shapes includes every 2d geometric shape and all 3d geometrical shapes. Web
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Printable Shapes 2D and 3D
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Printable Shapes Chart
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2D shapes Overview Studyladder Interactive Learning Games
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2D Shapes Planning Playtime
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Here You Will Find Our Range Of Free Printable Shape Sheets.
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Web Our 2D Shapes Worksheets Will Ensure They Are Prepared Every Step Of The Way.
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"# Solutions Q25 - 32 Q25 answer\n",
"(a) The vector $\\vec a$ and its projections onto the $x$- and $y$-axes are shown in figure 68. If $\\boldsymbol i$ and $\\boldsymbol j$ are unit vectors along the $x$- and $y$-axes, and as $\\vec a$ is described in the question as a unit vector, its length must be $| \\vec a | = 1$.\n",
"\n",
"By definition, vector $\\vec a$ has components which are multiples of the base vectors $\\boldsymbol i$ and $\\boldsymbol j$, and is therefore,\n",
"\n",
"$$\\displaystyle a = \\cos(\\alpha)\\boldsymbol i + \\sin(\\alpha)\\boldsymbol j$$\n",
"\n",
"because $\\cos(\\alpha)$ is the component on the $x$-axis, along which the base vector $\\boldsymbol i$ lies. The component along $\\boldsymbol j$ and the $y$-axis is $\\sin(\\alpha)$.\n",
"\n",
"\n",
"\n",
"Figure 68. Vectors and their projections.\n",
"________\n",
"\n",
"Similarly, vector $\\vec b = \\cos(\\beta)\\;\\boldsymbol i + \\sin(\\beta)\\;\\boldsymbol j$. Their dot product is \n",
"\n",
"$$\\displaystyle \\vec a\\cdot \\vec b = \\big(\\cos(\\alpha)\\;\\boldsymbol i + \\sin(\\alpha)\\;\\boldsymbol j\\big)\\big(\\cos(\\beta)\\;\\boldsymbol i + \\sin(\\beta)\\;\\boldsymbol j\\big) = \\cos(\\alpha)\\cos(\\beta) + \\sin(\\alpha)\\sin(\\beta)$$\n",
"\n",
"because $\\boldsymbol i\\cdot\\boldsymbol i = \\boldsymbol j\\cdot\\boldsymbol j = 1$ and $\\boldsymbol i\\cdot\\boldsymbol j = 0$. From the diagram, the angle _between_ the two vectors is $\\beta - \\alpha$, therefore the dot product is\n",
"\n",
"$$\\displaystyle \\vec a\\cdot \\vec b =|\\vec a||\\vec b|\\cos(\\beta-\\alpha)=\\cos(\\beta-\\alpha)$$\n",
"\n",
"again using the fact that $\\vec a$ and $\\vec b$ are unit vectors. Therefore, by equating the two dot product equations, the trigonometric identity is found:\n",
"\n",
"$$\\displaystyle \\cos(\\beta-\\alpha)=\\cos(\\alpha)\\cos(\\beta) + \\sin(\\alpha)\\sin(\\beta)$$\n",
"\n",
"Suppose that vector $\\vec b$ is changed, so that it points downwards in the figure making its angle $-\\beta$ to the x-axis, then the angle between the vectors becomes $\\alpha + \\beta$ and $\\vec b = \\cos(-\\beta)\\;\\boldsymbol i + \\sin(-\\beta)\\;\\boldsymbol j = \\cos(\\beta)\\;\\boldsymbol i - \\sin(\\beta)\\;\\boldsymbol j$ therefore \n",
"\n",
"$$\\displaystyle \\cos(\\beta+\\alpha)=\\cos(\\alpha)\\cos(\\beta) - \\sin(\\alpha)\\sin(\\beta)$$\n",
"\n",
"(b) Repeating the first part of the calculation in matrix - vector form, the two vectors are\n",
"\n",
"$$\\displaystyle \\vec a=\\begin{bmatrix}\\cos(\\alpha)\\\\ \\sin(\\alpha)\\end{bmatrix},\\quad \\vec b=\\begin{bmatrix}\\cos(\\beta)\\\\ \\sin(\\beta)\\end{bmatrix}$$\n",
"\n",
"the dot product of which is a little easier to calculate than $ijk$ vectors, viz,\n",
"\n",
"$$\\displaystyle \\vec a\\cdot\\vec b= \\begin{bmatrix}\\cos(\\alpha)& \\sin(\\alpha)\\end{bmatrix}\\begin{bmatrix}\\cos(\\beta)\\\\ \\sin(\\beta)\\end{bmatrix}=\\cos(\\alpha)\\cos(\\beta) + \\sin(\\alpha)\\sin(\\beta)$$\n",
"\n",
"## Q26 answer\n",
"(a) Adding together base vectors to form the $x$ and $y$ orbitals, and forming the dot product in matrix vector form, gives\n",
"\n",
"$$\\displaystyle \\vec \\psi_{2p_x}\\cdot \\vec \\psi_{2p_y}=\\frac{1}{2}\\begin{bmatrix}0&1&1\\end{bmatrix} \\begin{bmatrix} 0\\\\-1\\\\1\\end{bmatrix} = 0$$\n",
"\n",
"therefore the orbitals are orthogonal and they are also normalized because\n",
"\n",
"$$\\displaystyle \\vec \\psi_{2p_x}\\cdot \\vec \\psi_{2p_x}=\\frac{1}{2}\\begin{bmatrix}0&1&1\\end{bmatrix} \\begin{bmatrix} 0\\\\1\\\\1\\end{bmatrix} = 1$$\n",
"\n",
"and similarly for $\\psi_{2p_y}$. \n",
"\n",
"(b) If orthogonal, the dot product should be zero. It will contain nine terms\n",
"\n",
"$$\\displaystyle \\vec\\psi_1\\cdot\\vec \\psi_2 =\\begin{bmatrix} \\frac{1}{\\sqrt{3}}\\vec\\psi_s-\\cdot \\begin{bmatrix} \\frac{1}{\\sqrt{3}}\\vec\\psi_s+$$\n",
"\n",
"The product can be expanded out term by term and then the orthogonality of the base vectors used. This means that only $\\vec \\psi_{p_{+1}}\\vec\\psi_{p_{+1}}$ etc are not zero; hence the dot product is zero and these orbitals are orthogonal.\n",
"\n",
"Alternatively, the orbitals can be represented as combinations of base vectors in the _basis set_ of $(s, p_{+1} , p_{-1})$ orbitals:\n",
"\n",
"$$\\displaystyle \\begin{bmatrix} 1/\\sqrt{3} & -1/\\sqrt{2} &-1\\sqrt{6}0$$\n",
"\n",
"The orthogonality test for $p_1$ with $p_x$ and $p_y$ is\n",
"\n",
"$$\\displaystyle \\vec \\psi_{p_x}\\cdot \\vec \\psi_{p_1} =\\frac{1}{\\sqrt{2}}\\begin{bmatrix} 0 & 1 &\\frac{1}{\\sqrt{2}}\\left(\\frac{1}{\\sqrt{2}}+\\frac{1}{\\sqrt{6}}\\right)=-0.79$$\n",
"\n",
"$$\\displaystyle \\vec \\psi_{p_y}\\cdot \\vec \\psi_{p_1} =\\frac{1}{\\sqrt{2}}\\begin{bmatrix} 0 & 1 &-0.21$$\n",
"\n",
"so neither $p_x$ nor $p_y$ are orthogonal to the $p_1$ orbital. The same is true of the $p_2$ orbital. They would therefore not be of much use to study molecular properties.\n",
"\n",
"(c) The wavefunctions $\\psi_s, \\psi_{p_x}, \\psi_z$ are orthogonal to one another, and this makes the calculation easy. A basis set of these orbitals in the order, $(s, x, z)$ can be used. The $+xy$ and $-xy$ orbitals are orthogonal as shown by the dot product calculation;\n",
"\n",
"$$\\displaystyle \\vec \\psi_{sp^2(+xy)}\\cdot \\vec\\psi_{sp^2(-xy)}=\\begin{bmatrix} 1/\\sqrt{3}& 1/\\sqrt{6} &1\\sqrt{2}\\end{bmatrix}\\begin{bmatrix} 1/\\sqrt{3}\\\\ 1/\\sqrt{6} \\\\-1\\sqrt{2}\\end{bmatrix}=\\frac{1}{3}+\\frac{1}{6}-\\frac{1}{2}=0$$\n",
"\n",
"The similar calculation with the other orbitals shows that they are orthogonal to one another. The hybrid orbitals are not orthogonal to the $p_x$ or $p_y$ orbitals. The $p_x$ is represented with the vector $\\begin{bmatrix}0 &1& 0\\end{bmatrix}$ in the $(s, p_x, p_y)$ basis set. The $p_z$ orbital is orthogonal to both the $±\\pm xy$ orbitals. This can be understood because there is no $z$ component in these orbitals. By forming a basis set in $(s, p_x, p_y, p_z)$ the $xy$ orbitals will have a zero component for $p_z$, the $p_z$ orbital has $1$ here and zero elsewhere, therefore the dot product zero, e.g.\n",
"\n",
"$$\\displaystyle \\begin{bmatrix} -\\frac{1}{\\sqrt{3}} &\\pm\\frac{1}{\\sqrt{2}} &-\\frac{1}{\\sqrt{6}} & 0\\end{bmatrix} \\begin{bmatrix} 0\\\\0\\\\0\\\\1 \\end{bmatrix} = 0$$\n",
"\n",
"If you are not convinced, then the hybrid orbitals can be broken down into the $p_z, p_{+1}, p_{-1}$ orbitals and the calculation repeated.\n",
"\n",
"(d) The $p_x, p_y$ and $p_z$ orbitals are shown in figure 69, and the np$^2$ hybrid orbitals, which lie at $120^\\text{o}$ to one another in the $x-y$ plane as expected for the sp$^2$ hybridization of $p$ orbitals.\n",
"\n",
"\n",
"\n",
"Figure 69. The $2p$ orbitals (top) and the sp$^2$ hybrids (bottom).\n",
"_____________\n",
"\n",
"(e) To test for orthogonality of the base orbitals $p_0$ and $p_{+1}$, the calculation is\n",
"\n",
"$$\\displaystyle \\int_0^{2\\pi}\\int_0^\\pi \\psi_0^*\\psi_{+1}\\sin(\\theta)d\\theta d\\varphi =-\\left(\\frac{9}{32\\pi}\\right)^{1/2}\\int_0^{2\\pi}e^{+i\\varphi}d\\varphi \\int_0^\\pi\\cos(\\theta)\\sin^2(\\theta)d\\theta =0$$\n",
"\n",
"The $\\varphi$ integration is separated out because there is no term in both $\\varphi$ and $\\theta$. Both integrations are easy when transformed to exponential form but that in $\\varphi$ can be seen because $e^0=1$, as does $e^{2\\pi i}$ and so this integral is the same at both limits and hence zero. The same result is obtained for the $p_0$ and $p_{-1}$ orbital. \n",
"\n",
"Normalization for the $p_{+1}$ orbital is\n",
"\n",
"$$\\displaystyle \\int_0^{2\\pi}\\int_0^\\pi \\psi_{+1}^*\\psi_{+1}\\sin(\\theta)d\\theta d\\varphi =\\frac{3}{8\\pi}\\int_0^{2\\pi}d\\varphi\\int_0^\\pi\\sin^3(\\theta)d\\theta =1$$\n",
"\n",
"and the same integration occurs for $p{-1}$ as the $\\varphi$ term disappears with the complex conjugate. The $\\theta$ integral can be simplified by converting to the exponential form, or by using $\\sin^3(\\theta) = (3\\sin(\\theta) - \\sin(3\\theta))/4$."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Q27 answer\n",
"(a) The dot product is $\\displaystyle\\begin{bmatrix} 1&1&1&1\\end{bmatrix}\\begin{bmatrix} 1\\\\1\\\\-1\\\\-1\\end{bmatrix}=0$ that the vectors are orthogonal.\n",
"\n",
"(b) The $s$ orbital has spherical symmetry; the dot product of the $p$ orbitals is $\\displaystyle \\frac{1}{3}\\begin{bmatrix} 1&1&1\\end{bmatrix}\\begin{bmatrix} 1\\\\-1\\\\-1\\end{bmatrix}=-\\frac{1}{3}$ where the $3$ normalises. From this result $\\cos(\\theta) = -1/3$, making the angle $109.47^\\text{o}$.\n",
"\n",
"## Q28 answer\n",
"In this example, the s-orbital $\\psi_s$ is spherically symmetric and cannot affect in which direction the hybrid orbitals point and we need not include it in the calculation. Comparing the orbital coefficients $b=-1/\\sqrt{6},\\quad c=\\sqrt{3}/\\sqrt{ 6}=1/\\sqrt{2};\\quad d=0$ because there is no $\\psi_{p_z}$ wavefunction included in $\\psi_1$, and for $\\psi_2$, $b=-1/\\sqrt{6},\\quad c=-1/\\sqrt{2};\\quad d=0$.\n",
"\n",
"With these coefficients, two vectors can be made, \n",
"\n",
"$$\\displaystyle\\vec v_1=\\frac{1}{\\sqrt{6}}\\begin{bmatrix}-\\boldsymbol i+\\sqrt{3}\\;\\boldsymbol j\\end{bmatrix}, \\quad\\vec v_2=\\frac{1}{\\sqrt{6}}\\begin{bmatrix}-\\boldsymbol i-\\sqrt{3}\\;\\boldsymbol j\\end{bmatrix}$$\n",
"\n",
"and the $s$ orbital is ignored. The angle between them is found from the dot product $\\vec v_1\\cdot \\vec v_2 = |\\vec v_1 ||\\vec v_2 |\\cos(\\theta)$ and using equation 2, $\\cos(\\theta)=-1/2$ or $\\theta =120^\\text{o}$ so it seems reasonable that $\\psi_1$ and $\\psi_2$ represent sp$^2$ hybrid orbitals. If we did include the $s$ orbitals in a basis set of four elements, the $s$ and three $p$ orbitals, then these vectors are orthogonal which they have to be for any eigenfunctions that are solution of the Schroedinger equation,\n",
"\n",
"$$\\displaystyle \\frac{1}{6}\\begin{bmatrix} \\sqrt{2}& -1&\\sqrt{3}& 0\\end{bmatrix} \\begin{bmatrix} \\sqrt{2}\\\\-1\\\\ -\\sqrt{3}\\\\ 0\\end{bmatrix}=0$$\n",
"\n",
"(c) Substituting for the $p_{\\pm 1}$ orbitals gives\n",
"\n",
"$$\\displaystyle \\vec\\psi_1=\\frac{1}{\\sqrt{6}}\\left(\\sqrt{2}\\psi_2+\\frac{\\sqrt{3}-1}{\\sqrt{2}}\\psi_{-1}-\\frac{\\sqrt{3}+1}{\\sqrt{2}}\\psi_{+1} \\right)\\\\\n",
"\\vec\\psi_2=\\frac{1}{\\sqrt{6}}\\left(\\sqrt{2}\\psi_2-\\frac{\\sqrt{3}+1}{\\sqrt{2}}\\psi_{-1}+\\frac{\\sqrt{3}-1}{\\sqrt{2}}\\psi_{+1} \\right)$$\n",
"\n",
"Their dot product is found using the coefficients \n",
"\n",
"$$\\displaystyle \\vec\\psi_1\\cdot\\vec\\psi_2 =\\frac{1}{6}\\begin{bmatrix}\\sqrt{2}&\\frac{\\sqrt{3}-1}{\\sqrt{2}}& -\\frac{\\sqrt{3}+1}{\\sqrt{2}} \\end{bmatrix}\\begin{bmatrix}\\sqrt{2}\\\\-\\frac{\\sqrt{3}+1}{\\sqrt{2}}\\\\ \\frac{\\sqrt{3}-1}{\\sqrt{2}} \\end{bmatrix}=\\frac{1}{6}(2-1-1)=0$$\n",
"\n",
"therefore, these orbitals are orthogonal. To calculate the angle, the $s$ orbital part has to be ignored and the calculation repeated, remembering to re-normalize the vector, as it is now only two dimensional. In this case, the dot product is $-1/2$ and the angle $120^\\text{o}$. This confirms that sp$^{2}$ hybridization is involved as found in part (b).\n",
"\n",
"(d) The principal quantum number $n=2$, is good for all orbitals as only $2p$ or $2s$ orbitals are involved. The $p_x, p_y, p_1$ and $p_2$, have _undefined_ $m$ quantum numbers because they are hybrids of different base orbitals with different $m$ quantum numbers. The $p_1$ and $p_2$ orbitals are $s-p$ hybrids so that quantum number $l$ is also undefined.\n",
"\n",
"$$\\displaystyle \\begin{array}{l|ccc}\n",
"\\hline\n",
"& n& i & m\\\\\n",
"\\hline\n",
"p_0\\equiv p_z & 2&1&0\\\\\n",
"p_{+1}& 2 & 1 & +1\\\\\n",
"p_{-1}& 2 & 1 & -1\\\\\n",
"p_x& 2 & 1 & \\text{undefined}\\\\\n",
"p_y& 2 & 1 & \\text{undefined}\\\\\n",
"p_1& 2 & \\text{undefined} & \\text{undefined}\\\\\n",
"p_2& 2 & \\text{undefined} & \\text{undefined}\\\\\n",
"\\hline\\end{array}$$\n",
"\n",
"\n",
"## Q29 answer\n",
"(a) The body-centred cell: assuming,for simplicity,that the cube has length $1$, then by inspecting the body-centred cube the vectors to the points are\n",
"\n",
"$$ \\displaystyle \\begin{align}\\vec{oa} =& \\boldsymbol i/2 + \\boldsymbol j/2 + \\boldsymbol k/2, \\quad \\vec{ob} = k,\\quad \\vec{oc} = \\boldsymbol i\\\\\n",
"\\quad \\vec{od} =& 2\\boldsymbol i,\\quad \\quad \\vec{oe} = 2\\boldsymbol i +\\boldsymbol j +\\boldsymbol k,\\end{align}$$ \n",
"\n",
"In row matrix form $ \\vec{oa}=\\begin{bmatrix}1/2 &1/2 &1/2\\end{bmatrix},\\; \\vec{ob}=\\begin{bmatrix}0 &0& 1\\end{bmatrix}$.\n",
"\n",
"The distance $a \\to b$ is calculated by forming a vector $\\vec{ab}$ and finding its length; this vector is $\\vec{ab} = \\vec{ob} - \\vec{oa} = -\\boldsymbol i/2 - \\boldsymbol j/2 + \\boldsymbol k/2$ with length $\\sqrt{\\boldsymbol i\\cdot \\boldsymbol i/4-\\boldsymbol j\\cdot\\boldsymbol j/4+\\boldsymbol k\\cdot\\boldsymbol k/4}=\\sqrt{3/4}$ or $q\\sqrt{3/4}$ by accounting for the true size of a side.\n",
"\n",
"(i) Angle $\\angle bac$: by geometry the length $ac$ has to be the same as $ab$, but to find the angle $\\angle bac$ we need the vector $\\vec ca$, which is $\\vec{oc} - \\vec{oa} = +\\boldsymbol i/2 -\\boldsymbol j/2 -\\boldsymbol k/2$. The angle is calculated using\n",
"\n",
"$$\\displaystyle \\cos(\\theta)=\\frac{\\vec{ca}\\cdot\\vec{ab}}{|\\vec{ca}||\\vec{ab}|}=\\frac{4}{12}(\\boldsymbol i-\\boldsymbol j-\\boldsymbol k)\\cdot(-\\boldsymbol i-\\boldsymbol j+\\boldsymbol k)=-\\frac{1}{3}$$\n",
"\n",
"and therefore the angle $\\angle bac = 109.47^\\text{o}$. In matrix form this calculation is\n",
"\n",
"$$\\displaystyle \\frac{4}{3}\\begin{bmatrix}1/2&-1/2&-1/2\\end{bmatrix}\\begin{bmatrix}-1/2\\\\-1/2\\\\1/2\\end{bmatrix}=-\\frac{1}{3}$$\n",
"\n",
"The angle $\\angle bac$ is the same as in a tetrahedron, which is to be expected as a tetrahedron can fit inside a cube with vertices at opposite corners and across a diagonal.\n",
"\n",
"(ii) The angle $\\angle bad$ is found from the dot product of $\\vec{ad}$ with $\\vec{ab}$. Vector $\\vec{ad}$ is $\\vec{od}-\\vec{oa}=2\\;\\boldsymbol i-\\boldsymbol i/2-\\boldsymbol j/2- \\boldsymbol k/2=(3i-j-k)/2$ and has a length $q\\sqrt{11/4}$. The angle is calculated as \n",
"\n",
"$$\\cos(\\theta)=\\frac{1}{\\sqrt{33}}(3\\;\\boldsymbol i+\\boldsymbol j+\\boldsymbol k)\\cdot(-\\boldsymbol i-\\boldsymbol j+\\boldsymbol k)=-\\frac{3}{\\sqrt{33}}$$\n",
"\n",
"and the angle $\\angle bae= 121.4^\\text{o}$ which is the same as angle $bad$, as might have been spotted from the symmetry of the atoms.\n",
"\n",
"(b) The Python calculation for the face-centred cube is straightforward, once the matrices to hold the vectors from the origin, $Oa, Ob$, and so forth are defined. Dot products are used to find the lengths. Once the vectors are set up it is very easy to calculate any length or angle, whereas using geometry and repeatedly using Pythagoras' Theorem is far harder. If you want to repeat the calculation for the body-centred cube, change the initial vectors."angle ab-ac 73.22 degrees\n",
"angle ab-ad 80.41 degrees\n",
"angle ab-ae 97.42 degrees\n"
]
}
],
"source": [
"#--------------------------- # define routine to calculate angle \n",
"def get_angle(veca,vecb):\n",
" \n",
" len_veca = np.sqrt( np.dot(veca,veca))\n",
" len_vecb = np.sqrt( np.dot(vecb,vecb))\n",
" angle = np.arccos(np.dot(veca,vecb)/(len_veca*len_vecb) )*180/np.pi\n",
" return angle\n",
"#---------------------------\n",
"\n",
"Oa = np.array([1/2, 1 , 1/2]) # use np.array() to make vectors \n",
"Ob = np.array([0 , 0 , 1 ])\n",
"Oc = np.array([3/2, 0 , 1/2])\n",
"Od = np.array([3/2, 1/2, 1 ])\n",
"Oe = np.array([2 , 1/2, 1/2])\n",
"\n",
"ab = Ob - Oa # define new vectors\n",
"ac = Oc - Oa\n",
"ad = Od - Oa\n",
"ae = Oe - Oa\n",
"\n",
"print('{:s}{:8.2f}{:s}'.format('angle ab-ac ',get_angle(ab,ac) ,' degrees') )\n",
"print('{:s}{:8.2f}{:s}'.format('angle ab-ad ',get_angle(ab,ad) ,' degrees') )\n",
"print('{:s}{:8.2f}{:s}'.format('angle ab-ae ',get_angle(ab,ae) ,' degrees') )"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Q30 answer\n",
"Starting with the definition $\\vec v = |\\vec v|(\\cos(\\alpha)\\;\\boldsymbol i + \\cos(\\beta)\\;\\boldsymbol j+ \\cos(\\gamma)\\;\\boldsymbol k)$, \n",
"\n",
"the dot product of $\\vec v$ with itself is $\\vec v\\cdot\\vec v = |\\vec v|^2(\\cos^2(\\alpha) + \\cos^2(\\beta) + \\cos^2(\\gamma))$ \n",
"\n",
"where the rules $\\boldsymbol i\\cdot\\;\\boldsymbol i = 1$, and $\\boldsymbol i\\cdot\\;\\boldsymbol j = 0$ and so forth, were used. However, a vector is parallel to itself, therefore by definition, $\\vec v\\cdot\\vec v = |\\vec v |^2\\cos(0)$ and as $\\cos(0) = 1$ then $\\cos^2(\\alpha) + \\cos^2(\\beta) + \\cos^2(\\gamma) = 1$.\n",
"\n",
"## Q31 answer\n",
"(a) $\\vec A+\\vec B=2\\;\\boldsymbol i+2\\;\\boldsymbol j+6\\;\\boldsymbol k$, $|\\vec A|=\\sqrt{3}$, $|\\vec B|= \\sqrt{35}$, $\\vec A\\cdot\\vec B=3$, $\\cos(\\theta)=\\sqrt{3/35}$.\n",
"\n",
"(b) $\\vec A+\\vec B=5\\;\\boldsymbol i- 4\\;\\boldsymbol j+7\\;\\boldsymbol k$, $|\\vec A|=\\sqrt{26}$, $|\\vec B|=\\sqrt{38}$, $\\vec A\\cdot\\vec B=13$, $\\cos(\\theta) = 13/\\sqrt{26\\cdot 38} = 0.4135$.\n",
"\n",
"(c) $\\vec A+\\vec B=2\\;\\boldsymbol i+\\;\\boldsymbol j+\\;\\boldsymbol k$, $|\\vec A|=|\\vec B|\\sqrt{2}$, $\\vec A\\cdot\\vec B=1$, $\\cos(\\theta)=1/2$.\n",
"\n",
"## Q32 answer\n",
"Assume ideal geometries for the molecules,and use symmetry where possible to simplify the calculation, and then add all the vectorial components together. The resultant dipole could point in any direction, so it is necessary to define axes on the molecules, and calculate dipoles along each axis and then form the total. This definition based on symmetry, is effectively a principal axes transformation; see Chapter 7.13.8.\n",
"\n",
"(a) This molecule is flat,and if the $z-y$ plane is chosen to be the plane of the molecule, there is no z component. In benzene derivatives, each bond dipole is at $60^\\text{o}$ to the next one. If one chlorine atom is at position 1, the angle to the other atom in position 3, is $120^\\text{o}$. Drawing the molecule, shows that if atom 2 is along the y-axis, the molecule has $C_{2V}$ symmetry (has two mirror planes and a $C_2$ rotation axis) about this axis, which means that only half the bond dipoles need to be calculated, the total being twice this sum.\n",
"\n",
"\n",
"\n",
"Figure 70. Illustrating geometries for dipole moment calculation. The $z$ and $y$ are in the plane of the figure, $x$ points out.\n",
"_______\n",
"\n",
"The dipole is a projection of the bond dipoles on the $z$-axis, because by symmetry the $x$ and $y$ dipole components are zero. Only two Cl and two H atoms are needed, giving a dipole of\n",
"\n",
"$$\\displaystyle 2\\cos(60^\\text{o})\\mu_{CCl} + 2\\cos(120^\\text{o})\\mu_{CH} = -1.16\\; \\text{D}$$\n",
"\n",
"(b) In methylene chloride, the two H atoms are at $90^\\text{o}$ to the two Cl atoms because the molecule is tetrahedral. Using the same axes as in part (a), the H atoms are in the $z-y$ plane and the Cl atoms in the $z-x$ plane. By symmetry, the dipole component along the $x$- and $y$-axes is zero. If $\\theta$ is the bond angle, $\\cos^{-1}(-1/3)$, the dipole is the $z$ component which is\n",
"\n",
"$$\\displaystyle 2\\cos(\\theta/2)\\mu_{CH} + 2\\cos(\\pi - \\theta/2)\\mu_{CCl} = 1.34\\; \\text{D}$$\n",
"\n",
"(c) In chloroform,the positions of the chlorine atoms are quite difficult to calculate using geometry and the dipole can more easily be calculated using vectors.\n",
"\n",
"The geometry is tetrahedral, so that the atoms are at positions: carbon $(0, 0, 0)$ and the H and Cl atoms $(R)$ at\n",
"\n",
"$$\\displaystyle \\vec H = (1,1,1),\\quad \\vec R_1=(-1,-1,1),\\quad \\vec R_2=(-1,1,-1),\\quad \\vec R_3=(1,-1,1)$$\n",
"\n",
"The amount of the total dipole along each axis can be found by calculating the dot products with the base vectors $\\vec x=\\begin{bmatrix}1&0&0\\end{bmatrix},\\vec y=\\begin{bmatrix}0&1&0\\end{bmatrix},\\vec z=\\begin{bmatrix} 0& 0 & 1 \\end{bmatrix}$. The bond dipoles are $\\mu_H$ and $\\mu_{Cl}$ etc.\n",
"\n",
"The total dipole along the x axis is $\\vec \\mu_x = \\vec x\\cdot\\vec H\\mu_H + \\vec x\\cdot\\vec R_1\\mu_{R_1} +\\vec x\\cdot\\vec R_2\\mu_{R_2} +\\vec x\\cdot\\vec R_3\\mu_{R_3}$ and a similar calculation applies for the $y$ and $z$ axes and the vector of the total dipole, $\\vec mu=\\mu_x\\vec x+\\mu_y\\vec y+\\mu_z\\vec z$ of magnitude $\\mu\\sqrt{\\mu_x^2+\\mu_y^2+\\mu_z^2}$._{Cl} \\frac{\\mu_H-\\mu_{Cl}}{\\sqrt{3}}$$\n",
"\n",
"and the same result is found for the $y$ and $z$ components. The total dipole is $\\mu_H - \\mu_{Cl} = 1.16$ debye. The $\\sqrt{3}$ arises from normalizing the bond vector, because the dot product is used to produce direction cosines. The dipole points along the CH bond because all $x, y, z$ components are equal, and the H atom is at coordinate $(1, 1, 1)$ and the carbon at the origin.\n",
"\n",
"Repeating part (b) to calculate the dipole for methylene chloride using vectors produces_H 0$$\n",
"\n",
"and $\\mu_y = 0, \\mu_z = 2(\\mu_H - \\mu_{Cl})/\\sqrt{3}$ giving a dipole of $1.34$ D | 677.169 | 1 |
Practice problems assess your knowledge of this geometric theorem as well as the application of given information to determine that triangles are congruent. Quiz & Worksheet Goalstriangles are congruent, c. K ; we are given that DQG G corresponds with K . Since corresponding parts of congruent triangles are congruent, G K . MULTIPLE REPRESENTATIONS In this problem, you will explore the following statement. The areas of congruent triangles are equal . a. VERBAL Write a conditional statement to represent the relationship ... The angles of the triangles answer the key Glencoe geometry chapter 4 sheet answers 4-3 guide study and intervention congruent triangles answers 4-3 skills practice congruent triangles 4-5 Chapter 4. 25. Glencoe Geometry. A guide to study and intervention. Proof of The Triangles Congruent-SSS, SAS....Unit 4 Test Congruent Triangles Answer Key All Things … WebThis is just one of the solutions for you to be. We provide you all the answer keys for all the unit 4 congruent triangles homework 4 congruent triangles questions, as well as a wealth of extra study materials online. skills-practice-proving-triangles-congruent-answers 2/3 Downloaded from titleix.
Because this triangle has 3 angles with measures less than 90 8 and 2 congruent sides, it is an acute isosceles triangle. b. Because this triangle has a right angle and no congruent sides, it is a right scalene triangle. c. Because this triangle has one angle greater than 908 and no congruent sides, it is an obtuse scalene triangle. 95 8 40 8 ...May 28, 2023 · 4 5 skills practice proving triangles congruent [4EIX77] Practice 4-4 Using Congruent Triangles: CPCTC Explain how you can use SSS, SAS, ASA, or AAS with CPCTC to prove each statement true. Practice 4-4 Using... | 677.169 | 1 |
Angles – The Most Crucial Concept of the Subject Mathematics
One of the crucial parts of mathematics is geometry. Angles are the most common topic of geometry. In our daily life, we come across a lot of objects in our surroundings. When one will notice it, one will find that all the objects in our world are inclined at some angle with respect to any point or line. An angle is mostly measured in degrees or radians. The majority of shapes have a set angle, such as all three angles of an equilateral triangle being 60 degrees, and all sides of a rectangle and square being 90 degrees. When the angle between two lines is zero degrees, they are said to be parallel, and when the angle between the two lines is ninety degrees, they are said to be perpendicular. As previously said, there are numerous classifications of angles; however, we will focus on corresponding angles. They are categorized into two types. One is formed by parallel lines and a transversal, whereas the other one is formed by two lines that are not parallel and a transversal. Let us first discuss corresponding angles formed due to parallel lines and a transversal.
For understanding corresponding angles due to parallel lines and a transversal, it is necessary to know about parallel lines. One of the major conditions of parallel lines is that they never meet each other at any point. Two lines are said to be parallel lines when the angle between them is equal to zero degrees. The distance between two parallel lines is always constant and it cannot be zero. There are a lot of examples of parallel lines that we observe in our daily life. Railway tracks can be considered as an example for parallel lines. Now let us move our discussion to corresponding angles.
Corresponding angles (parallel lines and transversal): These angles are formed when two parallel lines are cut by the transversal. One can define them as when a third line intersects the two lines which are parallel to each other. The angles that are formed occupy the same relative position at all the intersected points. These angles are termed corresponding angles. Suppose there is a line that crosses both parallel lines. Now at the point where the transverse intersects both the parallel lines, there are a total of four angles formed. So, there will be a total of four pairs of corresponding angles formed with each parallel line. Now, one should also know that the corresponding angles that are formed by the parallel lines are always equal. When the transverse cuts the two parallel lines, not only corresponding angles are formed but few other angles are also formed like alternate interior and exterior angles.
Corresponding angles (non-parallel lines and transversal): In this case, when the transverse cuts the other two lines, then corresponding angles are formed. But in this case, there is no relation between the corresponding angles formed or we can say that the corresponding angles formed are not equal in this case as it was in parallel lines. Students need to have these concepts clear in their minds so that they can tackle tough mathematical problems more accurately and easily.
We attempted to cover all of the ideas related to corresponding angles in the preceding post. People can now discover a variety of platforms to obtain information thanks to the rise of online learning. Cuemath is one such platform. It is one of the most effective tools for making our math difficulties crystal plain. Its language is simple to comprehend. There are dozens of math-related subjects to read on it. Not only school or college students, but everyone of any age can benefit from this platform by obtaining access to the vast amount of information provided. In the last several years, the popularity of online learning has skyrocketed. Not only does studying from such online platforms save us energy, but it also saves us time. Such platforms should be used to their full potential | 677.169 | 1 |
A line PQ is drawn and with P as centre, a circle is drawn with radius greater than half of PQ but less than PQ. Again with same radius and Q as centre, draw another circle. If the two circles meet at X and Y, then what can be said about the points X and Y?
A
They will not be on the same line
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B
The midpoint of XY lies on PQ and they coincide
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C
The midpoint of XY lies on PQ and closer to P
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D
The midpoint of XY lies on PQ and closer to Q
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Open in App
Solution
The correct option is B
The midpoint of XY lies on PQ and they coincide
The above description gives the perpendicular bisector of PQ. Alternate solution: The quadrilateral formed by joining PXQY is a rhombus. In rhombus the diagonals are perpendicular bisectors. | 677.169 | 1 |
Basically, line segments' intersection is a mathematical concept. To detect the intersection of two line segments, find their intersection points. For 2D games, it is very helpful when an explosion animation appears at a position where the two lines intersect; for example, two laser shoots collide. The line segments' intersection can also help you to make a decision in pool games as a guideline, especially for beginners. In this recipe, you will learn how to detect the line segments' intersection.
Getting ready
To define the two line segments, we need the following two formulae:
If you put in 0 for U, you'll get the start point, if you put in 1, you'll get the end point.
With the two equations, if the intersection happens between the two line segments:
The equation could be rewritten as follows:
In order to get the Ua and Ub values, we need two equations. The previous equation could also be written using x and y factors of the points:
You can use the two equations to solve for Ua and Ub:
The denominator of both of the equations is the same. Solve it first. If it is zero, the lines are parallel. If both numerators are also zero, then the two line segments are coincident.
Since these equations treat the lines as infinitely long lines instead of line segments, there is a guarantee of having an intersection point if the lines aren't parallel. To determine if it happens with the segments we've specified, we need to see if U is between zero and one. Verify that both of the following are true:
If we've gotten this far, then our line segments intersect, and we just need to find the point at which they do and then we're done:
The following pseudo code describes the line segments' intersection algorithm coming from the previous description:
The method receives the four points of the two line segments and two flags for indicating the intersected and coincident states. The first two lines are to compute the numerators of ua and ub, then the following line is on calculating the denominator. After that, we begin to check the value of the denominator to see whether the two line segments intersected. If the denominator is almost zero, it means the two line segments are parallel. Meanwhile, if the numerators of Ua and Ub are both almost zero, the two line segments are coincident. Otherwise, if the denominator is greater than zero, it means the two lines where the two line segments lie intersect. In order to make sure the two line segments intersect, we need to check that the absolute values of ua and ub are both less than or equal to 1. If true, the intersection between them happens; now we can use ua or ub to calculate the intersectionPoint.
How to do it…
The following steps will show you how to master the practical method of using the line segments' intersection:
Create a Windows Phone Game project named LineSegmentIntersection, change Game1.cs to LineSegmentsIntersectionGame.cs. Then, add Line.cs to the project.
In the next step, we need to define the Line class in Line.cs. The class will draw the line between two points on the Windows Phone 7 screen.
Declare the indispensable variables of the class, and then add the following code to the class field:
[code]
// Line Texture
private Texture2D lineTexture;
// Origin point of line texture for translation, scale and
// rotation
private Vector2 origin;
// Scale factor
private Vector2 scale;
// Rotation factor
private float rotation;
// Axis X
Vector2 AxisX = new Vector2(1, 0);
// Dictance vector
Vector2 distanceVector;
// Line direction
Vector2 Direction = Vector2.Zero;
// The angle between the line and axis X
float theta = 0;
// Line thickness
private int Thickness = 2;
// Line color
private Color color;
[/code]
Create the constructor of the Line class:
[code]
public void Load(GraphicsDevice graphicsDevice)
{
// Initialize the line texture and its origin point
lineTexture = CreateLineUnitTexture(graphicsDevice,
Thickness, color);
origin = new Vector2(0, Thickness / 2f);
}
[/code]
Define the CalculateRotation() method called in the previous step in the Line class. This method calculates the angle between the line and the X-axis:
[code]
private void CalculateRotation(Vector2 distanceVector)
{
// Normalize the distance vector for line direction
Vector2.Normalize(ref distanceVector, out Direction);
// Compute the angle between axis X and line
Vector2.Dot(ref AxisX, ref Direction, out theta);
theta = (float)Math.Acos(theta);
// If the Y factor of distanceVector is less than 0
// this means the start point is lower than the end point,
// the rotation direction should be in the opposite
// direction.
if (distanceVector.Y < 0)
{
theta = -theta;
}
// return the angle value for rotation
rotation = theta;
}
[/code]
Implement the CalculateScale() method in the Line class. The method will calculate a scale represented as a Vector2 object. The X factor stores the number of textures while the Y factor stores the scale degree.
[code]
private void CalculateScale(Vector2 distanceVector)
{
// The Vector2 object scale determines how many textures
// will be drawn based on the input rotation and start
// point, X for the number, Y for the scale factor
float desiredLength = distanceVector.Length();
scale.X = desiredLength / lineTexture.Width;
scale.Y = 1f;
}
[/code]
Define the CreateLineUnitTexture() method, in the Line class, which creates the line unit texture according to the input line thickness.
[code]
// Create a unit texture of line, the texture will be used to
// generate a line with desired number
public static Texture2D CreateLineUnitTexture(
GraphicsDevice graphicsDevice,
int lineThickness, Color color)
{
// Initialize the line unit texture according to the line
// thickness
Texture2D texture2D = new Texture2D(graphicsDevice,
lineThickness, lineThickness, false,
SurfaceFormat.Color);
// Set the color of every pixel of the line texture
int count = lineThickness * lineThickness;
Color[] colorArray = new Color[count];
for (int i = 0; i < count; i++)
{
colorArray[i] = color;
}
texture2D.SetData<Color>(colorArray);
return texture2D;
}
[/code]
Define the Draw() method in the Line class that draws the line segment on Windows Phone 7.
[code]
// Draw the line
public void Draw(SpriteBatch spriteBatch, Vector2 startPoint,
Vector2 endPoint)
{
// Compute the distance vector between the line start
// point and end point
Vector2.Subtract(ref endPoint, ref startPoint,
out distanceVector);
// Calculate the rotation angle
CalculateRotation(distanceVector);
// Calculate the scale factor
CalculateScale(distanceVector);
// Draw the line texture on screen
spriteBatch.Draw(lineTexture, startPoint, null, color,
rotation, origin, scale, SpriteEffects.None, 0);
}
[/code]
From this step, we will begin to interact with the tap gesture and draw the line segments and the intersection point if intersection takes place on the Windows Phone 7 screen. First, add the following lines to the LineSegmentsIntersectionGame class field:
[code]
// Line Object
Line line;
// Circle Texture
Texture2D circleTexture;
// Points of two lines for intersection testing
Vector2 point1, point2, point3, point4, intersectionPoint;
// The flag for intersection
bool Intersection;
// The flag for coincidence
bool Coincidence;
[/code]
Initialize the four points of the two line segments. Insert the following lines to the Initialize() method:
[code]
// Initialize the points of two lines
point1 = Vector2.Zero;
point2 = newVector2(600, 300);
point3 = newVector2(0, 200);
point4 = newVector2(800, 200);
[/code]
Initialize the line and circleTexture objects. Add the following code to the LoadContent() method:
[code]
// Initialize the two line objects with white color
line = new Line(Color.White);
line.Load(GraphicsDevice);
// Initialize the texture of circle
circleTexture = CreateCircleTexture(GraphicsDevice, 5, Color.
White);
[/code]
Now, build and run the application. When you tap on the screen, it should run as shown in the following screenshots:
How it works…
We need to draw the line from a texture since there is no drawing method in XNA 4.0.
In step 2, the lineTexture holds the texture for the drawing line; the origin is the center point for rotation, translation, and scale. The scale, a Vector2 object, the X-factor tells SpriteBatch.Draw() method how many texture units will be drawn. The Y-factor stands for the scaling degree; rotation specifies the rotating degrees; AxisX is the vector represents the X-axis; distanceVector holds the vector between two designated points; Direction indicates the line segment direction; the theta variable shows the angle between the X-axis and the line segment; the Thickness means how thick the line segment should be; color defines the color of the line segment.
In step 4, the first line normalizes the distanceVector, which stores the vector between two points to Direction, a unit vector variable. Then, we use Vector2.Dot() and the Math.Acos() method to calculate the angle between Direction and distanceVector. When we get theta, if the distanceVector.Y is less than 0, this means the start point is lower than the end point, but the theta, which should be a negative angle, is still greater than 0 because in the XNA coordinate system, all the location coordinates are greater than 0, so the dot product is always greater than 0. Thus, we should negate the theta value to meet the actual angle value. Finally, return the theta value to the rotation variable.
In step 5, first distanceVector.Length() returns the length between the two end points of the given line segment. Then, we calculate the number of line unit textures based on the texture width and assign the value to scale.X. After that, we save the scale degree to scale.Y.
In step 6, this method first initializes the line unit texture, of which the size depends on the line thickness. Then, we set the input color to every pixel of the texture. Finally, we return the generated line unit texture.
In step 8, the line will be used to draw the line segments; circle is responsible for drawing the end points and the intersection points of the lines; the Intersection is the flag indicating whether the line segments' intersection happens; the Coincidence shows whether the line segments are coincident or not.
In step 10, the line has white color; the radius of circle textures is 5.
In step 11, the first line of the method is to detect the line segments' intersection. The DetectLineSegmentsIntersection() method uses the point1, point2, point3, and point4 to compute the intersection equation. If there is an intersection, the Intersection variable will be true and the intersectionPoint will return the intersected point. We have discussed a more detailed explanation of this method at the beginning of the recipe. The second part is to control the position of the first point to make an interactive manipulation on one of the line segments. If the tapped position is valid, the position of the first point will be changed to the current tapped position on screen.
In step 13, the method first computes diameter for the width and height of the circle textures. The center specifies the center point of the circle. After that, we initialize the circle texture and the texture color array of which the length is diameter*diameter and then we will iterate each pixel in the array. If the position is outside the region of the circle, the pixel color will be set to transparent, or the color what you want.
Implementing per pixel collision detection in a 2D game
In a 2D game, a general method for detecting collision is by using bounding box. This is the solution for a lot of situations where precision is not the most important factor. However, if your game cares whether two irregular objects collide with each other or overlap, the bounding box will not be comfortable with the. At this moment, per pixel collision will help you. In this recipe, you will learn how to use this technique in your game.
How to do it…
Create a Windows Phone Game project named PixelCollision2D, change Game1.cs to PixelCollision2DGame.cs. Then, add the PixelBall.png and PixelScene.png file to the content project.
Initialize the positions of ball and scene and enable the FreeDrag gesture. Insert the following code in to the Initialize() method:
[code]
// Initialize the position of ball
positionBall = new Vector2(600, 10);
// Initialize the position of scene
positionScene = new Vector2(400, 240);
TouchPanel.EnabledGestures = GestureType.FreeDrag;
[/code]
Now, build and run the application. It will run as shown in the following screenshots:
How it works…
In step 2, the font is used to draw the collision state; texScene loads the scene image; texBall holds the ball texture; textureDataScene and textureDataBall store their texture color array data; positionBall and positionScene specify the position of ball and scene textures; boundBall and boundScene define the bound around the ball and scene texture; Collided is the flag that shows the collision state; Selected indicates whether the ball is tapped.
In step 5, the IntersectPixels() method is the key method that detects the per pixel collision. The first four variables top, bottom, left, and right individually represent the top, bottom, left, and right side of the intersection rectangle of the two bound boxes around the two textures in the example. Then, in the for loop, we check whether the color alpha value of every pixel of both the textures within the intersection rectangle is completely transparent. If yes, the collision occurs; the method will return true, otherwise, it will return false.
In step 6, the first part is to check whether the ball is selected. If yes, then Selected will be true. The second part is about reading the on-going gesture; if the gesture type is FreeDrag, we will update the position of the ball and its bounding box. The third part calls the IntersectPixels() method to detect the pixel-by-pixel collision.
Implementing BoundingBox collision detection in a 3D game
Regardless of whether you are programming 2D or 3D games, collision detection based on bounding box is straightforward, i simple and easy to understand. You can imagine that every object is individually covered by a box. The boxes are moving along with the corresponding boxes; when the boxes collide, the objects collide too. The boxes are called the BoundingBox. To compose the BoundingBox, you only need to go through all the points or vertices, and then find the min and max ones. After that, the BoundingBox collision detection will depend on the min and max information of every BoundingBox to examine whether their min and max values are inside its own range to make the collision decision. Even in a more accurate collision detection system, bounding box collision detection will be taken first, before using the more precise, but costly method. In this recipe, you will learn how to apply the technique to a simple game.
How to do it…
The following steps will help you build your own BoundingBox information content processor and use the BoundingBox in your game:
Create a Windows Phone Game project named BoundingBoxCollision and change Game1.cs to BoundingBoxCollisionGame.cs. Then, create a Content Pipeline Extension Library named MeshVerticesProcessor and replace the ContentProcessor1.cs with MeshVerticesProcessor.cs. We create the content pipeline processor for processing and extracting the BoundingBox information from the model objects before running the game. This will accelerate the game loading speed because your application won't need to do this work again and again. After that, add the model file BigBox.FBX to the content project.
Next, we need to define the MeshVerticesProcessor class in MeshVerticesProcessor.cs of the MeshVerticesProcessor project. Extend the MeshVerticesProcessor class from ModelProcessor, because we need the model vertices information based on the original model.
[code]
[ContentProcessor]
publicclass MeshVerticesProcessor : ModelProcessor
[/code]
Define the FindVertices() method in the MeshVerticesProcessor class:
[code]
// Extracting a list of all the vertex positions in
// a model.
void FindVertices(NodeContent node)
{
// Transform the current NodeContent to MeshContent
MeshContent mesh = node as MeshContent;
if (mesh != null)
{
string meshName = mesh.Name;
List<Vector3> meshVertices = new List<Vector3>();
// Look up the absolute transform of the mesh.
Matrix absoluteTransform = mesh.AbsoluteTransform;
// Loop over all the pieces of geometry in the mesh.
foreach (GeometryContent geometry in mesh.Geometry)
{
// Loop over all the indices in this piece of
// geometry. Every group of three indices
// represents one triangle.
foreach (int index in geometry.Indices)
{
// Look up the position of this vertex.
Vector3 vertex =
geometry.Vertices.Positions[index];
// Transform from local into world space.
vertex = Vector3.Transform(vertex,
absoluteTransform);
// Store this vertex.
meshVertices.Add(vertex);
}
}
tagData.Add(meshName, meshVertices);
}
// Recursively scan over the children of this node.
foreach (NodeContent child in node.Children)
{
FindVertices(child);
}
}
[/code]
Build the MeshVerticesProcessor project. Add a reference to MeshVerticesProcessor.dll in the content project and change the Content Processor of BigBox.FBX to MeshVerticesProcessor, as shown in the following screenshot:
From this step, we will begin to draw the two boxes on screen and detect the bounding box collision between them in the BoundingBoxCollisionGame class in BoundingBoxCollisionGame.cs of the BoundingBoxCollision project. First, add the following lines to the class field:
[code]
// The sprite font for drawing collision state
SpriteFont font;
// Model box A and B
Model modelBoxA;
Model modelBoxB;
// The world transformation of box A and B
Matrix worldBoxA;
Matrix worldBoxB;
// BoundingBox of model A and B
BoundingBox boundingBoxA;
BoundingBox boundingBoxB;
// The bounding box stores the transformed boundingBox
BoundingBox boundingBox box A and the corresponding bounding box and detect the bounding box collision between box A and box B. Add the following code to the Update() method in the BoundingBoxCollisionGame class:
[code]
// Interact with tapping
TouchCollection touches = TouchPanel.GetState();
if (touches.Count > 0 && touches[0].State ==
TouchLocationState.Pressed)
{
Point point = new Point(
(int)touches[0].Position.X,(int)touches[0].Position.Y);
// If the tapped position is inside the left hit region,
// move the box A left
if (LeftHitRegion.Contains(point))
{
worldBoxA.Translation -= new Vector3(1, 0, 0);
}
// If the tapped position is inside the right hit region,
//move the box A right
if (RightHitRegion.Contains(point))
{
worldBoxA.Translation += new Vector3(1, 0, 0);
}
}
// Create a bounding box for the transformed bounding box A
boundingBox = new BoundingBox(
Vector3.Transform(boundingBoxA.Min, worldBoxA),
Vector3.Transform(boundingBoxA.Max, worldBoxA));
// Take the collision detection between the transformed
// bounding box A and bounding box B
if (boundingBox.Intersects(boundingBox 2, the [ContentProcessor] attribute is required. It makes the MeshVerticesProcessor class in to a content processor, which will show up in the content project when you change the model processor.
In step 3, the tagData receives the mesh name as the key and the corresponding mesh vertices as the value.
In step 4, the input—a NodeContent object—represents the root NodeContent of the input model. The key called method is the FindVertices() method, which iterates the meshes in the input model and stores the mesh vertices in the tagData with the mesh name.
In step 5, the first line transforms the current NodeContent to MeshContent so that we can get the mesh vertices. If the current NodeContent is MeshContent, declare meshName variable for holding the current mesh name, meshVertices saves the mesh vertices and stores the world absolute transformation matrix to the absoluteTransform matrix using MeshContent.AbsoluteTransform. The following foreach loop iterates every vertex of the model geometries and transforms it from object coordinate to world coordinate; it then stores the current vertex to the meshVertices. When all the vertices of the current mesh are processed, we add the meshVertices to the tagData dictionary with the meshName as the key. The last part is to recursively process the vertices of the child NodeContent objects of the temporary MeshContent.
In step 7, the font is responsible for drawing the collision state on screen; modelBoxA and modelBoxB hold the two box models; worldBoxA and worldBoxB represent the world transformation of the two boxes; boundingBoxA and boundingBoxB store the bound boxes individually around the two boxes; boundingBox will save the transformed bounding box A for collision detection; the cameraPosition, view, and projection will be used to initialize the camera; LeftHitRegion and RightHitRegion define the left and right hit regions on the Windows Phone 7 screen.
In step 9, in this method, we read the vertices of box A and B from the Model.Tag property. Then, we use BoundingBox.CreateFromPoints() to create the bounding box from the extracted vertices of the box model. Notice, so far, the generated bounding boxes are in the same place; we need to translate them to the place where the corresponding box model locates. Since we will use box A as the moving object, the position will be updated in real time. Now, we just translate the bounding box for box B.
In step 10, in the first part, we check whether the tapped position is in the left or right hit region and move box A. After that, we create a new boundingbox for representing the transformed bounding box A. Then, we take the bounding box collision detection between the boundingBoxA and boundingBoxB using the BoundingBox.Intersects() method. If a collision happens, the method will return true, otherwise it will return false.
Implementing BoundingSphere collision detection in a 3D game
Unlike the bounding box, bounding sphere based collision detection is faster. The technique just needs to compute the length between two points or vertices whether less, equal, or greater than the sum of radii. In modern games, bounding sphere based collision detection is preferred rather than the bounding box. In this recipe, you will learn how to use the technique in an XNA application.
How to do it…
Follow the steps below to master the technique of using BoundingSphere in your game:
Create a Windows Phone Game project named BoundingSphereCollision and change Game1.cs to BoundingSphereCollisionGame.cs. Then, create a Content Pipeline Extension Library named MeshVerticesProcessor and replace the ContentProcessor1.cs with MeshVerticesProcessor.cs. After that, add the model file BallLowPoly.FBX to the content project.
Define the MeshVerticesProcessor class in MeshVerticesProcessor.cs of the MeshVerticesProcessor project. The class is the same as the one mentioned in the last recipe Implementing BoundingBox collision detection in a 3D game. For a full explanation, please refer back to it.
Build the MeshVerticesProcessor project. Add a reference to MeshVerticesProcessor.dll in the content project and change the Content Processor of BallLowPoly.FBX to MeshVerticesProcessor, as shown in the following screenshot:
From this step, we will begin to draw the two balls on screen and detect the bounding sphere collision between them in the BoundingSphereCollisionGame class in BoundingSphereCollisionGame.cs of the BoundingSphereCollision project. First, add the following lines to the class field:
[code]
// The sprite font for drawing collision state
SpriteFont font;
// Model ball A and B
Model modelBallA;
Model modelBallB;
// The world transformation of ball A and B
Matrix worldBallA;
Matrix worldBallB;
// BoundingSphere of model A and B
BoundingSphere boundingSphereA;
BoundingSphere boundingSphereB ball A and the corresponding bounding sphere. Then, detect the bounding sphere collision between ball A and B. Add the following code to the Update() method:
[code]
// Check the tapped position
TouchCollection touches = TouchPanel.GetState();
if (touches.Count > 0 && touches[0].State ==
TouchLocationState.Pressed)
{
Point point = new Point((int)touches[0].Position.X,
(int)touches[0].Position.Y);
// If the tapped position is inside the left hit region,
// move ball A to left
if (LeftHitRegion.Contains(point))
{
worldBallA.Translation -= new Vector3(1, 0, 0);
}
// If the tapped position is inside the left right region,
// move the ball A right
if (RightHitRegion.Contains(point))
{
worldBallA.Translation += new Vector3(1, 0, 0);
}
}
// Update the position of bounding sphere A
boundingSphereA.Center = worldBallA.Translation;
// Detect collision between bounding sphere A and B
if (boundingSphereA.Intersects(boundingSphere 4, the font is responsible for drawing the collision state on screen; modelBallA and modelBallB hold the two box models; worldBallA and worldBallB represent the world transformation of the two boxes; boundingSphereA and boundingSphereB store the bound boxes individually around the two boxes; the cameraPosition, view, and projection will be used to initialize the camera; LeftHitRegion and RightHitRegion define the left and right hit regions on the Windows Phone 7 screen.
In step 6, the first part is to check whether the tapped position is in the left or the right hit region and to move ball A. After that, we update the center position of bounding sphere A with the newest position of ball A. Then, we take the bounding sphere collision detection between the boundingSphereA and boundingSphereB using the BoundingSphere. Intersects() method. If the collision happens, the method will return true, otherwise it will return false.
Implementing ray-triangle collision detection
Ray-triangle collision gives very accurate collision detection in games. Depending on the return value of the distance from the ray start position to the triangle, it is easy for you to decide whether a collision occurs. As you might know, all the models in 3D games are made of triangles, whether static or dynamic. The ray is like a bullet fired from a gun; here, you can consider the gun as another object, with a straight thin rope behind. Once the bullet hits the triangle—an object, the collision happens. A lot of methods on ray-triangle are available; in this recipe, you will learn how to implement the method which has the best time and space complexity to make your game run faster with less memory usage.
Getting ready…
The ray-triangle collision detection method provides more accurate data than other methods using BoundingBox or BoundingSphere. Before the best ray-triangle collision detection method was invented by Moller and Trumbore, most of the existing methods first compute the intersection point between the ray and the triangle's plane. After that, the intersection point will be projected to the axis-aligned plane to determine whether it is inside the 2D projected triangle. These kinds of methods need the plain equation of triangle based on the computed normal every frame, for a triangle mesh; to do this will cost considerable memory space and CPU resources. However, the method from Moller and Trumbore requires only two cross product computations and also gives us an intersection point.
As a detailed explanation, a point v in a triangle is represented by Barycentric coordinates and not Cartesian coordinates. Since the Barycentric coordinate is the most suitable coordinate system to describe a point position in a triangle, the point could be represented by the following formula:
The u and v coordinates—two of the Barycentric coordinates—are also used in texture mapping, normal interpolation like the Phong lighting algorithm, and color interpolation.
For a ray, a point on the ray is given by:
The intersection point between the ray and the triangle means the point is both on the ray and the triangle. To get the point, we have the formula:
We rearrange the previous equation to an expression in matrix notation:
The previous equation means the distance t from the ray origin to the intersection point and the Barycentric coordinate (u,v) can be found in the equation solution. If
is a matrix M, our job is to find the M-1. The equation will be:
Now, let
With the Cramer's rule, we find the following solution:
From linear algebra, the determinant is computed using the Triple Product:
The solution can be rewritten as follows:
The following pseudo code describing the algorithm comes from the solution of the previous equation:
As the parameters of the RayIntersectsTriangle() method, the vertex1, vertex2, and vertex3 are the three points of a triangle, ray is the object of the XNA built-in type ray, which specifies the origin point and the ray direction; the result will return the distance between the ray start point and the intersection point. In the body of the method, the first three lines compute the two triangle edges, then they use the ray.Direction and edge2 to compute the cross product directionCrossEdge1 that represents P, which equals P=D E2. Next, we use directionCrossEdge2 that takes the dot multiplication with edge1 to compute the determinate with the equation determinate=P E1. The following if statement is to validate the determinate. If the value tends to 0, the determinate will be rejected. Then, we use inverseDeterminant to represent the following fraction:
Now you have got the denominator of the fraction of Cramer's rule. With the value, the u, v, and t could be solved as the solution equation. Following the pseudo code, the next step is to calculate the P with equation S=O-P0. Here, ray.Position is o, vertex1 is P0, distanceVector is S. Based on the S value, you could get the u value from the equation u=P S*inverse, the code calls the Vector3.Dot() method between directionCrossEdge2 and distanceVector for the intermediate TriangleU, then the returned value multiplies with the inverseDeterminant for the final TriangleU. The v value triangleV comes from the equation v=Q D*inverse in which Q=S E1. Similarly, you could gain the t value rayDistance from the equation t=Q E2*inverse.
How to do it…
Now, let's look into an example for a direct experience:
Create a Windows Phone Game project named RayTriangleCollisionGame, change Game1.cs to RayTriangleCollisionGame.cs. Then, add the gameFont.spritefont file to the content project.
Draw the ray and triangle on the Windows Phone 7 screen. Insert the following code into the Draw()method:
[code]
GraphicsDevice.RasterizerState = Solid;
// Draw the triangle
DrawColoredPrimitives(vertexBufferTriangle,
PrimitiveType.TriangleList, 1, Matrix.Identity);
// Draw the line which visualizes the ray
DrawColoredPrimitives(vertexBufferLine, PrimitiveType.LineList,
1, worldRay);
spriteBatch.Begin();
spriteBatch.DrawString(font,
"Tap the Left or Right Part of nScreen to Move the ray",
new Vector2(0, 0), Color.White);
spriteBatch.End();
[/code]
Now, build and run the application. It runs as shown in the following screenshots:
How it works…
In step 2, the font object will draw the instruction about how to play with the example on screen; verticesTriangle is the vertex array of the testing triangle; vertexBufferTriangle is the vertex buffer that stores the triangle vertices; verticesLine holds the two points of a line, which visually represents the testing ray; the matrix worldRay stands for the ray world position. The following two matrices view and projection will be used to define the camera; the ray object will be the real tested ray; the distance indicates the actual distance from ray origin position to the ray-triangle intersection point; the LeftRectangle and RightRectangle are the hit regions for moving the ray to the left or to the right. The Solid variable specifies the render state of the graphics device.
In step 3, the LeftRectangle occupies the left half of the screen; the RightRectangle takes up the right half of the screen.
In step 4, the first part is to initialize three of the triangle vertices with the position and color. The original color is green, and when ray collides with the triangle, the color will be yellow. Then we set the triangle vertices to the triangle vertex buffer. The second part is about initiating the line that visualizes the ray and putting the data into the vertex buffer for line data. The final part defines the ray object with position and direction.
In step 5, before the RayIntersectsTriangle() method, the code is to check the tapped position to see whether it is in the LeftRectangle or the RightRectangle. When a valid tapping takes place, the ray will move along the X-axis by one unit, then we call the RayIntersectsTriangle() method to judge whether there is a collision between the ray and the triangle. If the returned distance is not null, this means that the collision happened, and we change the color of the triangle vertices to Color.Yellow. Otherwise, the color will be restored to Color.Green. The RayIntersectsTriangle() method has been discussed at the beginning of this recipe and inserts the definition of the RayIntersectsTriangle() method to the RayTriangleCollisionGame class.
In step 6, in the DrawColoredPrimitives() method, the effect receives the view and project matrices for the camera, the world matrix for the world position and transformation. The effect.VertexColorEnabled is set to true to make vertices have color. Then, we use the first pass of the current technique of BasicEffect and GraphicsDevice. The DrawPrimitives() method draws the primitives from the beginning of the vertex array in the vertex buffer.
In step 7, the DrawColoredPrimitives() method is used to draw the triangle that receives a parameter PrimitiveType.TriangleList, where 1 means the count of triangles. When drawing the line the PrimitiveType is LineList. | 677.169 | 1 |
Hint: In this question, we will first construct the given vectors and thereby it's resultant. Then we will further use the Pythagoras theorem and also we will use the basic trigonometry to solve and get the required result. Further, we will see the basics of vectors, for our better understanding.
Complete step by step solution: Let us assume P and Q is the given two vectors acting simultaneously at a given point. Also, both the given vectors are represented in magnitude and direction by two adjacent sides given as OA and OD of a parallelogram OABD as shown in figure below:
Now, let $\theta $ be the angle between the given vectors P and Q. Here, R is the resultant vector. So, as we know from the parallelogram law of vector addition, which gives that diagonal OB represents the resultant of P and Q.
Therefore, we get the required magnitude and direction of resultant vector R.
Additional information: As we know that, ordinary quantities that have a magnitude but not direction are called scalars. Example: speed, time.
Also, a vector quantity is known as the quantity having magnitude and direction. Vector quantities must obey certain rules of combination. These rules are:
VECTOR ADDITION: it is written symbolically as A + B = C. So, that it completes the triangle. Also, If A, B, and C are vectors, it should be possible to perform the same operation and achieve the same result i.e., C, in reverse order, B + A = C.
VECTOR MULTIPLICATION: This is the other rule of vector manipulation i.e., multiplication by a scalar- scalar multiplication. It is also termed as the dot product or inner product, and also known as the cross product.
If we take example: we already know the basic difference of speed and velocity i.e., speed is the measure of how fast an object can travel, whereas velocity tells us the direction of this speed. As we know speed is a scalar quantity that means speed has only magnitude not direction, whereas velocity is a vector quantity that means that velocity has both magnitude and direction as well. The S.I unit of velocity is meter per second (m/sec).
Note: We should remember that vectors have both magnitude and direction as well, whereas the scalar has only magnitude not direction. Also, we should know that vectors can be used to find the angle of the resultant vector from its parent vectors. | 677.169 | 1 |
This article will cover the definition of similarity in geometry and its applications.
Similarity in geometry definition
Similarity can be defined as an attribute exhibited by two or more figures when their shapes are the same.
An individual is up for a red-night game with his friends requiring them to blindfold each other and make a selection for a similar pair among 4 pastries; a doughnut, a burger bread, sliced bread, and a samosa. Determine the appropriate selection.
Solution:
A samosa is triangular in shape; a slice of bread is rectangular in shape; a burger bread is circular in shape, and a doughnut is circular in shape.
Hence the appropriate similar pair is the burger bread and doughnut.
Properties of Similarity
Two figures are said to be similar if they have the same shape but different sizes. Therefore, similar shapes have the following properties,
Corresponding angles are equal.
Corresponding sides are in the same ratio: this means that all sides in one figure must be multiplied by the same number to give the corresponding sides in the other figure.
To have a deep understanding of the application of similarity in geometry, present the similarity theorems for triangles.
Similarity Formulas
Many people (perhaps, you) get excited when there is a formula to solve problems in a topic. These formulas become the identity of these topics and serve to enhance memory retention. However, the concept of similarity lacks this approach. In clearer terms, there is hardly any formula(s) that is attributed to solving the similarity problems.
Nonetheless, problems with similarity are primarily reliant on the understanding and application of the properties of similarity which was discussed in the previous section. More so and more importantly, the understanding, as well as the application of the theorems which are discussed hereafter are indeed much more than having formulas to memorize.
The Similarity in geometry theorems
There are multiple ways in which we can determine whether or not two triangles are similar, by using one of the four triangle theorems.
Angle-Angle similarity
If two angles in a triangle are equal to two angles in another triangle, then these two triangles are similar.
Triangles ABC and DEF are similar, since
∠BCA=∠DFE
and
∠CAB=∠FDE.
Angle-Angle similarity, StudySmarter Originals
Side-Angle-Side similarity
If an angle in one triangle is equal to an angle in another triangle and the sides making up this angle are proportional, then these two triangles are similar.
What is meant by proportionality of sides, is that the two sides on triangle ABC must both be multiplied by the same number to give the sides of triangle DEF.
Side- Angle- Side similarity, StudySmarter Originals
The given sides in the above figure have a common ratio, that is,
DEAB=EFBC
and the respective angles formed by these corresponding sides are equal,
∠ABC=∠DEF.
Side-Side-Side similarity
Two triangles could also be classified as being similar in the event that their sides AC, AB, and BC which correspond to the sides of another triangle DF, DE and FE are indeed proportional.
Side-side-side similarity theorem (
In the diagram, all of the lines forming triangle DEF are the length of their respective side in triangle ABC multiplied by a constant factor r.
Right Angle - Hypotenuse - Side similarity
This theorem is valid only for right-angled triangles.
Two triangles are similar if the length of the hypotenuse and one other side in one triangle are proportional to the length of the hypotenuse and the other side in another triangle. That is
BCAC=EFDF
Right angle-hypotenuse-side similarity theorem (
When we use a side in a similarity theorem (for example in the SAS theorem), we do not mean that the sides are equal, but that the ratio between the triangle's sides is constant.
Symbol for Similarity
The symbol we use to show that two things are similar is ∼ . Suppose triangles ABC and DEF are similar, we could then write
Since we are only given sides, we want to use the SSS similarity theorem.
For us to be able to apply this theorem, we need to find a common ratio between the sides of triangle ABC and triangle DEF.
The ratio between the sides AB and DE is
ABDE=63=2:1
The ratio between sides AC and sides DF is
ACDF=42=2:1
The ratio between sides BC and EF is
BCEF=105=2:1
Since the ratio between the sides of triangle ABC to its respective sides on triangle DEF is constant, we can say that △ABC~△DEF
Similarity in polygons
Polygons are plane shapes that have three or more sides. This means that a triangle is also a polygon. The concept of similarity also occurs in other polygons other than triangles.
In fact, the similarity of triangles is a particular case of the similarity of polygons.
However, for similarity to occur between polygons, two conditions need to be met:
1. The corresponding angles of the pair in comparison must be equivalent.
2. The corresponding sides of the pair in comparison must have equivalent proportions.
Prove that these two rectangles are similar.
An illustration on polygons with similarity, StudySmarter Original
Solution:
Both rectangles have all their internal angles as right angles. This means that the first criterion which says that all corresponding angles must be equal has been met.
Next, we need to confirm that the ratio of their corresponding sides is equal.
The ratio of both widths is
93=3:1
and the ratio of both lengths is
155=3:1
Similarity in geometry examples
To further understand the concept of similarity, here are a few examples.
Determine similarity among the following pairs,
(a)
Using the angle-angle rule, StudySmarter Originals
(b)
Using the side-angle-side rule, StudySmarter Originals
(c)
Using the side-side-side rule, StudySmarter Originals
(d)
Using right angle-hypotenuse-side rule, StudySmarter Originals
Solution:
(a) Using the angle-angle rule, we can tell that both triangles in figure (a) are similar because knowing that the sum of angles in a triangle is 180º, hence the third angle in the first triangle is
180°-(63°+73°)=180°-136°=44°
This confirms that both triangles are similar since all corresponding angles are equal.
(b) The pair in figure (b) are not similar, although the ratio between corresponding sides is equal to 2:1, the corresponding angles between them are different and as such using the side-angle-side rule we can confirm that the triangular pair are not similar.
(c) The pair in figure (c) are not similar because the ratio of the two sides is 2:1 while the ratio of the third side is 5:3. By considering the side-side-side rule, the ratio of all corresponding sides must be equivalent, hence this pair of triangles are not similar.
(d) The pair in figure (d) is similar because they are both right triangles and the ratio of both corresponding hypotenuses and opposite sides is 1:4. This is in compliance with the right angle-hypotenuse-side rule of similarity.
Similarity - Key takeaways
Figures are similar if they have the same shape.
There are four similarity theorems for triangles: Angle-angle, side-angle-side, side-side-side, and right angle-hypotenuse-side.
If two triangles are similar, their respective sides are of proportionate length.
Learn with 12 Similarity relevance. | 677.169 | 1 |
Introduction
Have you ever wondered what a regular polygon is and what makes it unique? In this article, we will delve into the world of polygons, exploring their definition, properties, and examples. By the end,...
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Introduction
Have you ever wondered what a regular polygon is and what makes it unique? In this article, we will delve into the world of polygons, exploring their definition, properties, and examples. By the end, you'll have a clear understanding of this fascinating geometric concept and its applications.
What Is a Polygon?
A polygon is a two-dimensional enclosed figure formed by connecting three or more straight lines. These "flat figures" come in various shapes and sizes. For example, a square is a polygon with four equal-length sides. Take a look at the image below for more examples:
Parts of a Polygon
Sides
Sides are line segments that connect two vertices, forming the boundaries of the polygon.
Vertices
Vertices are the points where two sides meet. They define the corners of the polygon.
Angles
Polygons have both interior and exterior angles. The interior angle is formed within the enclosed surface of the polygon, while the exterior angle is the angle formed outside the polygon. Take a look at the image below for a visual representation:
Parts of a polygon
What Are Regular Polygons?
When all the sides and interior angles of a polygon are equal, we refer to it as a regular polygon. Examples of regular polygons include squares and equilateral triangles. Not only do regular polygons have congruent sides, but their angles are also equal, making them equiangular.
Regular polygon
Properties of Regular Polygons
The properties of regular polygons are as follows:
All sides of a regular polygon are equal in length.
All interior angles of a regular polygon are equal in measure.
The sum of the exterior angles of a regular polygon is always 360°.
Perimeter of a Regular Polygon
The perimeter of a regular polygon is the sum of all its sides. In a regular polygon with n sides, the perimeter is equal to n times the length of a side. For example, if the side of a regular polygon measures 6 cm and the polygon has 5 sides, the perimeter would be 30 cm.
Sum of Interior Angles of a Regular Polygon
For a regular polygon with n sides, the sum of its interior angles can be calculated using the formula (n - 2) × 180°. Let's take a regular polygon with 6 sides as an example. The sum of its interior angles would be (6 - 2) × 180° = 720°.
Measure of Each Interior Angle of a Regular Polygon
Since regular polygons are equiangular, the measure of each interior angle can be calculated by dividing the sum of interior angles by the number of sides. For instance, in an 8-sided regular polygon, each interior angle would be (8 - 2) × 180° / 8 = 135°.
Measure of Each Exterior Angle of a Regular Polygon
The measure of each exterior angle of a regular polygon is determined by dividing 360° (the sum of all exterior angles) by the number of sides. Therefore, for an n-sided regular polygon, each exterior angle measures 360° / n.
Number of Diagonals of a Regular Polygon
The number of diagonals in a polygon with n sides can be calculated using the formula n(n - 3) / 2. This accounts for the fact that each vertex connects to (n - 3) vertices, and to avoid double counting, we divide by two. For example, a regular polygon with 4 sides (a square) has 2 diagonals.
Number of Triangles of a Regular Polygon
The number of triangles formed by joining the diagonals from one corner of a regular polygon is equal to n - 2, where n represents the number of sides. For a polygon with 4 sides, there would be 2 triangles formed.
Lines of Symmetry of a Regular Polygon
A line of symmetry divides a shape or object into identical halves. In regular polygons, the number of lines of symmetry is equal to the number of sides (n). For example, a square has 4 sides, so it also possesses 4 lines of symmetry.
Order of Symmetry of a Regular Polygon
The order of rotational symmetry in a regular polygon is equal to the number of sides (n). This means that when a regular polygon is rotated a certain number of times (the order), it will align perfectly with its original position. The angle of rotational symmetry in a regular polygon can be calculated using the formula 360° / n. For example, a square has 4 sides, resulting in an order of rotational symmetry of 4 and an angle of 90°.
Different Regular Polygons
Here are some examples of regular polygons:
Fun Fact!
Regular polygons have played a significant role in art, architecture, and even nature. From the beautiful symmetry of snowflakes to the stunning patterns in Islamic geometric art, regular polygons continue to captivate and inspire us.
Solved Examples on Regular Polygon
Let's solve a couple of examples to solidify our understanding:
Example 1: Find the number of diagonals of a regular polygon with 12 sides.
Solution: The number of diagonals of an n-sided polygon can be calculated using the formula n(n - 3) / 2. For our case, the number of diagonals would be:
12 × (12 - 3) / 2 = 54 diagonals.
Example 2: If each interior angle of a regular polygon measures 120°, how many sides does it have?
Solution: We can use the formula (n - 2) × 180° / n to find the number of sides. Let's substitute the given angle:
120° = (n - 2) × 180° / n.
Solution: To determine if this is possible, we need to calculate the exterior angle of the polygon. The exterior angle is given by 180° - internal angle. So, the exterior angle in this case would be 180° - 100° = 80°.
Since the number of sides cannot be in decimal form, a regular polygon with internal angles of 100° each is not possible.
Practice Problems on Regular Polygon
Now that we have covered the fundamentals, why not put your knowledge to the test? Try solving these practice problems to enhance your understanding of regular polygons:
Find the measure of each interior angle of a regular polygon with 10 sides.
Determine the number of diagonals in a regular polygon with 7 sides.
Frequently Asked Questions on Regular Polygon
Question: What is a polygon?
Answer: A polygon is a two-dimensional enclosed figure formed by connecting three or more straight lines.
Question: What are regular polygons?
Answer: Regular polygons are polygons where all sides and interior angles are equal.
Question: How do you calculate the perimeter of a regular polygon?
Answer: The perimeter of a regular polygon is the sum of all its sides.
Question: What is the order of symmetry of a regular polygon?
Answer: The order of symmetry of a regular polygon is equal to the number of its sides.
Now that you have a solid grasp of regular polygons, you can appreciate their beauty and elegance in various aspects of life. From intricate architecture to intricate mathematical concepts, regular polygons continue to fascinate us with their symmetry and mathematical properties. So, the next time you come across a geometric shape, take a moment to observe its regularity and appreciate the wonders of mathematics in our | 677.169 | 1 |
Could a right triangle ever be an equilateral triangle?
This is a question our experts keep getting from time to time. Now, we have got the complete detailed explanation and answer for everyone, who is interested!
It is not possible for a right triangle to be an equilateral triangle. A right triangle is, by definition, a triangle that contains a right angle, where a right angle is defined as an angle that is…
Why should we use a right triangle? Can't be a triangle with equal sides, right?
In a triangle with equilateral sides, each of the sides is the same length. If we use the theorem of the longest side, which states that in a triangle, the side that crosses the acute angle is the longest, we get the following: If each of the sides has the same length, then each of the angles must also be the same. Hence, we cannot have an equilateral triangle with a right angle.
Is it possible for a right triangle to have three sides of identical length?
It is possible for a right triangle to take the form of an isosceles triangle, which describes a triangle with two sides that are identical in length. A right isosceles triangle has one angle that is 90 degrees and two angles that are 45 degrees. This is the only right triangle that may also be represented as an isosceles triangle… The triangle with angles of 30, 60, and 90 degrees is yet another intriguing example of a right triangle.
Is there a polygon with three sides that is not a triangle?
A triangle is a type of polygon that has three sides.
There are many distinct kinds of triangles (as seen in the diagram), including the following: Equilateral means that all of the sides have the same length, and that all of the internal angles are equal to sixty degrees. Isosceles is a shape that has two sides that are the same length and a third side that is a different length.
What do you call a triangle that has three sides that are all the same length?
Equilateral. An equilateral triangle is one that has three sides and angles that are all equal to one another. It is guaranteed to always have angles of sixty degrees in each corner.
Triangles – Equilateral, Isosceles and Scalene
We found 16 questions connected to this topic.
What is the name of a triangle with angles equal to 45 degrees?
For instance, the angles of a right triangle could create straightforward connections such as 45 degrees, 45 degrees, and 90 degrees. This type of right triangle is referred to as an "angle-based" right triangle. The lengths of the sides of a right triangle are said to be "side-based" if they form ratios of whole numbers, such as 3: 4: 5, or of other special numbers, such as the golden ratio.
Is it possible to draw an obtuse triangle with equal-sided equilateral sides?
An obtuse equilateral triangle can never be constructed. This is an impossibility. The term "equilateral" refers to a shape with equal side lengths.
Is it possible for an obtuse triangle to contain a right angle?
It is not possible for a triangle to have both a right angle and an obtuse angle at the same time. Because there is one right angle in a right-angled triangle, the other two angles in the triangle are acute angles. As a result, it is impossible for a triangle with an obtuse angle to have a right angle, and vice versa. In the triangle, the side that is perpendicular to the acute angle is the one that is the longest.
Is it possible for a triangle to have one angle that is right?
There can only be one right angle in a triangle since the total of all the angles in any triangle adds up to 180 degrees. If the other angles are to be greater than zero, then there can be no other angles.
Why is it that a right triangle can never be an obtuse triangle?
Because of the proportions of its internal angles, a right triangle can never be in an obtuse position. Every triangle has three sides, three angles, and three angles that are equal, regardless of the shape.
What is the maximum number of angles that are obtuse that can be found in a right triangle?
Johnson Z. One acute angle is the most that can be present in a triangle.
Why is it so difficult to construct a triangle with obtuse angles and equilateral sides?
In a triangle with equilateral sides, each angle measures exactly sixty degrees. Given that each of the three angles in a triangle is equal, divide the total number of degrees in the triangle by 180 and then by 3. A degree that is larger than 90 degrees is known as an obtuse angle. Because 60 and 90 are both divisible by the same number, an equilateral triangle cannot have an obtuse angle.
Is it possible to draw a triangle using two right angles as the angles of the triangle?
No, a triangle can never have two angles that are right angles at the same time. A triangle has exactly three sides, and the sum of the angles on the inside of the triangle adds up to 180 degrees. If a triangle contains two angles that are right angles, then the third angle must be 0 degrees, which indicates that the third side will overlap with the other side of the triangle.
Is it possible to draw a triangle with two angles that are 90 degrees from one another?
No, a triangle cannot have both acute and obtuse angles at the same time. An angle that has a measure that is larger than 90 degrees is known as an obtuse angle.
Do the numbers 4, 5, and 6 form right triangles?
A Pythagorean Triple is formed by the three digits 4, 5, and 6.
Can 30-60-90 angles produce a triangle?
The angles of a 30-60-90 triangle are always 30 degrees, 60 degrees, and 90 degrees. A right triangle is defined as any triangle that contains an angle of 90 degrees, hence a 30-60-90 triangle is a special kind of right triangle. Because it is a unique triangle, it also possesses side length values that are always in a consistent relationship with one another. These side length values are always in a relationship with one another.
What are the lengths of the sides of a triangle with angles of 45, 45, and 90?
A unique kind of right triangle known as a 45°-45°-90° triangle has two angles that are 45 degrees and one angle that is 90 degrees. The lengths of the three sides of this triangle are proportioned as follows: Side 1: Side 2: Side 3: Hypotenuse = n: n: n√2 = 1:1: √2. The right triangle formed by 45 degrees, 45 degrees, and 90 degrees is half of a square.
How many angles that are perfectly correct does a triangle have?
One triangle's interior can have no more than one right angle, which is defined as an angle with a measure of 90 degrees.
What do you call the side that is the longest in a triangle that has right angles?
The hypotenuse of a right-angled triangle is the side that is the longest overall. Always on the other side of the right angle is the hypotenuse.
Can you name the three angles that make up a right triangle?
Triangles are said to be right triangles when one of the interior angles is exactly 90 degrees, often known as a right angle. Because a triangle has three interior angles that sum up to 180 degrees, a right triangle has one angle that is always 90 degrees, hence the other two angles must always add up to 90 degrees for the triangle to be considered a right triangle.
Is it possible for a triangle to have right angles and be isosceles?
Is it possible for an isosceles triangle to also be a scalene or right angle triangle? It's possible for an isosceles triangle to also be a scalene or right angle triangle. A right triangle with isosceles sides has one angle that is exactly 90 degrees and two sides that are equal to each other. The angle that corresponds to these two sides is congruent since the two sides themselves are equal.
Is there a requirement that each triangle have an acute angle?
Absolutely, there are always at least two sharp angles in a triangle. Angles that measure fewer than 90 degrees are referred to as acute, while angles that measure more than 90 degrees are referred to as obtuse.
Does an acute triangle have equal sides?
A measure of an acute triangle's angle falls somewhere between zero and ninety degrees, regardless of the exact value. It is possible for the lengths of each of its three sides to be equal, unequal, or any two of the sides to be equal. An acute triangle is referred to as an equilateral acute triangle when all three of its angles measure less than ninety degrees and all three of its sides measure the same length.
How many angles that are exactly 90 degrees are there in an acute triangle?
The right triangle has one angle that is 90 degrees and two angles that are acute (less than 90 degrees). As a result of the fact that the sum of a triangle's angles is always 180 degrees…
How many triangles with an acute angle are there in total?
In any triangle, there can only be one angle that is considered obtuse. This is due to the fact that the sum of the degrees that make up a triangle's internal angles must always equal 180… | 677.169 | 1 |
8 1 Additional Practice Right Triangles And The Pythagorean Theorem Answers Integrated Arithmetic and Basic Algebra Bill E. Jordan 2004-08 A combination …
Did you know?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 Learn more at mathantics.comVisit for more Free math videos and additional subscription based content! InAngles. Triangles. Medians of triangles. Altitudes of triangles. Angle bisectors. Circles. Free Geometry worksheets created with Infinite Geometry. Printable in convenient PDF format.8: Pythagorean Theorem and Irrational Numbers. 8.2: The Pythagorean Theorem. 8.2.1: Finding Side Lengths of Triangles 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 ...
Practice using the Pythagorean theorem to solve for missing side lengths on right triangles. Each question is slightly more challenging than the previous. Pythagorean ……
Reader Q&A - also see RECOMMENDED ARTICLES & FAQs. 8-1 1. Plan What You'll Learn • To use the . Possible cause: Here is a right triangle, where one leg has a length of 5 units, the hypotenuse ha.
Practice. Find angles in isosceles triangl Jan 31, 2020 · 10. The length of one leg of a rigThe Pythagorean Theorem is an important math Here8-1 Additional PracticeRight Triangles and the Pythagorean TheoremFor Exercises 1-9, find the value of x. Write your answers in simplest radical … The remaining sides of the right triangle a Dec The Pythagorean Theorem is an important matQ Triangle J′K′L′ shown on the grid below is a dilation of The Pythagorean Theorem states that if a triangle is a right tria 8-1 Additional Practice Right Triangles and the Pythago Now I'll plug these into the Pythagorean Theorem, and[Sep 27, 2022 · In any right triangle, the 8.G.C.9. Know the formulas for the volumes of cones, cylinders, a | 677.169 | 1 |
CCSS.Math.Content.HSF-TF.A.3
(+) Use special triangles to determine geometrically the values of sine, cosine, tangent for π/3, π/4 and π/6, and use the unit circle to express the values of sine, cosine, and tangent for x, π + x, and 2π – x in terms of their values for x, where x is any real number | 677.169 | 1 |
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Welcome to cos 30°, our post aboutthe cosine of 30 degrees. For the cosine of 30 degrees we use the abbreviation cos for the trigonometric function together with the degree symbol °, and write it as cos 30°. If you have been looking for what is cos 30°, or if you have been wondering about cos 30 degrees in radians, then you are right here, too.Q. Evaluate sin60∘ cos30∘ +cos60∘sin30∘. Q. Evaluate each of the following. sin 60 cos 30° + cos 60° sin 30°. Q. Find the values of -. (i) 5 sin 30 ° + 3 tan 45 ° (ii) 4 5 tan 2 60 ° + 3 sin 2 60 ° (iii) 2sin 30 ° + cos 0 ° + 3sin 90°. (iv) tan 60 sin 60 + cos 60 (v) cos 2 45 ° + sin 2 30 ° (vi) cos 60 ° × cos 30 ° + sin ... Sec 30. The value of Sec 30 degrees is equal to 2/√3. In trigonometry, you may have learned about three main primary functions, such as sine, cosine and tangent along with them the other three trigonometric functions, such as secant, cotangent and cosecant. Here, you will find the value of sec 30 degrees along with the other secant degree values.Free trigonometric simplification calculator - Simplify trigonometric expressions to their simplest form step-by-step 當我們有一個三角形,邊長與角度如上圖所示時,則面積會等於一半的兩邊乘上夾角的 $ sin $ 值:$$ 面積 =\frac{1}{2}\cdot a\cdot b\cdot sin(\angle C) $$三邊長與對角的關係呈:$$ \frac{a}{sin\angle A} = \frac{b}{sin\angle B} = \frac{c}{sin\angle C} $$任意一邊長與另外兩邊的關係為:$$ c^2 = a^2 + b^2 - 2\cdot a\cdot b\cdot cos\angle C ...Simplify sin(30)+cos(30) Step 1. The exact value of is . Step 2. The exact value of is . Step 3. The result can be shown in multiple forms. Exact Form: Decimal Form:The Tan 30 Degrees. The value of tan 30 degrees is 1/√3. The value of tan π/6 can be evaluated with the help of a unit circle, graphically. In trigonometry, the tangent of an angle in a right-angled triangle is equal to the ratio of opposite side and the adjacent side of the angle. Tan 30 degrees is also represented by tan π/6 in terms of radians. Accurate trigonometric ratios for 0°, 30°, 45°, 60° and 90° The trigonometric ratios for the angles 30°, 45° and 60° can be calculated using two special triangles.Free math problem solver answers your trigonometry homework questions with step-by-step explanations.cos(30) 47: Find the Exact Value: cos(60) 48: Find the Exact Value: cos(0) 49: Find the Exact Value: cos(135) 50: Find the Exact Value: cos((5pi)/3) 51: Find the Exact Value: cos(210) 52: Find the Exact Value: sec(60 degrees ) 53: Find the Exact Value: sin(300 degrees ) 54: Convert from Degrees to Radians: 135: 55: Convert from Degrees to ...Sep 2, 2023 · The value of cos 30 is equal to the value of sin 60 since they complement each other within a triangle. Hence, the value of sin 30 can be used with cos 60 too, considering this pattern, we can conclude that sin (90-x) = cos x; where x can be any angle. Derivative Cos 30°: For deriving the value of cos 30°, let us take a look at the triangle ... InFind the Exact Value sin(30)+cos(60) Step 1. Simplify each term. Tap for more steps... Step 1.1. The exact value of is . Step 1.2. The exact value of is . Step 2 ... Free math problem solver answers your trigonometry homework questions with step-by-step explanationsSep 2, 2022 · The value of cos 30 degrees in decimals is 0.8660. Converting degree to radian, that is, θ in radians = θ × π/180° or θ × Pi/180°. Therefore, converting cos 30° in radians will give cos (30 × π/180°), and the final value in radians will become cos(π/6) or cos(0.5235). Following are the values of cos 30° in different forms, The
What is the exact value answers in degrees of two cos x minus radical three equals zero? 30 degrees explanation 2Cosx-radical 3=0 Then 2cosx=radical 3 and cos x= (radical 3)/2 Now remember that cos 300 is (radical 3)/2 from the 30/60/90 triangle. So the answer is 30 degreescos 1 cos 2 sin 1 sin 2 + i(sin 2 cos 1 + isin 1 cos 2) = r 1r 2 cos( 1 + 2) + isin( 1 + 2); where we used the sum formulas in Section 5.4 in the last line. To show that the division formula holds, you can use the multiplication formula and that z 1 = z 1 z 2 z 2. Examples Carry out each of the following operations: 1.3 cos ˇ 3 + isin ˇ 3 4 ... …
Step 4: Determine the value of tan. The tan is equal to sin divided by cos. tan = sin/cos. To determine the value of tan at 0° divide the value of sin at 0° by the value of cos at 0°. See the example below. tan 0°= 0/1 = 0. Similarly, the table would be. Angles (In Degrees) 0°. 30°. What is the value of cos 75°? | 677.169 | 1 |
Description. This Relationships in Triangles Unit Bundle contains guided notes, homework assignments, three quizzes, a study guide and a unit test that cover the following topics: • Midsegments of Triangles (includes reinforcement of parallel lines) • Inequalities in Triangles: Determine if three sides can form a triangle.Homework 1 Triangle Midsegments Answer Key. In triangle geometry, midsegments play an important role in understanding the properties and relationships between triangle sides and angles. A midsegment of a triangle is a line segment that connects the midpoints of two sides of the triangle. It is also known as a midline. Homework 1 focuses on ... 11+ chapter 5 relationships in triangles reply key. Unit 5 take a look at relationships in triangles reply key beneath is one of the best info and information about unit 5 relationships in triangles reply key compiled and compiled by the. Unit 3 relations and capabilities homework 6. Supply: razanac-tzo.internetTherefore the angles from least to greatest are: ∠N, ∠R, ∠Q. 13) The sides are 6 in, 7 in, 10 in (least to greatest) So the angles opposite to those sides (in the same order) are also …
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First time poster so please be gentle! I really should remember this solution from high school maths, but I'm afraid that was a very long time ago :(
I am looking for a formula to determine if a given point lies within the locus of all points equidistant from a line segment marked on a sphere. What I am ultimately trying to do is this:
I have a MySQL database with two tables: Buildings and People
Every building and person has a location described in real-world co-ordinates as their lat/lng pair
Buildings are always in a fixed position: their lat/lng pair never changes
People can move, and for the sake of simplicity I assume they always move in a straight line from point A to point B (noting that it's not really a straight line, as it must follow the curvature of the earth)
What I want to determine is for a given journey a person takes (from A to B), assuming their range of vision is 300 metres, what are all the buildings they would see along their route?
At the moment, I can successfully determine all buildings a person can see from one point (i.e., assuming the person is standing still). This is basically similar to determining all points within a circle given a certain radius described on a sphere (Haversine, I believe), and the SQL query looks like this:
I could also implement a point-in-polygon algorithm probably using ray casting. So to solve the problem (albeit inefficiently and inelegantly) I could:
Find the points within a rectangle of "side" edges at A and B, and "top" edges 600 metres apart and perpendicular to the "side" edges, using point-in-polygon (not sure if this will be accuracte given the curvature of the earth, but I could possibly live with that)
Find the points within the two circles described by points A and B with radii 300 metres (because I think it's easier to find points in circles rather than points in semi-circles)
Discard any duplicate points found (two buildings can't have exactly the same lat/lng pair)
But there must be a more correct, accurate, efficent and elegant way. I'd be very keen to see a mathematical, formulaic approach to this proble, with extra points for the algorithm written in SQL!
1 Answer
1
I think the formula you are looking for is:
$$(x-x_m(t))^2 + (y-y_m(t))^2 \leq r^2 $$
where $x_m(t)$ and $y_m(t)$ are functions of x-coordinates and y-coordinates of the man under consideration. In you case it can have the sequence of points you have stored in table.
This equation governs the area inside a circle whose center is moving as a function of time. You need to put all the building location coordinates for every new coordinate of man and check which buildings satisfy the inequality.
$\begingroup$Hi there Aniket. If I understand your formula correctly, you are suggesting sampling the line every t seconds and running the formula at every point, is that correct? So that if a person travels at 10 m/s for 100m, I would have to run the formula at every 10m along the path and get all the points within the circle's radius 10 times, is that right? If so, how does this account for excluding overlaps so that the same building isn't "seen" multiple times? And does your formula take into account the curvature of the earth? Thanks, Arj.$\endgroup$
$\begingroup$Yes, your interpretation is right. In fact, if you are working on a machine that can handle functions (like MATLAB) then you can directly generate an end-function that will provide area covered by the person walking in any arbitrary continuous function of path (maybe straight line, curved line, sinusoid but in 2D only, not the curvature of Earth) between distinct endpoints. For excluding multiple times of the same building you can always remove the duplicate coordinates found in every iteration. I can try to edit this formula to incorporate third dimension (curvature of Earth).$\endgroup$
$\begingroup$Hi Aniket, thanks for your reply. No need to edit your formula - I understand the premise and have the Haversine formula in SQL and PHP to handle the calculations. I guess I was looking for a formula that didn't use time or distance sampling, and also one that automatically excluded duplicates. However if this is not mathematically possible, then I guess I will go with your approach. Arj$\endgroup$
$\begingroup$Yeah I think the solution I provided cannot work in discrete sense because if time steps are not small enough then the actual area is a sequence of circles instead of a rectangle with circular ends. The problem in finding a single equation is that a rectangle in itself is a discontinuous shape, behaves differently in different domains. So adding semicircular ends does no good. So it is hard to find an equation that directly draws the area you require. Please do let me know if you find such an equation. I would love to know it.$\endgroup$
$\begingroup$Thanks for your reply Aniket. There was definitely such a formula when I was in high school. However it's in a very old text book which I don't have any more. And I can't find it on the web :($\endgroup$ | 677.169 | 1 |
PlatonicSolids
Platonic solids are regular, convex polyhedrons. They are constructed by congruent (identical in shape and size) regular (all angles equal and all sides equal) polygonal faces with the same number of faces meeting at each vertex.
Five solids satisfy the above criteria:
Figure
Tetrahedron
Cube
Octahedron
Icosahedron
Dodecahedron
Vertices
4
8
6 (2 × 3)
12 (4 × 3)
20 (8 + 4 × 3)
Edges
6
12
12
30
30
Faces
4
6
8
20
12
The relationship between vertices, edges and faces is given by Euler's formula:
#!/usr/bin/env pythonfromcollectionsimportnamedtupleFiltersSourcesimportvtkPlatonicSolidSourcefromvtkmodules.vtkRenderingCoreimport(vtkActor,vtkActor2D,vtkPolyDataMapper,vtkRenderWindow,vtkRenderWindowInteractor,vtkRenderer,vtkTextMapper,vtkTextProperty)defmain():colors=vtkNamedColors()mappers=list()actors=list()text_mappers=list()text_actors=list()renderers=list()# Create a common text property.text_property=vtkTextProperty()text_property.SetFontSize(16)text_property.SetJustificationToCentered()# Create the render window and interactor.ren_win=vtkRenderWindow()ren_win.SetWindowName('PlatonicSolids')iren=vtkRenderWindowInteractor()iren.SetRenderWindow(ren_win)name_orientation=get_name_orientation()lut=get_platonic_lut()platonic_solids=list()foriinrange(0,len(name_orientation)):platonic_solids.append(vtkPlatonicSolidSource())platonic_solids[i].SetSolidType(i)mappers.append(vtkPolyDataMapper())mappers[i].SetInputConnection(platonic_solids[i].GetOutputPort())mappers[i].SetLookupTable(lut)mappers[i].SetScalarRange(0,19)actors.append(vtkActor())actors[i].SetMapper(mappers[i])text_mappers.append(vtkTextMapper())text_mappers[i].SetInput(name_orientation[i].name)120,16)renderers.append(vtkRenderer())renderers[i].AddActor(actors[i])renderers[i].AddViewProp(text_actors[i])ren_win.AddRenderer(renderers[i])# Set up the viewports.grid_dimension_x=3grid_dimension_y=2renderer_size=300ren_win.SetSize(renderer_size*grid_dimension_x,renderer_size*grid_dimension_y)forrowinrange(0,grid_dimension_y):forcolinrange(0,grid_dimension_x):index=row*grid_dimension_x+col# (x_min, y_min, x_max, y_max)viewport=[float(col)/grid_dimension_x,float(grid_dimension_y-(row+1))/grid_dimension_y,float(col+1)/grid_dimension_x,float(grid_dimension_y-row)/grid_dimension_y]ifindex>(len(actors)-1):# Add a renderer even if there is no actor.# This makes the render window background all the same color.ren=vtkRenderer()ren.SetBackground(colors.GetColor3d('SlateGray'))ren.SetViewport(viewport)ren_win.AddRenderer(ren)continuerenderers[index].SetViewport(viewport)renderers[index].SetBackground(colors.GetColor3d('SlateGray'))renderers[index].ResetCamera()renderers[index].GetActiveCamera().Azimuth(name_orientation[index].azimuth)renderers[index].GetActiveCamera().Elevation(name_orientation[index].elevation)renderers[index].GetActiveCamera().Zoom(name_orientation[index].zoom)renderers[index].ResetCameraClippingRange()iren.Initialize()ren_win.Render()iren.Start()defget_name_orientation():""" Get the platonic solid names and initial orientations. :return: The solids and their initial orientations. """# [[name, azimuth, elevation, zoom] ...]res=[['Tetrahedron',45.0,30.0,1.0],['Cube',-60.0,45.0,0.8],['Octahedron',-15.0,10.0,1.0],['Icosahedron',4.5,18.0,1.0],['Dodecahedron',171.0,22.0,1.0]]platonic_solids=namedtuple('platonic_solids',('name','azimuth','elevation','zoom'))# Convert res to a list of named tuples.res=[platonic_solids(*row)forrowinres]returnresdefget_platonic_lut():""" Get a specialised lookup table for the platonic solids. Since each face of a vtkPlatonicSolidSource has a different cell scalar, we create a lookup table with a different colour for each face. The colors have been carefully chosen so that adjacent cells are colored distinctly. :return: The lookup table. """lut=vtkLookupTable()lut.SetNumberOfTableValues(20)lut.SetTableRange(0.0,19.0)lut.Build()lut.SetTableValue(0,0.1,0.1,0.1)lut.SetTableValue(1,0,0,1)lut.SetTableValue(2,0,1,0)lut.SetTableValue(3,0,1,1)lut.SetTableValue(4,1,0,0)lut.SetTableValue(5,1,0,1)lut.SetTableValue(6,1,1,0)lut.SetTableValue(7,0.9,0.7,0.9)lut.SetTableValue(8,0.5,0.5,0.5)lut.SetTableValue(9,0.0,0.0,0.7)lut.SetTableValue(10,0.5,0.7,0.5)lut.SetTableValue(11,0,0.7,0.7)lut.SetTableValue(12,0.7,0,0)lut.SetTableValue(13,0.7,0,0.7)lut.SetTableValue(14,0.7,0.7,0)lut.SetTableValue(15,0,0,0.4)lut.SetTableValue(16,0,0.4,0)lut.SetTableValue(17,0,0.4,0.4)lut.SetTableValue(18,0.4,0,0)lut.SetTableValue(19,0.4,0,0.4)returnlutif__name__=='__main__':main() | 677.169 | 1 |
Right triangle ratio crossword
The Crossword Solver found 30 answers to "trig ratios found 2 answers for the crossword clue Trigonometric ratios. If you haven't solved the crossword clue Trigonometric ratios Search for crossword answers and clues. Word. Letter count. ... Right-triangle ratio, for short. Answer for the clue "Right-triangle ratio, for short ", 5 letters: cotan. Alternative clues for the word cotan . Ratio of the adjacent to the opposite side; Trig function: Abbr; Trig ratio; Word definitions for cotan in dictionaries.Our crossword solver found 10 results for the crossword clue "right triangle ratio".5. 6. jump to corrections. All crossword answers with 3-6 Letters for Trig ratio found in daily crossword puzzles: NY Times, Daily Celebrity, Telegraph, LA Times and more. Search for crossword clues on crosswordsolver.com. triangle ratio Crossword Clue. The Crossword Solver found 30 answers to "triangle ratio Right triangle ratio, or a kind of wave. We will try to find the right answer to this particular crossword clue. Here are the possible solutions for "Right triangle ratio, or a kind of wave" clue. It was last seen in American quick crossword. We have 1 possible answer in our database.The Crossword Solver found 30 answers to "Ratio of acute side length angle in a right angle to length of opposite side Ratio. We will try to find the right answer to this particular crossword clue. Here are the possible solutions for "Ratio" clue. It was last seen in The Independent quick crossword. We have 1 possible answer in our databaseLet me guess, you have been playing The Denver Post crossword and got stuck on the clue Basketball's Wilt "The ___" Chamberlain. Well, you have come to the right place to find the answer to this clue. ... Right-triangle ratio: Business card datum: Reduce to rubble: In a 22-Down: Green kind of energy: With 11-Down, 1970 Perry Como hit: Tourney ... Cl The Crossword Solver found 30 answers to "Airspeed ratio indicatorRight triangle ratio" 18 letters crossword answer - We have 2 clues. Solve your "Right triangle ratio" crossword puzzle fast & easy with the-crossword-solver.comWhen facing difficulties with puzzles or our website in general, feel free to drop us a message at the contact page. We have 1 Answer for crossword clue Ratio Of An Angles Opposite Side To The Hypotenuse of NYT Crossword. The most recent answer we for this clue is 4 letters long and it is Sine.We have got the solution for the Triangle ratio crossword clue right here. This particular clue, with just 4 letters, was most recently seen in the Wall Street Journal on July 7, 2023. And below are the possible answer from our database. Triangle ratio Answer is: SINE.The Crossword Solver found 30 answers to "photography ratio photography (filmmaking technique ...For the word puzzle clue of a ___ ratio is the ratio of the lengths of two sides in a right triangle, the Sporcle Puzzle Library found the following results.Explore more crossword clues and answers by clicking on the results or quizzes.
The Crossword Solver found 30 answers to "Words in a ratioWeight to height ratio total the crossword has more than 80 questions in which 40 across and 40 down. If you can't find the answers yet please send as an email and we will get back to you with the solution. Right triangle ratio
Upon examining the given clues, we have managed to identify a total of 2 possible solutions for the crossword clue "Math ratio ... Right triangle ratio; Trig ratio: Abbr. *Kind of ratio; Light ratio in astronomy; The golden ratio; Speed ratio; Ratio symbols; Betting ratio; Ratio of AB to BC, say;The Crossword Solver found 30 answers to "Right triangle ratio: Abbr./51883 ratios of the sides of a right triangle are called trigonometric . Possible cause: Below are possible answers for the crossword clue the reciprocal of a cosi.
A qualification ratio is actually two ratios that banks use to determine whether a borrower is eligible for a mortgage. A qualification ratio is actually two ratios that banks use ...
Below are possible answers for the crossword clue Type Of Ratio In Trigonometry. Clue. Length. Answer. Type Of Ratio In Trigonometry. 4 letters. sine. Definition: 1. ratio of the length of the side opposite the given angle to the length of the hypotenuse of a right-angled triangle. View more information about sine.Answers for Hypotenuse to adjacent side ratio crossword clue, 6 letters. Search for crossword clues found in the Daily Celebrity, NY Times, Daily Mirror, Telegraph and major publications. Find clues for Hypotenuse to adjacent side ratio or most any crossword answer or clues for crossword answers.The Crossword Solver found 30 answers to "Adjacent side/ hypotenuse, in trig", 3 letters crossword clue. The Crossword Solver finds answers to classic crosswords and cryptic crossword puzzles. Enter the length or pattern for better results. Click the answer to find similar crossword clues . A clue is required. | 677.169 | 1 |
So I have this problem involving astronomy, but because astronomy uses all sorts of fancy words I'm going to make it more simple by using an analogy of the earth. The process, mathematically would be exactly the same.
There are two cities on the earth with the lat/long coordinates (40W,20S) (which we'll call G) and (50E,6N) (Which we'll call H). What point on the equator is equidistant from both G and H?
However: I don't want the straight distance from G or H to this mystery point on the equator. I want to take into account the spherical shape of the earth, so it would require some spherical geometry.
I need the answer in latitude/longitude, not cartesian coordinates. Because I plan on doing several of these, please explain how you calculated this.
$\begingroup$While I've voted to reopen on the basis of recent edits, it seems curious that you seem on the one hand comfortable abstracting your actual astronomical problem to a setting involving a "spherical earth" with longitude and latitude, but on the other hand uncomfortable with converting between Cartesian coordinates and longitude/latitude. The spherical geometry presents some interesting parallels and contrasts with plane geometry, which I suspect you have already spotted.$\endgroup$
1 Answer
1
Let's forget about the $r$ coordinate because we'll be solely working on the surface of the sphere, $r = R$.
First convert your coordinates that are in degrees to radians for discussion. Let $\theta = $ a general point on the equator.
The spherical surface distance between $(\phi, \theta)$ and $(a,b)$ can be found by converting to Cartesian coordinates, taking the dot product to get the $\cos ($ the angle between the two vectors $)$, calculating the length of an arc with that angle.
Now you can come up with the formula for Spherical-to-Cartesian coordinates by drawing the system out on paper, or just refer here:
$\begingroup$when you say "cosθsinacosb+sinθsinasinb=cosθsina′cosb′+sinθsina′sinb′cosθsinacosb+sinθsinasinb=cosθsina′cosb′+sinθsina′sinb′" do you mean that a is the longitude of one and b is its latitude, while a' and b' are the other's coordinates, respectively? Will I need to convert to cartesian to do this particular method? What is the purpose of the arctan (theta) at the end? Is it necessary?$\endgroup$ | 677.169 | 1 |
A landscaper wants to plant begonias along the edges of a triangular plot of land in Winton Woods Park. Two of the angles of the triangle measure 95⁰and 40⁰. The side between the two angles is 80 feet long. What is the perimeter of this triangular plot of land 18:55:172023-03-20 18:55:17A landscaper wants to plant begonias along the edges of a triangular plot of land in Winton Woods Park. Two | 677.169 | 1 |
How To Jk kl and lj are all tangent: 4 Strategies That Work
Since JK, KL, and LJ are all tangent to circle O, we know that JL and JK are radii of the circle O. Let's represent the radius as r. Step 3: Identifying lengths of JL and JK We have the radii JL and JK, and we have a right triangle formed with AL, r, and LJ. Using the Pythagorean theorem, we can write the equation: AL^2 + LJ^2 = AJ^2 J4. m∠O = 111. 5. m∠P = 12. 6. BC is tangent to circle A at B and to circle D at C (not drawn to scale). AB = 7, BC = 18, and DC = 5. Find AD to the nearest tenth. 7. AB is tangent to circle O at B. Find the length of the radius r for AB = 5 and AO = 8.6. Round to the nearest tenth if necessary. O is the center of the circle. 69. ab is tangent to a=24 and BC=50 what is ab. 70. ab is tangent circle O at B find the length of the radius r for ab=5 and ao= 8.6 round to the nearest tenth necessary. 7. jk and kl and lj are all tangent to 0 not drawn to scale ja=9, al=10 and ck=14 find the perimeter of JKL. 66 JK, KL, and LJ are all tangent to O (not drawn to scale). JA = 14, AL = 15, and CK = 13. Find the perimeter of ^JKL (^ is in place for the triangle symbol thing) D. 84. All answers for Connexions Academy Geometry B - Inscribed Angles Learn with flashcards, games, and more — for free. Lines JK, KL, and LJ are tangent to the circle. Find the perimeter of triangle JKL. 4. Find x. 5. Find y. 6. Find x. 7. Find x. 8. Find the volume of the rectangular based pyramid. 9. Find the volume of the cone. 10. Find the volume of the figure. 11. Name the cross sections. 12. What 3D object is produced by the rotating cross ... JJK, KL and LJ are all tangent to O (not drawn to scale). JA = 13, AL = 9, and CK = 11. Find the perimeter of JKL. a.)66 b.)44 c.)33 d.)48 J Math D. Circles P and Q are externally tangent. AB is a commNon external tangent. B 7. If the radii of circles P and Q are 8 cm and 12 cm, respectively, find the length of the tangent segment AB. 8. If the radii of circle P and Q are 5 cm and 9 cm, respectively, find the length of the tangent segment AB. 40. D. Circles P and Q are externally tangent.jk kl and lj are all tangent to o ja = 14 al = 15 and ck=13 find the perimerter; admin. Mathematics High School..JA =14, AL=15, and CK=13.Solution for 40.) A line in the plane of a circle that intersects the circle at two points is the a. tangent b. secant c. radius d. diameter. Skip to main content. close. Start your trial now! First week only $4.99! arrow ... JK, KL, and LJ are all tangent to circle O ...Find an answer to your question jk,kl, and lj are all tangent to circle o, ja=13,al=7, and ck=10 what is the perimeter of angle JKL See what teachers have to say about Brainly ... two tangent segments drawn from …The tangent segment theorem states that if a line is tangent to a circle, then the lengths of the two tangent segments drawn from an external point to the point of tangency are equal. So, JK = JA = 13 and LJ = AL = 19. Now we can add the lengths of all three segments to find the perimeter: Perimeter = JK + KL + LJ Perimeter = 13 + KL + 19answered • expert verified. JK, KL, and LJ are all tangent to circle O. JA = 12, AL = 11, CK = 13. What is the perimeter of triangle JKL? A. 46 units. B. … Lines JK, KL, and LJ are all tangent to circle O. The Diagram is not drawn to scale. If JA = 10, AL = 23 and CK = 9, what is the perimeter of triangle JKL? a. 84 b. 66 c. 42 d. 38 Is the answer 84? Please explain how you got the answer. 2 months ago. Solution 1. Guest #9411870. 2 months ago JK, KL, and LJ are all tangent to circle O (not drawn to scale). JA = 7 cm, AL = 6 cm, and CK = 9 cm. Find the perimeter of AJKL. 04:29. 7. In the figure below; M is …Feb 7, 2021 · JK,KL, and LJare all tangent to circle O (not drawn to scale), and JK≅LJ.JA= 9, AL= 10, and CK= 14. Find the perimeter of ΔJKL.A.66B.38C.46D.33 Background image We store cookies data for a seamless user experienceDAis tangent to the circle atAandDC is tangent to the circle atC. Findm∠D form∠B= 56. (The figure is not drawn to scale.) A. 136 B. 124 C. 68 D. 62 112 A b ____ 28. A park maintenance person stands 16 m from a circular monument. Assume that her lines of sight form tangents to the monument and make an angle of 56°.Math 7. In the diagram shown below, QT and QR are tangent to circle O at points T and R, respectively. If mTR =122°, then which of the following is the measure of ZRỌT ? R (1) 61° (2) 58° (3) 177° (4) 119°. 7. In the diagram shown below, QT and QR are tangent to circle O at points T and R, respectively.The KL gene provides instructions for making the protein alpha-klotho, which is found primarily in kidney cells. Learn about this gene and related health conditions. The KL gene pr...AB and AC are tangent to circle O at B and C respectively. If the circle's radius is 2 and AB has a…. A: AB=AC=9 So option ( B) is correct. If we draw two tangent lines from same point then tangent lines…. . Given segment EB is tangent to circle A at point B. Find the distance from E to Aif the radius of….JClickSince JK, KL, and LJ are all tangent to circle O, they are all perpendicular to the radii drawn to the points of tangency. This means that JK, KL, and JL bisect angles J, K, and L respectively. Let x be the length of JK, y be the length of KL, and z be the length of LJ.Draw three non collinear points, J, K and L. Then draw the lines JK, KL and LJ. Problem 3 : Draw two lines, label points on the lines and name two pairs of opposite rays. Problem 4 : Sketch the figure described. "A line that intersects a plane in one point" Problem 5 : Sketch the figure described. "Two planes that intersect in a line"How to Raise a Reckless, Glorious, Girl— According to Merriam-Webster Dictionary, the definitions of the words: Reckless and glorious, are the following: reckless reck&a...QuestionGet the detailed answer: 3. JK, KL, and LJ are all tangent to O. JA = 6, AL = 11, and CK = 13. Find the perimeter of triangle $$\triangle$$JKL. A) P=40 B)Jun 8, 2023 · 1. JK, KL, and LJ are all tangent to circle O. The diagram is not drawn to scale. If JA = 14, AL = 12, and CK = 8, what is the perimeter of ΔJKL? 2. The farthest distance a satellite signal can directly reach is the length of the segment tangent to the curve of Earth's surface. The diagram is not drawn to scale. Oct 27, 2021 · Jk, kl, and lj are all tangent to circle o. ja = 12 Study with Quizlet and memorize flashcards containing ter May 11, 2018 ·. Oct JK, KL, and LJ are all tangent to circle O (not; drawn to JK, KL, and LJ are all tangent to circle O (not; drawn to scale), and JK LJ. JA = 7, AL = 12, and. CK = 10. Find the perimeter of JKL. a. 34. b. 38. c. 58. d. 29. Assume that lines that appear to be tangent are tangent. O is the center of the circle. Find the value of x.The tangent line for a graph at a given point is the best straight-line approximation for the graph at that spot. The slope of the tangent line reveals how steep the graph is risin... This problem has been solved! You'll get ... | 677.169 | 1 |
STATEMENT-1: The locus of the centre of a circle which touches two given circles with different radius and centre will be a hyperbola.
and
STATEMENT-2: If the differnece of distance of a variable point to two given is always constant and less than the distnace between the points then the locus will be a hyperbola
A
Statement -1 is True, Statement-2 is True; Statement-2 is NOT a correct expalanation for Statement-1
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B
Statement-1 is True, Statement-2 is False
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C
Statement -1 is True, Statement-2 is True; Statement-2 is a correct expalanation for Statement-1
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D
Statement-1 is False, Statement-2 is True.
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Open in App
Solution
The correct option is C Statement -1 is True, Statement-2 is True; Statement-2 is a correct expalanation for Statement-1 Let r1,r2 be the radii of the given circles and r be the radius of variable circle, Let C−1,C−2,C be the centre of the given and the variable circles respectively, then CC1=r+r1 CC2=r+r2 ⇒CC1−CC2=r1−r2 | 677.169 | 1 |
45 deg. 45 deg. b/c. a/c. b/a. a/c. b/c. a/b. phi. theta. theta. 0.866. Note: if you are using a calculator to find these ratios, make sure you have set the mode to DEG for degrees. if you are using microsoft excel, (I use it like a calculator at work) it calculates angles in radians. So, using excel to calculate the cos of 30 degrees you would
For any problem involving a 30°-60°-90° triangle, the student should not use a table. The student should sketch the triangle and place the ratio numbers. Since the cosine is the ratio of the adjacent side to the hypotenuse, you can see that cos 60
2015-08-05
sin(60 degrees) = cos(30 degrees) cos(60 degrees) = sin(30 degrees) We will not do this computation, but if we replace 30 degrees with 20 degrees and 60 degrees with 70 …
What is {eq}\cos 30° {/eq} in the simplest radical form? Trigonometric function A function of form f(x)=sinx, cosx, tanx etc are called the trigonometric function. How to solve: Use a special right triangle to write cos 30° as a fraction. By signing up, you'll get thousands of step-by-step solutions to
2019-10-07
Exact values of sin(30), cos(30), tan(30), csc(30), sec(30), cot(30), Find exact values of all trigonometric functions when the angle is 30 degrees,blackpenr
$\cos{(30^°)} \,=\, \dfrac{\sqrt{3}}{2}$ The value of cos of angle $30$ degrees is equal to $\dfrac{\sqrt{3}}{2}$ in fraction form exactly.
Find the resultant of the following two displacements: 2 m at 30 deg and 4 m
2019-03-28
Without using calculator,evaluate Sin15 deg,Cos 15 deg,Sin 75deg. O need the answer urgently pls. Prince. Nov 14, 2011 . sin 15° = sin(45-30) = sin45cos30 - cos45sin30 = (1/√2)(√3/2) - (1/√2)(1/2) = (√3 - 1)/(2√2) so cos 15° the same way and for the
2008-06-02
The cos of 37 degrees is 0.79864, the same as cos of 37 degrees in radians. | 677.169 | 1 |
Vector Projection Calculator is a step-by-step online calculator that calculates the projection of one vector onto another. It utilizes the standard vector projection formula, and regardless of your math skills, it is pretty easy for everyone. Importantly, our Vector Projection Calculator can be used to find the project of both two-dimensional and three-dimensional vectors. Besides calculating the projection of one vector onto another, our calculator also shows the projection factor.
Additionally, do you want to learn about cross (vector) product? You should check out comprehensive Cross Product Calculator.
What is a vector?
A vector is an object composed of two parts: magnitude and direction. If we look at a vector geometrically, we consider it a straight line with the direction shown by the pointing arrow. The magnitude of the vector is equivalent to the line's length, whereas the direction is expressed by the point in which the line's arrow is pointing.
Generally, two vectors are the same if we translate them or change their position without affecting their direction and magnitude. On the opposite, when we have two vectors, and we stretch its tail or head, we can no longer consider those vectors the same. However, regardless of how distant vectors are in the coordinate system, as long as their direction (arrow) and magnitude (line length) remain the same, we consider them as the two equal vectors.
Type of vectors
Considering the magnitude and direction of the vectors, we classify them into six major groups:
Zero or null vector
A zero or null vector is a vector with a magnitude of 0. In other words, the zero vector does not have direction, and its starting point is equal to the ending point.
Unit vector
A unit vector has a magnitude of 1 unit in length. For example, a vector x with the magnitude x is written as x̂.
Position vector
A position vector indicates nothing but a particular point in the plane.
Co-Initial Vectors
A co-initial vector is self-explanatory. Two or more vectors with the same starting point are considered co-initial vectors.
Like and Unlike Vectors
When vectors are pointing in the same direction, we call them like vectors. As opposed to that, when two or more vectors have the opposite direction, we consider them unlike vectors.
Coplanar Vectors
If two vectors lie in the same plane, we consider them coplanar vectors.
Collinear or parallel vectors
If we have two vectors that lie in the plane parallel to each other, we call them collinear or parallel vectors.
Equal Vectors
As long as two vectors have the same magnitude and direction, they are equal vectors regardless of their starting and end points.
Displacement Vector
If a point is displaced from point A to B, AB is considered a displacement vector.
Negative Vector
Suppose we imagine vectors a and b, assumably. In that case, they have the same magnitude but point in the opposite directions – we call them negative vectors. In other words, vector b is the negative of vector a.
Angle Between Two Vectors
If we consider two vectors and their positions in the coordinate system, they relate to each other. Additionally, we can calculate or measure the angle between them using the following formula:
cos \theta = \vec{a}.\vec{b} \div \left| a\right|.\left| b\right|
The angle between two vectors equals the cosine of the angle between them.
Dot Product of Two Vectors
Dot product, or in simpler terms (multiplication of two vectors), is a math calculation that we perform using the formula below:
\vec{a}.\vec{b} = a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3}
This calculation results in a scalar unit or a number, so we often call dot product the scalar product of two vectors.
Projection of two vectors
The projection of one vector onto another is the product of their magnitudes and the cosecant of the angle between them. For example, suppose vector a projected onto vector b is equivalent to the product of vector a and the cosecant of the angle between vectors a and b. The value we get from the projection is scalar.
It's important to emphasize that the projection vector has magnitude as a part of the magnitude of vector a. However, the direction of the projection vector is the same as the direction of vector b.
Vector projection formula
This is the standard vector projection formula used to calculate the projection of two-dimensional or three-dimensional vectors. Since an orthogonal approach is used in this formula, theoretically, it wouldn't be incorrect to call it orthogonal projection formula, as well.
Calculate vector projection using our Vector Projection Calculator
Okay, we saw how the vector projection formula works under the hood, but there is an easier way to calculate the projection of two vectors. Instead of calculating it manually, you can use our free step-by-step online vector projection calculator to find the projection of one vector onto another. You have our calculator in front of you, but it seems complicated or unclear to you? Don't worry because, in this section, I will explain to you step by step the process of entering the coordinate values of the two vectors and calculating the projection of a vector a onto vector b. So let's get hands dirty!
Projection of vectors calculator steps:
1) Determine whether you have two-dimensional or three-dimensional vectors 2) Let's suppose you have two 2D vectors (vector a [2, 8] and vector b [1, 5] 3) Enter the values in the respective calculator fields 4) You can choose to enter values of vectors as coordinates or points (whatever you prefer) 5) Have a look at the scalar result
Projection of a onto b = [1.6154, 8.077] Projection factor = 1.6154
1.6154 – projection on the x-axis 8.077 – projection on the y-axis
Difference between scalar and vector projection
The difference between scalar and vector projections is in the formula that we use in their calculations. Scalar is a value (number), whereas a vector is not only the number (magnitude) but a composition of the magnitude and direction. Direction is something that differentiates vectors from scalars.
Having this in mind, let's set the formula for both scalar and vector projections and comment on their differences in approach.
For scalar projections, we first find the dot product of the vectors a & b and then divide that value by the length of the vector b.
3D vector projection
A three-dimensional projection of one vector onto another uses the same approach as 2D vectors. However, the only difference is in the number of axis involved. This is because 3D vectors lie in the 3D coordinate system and have x,y,z properties. In the example above, we showed you how to calculate vector projects but of 2D vectors. Therefore, let's use our step-by-step vector projection calculator again and calculate the vector projection of two 3D vectors.
Real-world applications of vector projection
One of the frequent uses of vector projection we can find in physics for finding the projection of a force in the specified direction. However, civil engineers found it helpful for measuring the strength of materials and structures. Additionally, the Vector projection formula is crucial in spacecraft and aircraft navigation systems. Basically, anything that includes linear algebra indirectly or directly use the vector projection formula, such as machine learning & data science. Furthermore, software engineers use vector projections when dealing with matrices and their multiplications.
FAQ
What is a vector projection?
When we project a vector onto another, we break it into two component vectors, having one parallel to the second vector and one perpendicular to the second vector.
How to calculate vector projection?
We calculate the projection of one vector onto another by multiplying their magnitudes by the cosecant of the angle between them. In other words, vector projection of vectors a & b is equivalent to the product of vector a and the cosecant of the angle between vectors a and b.
Vector projection formula: P = (a \cdot b \div b \cdot b) \times b!
How to find scalar and vector projections
Scalar projections can be found using the following formula: Scalar projection of vector b onto a = (dot product of a & b) \div (length of vector b)
Vector projections can be found using the following formula: Vector projection of vector b onto a = Scalar projection of the vector b \times (Unit vector of a) | 677.169 | 1 |
All Angle Converters
The angle converters below provide more detail about converting between the individual angle units. Each one includes a definition of the individual angle units, step-by-step instructions on performing the conversion, conversion examples, together with conversion charts and other visualisations.
What Angle Units Are Supported?
Name
Symbol
Measurement System
Description
circle
circle
Non-SI (International)
A circle is not typically used as a unit of measure in the same way that units such as meters or seconds are used. However, there are some situations where the concept of a circle may be used as a unit of measure.
degree
°
Non-SI (International)
A degree is a unit of measure for angles, which is used to quantify the size of an angle. One degree is defined as 1/360th of a full rotation, which is a complete revolution around a circle.
gon
gon
Non-SI (International)
The gon is a unit of measure for angles, which is used to quantify the size of an angle. One gon is defined as 1/400th of a full rotation, which is a complete revolution around a circle.
gradian
grad
Non-SI (International)
The gradian is a unit of measure for angles, which is used to quantify the size of an angle. One gradian is defined as 1/400th of a full rotation, which is a complete revolution around a circle.
angular mil
µ mil
Non-SI (International)
An angular mil, also known as a milliradian or mil, is a unit of angular measurement commonly used in military and long-range shooting applications. One mil is equal to 1/6400th of a circle or 1/1000th of a radian.
arcminute
arcmin
Non-SI (International)
An arcminute (or minute) is a unit of angular measurement that is often used as a subdivision of a degree. One degree is equal to 60 minutes.
quadrant
quadrant
Non-SI (International)
A quadrant is an angular measure that is equal to one quarter of a circle or 90 degrees. The term "quadrant" comes from the Latin word "quadrans" which means "a quarter."
radian
rad
International System of Units (SI) / Metric Systemrevolution
rev
Non-SI (International)
A revolution is a unit of angular measure that is equal to one complete turn around a circle, which is 360 degrees or 2π radians.
right angle
right angle
Non-SI (International)
A right angle is not a unit of angular measure, but rather a specific type of angle that measures exactly 90 degrees, or one quarter of a full circle. A right angle is formed when two lines or segments intersect each other in such a way that they form two adjacent angles that are equal to each other and measure exactly 90 degrees.
arcsecond
arcsec
Non-SI (International)
An arcsecond (or second) is a unit of angular measure that is used to measure very small angles, particularly in astronomy and geodesy. One second is defined as 1/3600th of a degree or 1/60th of a minute of arc.
sextant
sextant
Non-SI (International)
A sextant is not a unit of angular measure, but rather a navigational instrument used to measure the angle between two objects, such as the horizon and a celestial object like the sun or a star. The sextant was originally developed for use in navigation at sea, and is still used today by sailors, aviators, and astronomers.
sign
sign
Non-SI (International)
Sign is a unit of angle used in astrology and defined as 1/12 of a turn or 30 degrees. This is so because astrologers divide the Sun's annual path through the sky into 12 parts, which they call signs. Each sign corresponds to one of the twelve zodiacal constellations.
turn
tr
Non-SI (International)
A turn is a unit of angular measure that represents a full rotation around a point or axis. It is equivalent to 360 degrees or 2π radians. | 677.169 | 1 |
Exploring the Significance and Application of Diagonals in Various Fields
Exploring the Significance and Application of Diagonals in Various Fields
Diagonals, in their simplest essence, are lines that connect two opposite corners or points within a shape, often forming angles. While this geometric concept may seem straightforward, its applications extend far beyond the realm of basic geometry. Diagonals play a crucial role in various disciplines, from mathematics and architecture to art and sports. In this article, we delve into the significance and application of diagonals across different fields, highlighting their diverse functionalities and implications.
Understanding Diagonals in Geometry:
In geometry, diagonals hold fundamental importance in defining shapes and understanding their properties. Take, for instance, the square and the rectangle. The diagonal of a square divides it into two congruent right triangles, while in a rectangle, diagonals are equal in length and bisect each other. This simple geometric concept forms the basis for more complex calculations and constructions in mathematics.
Diagonals also feature prominently in polygons. In a polygon with n sides, the number of diagonals can be calculated using the formula: �=�(�−3)2D=2n(n−3), where D represents the number of diagonals. This formula underscores the relationship between the number of sides and the diagonals within a polygon, showcasing the interconnectedness of geometric elements.
Moreover, diagonals play a crucial role in determining the properties of three-dimensional shapes such as cubes, rectangular prisms, and pyramids. They aid in visualizing spatial relationships and understanding the structure of these objects, making them indispensable in fields like architecture, engineering, and computer graphics.
The use of diagonals in architectural design can also evoke a sense of movement and fluidity. Buildings like the Sydney Opera House and the Burj Khalifa incorporate diagonal elements in their facades, adding a modern and dynamic touch to their architectural language.
Furthermore, diagonals play a crucial role in urban planning and landscape design. They help in defining pathways, sightlines, and zones within urban spaces, contributing to the functionality and aesthetics of the environment. By strategically incorporating diagonals into designs, architects and urban planners can create harmonious and visually engaging landscapes.
Diagonals in Art and Photography:
In the realm of art and photography, diagonals are powerful compositional tools that guide the viewer's eye and create visual tension. The diagonal rule, a principle derived from the golden ratio, suggests that placing key elements along diagonal lines enhances the overall balance and harmony of an image.
Artists often use diagonals to convey a sense of movement, depth, and perspective in their compositions. Paintings like Leonardo da Vinci's "The Last Supper" and Edvard Munch's "The Scream" employ diagonal lines to draw the viewer's attention to focal points and create a dynamic sense of space.
Similarly, photographers utilize diagonals to add drama and interest to their images. Whether capturing architectural structures, landscapes, or portraits, the strategic placement of diagonal lines can transform an ordinary scene into a captivating visual narrative. Diagonals also play a crucial role in leading the viewer's gaze through the frame, enhancing the overall impact of the photograph.
Diagonals in Sports and Athletics:
In sports and athletics, diagonals manifest in various forms, influencing gameplay, strategy, and performance. In sports such as basketball and soccer, players often utilize diagonal passes and movements to create scoring opportunities and break through defensive lines. Diagonal runs and cuts are fundamental tactics employed by athletes to outmaneuver opponents and create space on the field or court.
Moreover, diagonals play a significant role in biomechanics and athletic training. Coaches and trainers analyze the body's movement patterns to optimize performance and prevent injuries. Understanding the role of diagonals in muscle activation and joint mechanics helps athletes improve their agility, coordination, and overall athletic prowess.
Diagonals in Mathematics and Science:
Beyond geometry, diagonals find applications in various branches of mathematics and science. In graph theory, diagonals represent connections or edges between vertices in a graph, facilitating the analysis of networks, circuits, and relationships between data points. Diagonal matrices, characterized by nonzero elements only along the main diagonal, are utilized in linear algebra and numerical analysis for solving systems of equations and performing matrix operations efficiently.
In physics, diagonals appear in the context of vectors and tensors, representing components along specific directions within a multidimensional space. Diagonalization, a process of transforming a matrix into a diagonal form, simplifies complex calculations and enables the study of eigenvalues and eigenvectors in quantum mechanics, quantum chemistry, and other fields of theoretical physics.
Conclusion:
Diagonals, though simple in concept, hold immense significance and application across various disciplines. From geometry and architecture to art, sports, mathematics, and science, diagonals serve as versatile tools for visualization, analysis, and expression. Whether in shaping the built environment, guiding the viewer's gaze, or optimizing athletic performance, the influence of diagonals permeates through countless aspects of our lives, underscoring their enduring relevance and importance in the modern world. | 677.169 | 1 |
Which Best Describes The Triangle Or Triangles
Triangles are one of the most fundamental geometric shapes, and they come in a variety of different types. In this blog post, we're going to take a closer look at the different types of triangles and their properties. We'll also discuss some of the common applications of triangles in math, science, and engineering.
Pain Points
When it comes to triangles, there are a few common pain points that people often experience. These include:
Not being able to identify the different types of triangles
Not understanding the properties of different types of triangles
Not being able to apply triangles to real-world problems
Solution
This blog post will help you to overcome these pain points by providing you with a clear and concise overview of the different types of triangles and their properties. We'll also provide you with some examples of how triangles are used in real-world applications.
Summary
In this blog post, we have discussed the different types of triangles and their properties. We have also provided some examples of how triangles are used in real-world applications. This information will help you to better understand triangles and their many uses.
The Enigmatic Triangle: Unveiling Its Geometrical Wonders
Introduction
Triangles, the fundamental building blocks of geometry, have captivated mathematicians, scientists, and artists for centuries. Their unique properties and versatile applications have made them indispensable in a myriad of fields. This comprehensive article delves into the fascinating world of triangles, exploring their various types, properties, and intriguing applications.
Types of Triangles
Triangles are classified into several types based on their side lengths and angle measures:
Equilateral Triangles: All three sides are of equal length, with angles of 60 degrees.
Isosceles Triangles: Two sides are of equal length, forming two congruent angles.
Scalene Triangles: All three sides are unequal, and all angles differ.
Properties of Triangles
Triangles possess several fundamental properties:
Sum of Interior Angles: The sum of the interior angles of a triangle is always 180 degrees.
Triangle Inequality Theorem: In any triangle, the sum of the lengths of any two sides is always greater than the length of the third side.
Perpendicular Bisector Theorem: The perpendicular bisector of a triangle's side passes through the triangle's vertex opposite to that side.
Special Triangles
Certain triangles exhibit distinctive properties:
Right Triangles: One angle is 90 degrees, forming a right angle.
Pythagorean Theorem: In a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.
Triangles in Art
Symbolism: Triangles have been used to represent various concepts, such as stability, balance, and the Holy Trinity.
Perspective: Triangles are used in perspective drawing to create the illusion of depth and three-dimensionality.
Triangles in Mathematics
Triangles are fundamental in mathematical concepts:
Trigonometry: The study of triangles and their angle relationships forms the basis of trigonometry.
Geometry: Triangles are used to solve problems involving area, perimeter, and angle measures.
Calculus: Triangles are used to approximate curves, calculate integrals, and derive derivatives.
Isosceles Triangle
An isosceles triangle has two equal sides and two equal angles. The third side, called the base, is unequal to the other two sides. The angles opposite the equal sides are also equal.
Scalene Triangle
A scalene triangle has no equal sides or angles. All three sides and all three angles are different.
Right Triangle
A right triangle has one angle that measures 90 degrees. The side opposite the right angle is called the hypotenuse. The other two sides are called the legs.
Equilateral Triangle
An equilateral triangle has three equal sides and three equal angles. All three angles measure 60 degrees.
Conclusion
Triangles, with their diverse properties and applications, are a testament to the beauty and versatility of geometry. Their ubiquity in nature, art, and mathematics highlights their fundamental role in shaping our understanding of the world around us. From towering structures to intricate designs, triangles continue to inspire and intrigue, serving as a cornerstone of scientific, artistic, and technological advancements.
Frequently Asked Questions
What is the most common type of triangle?
Scalene triangle
What is the sum of the interior angles of a triangle?
180 degrees
What is the Pythagorean Theorem?
In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.
Where can triangles be found in nature?
Leaf veins, flower petals, animal skeletons, and crystallography
How are triangles used in art?
To create dynamic compositions, convey symbolism, and enhance perspective | 677.169 | 1 |
The party hat II
An equilateral triangle with a circular arc through its red centre and two vertices. If the yellow area equals the green one, what's the angle α?
Scroll down for a solution to this problem.
Solution
The angle α is 60°.
Extend the circle and draw the inscribed regular hexagon. It is easy to see that ABCD is a parallelogram. From the condition yellow is green it follows that the extended common base DE cuts the diagonal AC in half. So it must be the other diagonal DB. Now, ∠BEC=∠BAC=120°, making α=60°. | 677.169 | 1 |
Calculate a point that is tangent to the circle and passes through the origin.
Possible Answers:
Correct answer:
Explanation:
To construct a line that is tangent to a point on the circle and passes through the origin, origin results in the followingLooking at the image, it is seen that the line only touches the circle once therefore, the line is tangent to the circle. Thus this statement is trueGiven a circle, a tangent line to the circle can be constructed if it intersects the circle at two points.
Possible Answers:
True
False
Correct answer:
False
Explanation:Therefore, looking at the graph
it is seen that the line only intersects the circle once. Thus, the statement "The line is tangent to the circle." is trueTherefore, the point on the circle that creates a tangent line with the point given is .Using the plotted circle one possible line would touch the circle | 677.169 | 1 |
Question Video: Identifying the Number of Polygons in a Figure
Mathematics • Third Year of Primary School
How many polygons are there in the following figure?
01:09
Video Transcript
How many polygons are there in the following figure?
A polygon is a plane shape with straight sides. How many of those plane shapes with straight sides are in this figure? Well here's a rectangle. It's a plane shape with straight sides. And here's a second rectangle.
What about this shape? Does this shape qualify as a polygon? No, these four shapes are not polygons. They can't qualify as polygons, because they have a curved side or a curved edge. However, there is actually one more polygon in this figure. If you take the two smaller rectangles and put them together, they make one larger rectangle, which accounts for the third polygon in this figure. Three polygons are in the figure. | 677.169 | 1 |
Practice (41)
Point $B$ lies on line segment $\overline{AC}$ with $AB=16$ and $BC=4$. Points $D$ and $E$ lie on the same side of line $AC$ forming equilateral triangles $\triangle ABD$ and $\triangle BCE$. Let $M$ be the midpoint of $\overline{AE}$, and $N$ be the midpoint of $\overline{CD}$. The area of $\triangle BMN$ is $x$. Find $x^2$.
Point $A,B,C,D,$ and $E$ are equally spaced on a minor arc of a cirle. Points $E,F,G,H,I$ and $A$ are equally spaced on a minor arc of a second circle with center $C$ as shown in the figure below. The angle $\angle ABD$ exceeds $\angle AHG$ by $12^\circ$. Find the degree measure of $\angle BAG$.
Triangle $ABC$ has positive integer side lengths with $AB=AC$. Let $I$ be the intersection of the bisectors of $\angle B$ and $\angle C$. Suppose $BI=8$. Find the smallest possible perimeter of $\triangle ABC$.
Triangle $ABC$ has side lengths $AB = 12$, $BC = 25$, and $CA = 17$. Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$, vertex $Q$ on $\overline{AC}$, and vertices $R$ and $S$ on $\overline{BC}$. In terms of the side length $PQ = w$, the area of $PQRS$ can be expressed as the quadratic polynomial Area($PQRS$) = $\alpha w - \beta \cdot w^2$.
Then the coefficient $\beta = \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
The circumcircle of acute $\triangle ABC$ has center $O$. The line passing through point $O$ perpendicular to $\overline{OB}$ intersects lines $AB$ and $BC$ and $P$ and $Q$, respectively. Also $AB=5$, $BC=4$, $BQ=4.5$, and $BP=\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$, respectively, and are externally tangent at point $A$. Point $B$ is on $mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$. Points $B$ and $C$ lie on the same side of $\ell$, and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
A disk with radius $1$ is externally tangent to a disk with radius $5$. Let $A$ be the point where the disks are tangent, $C$ be the center of the smaller disk, and $E$ be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of $360^\circ$. That is, if the center of the smaller disk has moved to the point $D$, and the point on the smaller disk that began at $A$ has now moved to point $B$, then $\overline{AC}$ is parallel to $\overline{BD}$. Then $sin^2(\angle BEA)=\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
On square $ABCD$, points $E,F,G$, and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34$. Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P$, and the areas of the quadrilaterals $AEPH, BFPE, CGPF,$ and $DHPG$ are in the ratio $269:275:405:411.$ Find the area of square $ABCD$.
A rectangle has sides of length $a$ and 36. A hinge is installed at each vertex of the rectangle, and at the midpoint of each side of length 36. The sides of length $a$ can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with the sides of length $a$ parallel and separated by a distance of 24, the hexagon has the same area as the original rectangle. Find $a^2$.
Circle $C$ with radius 2 has diameter $\overline{AB}$. Circle D is internally tangent to circle $C$ at $A$. Circle $E$ is internally tangent to circle $C$, externally tangent to circle $D$, and tangent to $\overline{AB}$. The radius of circle $D$ is three times the radius of circle $E$, and can be written in the form $\sqrt{m}-n$, where $m$ and $n$ are positive integers. Find $m+n$.
In $\triangle RED$, $\angle DRE=75^{\circ}$ and $\angle RED=45^{\circ}$. $|RD|=1$. Let $M$ be the midpoint of segment $\overline{RD}$. Point $C$ lies on side $\overline{ED}$ such that $\overline{RC}\perp\overline{EM}$. Extend segment $\overline{DE}$ through $E$ to point $A$ such that $CA=AR$. Then $AE=\frac{a-\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer. Find $a+b+c$.
In $\triangle{ABC}, AB=10, \angle{A}=30^{\circ}$, and $\angle{C=45^{\circ}}$. Let $H, D,$ and $M$ be points on the line $BC$ such that $AH\perp{BC}$, $\angle{BAD}=\angle{CAD}$, and $BM=CM$. Point $N$ is the midpoint of the segment $HM$, and point $P$ is on ray $AD$ such that $PN\perp{BC}$. Then $AP^2=\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Let $ABCD$ be a square, and let $E$ and $F$ be points on $\overline{AB}$ and $\overline{BC},$ respectively. The line through $E$ parallel to $\overline{BC}$ and the line through $F$ parallel to $\overline{AB}$ divide $ABCD$ into two squares and two nonsquare rectangles. The sum of the areas of the two squares is $\frac{9}{10}$ of the area of square $ABCD.$ Find $\frac{AE}{EB} + \frac{EB}{AE}.$
A paper equilateral triangle $ABC$ has side length $12$. The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$. The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$, where $m$, $n$, and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p$.
Let $\bigtriangleup PQR$ be a triangle with $\angle P = 75^\circ$ and $\angle Q = 60^\circ$. A regular hexagon $ABCDEF$ with side length 1 is drawn inside $\triangle PQR$ so that side $\overline{AB}$ lies on $\overline{PQ}$, side $\overline{CD}$ lies on $\overline{QR}$, and one of the remaining vertices lies on $\overline{RP}$. There are positive integers $a, b, c,$ and $d$ such that the area of $\triangle PQR$ can be expressed in the form $\frac{a+b\sqrt{c}}{d}$, where $a$ and $d$ are relatively prime, and c is not divisible by the square of any prime. Find $a+b+c+d$.
Triangle $AB_0C_0$ has side lengths $AB_0 = 12$, $B_0C_0 = 17$, and $C_0A = 25$. For each positive integer $n$, points $B_n$ and $C_n$ are located on $\overline{AB_{n-1}}$ and $\overline{AC_{n-1}}$, respectively, creating three similar triangles $\triangle AB_nC_n \sim \triangle B_{n-1}C_nC_{n-1} \sim \triangle AB_{n-1}C_{n-1}$. The area of the union of all triangles $B_{n-1}C_nB_n$ for $n\geq1$ can be expressed as $\tfrac pq$, where $p$ and $q$ are relatively prime positive integers. Find $q$.
In equilateral $\triangle ABC$ let points $D$ and $E$ trisect $\overline{BC}$. Then $\sin(\angle DAE)$ can be expressed in the form $\frac{a\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is an integer that is not divisible by the square of any prime. Find $a+b+c$.
A hexagon that is inscribed in a circle has side lengths $22$, $22$, $20$, $22$, $22$, and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$, where $p$ and $q$ are positive integers. Find $p+q$.
Given a circle of radius $\sqrt{13}$, let $A$ be a point at a distance $4 + \sqrt{13}$ from the center $O$ of the circle. Let $B$ be the point on the circle nearest to point $A$. A line passing through the point $A$ intersects the circle at points $K$ and $L$. The maximum possible area for $\triangle BKL$ can be written in the form $\frac{a - b\sqrt{c}}{d}$, where $a$, $b$, $c$, and $d$ are positive integers, $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$.
In $\triangle ABC$, $AC = BC$, and point $D$ is on $\overline{BC}$ so that $CD = 3\cdot BD$. Let $E$ be the midpoint of $\overline{AD}$. Given that $CE = \sqrt{7}$ and $BE = 3$, the area of $\triangle ABC$ can be expressed in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n$.
Let $\triangle ABC$ be a right triangle with right angle at $C.$ Let $D$ and $E$ be points on $\overline{AB}$ with $D$ between $A$ and $E$ such that $\overline{CD}$ and $\overline{CE}$ trisect $\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m$ and $n$ are relatively prime positive integers, and $p$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$
Three concentric circles have radii $3,$ $4,$ and $5.$ An equilateral triangle with one vertex on each circle has side length $s.$ The largest possible area of the triangle can be written as $a + \tfrac{b}{c} \sqrt{d},$ where $a,$ $b,$ $c,$ and $d$ are positive integers, $b$ and $c$ are relatively prime, and $d$ is not divisible by the square of any prime. Find $a+b+c+d.$
Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$, $BC=7$, and $AC=3$. The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$. Let $\gamma$ be the circle with diameter $\overline{DE}$. Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$. Then $AF^2 = \frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
In rectangle $ABCD$, $AB=12$ and $BC=10$. Points $E$ and $F$ lie inside rectangle $ABCD$ so that $BE=9$,$DF=8$,$\overline{BE}||\overline{DF}$,$\overline{EF}||\overline{AB}$, and line $BE$ intersects segment $\overline{AD}$. The length $EF$ can be expressed in the form $m\sqrt{n}-p$, where $m$,$n$, and $p$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n+p$. | 677.169 | 1 |
Teach your students to identify and calculate supplementary angles with this one-page maths worksheet.
Looking for a Supplementary Angles Worksheet?
Supplementary angles add up to 180°. This means that if you were to divide a straight angle into several smaller angles, the total sum of the supplementary angles would be 180°.
This worksheet provides students with the opportunity to calculate the size of the missing supplementary angle on a variety of straight angles.
An answer key is included with your download to make grading fast and easy!
Tips for Differentiation + Scaffolding
In addition to independent student work time, use this worksheet as an activity for:
Whole-class review (via smartboard)
Guided maths groups
Independent practice
Homework assignment
Summative assessment
Some students may need a reminder of what complementary angles are when completing the worksheet. Our Complementary and Supplementary Angles Poster is the perfect resource to paste into your students' workbooks as a visual guide to refer back to while completing the worksheet.
Download This Printable Angles Worksheet
Use the Download button to access the easy-print black-and-white PDF.
Because this resource includes an answer sheet, we recommend you print one copy of the entire file, then make photocopies of the blank worksheet for students to complete.
Looking for a more sustainable way to use this resource? Project the worksheet onto a screen and work through it as a class by having students record their answers in their notebooks.
Click below for more time-saving angles resources to cut down on your planning time! | 677.169 | 1 |
Math Games
Square Graph Paper
Download and print all the square graph paper you want.
By Michael Hartley
There are lots of ways to tessellate the plane. The most regular tessellations are with squares, hexagons or triangles. The tessellations with triangles and hexagons form a pair - if you draw a dot in the center of each hexagon, and join neighboring dots, you get the tessellation with triangles, and vice-versa. The tessellation with squares stands alone - of you draw a dot at the center of each square and join neighboring dots, you get the same square tessellation, shifted up and across by half an edge length!
It's easy to see why these are the only tessellations with regular figures - the angles inside a regular polygon with n sides are 180-360/n degrees each. For the triangle, with n=3, these angles are 180-120=60 degrees, and six of these angles fit around a point. For the square, the angles are 90 degrees, and you get four squares around a point because 4 times 90 degrees is 360 degrees. Likewise, you can fit three regular hexagons around a point. For any other regular polygon, the angles at each corner don't go evenly into 360 degrees, so they can't be arranged to fit around even a single corner, let alone a whole plane.
If you let yourself use more than one type of regular polygon, you can fill the plane in many more ways. You can download and print out some of these at my page of archimedean graph paper. These patterns are very pretty.
There are all sorts of ways you can use square graph paper - obviously, you can use it to draw graphs! When moving house once, I drew up the plans for our new house on graph paper and made cut-out cardboard scale copies of our furniture. Sitting down with a supply of Blu Tack, my wife and I planned where all our furniture would go. It made moving day so much simpler! Many games are played on a square grid. You could use these printable square graph paper downloads to print yourself some game boards.
In fact, I'm sure you already have an idea in mind, so without further ado, here's your download links : | 677.169 | 1 |
Number of edges formed inside a right triangle by the straight line segments mutually connecting all vertices and points on the two shorter edges whose positions on one edge equal the Farey series of order n while on the other they divide its length into n equal segments. | 677.169 | 1 |
Dentro del libro
Resultados 1-5 de 100
Pįgina 16 ... third sides , equal ; and the two triangles shall be equal ; and their other angles shall be equal , each to each , viz . those to which the equal sides are opposite . Let ABC , DEF be two triangles which have the two sides AB , AC ...
Pįgina 28 ... third side . Let ABC be a triangle ; any two sides of it together are greater than the third side , viz . the sides BA , AC greater than the side BC ; and AB , BC greater than AC ; and BC , CA great- er than AB . Produce BA to the point ...
Pįgina 29 ... third . " Let A , B , C be the three given straight lines , of which any two whatever are greater than the third , viz . A and B greater than C ; A and C greater than B ; and B and C than A. It is required to make a triangle of which ...
Pįgina 32 ... third angle of the other . A Let ABC , DEF be two triangles which have the angles ABC , BCA equal to the angles DEF ... third angle BAC to the third angle EDF . G B C E F For , if AB be not equal to DE , one of them must be the greater ...
Pįgina 33 ... third angle BAC to the third angle EDF . Next , let the sides A which are opposite to equal angles in each triangle be equal to one another , viz . AB to DE ; likewise in this case , the other sides shall be equal , AC to D DF , and BC ...
Pasajes populares
Pįgina 17 - FG; then, upon the same base EF, and upon the same side of it, there can be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise their sides terminated in the other extremity: But this is impossible (i.
Pįgina 67 - Ir any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle. Let ABC be a circle, and A, B any two points in the circumference ; the straight line drawn from A to B shall fall within the circle.
Pįgina 92 - IF a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle, the angles made by this line with the line touching the circle, shall be equal to the angles which are in the alternate segments of the circle.
Pįgina 26 - If from the ends of a side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the triangle, but shall contain a greater angle.
Pįgina 55 - If a straight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the square of the other part, is equal to the square of the straight line, which is made of the whole and that part.
Pįgina 318 - 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.
Pįgina 22 161Pįgina 21 - When a straight line standing on another straight line, makes the adjacent angles equal to one another, each of the angles is called a right angle ; and the straight line which stands on the other is called a perpendicular to it.
Información bibliogrįfica
Tķtulo | 677.169 | 1 |
Categories
The Basics of Quadrilaterals: Understanding Shapes in Mathematics
Introduction to Quadrilaterals
Quadrilaterals, a fundamental concept in geometry, refer to a class of shapes known for having four sides. These shapes are ubiquitous, not only in mathematical textbooks but also in our daily surroundings, from the design of a window pane to the layout of a piece of land. Understanding quadrilaterals is not only a stepping stone in the world of geometry but also a critical skill in various practical applications. This article delves into the basics of quadrilaterals, their properties, and the diverse types that exist within this category.
Defining Quadrilaterals
A quadrilateral is any polygon with four edges (or sides) and four vertices (or corners). The sum of the interior angles in any quadrilateral is always 360 degrees. This characteristic is essential for solving numerous geometrical problems and is a key attribute that helps in identifying various properties of different quadrilaterals.
Properties of Quadrilaterals
While all quadrilaterals share some basic properties, such as having four sides and the sum of internal angles equalling 360 degrees, they can be further categorized by additional specific features. These include the lengths of sides, whether or not sides are parallel or equal, and the types of angles present. Based on these properties, quadrilaterals are divided into several types, each with unique characteristics.
Parallel Sides
Parallel sides are a significant property for classifying quadrilaterals. For instance, in parallelograms, both pairs of opposite sides are parallel. This property is crucial for identifying rectangles, squares, and rhombi, all of which are types of parallelograms.
Angle Measures
The measurement of angles is another way to categorize quadrilaterals. Right angles are a notable feature in rectangles and squares, determining their shape and properties.
Types of Quadrilaterals
Quadrilaterals are categorized into several types, primarily based on their sides, angles, and other geometric properties. Some of the most common types include:
Parallelograms
Characterized by pairs of parallel sides, parallelograms include several well-known shapes such as rectangles, rhombi (pl. for rhombus), and squares. Each shape within this category has its unique properties. For example, all angles in a rectangle are right angles, whereas all sides in a rhombus are equal in length.
Trapezoids (or Trapezia)
Trapezoids are defined by having at least one pair of parallel sides. They differ from parallelograms in that only one pair of sides needs to be parallel. The distinguishing feature between different types of trapezoids (like isosceles trapezoid) revolves around the length of the non-parallel sides and the angles they form.
Kites
Kites are quadrilaterals with two pairs of adjacent sides that are equal. A notable property is that one pair of opposite angles is equal, which is directly attributable to its distinct side lengths. Kites are easily recognizable by their typical 'kite' shape, from which they derive their name.
Conclusion
The study of quadrilaterals opens up a fascinating world of geometry, revealing the diversity and complexity of shapes that can be created with just four sides. From practical applications in engineering and design to solving complex mathematical problems, the understanding of quadrilaterals is foundational. By grasping the basic properties and types of quadrilaterals, learners can build upon their geometrical knowledge, paving the way for more advanced studies in mathematics. | 677.169 | 1 |
If AA', BB' are two intersecting at P chords of a circle, then the common
chord of the circles with diameters AA', BB'
passes through P.
Discussion. [FvL, 2/7/02]: The
power of P w.r.t. the first circle is PA*PA'=PB*PB'. From that we see that
the powers of P w.r.t. the circles with diameters AA' and BB' are equal.
So P lies on the radical axis of these two circles, which is the line containing
their common chord. Since P lies in the interior of both circles, P lies
on the chord itself. | 677.169 | 1 |
Standard K.G.B.4 - Practice identifying three dimensional shapes based on their name/'corners') and other attributes (e.g., having sides of equal length). | 677.169 | 1 |
Eureka Math Precalculus Module 2 Lesson 19 Answer Key
Engage NY Eureka Math Precalculus Module 2 Lesson 19 Answer Key
Eureka Math Precalculus Module 2 Lesson 19 Example Answer Key
Example: The Parallelogram Rule for Vector Addition
When the initial point of a vector is the origin, then the coordinates of the terminal point will correspond to the horizontal and vertical components of the vector. This type of vector, with initial point at the origin, is often called a position vector.
a. Draw arrows to represent the vectors v = 〈5, 3〉 and u = 〈1, 7〉 with the initial point of each vector at (0, 0).
b. Add v + u end – to – end. What is v + u? Draw the arrow that represents v + u with initial point at the origin.
c. Add u + v end – to – end. What is u + v? Draw the arrow that represents u + v with an initial point at the origin.
Answer:
Exercise 3.
Write a rule for the component form of the vector v shown in the diagram. Explain how you got your answer.
Answer:
The component form is v = 〈x2 – x1, y2 – y1〉. The components of the vector are the distance A is translated vertically and horizontally to get to B. To find the horizontal distance from A to B, subtract the x – coordinates. To find the vertical distance from A to B, subtract the y – coordinates.
When we use the initial and terminal points to describe a vector, we often refer to the vector as a directed line segment. A vector or directed line segment with initial point A and terminal point B is denoted \(\overrightarrow{A B}\).
Exercise 5.
Consider points P(2, 1), Q( – 3, 3), and R(1, 4).
a. Compute \(\overrightarrow{P Q}\) and \(\overrightarrow{Q P}\) and show that \(\overrightarrow{P Q}\) + \(\overrightarrow{Q P}\) is the zero vector. Draw a diagram to show why this makes sense geometrically.
Answer:
\(\overrightarrow{P Q}\) = 〈 – 5, 2〉 and \(\overrightarrow{Q P}\) = 〈5, – 2〉
\(\overrightarrow{P Q}\) + \(\overrightarrow{Q P}\) = 〈 – 5 + 5, 2 – 2〉 = 〈0, 0〉
This makes sense because one vector translates P to Q, and then the other vector translates the point back from Q to P. The sum would be the zero vector because the original point is returned to its starting location.
b. Plot the points P, Q, and R. Use the diagram to explain why \(\overrightarrow{P Q}\) + \(\overrightarrow{Q R}\) + \(\overrightarrow{R P}\) is the zero vector. Show that this is true by computing the sum \(\overrightarrow{P Q}\) + \(\overrightarrow{Q R}\) + \(\overrightarrow{R P}\).
Answer:
The diagram shows that the sum should be the zero vector because point P is translated back to its original coordinates when you add the three vectors.
\(\overrightarrow{P Q}\) + \(\overrightarrow{Q R}\) + \(\overrightarrow{R P}\) = 〈 – 5 + 4 + 1, 2 + 1 – 3〉 = 〈0, 0〉
b. Draw a diagram representing the vectors as arrows placed end – to – end to support why their sum would be the zero vector.
Answer:
If we position v to have an initial point at (0, 0) then v translates the point (0, 0) to (2, – 3). Each additional vector translates the point by its components. The result of all four translations returns the point back to the origin.
i. How do the magnitudes of these vectors compare to one another and to that of v?
Answer:
The magnitude of 3v and – 3v are the same. The magnitude of these vectors is 3 times the magnitude of v. You can see that because scalar multiplication dilates the vector by the scalar multiple.
ii. How do the directions of 3v and – 3v compare to the direction of v?
Answer:
When the scalar is positive, the direction of the scalar multiple is the same going along v. When the scalar is negative, the direction is opposite the direction of v.
Question 6.
Consider Example 5, part (b) in the lesson and Problem 5, part (a) above. What can you conclude about three vectors that form a triangle when placed tip – to – tail? Explain by graphing.
Answer:
If three vectors form a triangle when placed tip – to – tail, then the sum of those three vectors is zero.
b. Is vector u equal to vector c?
Answer:
Yes, u = 〈2, 1〉 = c, they have the same components and direction.
c. Jens says that if two vectors u and v have the same initial point A and lie on the same line, then one vector is a scalar multiple of the other. Do you agree with him? Explain how you know. Give an example to support your answer.
Answer:
Yes. Suppose that the terminal point of u is C and the terminal point of v is B. Then u has magnitude ‖AC‖, where ‖AC‖ ≠ 0, and v has magnitude ‖AB‖, where‖AB‖≠0. If u and v point in the same direction, then v = \(\left(\frac{\| A B}{\|A C\|}\right) \mathrm{u}\). If u and v point in opposite directions, then v = – \(\left(\frac{\| A B}{\|A C\|}\right) \mathrm{u}\). For example in the diagram above, vector u = \(\overrightarrow{A C}\) = 〈2, 1〉 and vector v = \(\overrightarrow{A B}\) = 〈8, 4〉, and we have v = \(\frac{\sqrt{80}}{\sqrt{5}} \mathrm{u}\) = 4u.
Question 2.
For vectors u and v as in Question 1, explain how to find u + v using the parallelogram rule. Support your answer graphically below.
Answer:
To find u + v using the parallelogram rule, form a parallelogram with vectors u and v as sides. The diagonal of the parallelogram is the sum of the two vectors. | 677.169 | 1 |
Question Category: Practice Set Part 3.2
(i)Circles touching externally If the circles touch each other externally, distance between their centres is equal to the sum of their radii.Distance between the centres = 4+2.8 = 6.8cm(ii)Circles touching internally The distance between the centres of the circles touching internally is equal to the difference of their radii.Distance between the centres = 4-2.8 = […]
The distance between the centres of the circles touching internally is equal to the difference of their radii.We have r1 = 3.5 and r2 = 4.8Distance between their centres = 4.8-3.5 = 1.3 | 677.169 | 1 |
MATHEMATICS O LEVEL(FORM THREE) NOTES – THE EARTH AS A SPHERE
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THE EARTH AS A SPHERE
The earth surface is very close to being a sphere. Consider a sphere representing the shape of the earth as below.
NS is the Axis of the earth in which the earth rotates once a day.
O is the center of the earth.
The radius of the Earth is 6370 Km.
GREAT CIRCLES.
Is that which is formed on the surface of the earth by a plane passing through the center of the Earth. Its radius is equal to the radius of the Earth.
Examples; ARB and ANB.
The equator is also a great circle.
NDBS, NGS and NPRS all are meridians. (Longitude).
EQUATOR: Is the line that circles the Earth midway between the North and the South Pole.
PRIME MERIDIAN (: is a meridian (longitude) which passes through Greenwich, England.
SMALL CIRCLES : is one formed on the surface of the Earth by a plane that cuts through the Earth but does not pass through the centre of the Earth. Eg. ARB.
LATITUDE: The angular distance of a place north to south of the earth's equator or of a celestial object north or south of the celestial equator. It is measured on the meridian of the point.
Example 1. CASE 1
1.1. Points A and B have the same longitudes but different latitudes. Point A (66o N, 400E) B, Point B ( 25oN ,400E)A. Find the angle subtended at the center of the Earth by arc AB if A is (250N,400E) and B is (66os ,400E)
2
AOB =BOP- AOP
AOB =660– 25o =410
Exercise 1
In questions 1 to 4 consider the town and cities indicated an d the answer the questions that follow.
4. Find the angle subtended at the center of the Earth by arc AB if A is Mwanza and B is Mbeya. 5. Find the angle subtended at the centre of the Earth by arc XY if X is Nakuru and Y is Kampala Solution
1. A Mwanza (30 S, 330 E)
B Mbeya (90 S, 330 E)
SOB = BOA- AOS
SOB= 90 – 30
SOB = 60
5. c.X Nakuru ( 00 , 360 E)
d.Y Kampala ( 00 , 330 E)
e.TOX = XOY –TOY
f.TOX= 360 – 33 0
g.TOX = 30
CASE 2
Point A and B are on the same latitude but different longitudes.
Examples
1.Find the angle subtended at the center of the Earth by arc XY if X is Nakuru (00 , 360 E) and Y is Kampala ( 00 , 330 E)
XOY = XOP- YOP
= 360 – 330
= 30
2.Two towns A and B are on the equator.The longitude of A is 350 E and that of B is 720 W. Find the angle subtended by the arc AB at the center of the Earth. Solution
AOB = AOP + POB
= 350 + 720
= 1070
3.Two towns A and B in Africa are located on the Equator. The longitude of A is 100 E and that of B is 420 E . Find the angle subtended by the arc AB at the center of the Earth. Solution
AOB = SOB – SOA
= 420 – 100
= 320
The angle formed by arc AB is 320
Exercise 2
1. In the figure below, if the center through N,G,S is the prime meridian , the center of the Earth and the Equator passes through B and G , the longitude and latitude of A.
Longitude of A = ( 00 , 300 W)
Latitude of A = ( 500 N, 00 )
There after draw a figure similar to that of question 1 to illustrate the position of point H (600S,450E).
2. P and Q are towns on latitude 00 . if the longitude of P is 1160 E and that of Q is 1050 W, find the angle subtended by the arc connecting the two places at the center of the Earth . Draw a figure to illustrate their positions.
LENGTH OF A GREAT CIRCLE.
Distance on the surface of the Earth are usually expressed in nautical miles or in kilometers. A nautical mile is the length of an arc of a great circle that subtends an angle of 1 minute at the center of the Earth.
1.If the latitude of Nakuru is O0 , find the distance ( length) in nautical miles from this town to the North Pole.
Solution.
From O0 to North Pole , the angle is 900
10 = 600 nautical miles.
900 = ?
900 x 600 = 5400 nautical miles
1 nautical mile = 1.852 km 5400 nautical mile = ? km
Distance = 1. 852 x 5400 = 10000.8km
2.calculate the distance of the prime meridian from south to North pole in
a. nautical mile
b. kilometers.
Solution
a.10 = 60 nautical miles.
1800 = ? nautical miles
1800 x 600 = 10800 nautical miles. ... The distance of prime meridian from south to north pole is 10800Nm
b.1 nautical mile = 1.852 km
10800 Nautical miles =?km
10800x 1.852= 20001.6 km ... The distance of prime meridian from south to north pole is 20001.6 km
3.Calculate the distance of the equator from east to West in Nautical Miles.
10 = 60nautical miles.
180o ?
1800 x 60 0= 10800 nautical miles.
... The distance of the equator from East to West in Nm is given by 10800Nm
Length of small circle.
Let P be any point on the surface of the earth through this point a small circle is drawn with parallel of latitude θº as shown above. The radius of the earth as R and the radius of the parallel latitude (r) are both perpendicular to the polar axis. Note: SP is parallel to OQ (Both are perpendicular to NS) ...θº = OPS Then we have From trigonometrical ratios r= R cos θ where R is the radius of the earth and r is the radius of the small circle of latitude θ. ... Distance of parallel of latitude θ = 2πr = 2πRcos θ
Example 1.
1.Calculate the circumference of a small circle in kilometers along the parallel of latitude 100 S.
Soln
C = 2πR cos θ
= 2x 3.14x 6370x cos 100
=39395.54528km
2.Calculate the length of the parallel of the latitude through Bombay If Bombay is located 190 N, 730 E
C= 2πR cos θ
= 2×3.14x6370xcos190
=37823.40km
In nautical miles.
3782km/1.852km/miles = 20423 nautical miles.
Exercise
In the questions below , take the radius of the Earth , R= 6370km and π = 3.14
1. The city of Kampala lies along the equator. Calculate the distance in kilometers from the city of Kampala to the South Pole 2. How far is B from A if A is 00 ,00 and B is 00, 1800 E. 3. What is the latitude of a point P north of the Equator if the length of the parallel of the latitude through p is 28287 kilometers.( give your answer to the nearest degree. 4.What is the radius of a small circle parallel to the equator along latitude 700 N
Consider the figure below, points A and B are two points having the same latitude 00 , since both lie on the parallel of latitude but they are different in their longitudes i.e. That point A is on a different longitude from that of point B. the difference between their longitudes is θ.
3600 = 2πr θ = ?
Example.
1. A ship is streaming in a western direction from Q and P. if the position of P is ( 400 S, 1780E) and that of Q ( 400 S, 1720E). how far does the ship move from Q to P?
Solution
Difference in longitude = 178-172= 60
Exercise
1. Two points on latitude 50 0 N lie on longitudes 350E and 40 0W. what is the distance between them in nautical miles. 2. An airplane flies westwards along the parallel of latitude 200 N from town A on longitude 400 E to town B on a longitude 100W. find the distance between the two towns in kilometers. 3. An aeroplane flies from Tabora ( 50S, 330E) to Tanga ( 50S, 390E) at 332 kilometers per hour along a parallel of latitude. If it leaves Tabora at 3 pm, find the arrival time at Tanga airport? 4. The location of Morogoro is (70 S , 380E) and that of Dar-Es-Salaam is ( 70S, 390E). find the distance between them In kilometers. 5. A ship after sailing for 864 nautical miles eastwards find that her longitude was altered by 300. What parallel of the latitude is the ship sailing? 6. An aeroplane takes off from B (550 S, 330E) to C ( 550 S, 390E) at a speed of 332 km/ hr . if it leaves B at 3:00 pm , at what time will it arrive at C airport? 7. A ship sails due North from latitude 200 S for a distance of 1440km. find the latitude of the point it reaches. | 677.169 | 1 |
Ex 5.1, 1 (v) - Chapter 5 Class 9 Introduction to Euclid's Geometry
Last updated at April 16, 2024 by Teachoo
Transcript
Ex 5.1, 1
Which of the following statements are true and which are false? Give reasons for your answers.
(v) In the following figure, if AB = PQ and PQ = XY, then AB = XY.
According to Euclid's first axiom , things which are equal to the same thing are equal to one another
So, the given statement is true | 677.169 | 1 |
What is the measure of this angle?
Find an answer to your question ✅ "What is the measure of this angle? ..." in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions. | 677.169 | 1 |
Properties Of Quadrilaterals Worksheet
Showing top 8 worksheets in the category properties of quadrilaterals. Perimeter of quadrilaterals worksheets calculate the perimeter of various special quadrilaterals like squares rectangles parallelograms rhombuses kites and trapezoids with this array of worksheets with dimensions presented as integers and decimals comprehend the congruent property and solve algebraic expressions to find the side length.
Number of parallel sides.
Properties of quadrilaterals worksheet. Rectangle parallelogram trapezoid rhombus square and kite. Showing top 8 worksheets in the category properties of quadrilaterals. In this worksheet you will be asked about the following quadrilaterals.
Classifying quadrilaterals worksheets quadrilaterals are classified by their properties e g. There are many different types of quadrilaterals which are classified by their sides and angles. Some
In these worksheets students describe the properties of different types of quadrilaterals. Some of the worksheets for this concept are properties of quadrilaterals classifying quadrilaterals date period polygons quadrilaterals and special parallelograms geometry work quadrilaterals performance based learning and assessment task properties properties of quadrilaterals fifth grade. Quadrilaterals are figures with four sides and four angles.
Some Displaying top 8 worksheets found for properties of quadrilater | 677.169 | 1 |
What Does Negative Longitude Mean
What Does Negative Longitude Mean?
Note that negative latitudes represent the southern hemisphere and negative longitudes represent the western hemisphere. … In our example 0.3069° times 60 gives us 18.42 minutes latitude while 0.8583° times 60 gives us 51.50 minutes of longitude.Nov 25 2014
Can longitude lines be negative?
Each line of longitude runs north and south and measures the number of degrees east or west of the Prime Meridian. Values range from positive 180 to negative 180 degrees.
Is north longitude positive or negative?
Latitude/longitude coordinates follow a regular Cartesian system starting from the Equator and the Prime Meridian. Positive coordinates are to the right (east) and up (north). This means that the Western Hemisphere has negative longitudes.
How do you know if longitude is east or west?
To measure longitude east or west of the Prime Meridian there are 180 vertical longitude lines east of the Prime Meridian and 180 vertical longitude lines west of the Prime Meridian so longitude locations are given as __ degrees east or __ degrees west.
Is north latitude or longitude?
If you take a look at a map or globe of the world you may notice lines running east-west and north-south. The lines run east-west are known as lines of latitude. The lines running north-south are known as lines of longitude.
See alsowhen meters are longer and more complex we use the term
Why is longitude a negative number?
If the number is negative it represents south of the Equator. The line of longitude is read as 2 degrees (2) 10.4418 minutes (10.4418) east. The coordinate for the line of longitude represents east of the Prime Meridian because it is positive. If the number is negative it represents west of the Prime Meridian.
How do you write negative longitude?
If the number is negative such as -73.9917° (the minus is always shown if the number is negative) then the longitude is WEST!" "OK!"
What is negative latitude?
Positive latitudes are north of the equator negative latitudes are south of the equator. Positive longitudes are east of the Prime Meridian negative longitudes are west of the Prime Meridian. Latitude and longitude are usually expressed in that sequence latitude before longitude.
Does latitude go first?
Handy tip: when giving a co-ordinate latitude (north or south) always precedes longitude (east or west). Latitude and longitude are divided in degrees (°) minutes (') and seconds ("). There are 60 minutes in a degree and 60 seconds in a minute (similar to measuring time).How do you know if longitude is north or south?
2 Answers. Latitude is positive in north and negative in south. Similarly Longitude is positive in east and negative in west.
What is difference latitude andWho bisects earth?
The Equator or line of 0 degrees latitude divides the Earth into the Northern and Southern hemispheresSee alsohow do you break up hardened rock salt
Is 0 degrees north or south?
The latitude which most people are familiar with is the equator. This is 0 degrees latitude and it divides the world into the Northern and Southern hemispheres. The North Pole is referred to as latitude 90 degrees north. Lines of Longitude are referred to as Meridians of Longitude.
Which is negative longitude or latitude
In which direction latitude of line is negative?
In which direction latitude of the line is negative? Explanation: The latitude of the line is negative when measured southward. It is termed as southing.
Can minutes be negative?
Negative minutes and Negative seconds seems the most appropriate path placing the negative on the degrees then the minutes and then the seconds. Because putting a negative on a zero is wrong put it in on the minute if it is not zero first and then the second if both degree and minute are zero.
What is latitude and longitude format?
Latitude and longitude are a pair of numbers (coordinates) used to describe a position on the plane of a geographic coordinate system. The numbers are in decimal degrees format and range from -90 to 90 for latitude and -180 to 180 for longitude. For example Washington DC has a latitude 38.8951 and longitude -77.0364 .
How many longitudes are there?
360
Lines of latitude are known as parallels and there are 180 degrees of latitude in total. The total number of latitudes is also 180 the total number of longitudes is 360.
What is the meaning of 8 4 N?
It means 8 degrees 4 minutes North Latitude. The terminology is somewhat archaic and dates back to the days of sailing ships.
What is an example of longitude?
An example of longitude is that New York City is at 74° degrees west and local time is 5 hours less than Greenwich mean time. Angular distance measured west or east of the prime meridian.
How was longitude discovered?
Eratosthenes in the 3rd century BCE first proposed a system of latitude and longitude for a map of the world. … He also proposed a method of determining longitude by comparing the local time of a lunar eclipse at two different places to obtain the difference in longitude between them.
Is latitude north or west?
Lines of latitude run east and west. Lines of latitude run horizontally around the Earth and tell you how far north or south you are from the Equator.
Why do lines of latitude never intersect?
Circles of latitude are often called parallels because they are parallel to each other that is planes that contain any of these circles never intersect each other. A location's position along a circle of latitude is given by its longitude. … A circle of latitude is perpendicular to all meridians.
See alsohow many miles wide is the us
How many longitudes are there in India?
Therefore the total number of latitudes are 181 and the total number of longitudes are 360.
How do you speak longitude?
What is latitude on a map?
Latitude is a measurement on a globe or map of location north or south of the Equator. … As aids to indicate different latitudinal positions on maps or globes equidistant circles are plotted and drawn parallel to the Equator and each other they are known as parallels or parallels of latitude.
Which way does latitude run?
Have students look at the U.S. map and find the lines running across and up and down the page. Tell students that the lines running across the page are lines of latitude and the lines running up and down the page are lines of longitude. Latitude runs 0–90° north and south. Longitude runs 0–180° east and west.
What is 0 degree meridian is called?
The prime meridian is the line of 0° longitude the starting point for measuring distance both east and west around the Earth
What is meant by great circle?
A great circle is the largest possible circle that can be drawn around a sphere. … The Equator is another of the Earth's great circles. If you were to cut into the Earth right on its Equator you'd have two equal halves: the Northern and Southern Hemispheres. The Equator is the only east-west line that is a great circle.
Why do we have 360 longitudes and only 181 latitudes?
Longitude lines runs from North to south pole means a complete circles and hence covers 360 degrees and that is why there are 360 longitudes. … Each section from equator to pole is 90 degrees and two poles have 2 quarters of circle/globe hence 90X2 180 lattitudes. adding the Equator it becomes 181 latitudes. | 677.169 | 1 |
Page 1 ... right triangle is one which has a right angle , all other spherical triangles are called oblique . We shall in spherical trigonometry , as we did in 1 plane trigonometry , attend first to the solution of right SPHERICAL TRIGONOMETRY. ...
Page 2 ... angles of a spherical right tri- angle . Solution . The importance of this problem is obvi- ous ; for , unless some relations were known between the sides and the angles , they could not be determin- ed from each other , and there could ...
Page 3 ... right ...
Page 5 ... angle A , and the opposite side a , leads to the following corresponding one between h , the angle B , and the ... right trianGLES . 5 B'A' | 677.169 | 1 |
Exploring the Essence of Sine, Cosine, and Tangent Functions
Sine, cosine, and tangent are three fundamental trigonometric functions that play a crucial role in mathematics, physics, engineering, and many other disciplines. In this comprehensive guide, we will delve into the essence of these functions, exploring their definitions, properties, and applications. Join us on a journey to uncover the beauty and significance of sine, cosine, and tangent in the world of mathematics and beyond.
Understanding Sine, Cosine, and Tangent
Defining Trigonometric Functions
Sine, cosine, and tangent are trigonometric functions that relate the angles of a right triangle to the lengths of its sides. These functions are derived from the ratios of the sides of a right triangle and are fundamental tools for solving geometric problems involving angles and distances.
Sine Function (sinθ)
The sine function, denoted as sinθ, is defined as the ratio of the length of the side opposite the angle (opposite side) to the length of the hypotenuse in a right triangle. In other words, sinθ = opposite/hypotenuse. The sine function varies between -1 and 1 as the angle θ changes from 0 to 90 degrees (or 0 to π/2 radians).
Cosine Function (cosθ)
The cosine function, denoted as cosθ, is defined as the ratio of the length of the adjacent side to the length of the hypotenuse in a right triangle. In mathematical terms, cosθ = adjacent/hypotenuse. Like the sine function, the cosine function also varies between -1 and 1 as the angle θ changes.
Tangent Function (tanθ)
The tangent function, denoted as tanθ, is defined as the ratio of the length of the side opposite the angle to the length of the adjacent side in a right triangle. Symbolically, tanθ = opposite/adjacent. Unlike sine and cosine, the tangent function is unbounded and can take on any real value as the angle θ changes.
Properties and Graphs of Trigonometric Functions
Periodicity and Symmetry
All three trigonometric functions—sine, cosine, and tangent—are periodic, meaning their values repeat in a regular pattern as the angle varies. The period of sine and cosine functions is 360 degrees (or 2π radians), while the period of the tangent function is 180 degrees (or π radians). Additionally, the cosine function is an even function, whereas the sine and tangent functions are odd functions.
Graphical Representation
The graphs of sine, cosine, and tangent functions exhibit distinct patterns that reflect their respective properties. The sine function graph oscillates between -1 and 1, forming a wave-like pattern with peaks and troughs. The cosine function graph is similar to the sine function graph but shifted horizontally by π/2 radians (or 90 degrees). The tangent function graph consists of repeating vertical asymptotes and periodic branches that extend to positive and negative infinity.
Applications of Trigonometric Functions
Geometry and Trigonometry
Trigonometric functions are extensively used in geometry and trigonometry to solve problems involving angles, distances, and triangles. By applying trigonometric ratios and identities, mathematicians can determine unknown side lengths, angles, and area in geometric figures.
Physics and Engineering
In physics and engineering, trigonometric functions are used to analyze and model various phenomena, including motion, waves, and oscillations. From calculating the trajectory of a projectile to designing electrical circuits, engineers rely on trigonometry to solve complex problems and optimize designs.
Navigation and Astronomy
Trigonometric functions play a crucial role in navigation and astronomy, where they are used to determine the positions, distances, and trajectories of celestial objects and spacecraft. Navigators and astronomers use trigonometric principles to navigate ships, aeroplanes, and satellites and to study the motion of planets and stars.
Art and Music
Trigonometric functions also find applications in art and music, where they are used to create visual patterns, sculptures, and sound waves. Artists and musicians leverage the mathematical properties of trigonometric functions to generate intricate designs, harmonious compositions, and immersive experiences for audiences.
FAQs about Sine, Cosine, and Tangent Functions
How are sine, cosine, and tangent related to each other?
Sine, cosine, and tangent are related to each other through trigonometric identities and ratios. For example, the tangent function is defined as the ratio of sine to cosine (tanθ = sinθ/cosθ), while the reciprocal identities (cosecant, secant, and cotangent) express the inverse ratios of sine, cosine, and tangent.
What are the domains and ranges of sine, cosine, and tangent functions?
The domains of sine, cosine, and tangent functions are all real numbers, as they are defined for all angles in the unit circle. The ranges of sine and cosine functions are between -1 and 1, while the range of the tangent function is all real numbers except at the points where it is undefined due to vertical asymptotes.
How do trigonometric functions extend beyond right triangles?
While trigonometric functions are initially defined in the context of right triangles, they can be extended to any angle in the Cartesian coordinate system using the unit circle or the definitions of sine, cosine, and tangent in terms of exponential functions. This allows for the application of trigonometric functions to non-right triangles and circular motion.
What are some practical applications of trigonometric functions in everyday life?
Trigonometric functions are used in various real-world applications, including architecture, engineering, physics, astronomy, navigation, art, and music. For example, architects use trigonometry to design buildings and calculate structural loads, while astronomers use trigonometry to study the motion of celestial bodies and navigate spacecraft.
How can I learn more about trigonometric functions and their applications?
There are numerous resources available for learning about trigonometric functions, including textbooks, online courses, tutorials, and educational websites. Additionally, practising solving trigonometric problems and applying trigonometric principles to real-world scenarios can help deepen your understanding of these fundamental concepts.
Why are trigonometric functions important in mathematics and science?
Trigonometric functions are essential tools in mathematics and science due to their ability to model and analyze periodic phenomena, solve geometric problems, and describe complex relationships in nature. From calculating angles and distances to analyzing waveforms and oscillations, trigonometry provides a powerful framework for understanding the world around us.
Conclusion:
In conclusion, sine, cosine, and tangent functions are fundamental components of trigonometry that play a vital role in mathematics, science, engineering, and various other fields. By understanding the definitions, properties, and applications of these functions, we can gain insights into the underlying principles of geometry, physics, and the natural world. Whether exploring the depths of space or designing innovative technologies, the power of trigonometry continues to shape our understanding and advance human knowledge. | 677.169 | 1 |
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