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Different Shapes Names with pictures | Shape List and Types
Learning the Different Shapes Names in English can grow your English vocabulary for daily use. There are many shapes in the world, for example, circles, squares, triangles, and rectangles as well as the more intricate ones trapezoids, octagons, and diamonds, ETC. in this article we have covered Different Shape Names with pictures to recognize how they look | 677.169 | 1 |
Best 13 Problem-solving tips for trigonometry
Problem-solving tips for trigonometry – Trigonometry is a branch of mathematics that deals with the relationships between angles and sides of triangles. It is a fundamental subject in physics, engineering, and navigation. However, many students find trigonometry challenging due to its abstract nature and complex problem-solving requirements.
Let me help you explore valuable problem-solving tips that will help students navigate the complexities of trigonometry and enhance their proficiency in this crucial area of mathematics.
1. Understand Trigonometric Functions and Identities
Before diving into problem-solving, it is crucial to have a solid understanding of trigonometric functions and identities. Please familiarize yourself with basic trigonometric ratios such as sine, cosine, and tangent and their reciprocal functions. Review the Pythagorean identities, sum and difference identities, double-angle identities, and other essential trigonometric identities. A thorough understanding of these concepts will lay a strong foundation for effective problem-solving in trigonometry.
2. Visualize Trigonometric Relationships
Trigonometry involves working with angles and triangles. Visualizing these relationships can greatly aid problem-solving. Draw diagrams or use interactive software to represent triangles and angles. Label the sides and angles appropriately. Visualizing the information will help you identify relevant trigonometric functions and determine how they relate to the problem.
3. Identify the Type of Problem
Trigonometry encompasses many problem types, including finding missing side lengths, determining unknown angles, solving word problems, and applying trigonometric identities. It is crucial to identify the type of problem you are dealing with to choose the most appropriate problem-solving approach. Read the problem carefully and recognize the specific information being sought.
4. Utilize Trigonometric Ratios and Formulas
Trigonometric ratios and formulas play a vital role in solving trigonometry problems. Use the appropriate trigonometric ratio (sine, cosine, or tangent) to calculate the desired value based on the given information. Remember the mnemonic SOH-CAH-TOA (Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent) to recall the trigonometric ratios for right triangles. For non-right triangles, employ the Law of Sines and the Law of Cosines to solve for unknown angles and side lengths.
5. Break Down Complex Problems into Smaller Steps
Complex trigonometry problems can appear overwhelming at first glance. Break them down into smaller, more digestible steps to make them more manageable. Identify the individual components and consider each one separately. By solving one step at a time, you can gradually build up to the final solution, ensuring accuracy and understanding.
6. Practice Applying Trigonometry to Real-World Scenarios
Trigonometry is not confined to abstract mathematical problems—it has numerous practical applications in real-world scenarios. To enhance your problem-solving skills, practice applying trigonometry to real-life situations. Solve word problems involving angles of elevation and depression, distance and height calculations, navigation problems, and more. By connecting trigonometry to real-world contexts, you can strengthen your problem-solving abilities and appreciate its relevance in various fields.
7. Verify Solutions and Check for Reasonableness
Once you have obtained a solution, verifying its accuracy and checking for reasonableness is essential. Substitute your calculated values into the original problem to confirm that the solution satisfies all the given conditions. Ensure that the angles and side lengths make sense within the context of the problem. This verification step helps identify any errors or miscalculations and ensures the validity of your solution.
8. Review Trigonometry Concepts Regularly
Regularly reviewing trigonometry concepts is essential for maintaining a strong foundation and deepening understanding. Concepts such as trigonometric ratios, identities, graphs, and equations are interconnected, and consistent review helps make connections and prevent knowledge decay.
You can reinforce your skills and confidently approach trigonometry problem-solving by dedicating time to reviewing materials, practicing problems, and engaging in active learning. Keep your trigonometry knowledge fresh and sharpened to tackle more complex problems and excel in this crucial branch of mathematics.
9. Utilize Trigonometric Identities and Equations
Trigonometric identities and equations are powerful tools in trigonometry problem-solving. Familiarize yourself with common identities such as the reciprocal identities, Pythagorean identities, and the sum and difference identities. Additionally, be comfortable with solving trigonometric equations by applying algebraic techniques. These identities and equations can help simplify expressions, manipulate equations, and find solutions more efficiently.
10. Work with Reference Angles
Reference angles are an essential concept in trigonometry. They are the acute angles formed between the terminal side of an angle and the x-axis. Finding the reference angle can simplify calculations when dealing with angles outside the primary range (0 to 360 degrees or 0 to 2π radians). Utilize reference angles to determine trigonometric functions' positive or negative signs and work with special angles.
11. Understand Trigonometric Graphs
Graphing trigonometric functions can provide valuable insights for problem-solving. Familiarize yourself with the graphs of sine, cosine, and tangent functions and their key characteristics such as amplitude, period, and phase shift. Understanding how changes in the values of these parameters affect the graph will help you interpret and analyze trigonometric equations and apply them effectively to problem-solving situations.
12. Seek Help and Practice Collaboratively
Trigonometry can be challenging, and seeking help when needed is essential. Collaborate with classmates, join study groups, or seek guidance from your teacher or tutor. Exploring problems together offers different perspectives, alternative solutions, and valuable insights. Engaging in collaborative problem-solving can enhance your understanding of trigonometry concepts and strengthen your problem-solving skills.
13. Practice Time Management during Exams
Time management is crucial during trigonometry exams or timed assessments. Practice solving trigonometry problems within specific time constraints to improve speed and efficiency. Allocate time strategically, focusing on easier or higher-scoring problems and leaving more challenging ones for later. Regular practice under time pressure will help you adapt to the exam environment and improve your overall performance.
Regularly review concepts, utilize trigonometric identities and equations, and work with reference angles and graphs. Seek help and collaborate with others, and practice time management during exams. With dedication and perseverance, you can overcome the challenges of trigonometry and excel in this essential branch of mathematics. | 677.169 | 1 |
How many edges are there. Each of the vertices intersects with three faces and three edges. C...
If you're looking for a browser that's easy to use and fast, then you should definitely try Microsoft Edge. With these tips, you'll be able to speed up your navigation, prevent crashes, and make your online experience even better!By When it comes to browsing the internet, having a reliable and efficient web browser is essential. With a plethora of options available, it can be challenging to determine which one is right for you cube has 12 edges, 24 angles, eight vertices and six faces. A cube is a regular solid made up of six equal squares. Additionally, all angles within the cube are right angles and all sides are the same lengthOct 14, 2020 · We know for any graph G, the sum of the degrees of its vertices is twice its number of edges. In this case, the sum of degrees is: 5(4)+2(2)=20+4=24. According to our fact, 24=2 times number of edges. Therefore, number of edges=24/2= 12. Does this seem correct? Is there a better, more detailed way of explaining this? WhereasHere's the Solution to this Question. Let m be the the number of edges. Because the sum of the degrees of the vertices is. 15 \times8 = 120 15×8 = 120 , the handshaking theorem tells us that 2m = 120\implies m=60 2m = 120 m = 60 . So the number of edges m = 60.How many sides does a rectangle have? A rectangle is a 2D shape in geometry, having 4 sides and 4 corners. Its two sides meet at right angles. Thus, a rectangle has 4 angles, each measuring 90 ̊What should our position be in the USA by Chris Sanders - Amendment #1 "Congress shall make no law respecting an establishment of religion." This is a...InLooking to maximize your productivity with Microsoft Edge? Check out these tips to get more from the browser. From customizing your experience to boosting your privacy, these tips will help you use Microsoft Edge to the fullest.Nov 7, 2021 · A In today's digital age, having a reliable and efficient web browser is essential. With the vast array of options available, it can be challenging to choose the right one for your needs. One browser that has gained popularity in recent years...If you're in the market for a reliable and stylish SUV, look no further than a used Ford Edge. Known for its exceptional value and reliability, the Ford Edge has become a popular choice among car buyers.There are many types of Fantasy leagues out there, but finding the right one for you could be tricky. Our Jamey Eisenberg tells you the types of leagues that are out there and helps you find theHow many edges are there in the graph? a. b. 6 с. 8 d. 10 е. 12 12. How many vertices are there in the graph? a. 1 b. 2 C. 3 d. 4 е. 5 13. Which of the following describes the graph? All vertices have degree. b. The graph is not connected. a. Each vertex has 3 degrees d. Each edge has 3 degrees.$\ The maximum number of edges is clearly achieved when all the components are complete. Moreover the maximum number of edges is achieved when all of the components except one have one vertex.Major League Baseball has suspended Houston Astros right-hander Bryan Abreu for two games after determining that he intentionally plunked Texas Rangers outfielder Adolis García during Friday's ...Find step-by-step Discrete math solutions and your answer to the following textbook question: A connected, planar graph has nine vertices having degrees 2, 2, 2, 3, 3, 3, 4, 4, and 5. How many edges are there? How many faces are there?.Faces in appearance and properties.TheA Two edges are parallel if they connect the same pair of vertices.Dec Next we'll work out how many edges the sphere has, which are where two faces meet. A shere has 0 edges. Next we'll count the corners of the sphere (the corners). No surprises, a …This problem has been solved! You'll get a detailed solution from a subject matter expert that helps you learn core concepts. Question: Consider the following undirected graph. (a) How many edges are there in this graph? (b) Give the degree of each vertex. (c) Do these numbers agree with Euler's first observation? Q candidate.Major League Baseball has suspended Houston Astros right-hander Bryan Abreu for two games after determining that he intentionally plunked Texas Rangers outfielder Adolis García during Friday's ...2. (F) Let G have n vertices and m edges. How many induced subgraphs are there? How many spanning subgraphs are there? There are 2n induced subgraphs (all subsets of vertices) and 2m spanning subgraphs (all subsets of edges). 3. How many spanning subgraphs of K n are there with exactly m edges? n m , since we x all of the vertices and pick m ...Hyundai has long been known for its commitment to innovation and cutting-edge technology in their vehicles. With each new release, they continue to push the boundaries of what is possible in terms of performance, safety, and convenience.How Many Faces, Edges And Vertices Does A Hexagonal Prism Have? Here we'll look at how to work out the faces, edges and vertices of a hexagonal prism. We'll...Aug 16, 2023 · Edges are the lines of a 2D or 3D shape. They are the lines that join the vertices (corner points) up to form shapes and faces. Although many shapes have straight lines and straight edges, there are shapes which have curved edges, such as a hemisphere. A cube will have 12 straight edges as seen below; 9 are visible and 3 are hidden. In a complete graph with $n$ vertices there are $\frac{n−1}{2}$ edge-disjoint Hamiltonian cycles if $n$ is an odd number and $n\ge 3$. What if $n$ is an even number? In today's digital age, having a reliable and efficient web browser is essential. With so many options available, it can be overwhelming to choose the right one for your needs. One popular choice among users is Microsoft Edge.There are many types of Fantasy leagues out there, but finding the right one for you could be tricky. Our Jamey Eisenberg tells you the types of leagues that are out there and helps you find the ...Harassment is any behavior intended to disturb or upset a person or group of people. Threats include any threat of suicide, violence, or harm to another.The four-time Pro Bowler - owner of PFF's highest pass-rushing grade among edges - has generated 26 pressures in five games while boasting the highest pass-rush win rate in the NFL (29.8%).The maximum number of edges is clearly achieved when all the components are complete. Moreover the maximum number of edges is achieved when all of the components except one have one vertex. (Hint: Don't try to draw the graph and count!) the handshake theorem, this is twice the number of edges, so there are 90/2 = 45 edges.) How many edges a complete graph on 10 vertices must have? No a complete graph with10 vertices has 10×9/2=45 edges.23-May-2018. How many vertices does a complete graph have with 21 edges?Edge Connectivity Let 'G' be a connected graph. The minimum number of edges whose removal makes 'G' disconnected is called edge connectivity of G. In other words, the number of edges in a smallest cut set of G is called the edge connectivity of G. If 'G' has a cut edge, then λ(G) is 1. How many edges are there in the graph?. Christmas Vacation Care - 9am- 3pm Ages 5 year- 13 years old (pAnswer and Explanation: Become a Study.com member to unlock Thus the number of edges is 2 less than the sum of the numbers of vertices and faces. For example, a cube has 8 vertices and 6 faces, and hence 12 edges. Incidences with other faces. In a polygon, two edges meet at each vertex; more generally, by Balinski's theorem, at least d edges meet at every vertex of a d-dimensional convex polytope. Faces Edges and Vertices. Faces, edges, and vertices are Here Find step-by-step Discrete math solutions and your... | 677.169 | 1 |
Message #1480
Hi Andy,
"Movements:" give us description of the symmetry group of the puzzle body. Vectors from this set define reflections around their orthogonal hyperplanes. Movements are used to define the set of axes and their order, but vectors fr0om this set don't have to be vectors of multi-axes, or moving planes or so on (but usually they are).
"Vector:" is a part of description of the cutting hyperplane. It is orthogonal to a set of parallel hyperplanes, and it always belongs to multi-axes of all twists of all layers cutting by these hyperplanes.
"Planes:" it's desription of moving planes of the layer orthogonal to "Vector". Each moving plane is defined by a pair of vectors. Both of them are perpendicular to "Vector". Angle between vectors is a half of the minimal twist angle (so we are able to define 180-deg twists by selecting of two perpendicular vectors).
"Cuts:" distances from the center to cutting hyperplanes (in the direction of "Vector"). For the symmetrical bodies set of distances is usually symmetrical, but for simplex and some other bodies like (4,5)-duoprisms positive and negative directions of the axis are different.
And yes, multi-axis is (D-2)-dimensional space.
Andrey
— In [email protected], Andrew James Gould <agould@…> wrote:
>
> Hi Andrey,
> I forgot both are planes of rotation in 4D. I'll used the words 'fixed' and 'moving.'
>
> Does 'Movement:' = R1 vectors and
> 'Vector:' = R vector
> from your previous email? They make the fixed multi-axis [R,R1]?
>
> Consider an N-dimensional simplex (N>2). For any 120-degree rotation that can do a 3-cycle on 3 vertices (and the 2D triangle between them), the moving plane of rotation is determined as well as the fixed multi-axis. Orient the simplex so that the moving plane of rotation is an x-y rotation and move the simplex so that the center of the simplex is at the origin. Then since the moving plane of rotation is orthogonal to the multi-axis, (1) the multi-axis is the restricted set x=0 and y=0. Also, since the remaining fixed vertices are equidistant from the moving verticies, (2) the x and y coordinates of the fixed vertices are restricted to 0. (1) and (2) imply that a multi-axis for such a 120-degree rotation contains all fixed vertices. (It also contains the center of the moving triangle.) That help?
>
> I assumed your multi-axis is everything spanned by whatever axes you list in it.
>
> –
> Andy
> | 677.169 | 1 |
Holt Mathematics Worksheets With Answers
Holt Mathematics Worksheets With Answers - A scientist has 1,050 samples to observe. Name a right angle in the figure. Holt mcdougal mathematics course 3, publisher: We have leap years because the. Holt mcdougal mathematics course 2, publisher: Name two acute angles in.
Holt mathematics homework and practice workbook course 2 answers
Holt mcdougal mathematics course 2, publisher: Web what is a leap year? Web the expression 2x1 2yrepresents the total amount of molding you need where xis the width of the room (in feet) and yis the length of the room (in feet). Use the table below to find videos, mobile apps, worksheets and lessons that supplement holt mcdougal 6th grade.
Holt mcdougal 7th grade math book answers
Web the expression 2x1 2yrepresents the total amount of molding you need where xis the width of the room (in feet) and yis the length of the room (in feet). A scientist has 1,050 samples to observe. 2024 is a leap year, which means we have an extra day on february 29. Web use the table below to find videos,.
Web over 1000 online math lessons aligned to the holt textbooks and featuring a personal math teacher inside every lesson! Leap years happen every four years. Find the total amount of. Understanding points, lines, and planes.
Web holt mcdougal mathematics 6 common core grade 6 workbook & answers help online. Name two acute angles in. Name four rays in the figure. | 677.169 | 1 |
Hint: To solve this question we will see what are the basic requirements for two triangles to be called similar or congruent and on that basis we will judge whether the statement given in the question is true or not.
Complete step-by-step answer:
In question, it is asked that if there are two triangles and their corresponding angles are equal, then these triangles will be congruent or not. Now, let us first see the difference between congruence and similarity of triangles. Two triangles are said to be congruent if they both are identical in shape, size and measure, that is if we have two triangles, then if their sides are in ratio of 1:1, there all three angles are same then triangles are congruent. Two triangles are said to be similar if they both are in the same shape and same measure but not have similar size, that is, two triangles are said to be similar if all three corresponding sides are in the same ratio, say m:n and angle measure are the same but size may be different. Now, the question is that if corresponding angles are equal then those two triangles will be congruent or not. Answers may be true for congruence but always true for similarity. The reason is simple that two triangles have equal corresponding angles but size is not mentioned here and corresponding angles are equal and satisfies all the conditions of similarity but not of congruence as by definition of congruence, shapes must be identical. $ \text{In figure, }\vartriangle \text{ABC and }\vartriangle \text{EFG}$ are similar triangles with same So, the correct answer is "Option C".
Note: These questions are theoretical and this needs deep knowledge of chapter. So, one must know all the properties, theorems and statements on similarity and congruence of two triangles. Always use counter examples wherever you need to disprove something if possible. | 677.169 | 1 |
Investigation: Proof of the triangle inequality
We will now present a complete paragraph proof of the triangle inequality theorem.
Tips for reading mathematical proofs
Reading proofs takes much longer than reading text written in everyday language, and that's okay! Follow these tips for understanding and reading proofs:
Always stop reading if you don't understand something. It's okay to go back and re-read a statement several times.
If a writer draws something in a proof, try drawing it on your own with pencil and paper.
After reading a few sentences, see if you can state back the meaning in your own words.
Consider a general triangle, \Delta PQR:
What we want to prove now is that PQ+QR>PR, PQ+PR>QR, and QR+PR>PQ.
It seems daunting to prove three things! But we begin by assuming that \overline{PQ} is the longest side (as it is in the diagram above). Even if two or more of the sides have the same length as \overline{PQ}, we are only assuming PQ\geq QR and PQ\geq PR.
This gives us two of the inequalities essentially for free. PQ\geq PR becomes PQ+QR\geq PR+QR by the addition property of (in)equality, and since QR is a nonzero amount, we can be sure that PQ+QR>PR on its own. A similar calculation establishes the second inequality.
So now we are left with the third inequality, which is going to take the most work. We are going to describe a construction and use it on a particular diagram, but we need to make sure it will work for any triangle and be careful as we proceed.
We perform the following steps:
Draw a circle of radius QR centered at R.
Extend \overline{PR} to \overrightarrow{PR}.
The ray from step 2 intersects the circle from step 1 at a point - we mark it and call X.
Here is the results of the process with our triangle:
As indicated on the diagram, the radii \overline{RX} and \overline{QR} are congruent, since they're radii of the same circle, so they have the same measure. By substitution, QR+PR>PQ is equivalent to PR+RX>PQ, and proving this second statement is the same as finishing the proof altogether.
Remove the ray and the circle and create \overline{XQ} to form the triangle \Delta RQX:
As we noted before, two of the sides of this triangle are congruent, and so this triangle is isosceles. Since base angles of an isosceles triangle are equal, we can write \angle RXQ\cong\angle RQX.
Now we consider the other triangle we formed, \Delta PQX:
The angle addition postulate tells us that m\angle PQR + m\angle RQX = m\angle PQX. If we subtract m\angle PQR from the left-hand side only, we break equality, and have the inequality m\angle RQX < m\angle PQX.
Using the established congruence \angle RXQ\cong\angle RQX, we can substitute the measure of \angle RXQ for the measure of \angle RQX and produce m\angle RXQ < m\angle PQX. Now \angle PXQ=\angle RXQ since P, R, and X all lie on the ray \overrightarrow{XR}, and so by substitution again we can write m\angle PXQ < m\angle PQX. We can then conclude that PQ < PX, as an angle of smaller measure is opposite the shorter side.
By the segment addition postulate we know that PX = PR + RX, and by substituting this into the last inequality we found in the previous paragraph we know that PQ < PR + RX. We once again use the fact that RX=QR and by substitution we have PQ < PR + QR. We can then rearrange this inequality to produce QR+PR>PQ, QED.
Extra Challenge
Now that you've read the paragraph proof of the triangle inequality theorem, how can you be sure you really understand it?
Try re-creating the proof in a new medium. For example, by explaining it in a video, cartoon strip, or as a two-column proof.
Outcomes
CC.2.3.HS.A.3
Verify and apply geometric theorems as they relate to geometric figures. | 677.169 | 1 |
Kites, Trapezoids, And Midsegments use your Geometry STAAR Reference Chart (green one), a graphing calculator, and your Geometry Tool Box on this assessment. Additionally, you may wish to use a scratch piece of paper or a dry erase sleeve to avoid doing too much work totally in your head.
Questions and Answers
1.
The figure below is a kite.
What is the angle measure for x ?
Explanation The angle measure for x in a kite is equal to 15 degrees.
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2.
The figure below is a kite.
What is the angle measure for y ?
Explanation The angle measure for y in the given figure is 42.
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0
3.
The figure below is a kite.
What is the angle measure for x ?
Explanation The angle measure for x in the given figure is 63 degrees. This can be determined by recognizing that the opposite angles in a kite are congruent. Since the figure is a kite, the angle opposite to x must also measure 63 degrees.
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4.
The figure below is a kite.
What is the angle measure for y ?
Explanation The angle measure for y in a kite is equal to 73 degrees.
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5.
The figure below is an isosceles trapezoid.
What is the angle measure for x ?
Explanation The angle measure for x in an isosceles trapezoid is equal to the measure of the base angles. In an isosceles trapezoid, the non-parallel sides are congruent, which means the base angles are also congruent. Since the given answer is 59, it implies that the measure of angle x is 59 degrees.
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6.
The figure below is an isosceles trapezoid.
What is the angle measure for y ?
7.
The figure below is an isosceles trapezoid. The perimeter of the trapezoid is 105 cm.
What is the value of x ?
8.
What is the angle measure for x ?
Explanation The angle measure for x is 60 because the given answer is 60.
Rate this question:
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9.
What is the angle measure for y ?
Explanation The angle measure for y is 140 because the given answer matches the value provided in the question.
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10.
What is the length of z ?
Explanation The length of z is 13.
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11.
The figure below is a trapezoid.
What is the length of q ?
Explanation The length of q is 35 because the figure given is a trapezoid, and the sides of a trapezoid are parallel. Therefore, the opposite sides of the trapezoid are equal in length. Since the length of one side is given as 35, the length of the opposite side, q, must also be 35. | 677.169 | 1 |
Midpoint And Distance Formula Worksheet With Answers is an accumulation tips and techniques from teachers, doctoral philosophers, and professors, for you to use worksheets in class. Midpoint And Distance Formula Worksheet With Answershas been used in schools in several countries to improve Cognitive, Logical and Spatial Reasoning, Visual Perception, Mathematical Skills, Social Skills and Personal Skills.
Midpoint And Distance Formula Worksheet With Answers is supposed to provide guidance teaching how to integrate worksheets into cannot curriculum. Because we receive additional material from teachers throughout america, we desire to continue to be expanded Midpoint And Distance Formula Worksheet With Answers content. Please save the many worksheets that we provide on this internet site based on your entire needs in class possibly at home.
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What does perpendicular sides look like?
Perpendicular lines are two or more lines that intersect at a 90-degree angle, like the two lines drawn on this graph. These 90-degree angles are also known as right angles. When the lines are parallel or perpendicular, text will appear to let you know you've done it! o Look at the slopes of the two parallel lines.
What is perpendicular sides mean?
In elementary geometry, the property of being perpendicular (perpendicularity) is the relationship between two lines which meet at a right angle (90 degrees). A line is said to be perpendicular to another line if the two lines intersect at a right angle.
Is a rectangle have perpendicular sides?
Rectangles have four straight sides. Each pair of opposite sides is parallel, and adjacent sides are perpendicular. This means that every angle in a rectangle is a right (90∘) angle.
How do you find perpendicular sides?
If two lines are perpendicular to the same line, they are parallel to each other and will never intersect.
Adjacent sides of a square and a rectangle are always perpendicular to each other.
Sides of the right-angled triangle enclosing the right angle are perpendicular to each other.
Can a parallelogram have perpendicular sides?
No. A parallelogram is a four sided shape with opposite sides parallel and the same length. Not all parallelograms have perpendicular sides.
What is perpendicular parallelogram?
A Parallelogram with Perpendicular Diagonals is a Rhombus A rhombus is a special kind of parallelogram, in which all the sides are equal. We've seen that one of the properties of a rhombus is that its diagonals are perpendicular to each other.
What parallelogram has perpendicular diagonals?
If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. If one diagonal of a parallelogram bisects a pair of opposite angles, then the parallelogram is a rhombus. To prove a square, you must prove it is both a rectangle and a rhombus.
Why diagonals of kite are perpendicular?
The intersection of the diagonals of a kite form 90 degree (right) angles. This means that they are perpendicular. The longer diagonal of a kite bisects the shorter one. This means that the longer diagonal cuts the shorter one in half.
How do you prove diagonals are perpendicular?
To prove that two lines are perpendicular, when all we have are those two lines, we can use the Linear Pair Perpendicular Theorem – If two straight lines intersect at a point and form a linear pair of equal angles, they are perpendicular.
What part of a kite is perpendicular?
Diagonals
Do kites perpendicular bisector?
The angles opposite the axis of a kite are equal. The axis of a kite bisects the vertex angles through which it passes. The axis of a kite is the perpendicular bisector of the other diagonal.
Are the diagonals of a trapezoid perpendicular?
The diagonals in an isosceles trapezoid will not necessarily be perpendicular as in rhombi and squares. They are, however, congruent. Any time you find a trapezoid that is isosceles, the two diagonals will be congruent.
What happens if the diagonals of a trapezoid are perpendicular?
Diagonals of an isosceles trapezoid are perpendicular to each other and the sum of the lengths of its bases is 2a. Diagonals of an isosceles trapezoid are perpendicular to each other and the sum of the lengths of its bases is 2a.
Do rectangles have perpendicular diagonals?
The diagonals of the rectangle are equal i.e. AC = BD and they bisect each other. If in case of square and rhombus, the diagonals are perpendicular to each other. But for rectangles, parallelograms, trapeziums the diagonals are not perpendicular. ∴ The diagonals of a rectangle are not perpendicular to each other.
What Quadrilaterals have diagonals that are perpendicular?
Quadrilaterals
A
B
in these quadrilaterals, the diagonals are perpendicular
rhombus, square
a rhombus is always a…
parallelogram
a square is always a…
parallelogram, rhombus, and rectangle
a rectangle is always a…
parallelogram
Which Quadrilaterals diagonals are not perpendicular?
1 Answer. Rectangle is the quadrilateral, which has diagonals that are congruent and bisect but are not perpendicular. | 677.169 | 1 |
Is found at 0 longitude?
Is found at 0 longitude?
The prime meridian is the line of 0° longitude, the starting point for measuring distance both east and west around the Earth. The prime meridian is arbitrary, meaning it could be chosen to be anywhere.
What is the line of latitude called?
parallels
Lines of latitude, also called parallels, are imaginary lines that divide the Earth. They run east to west, but measure your distance north or south. The equator is the most well known parallel. At 0 degrees latitude, it equally divides the Earth into the Northern and Southern hemispheres.
What is the 0 degree line of latitude called quizlet?
Zero degrees (0°) latitude is the equator, the widest circumference of the globe. Latitude is measured from 0° to 90° north and 0° to 90° south—90° north is the North Pole and 90° south is the South Pole. Imaginary lines, also called meridians, running vertically around the globe.
What is 0 latitude called?
The Equator
The Equator is the line of 0 degrees latitude. Each parallel measures one degree north or south of the Equator, with 90 degrees north of the Equator and 90 degrees south of the Equator.
What is the 0 longitude line?
The prime meridian is the line of 0° longitude, the starting point for measuring distance both east and west around the Earth. The prime meridian is arbitrary, meaning it could be chosen to be anywhere. 6 – 12+
What is another name for 0 latitude?
the Equator
If you look at the intersection of 0 degrees latitude (known as the Equator) and 0 degrees longitude (known as the Prime Meridian) on a map, you will see that the confluence falls in the middle of the Atlantic Ocean, in the Gulf of Guinea off the coast of western Africa.
How do you write 0 degrees latitude?
The equator's line of latitude is marked by 0 degrees. When writing latitude and longitude, use the symbol "°" to indicate degrees. As you move north of the equator, lines of latitude increase by one degree until they reach 90 degrees. The 90 degree mark is the North Pole.
What does 0 latitude represent on the map quizlet?
in degrees north or degrees south the equator is 0 degree latitude . -the ends of that axis are the North Pole and the south pole: they mark the opposite positions of Earth rotational axis, around which turns into 24 hrs. -Latitude of a point is the angular distances between the point and the equator.
What are latitude lines on a map called quizlet?
All meridians meet at the North and South Poles. Longitude is related to latitude, the measurement of distance north or south of the Equator. Lines of latitude are called parallels. Maps are often marked with parallels and meridians, creating a grid.
Where are the lines of latitude?
Latitude lines are geographical coordinates that are used to specify the north and south sides of the Earth. Lines of latitude, also called parallels, run from east to west in circles parallel to the equator. They run perpendicular to the lines of longitude, which run from the north to the south.
Where is longitude 0 and latitude 0?
Gulf of Guinea
Location of 0 Latitude, 0 Longitude To be exact, the intersection of zero degrees latitude and zero degrees longitude falls about 380 miles south of Ghana and 670 miles west of Gabon. 1 This location is in the tropical waters of the eastern Atlantic Ocean, in an area called the Gulf of Guinea.
Are lines of longitude while are lines of latitude quizlet?
Unlike latitude lines, longitude lines are not parallel. Meridians meet at the poles and are widest apart at the equator. Zero degrees longitude (0°) is called the prime meridian. The degrees of longitude run 180° east and 180° west from the prime meridian | 677.169 | 1 |
Three circles in a plane, exterior to each other, are given in
position. Find the triangle with minimum perimeter that has one vertex
on each circle.
From the contents of the chapter it is obvious (using light reflections on three circular mirrors and rubber band methods) that the two sides of the required triangle that meet in a vertex on a given circle include equal angles with the radius.
But how can we construct (with the compass and straightedge) these vertices (A,B,C)?
UPD
Let one of the circles be an infinite radius (a straight line):
Looks like the same solution... And no idea about construction.
So let all of the circles be an infinite radius:
And we get Fagnano's problem with clear construction.
Hope this will be useful (?)
$\begingroup$Setting it differently, is the common point of $AD, BE, CF$ constructible by straightedge and compass ? You must know that there are many cases of remarkable points that are not contructible by straightedge and compass...$\endgroup$
$\begingroup$@lesobrod: I have already drawn the triangle $PQR$ in my comment and it applies to all circles with arbitrary radii. So I am certain that this generalization with Fagnano's triangle (when the the straight lines are replaced by an arena of 3 circles) would hold good in the sought future solution here.$\endgroup$
4 Answers
4
Solution minimum triangle is determined by the given circles and angle bisectors.The following states that triangle $ABC$ has minimum perimeter for all triangles such as $EDF.$ Co-tangential circles are also included to suggest a set of problems with same inner minimum perimeter triangle $ABC$.
EDIT1/2:
Intersections of angle bisectors from vertices $(A,B,C) $ with given circles form triangle $PQR$ vertices of minimum perimeter length. This results from the ellipse property... constancy of major axis length between shown ellipse foci with their mirror reflective property considered pairwise among $(P,Q,R)$.
$\begingroup$Thank you for your diagram. I don't understand your construction, though: how are $A, B$ and $C$ determined? Or, to put it another way, how is $I$ determined? (Once $I$ is found, it's easy to find $A, B$ and $C$.)$\endgroup$
$\begingroup$Thank you for your second diagram. That answers the question in my earlier comment. (Note that in the first diagram and my comment, the centres are $D, E, F$ and the vertices are $A, B, C$, but in the second diagram, the centres are $A, B, C$ and the vertices are $P, Q, R$.) Further question: do the angle-bisectors in the second diagram also show the vertices of that triangle of greatest perimeter with one vertex on each circle?$\endgroup$
Consider $A$ and $C$ fixed.
Then for the perimeter to be minimum, $B$ shall lie on the ellipse with foci
in $A$ and $C$, minimal sum of the distances from them, so on the one tangent to the
circle through $B$.
By the property of elliptical mirror, the normal to the tangent in $B$, shall halve the angle $ABC$.
Then the proof of Narasimham's answer follows.
Moreover, with $A$ and $C$ fixed, we are minimizing $p$, and $p-b$ as well.
By symmetry, the triangle has also minimal area.
As easy to understand the construction boils down to finding the in-center $I$ of the triangle $ABC$.
An algebraic way for this could be the following. Let the coordinates and the radius of $i$-th circle ($i=1,2,3$) be $(x_i,y_i)$ and $r_i$, respectively, and the coordinates of the point $I$ be ($x,y$).
Then the coordinates of the vertex $A$ read:
$$
(x_A,y_A)=(x_1,y_1)+r_1\frac{(x-x_1,y-y_1)}{\sqrt{(x-x_1)^2+(y-y_1)^2}},\tag1
$$
and similarly for the vertices $B$ and $C$.
From the condition that $I$ is the in-center of $\triangle ABC$ we have the following equations to determine $(x,y)$:
$$
\frac{\frac{x-x_A}{x_B-x_A}-\frac{y-y_A}{y_B-y_A}}
{\sqrt{\frac1{(x_B-x_A)^2}+\frac1{(y_B-y_A)^2}}}=
\frac{\frac{x-x_B}{x_C-x_B}-\frac{y-y_B}{y_C-y_B}}
{\sqrt{\frac1{(x_C-x_B)^2}+\frac1{(y_C-y_B)^2}}}=
\frac{\frac{x-x_C}{x_A-x_C}-\frac{y-y_C}{y_A-y_C}}
{\sqrt{\frac1{(x_A-x_C)^2}+\frac1{(y_A-y_C)^2}}},\tag2
$$
where $x_A,y_A,x_B,y_B,x_C,y_C$ are to be substituted using (1).
After expanding the equations one should end up with a system of two polynomial equation for $(x,y)$ of a rather high order. It is probable that the solution to this equation is not constructible.
$\begingroup$From the construction it is obvious that $F$ is Fermat point both for $\triangle ABC$ and $\triangle PQR$. But we are looking for the in-center of $\triangle PQR$, which generally does not coincide with the Fermat point (except for equilateral triangle).$\endgroup$ | 677.169 | 1 |
Elements of Geometry: Containing the First Six Books of Euclid: With a Supplement on the Quadrature of the Circle and the Geometry of Solids 50.
УелЯдб 43 ... parallelogram are equal to one ano- ther , and the diameter bisects it , that is , divides it into two equal parts . N. B. A Parallelogram is a four - sided figure , of which the opposite sides are parallel ; and the diameter is the ...
УелЯдб 44 ... parallelogram ACDB into two equal parts . Therefore , & c . Q. E. D. PROP . XXXV . THEOR . Parallelograms upon the same base and between the same parallels , are equal to one another . M ( SEE THE 2d and 3d figures . ) Let the ...
УелЯдб 45 ... parallelogram ABCD is equal to the parallelogram EBCF . Therefore , parallelograms upon the same base , & c . Q. ED . PROP . XXXVI . THEOR . Parallelograms upon equal bases , and between the same parallels , are equal to one another . A ...
УелЯдб 46 ... parallelogram EBCA , because the diameter AB bisects ( 34. 1. ) it ; and the triangle DBC is the half of the parallelogram DBCF , because the diameter DC bisects it : And the halves of equal things are equal ( 7. Ax . ) ; therefore the ...
УелЯдб 47 ... parallelogram is double of the triangle . Let the parallelogram ABCD and the triangle EBC be upon the same base BC , and between the same pa- rallels BC , AE ; the parallelogram ABCD is double of the triangle EBC . A Join AC ; then ...
ДзмпцйлЮ брпурЬумбфб
УелЯдб 153 - If from the vertical angle of a triangle a straight line be drawn perpendicular to the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the perpendicular and the diameter of the circle...
УелЯдб 19УелЯдб 18 - 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.
УелЯдбУелЯдб | 677.169 | 1 |
Transformations Unit: Geometry Interactive Notebook
Geometry – Unit 2 Review – Transformations
Geometry – Unit 2 Review – Transformations
After studying the basics of geometry and its basic relationships among lines and angles, we move on to our transformations unit. Specifically, congruence transformations. We will not learn about dilations until we reach our similarity unit. This unit has one big advantage: Much of what we learn here is actually repeated from 8th grade. Keep reading to see how I teach my high school geometry transformations unit.
Introduction to Transformations Unit
To kick off our transformations unit, we start with basic vocabulary related to transformations. Some of the vocabulary, such as pre-image, image, and rigid motion, is new for students. We also talk about the notation of transformations and how a prime will appear after a letter for each transformation. After vocabulary, we look at four transformations and students identify them as a translation, reflection, rotation, and dilation using their prior knowledge from 8th grade. We use the dilation as a counterexample for the rigid motions. Then, students practice identifying translations, reflections, and rotations with a card sort. (Paper Card Sort, Digital Card Sort)
Translations
After the introduction to transformations, we learn about the 3 rigid motions one day at a time, starting with translations. To begin the lesson, we define translations and look at the different notations for them. We practice determining rules of translations using graphs, graphing translations, and determining the preimage given the vertices of the image and rule.'
Transformations Unit: Reflections
For reflections, we start with not just vocabulary and notation, but also the properties of a line of reflection. To practice, we determine the line of reflection given a graph of reflections, and then we graph a few reflections. We also practice applying the rules for the special reflections (x-axis, y-axis, y = x, y = -x). I always enjoyed teaching the special reflections, but they are no longer used as frequently in our state assessments
Rotations
For the sake of consistency, we follow the same format as we did for translations and reflections when we learn rotations. In addition to the vocabulary and notation, we spend time discussing clockwise and counterclockwise rotations, and how to determine the angle of rotation. (We only use 90, 180, and 270 degree angles of rotation.) For practice, we identify the angles of rotation given graphs. Next, we talk about the rules for special rotations (90, 180, and 270) before practicing applying and graphing them. But again, special rotations are barely referenced on the state exam. I only teach these because I enjoy teaching them.
Transformations Unit: Line & Rotational Symmetry
Line and rotational symmetry is my favorite lesson of the unit! Most students have a knowledge of symmetry before geometry from art classes, and possibly elementary school. For this lesson, students complete a table as they complete a hands-on discovery activity. Students use patty paper to trace shapes and determine how many lines of symmetry they have. They also use folders that I attach shapes and a 360 degree protractor to so they can actually rotate the shapes to determine their rotational symmetry.
Prior to beginning the activity, I demonstrate what to do using a rectangle. I show how I fold the patty paper that I traced a rectangle onto to test its line symmetry. And, I show how I spin a rectangle around the 360 degree protractor to test its rotational symmetry. After this activity, students understand line and rotational symmetry very well, and are able to determine the symmetry without using any tools.
Sequence of Transformations
In the past, this lesson was all about performing sequences of transformations, especially using the rules for special reflections and rotations. Since we started using common core standards, however, the emphasis is on analyzing sequences. Students are asked to determine which two transformations were used given a graph of two figures. Also, they are often given a graph depicting 3 figures, and asked what one transformation would map the preimage onto the final image.
In this lesson, we focus on the aforementioned concepts, but still practice graphing sequences of transformations.
Congruence
Our final lesson of the transformations unit focuses on congruence. These are all congruence transformations after all. We define congruent and correspondence before analyzing two examples. In those examples, we are looking for which angles and which line segments are corresponding and congruent. After that, we practice using transformations to justify the congruence of two figures. This lesson directly sets us up for our next unit on triangles.
Are you preparing to teach your transformations unit? Whether you are teaching in person or online, you can use interactive notebooks to teach transformations. What I love about interactive notebooks is that students are able to refer to their notes anytime they need – which is very helpful with all of those rules for special reflections and rotations. You can get everything I use to teach the transformations in one convenient bundle. A digital interactive notebook of this unit is also available. | 677.169 | 1 |
How to Find the Center of a Circle | Finding the Exact Center of a Circle
Often, when it comes to DIY projects, we create our own circles by starting from a center point. Whether a compass, string guide, or specialized cutting jig, a circle emerges from our pencils or saws precisely because we've created an established and consistent distance from a single origin.
But, what happens when the shape already exists, and you need to know how to find the center of a circle? You can do it in less than a minute without any specialized math, memorizing a formula, or even knowing what pi is.
How to Find the Center of a Circle
Lay your ruler across the circle. You can do this anywhere, other than the exact diameter of the circle. Place one side on the edge, and rotate the ruler until you hit a nice round number on the other side. Note in the photo that the ruler hits the edges at the exact 0″ and 9″ marks. Scribe a line along the ruler.
Perhaps you'll remember from geometry class that what you've drawn here is a chord – a straight line whose endpoints both lie on a circle. Find the center point of your chord, and make a small mark. In this example, it's 4 1/2″.
Then, use your square to line up one side along your chord, and the 90° corner at the center tick mark. Draw a line along the opposite edge.
Your circle will look like this. In my case, I have a 9″ chord, and a perpendicular line intersecting at half its distance, 4 1/2″.
Now, just repeat this process somewhere else on the circle. It doesn't matter where it is in relationship to the first chord, but you should choose an easily divisible number again to keep things simple.
Ta-da! The place where the two perpendicular lines meet is the exact center of the circle. Because that's how geometry works.
You can repeat the process a few more times, just in case your work wasn't perfect the first time. This will allow you to take an average of all the cross points to find the center.
If your measurements are careful, you'll find all the line will cross at the exact same point. Hooray, math! | 677.169 | 1 |
Two points A and B are on the same side of a tower and in the same straight line with its base. The angle of depression of these points from top of the tower are ${{60}^{\circ }}$ and ${{45}^{\circ }}$ respectively. If the height of the tower is 15 m, then find the distance between these points.
Hint: Draw a diagram by carefully reading the question and applying the constraints given in it. Assume two variables (say 'x' and 'y') for distance of point 'A' and 'B' respectively from the base of the tower. Then use the formula $\text{tan }\!\!\theta\!\!\text{ =}\dfrac{\text{perpendicular}}{\text{base}}$ in two triangle separately to find 'x' and ;y; and then calculate distance between 'A' and 'B' by substituting 'x' from 'y'.
Complete step by step answer: According to question, we can draw the following diagram- | 677.169 | 1 |
Parallels and Parallelograms Maths Investigation.
Definitions:
Transversal: A that cuts across a set of lines or the of a . Transversals often cut across .
Parallel line: Two that do not intersect. Note: Parallel lines have the same .
Parallelogram: A with two pairs of .
A1 :
ᴗ :
This investigation aims at finding a relationship between the numbers of horizontal parallel lines and the transversals. When these lines intersect they form parallelograms. The aim of this investigation is examine and determine a general statement for transversals and horizontal lines and how they affect the number of parallelograms formed within the figure. A diagram of a parallelogram and a transversal is shown below.
Figure 1: two transversals
Figure 1 below shows a pair of horizontal parallel lines and a pair of parallel transversals. One parallelogram (A1) is formed. A1
Adding a third transversal
A third parallel transversal is added to the diagram as shown in Figure 2. Three parallelograms are formed: A1 , A2 , and A1ᴗA2.
A2
A1
Figure 2: three transversals
Adding a third transversal gives us a total of three parallelograms.
Adding a fourth transversal
Figure 3: four transversals
A1 , A2 , A3,A1ᴗA2,A2ᴗA3 and A1ᴗA3.
Adding a fourth transversals gives us a total of six (6) parallelograms.
Adding a seventh transversals gives us a total of twenty-one parallelograms.
Table 1: Side by Side view of Corresponding Transversals to Parallelograms
The general formula needs to be deduces from the patterns that are seen in the table and previously discovered and previously discovered maths formula. To discover the relationship between parallelograms and the number of vertical transversals a similar sequence of numbers needs to be investigated.
Graph 1
--> Let n = number of transversals and let p = number of parallelograms
This is a preview of the whole essay
2 3 4 5 6
1 1 1 1
Table 2: Two Horizontal Parallel Lines
General statement
Since the number of parallelograms created as the number of transversals increased each had a Second Order difference of 1, it was immediately known that the general formula must be a quadratic equation.
If there are n transversals and two horizontal lines, then p = sum of all integers from 1 to (n - 1).
Adding a third horizontal parallel line with seven transversal lines gives us a total of sixty-three parallelograms.
Table 3: Side by Side view of Corresponding Transversals to Parallelograms
Paralleograms:
3, 9, 18, 30, 45, 63, etc.
9-3 18-9 30-18 45-30 63-45
6 9 12 15 18
3 3 3 3
Table 4: Three Horizontal Parallel Lines
Now the second order difference is 3- triple the first set of parallelograms. Due to the second order being three, I deducted and found that the number of total parallelograms was increasing in multiples of three as shown by the fourth column of the table above.
General statement
Since the number of parallelograms created as the number of transversals increased each had a Second Order difference of 3, it was immediately known that the general formula must be a quadratic equation.
If there are n transversals and three horizontal lines, then p = sum of all integers from 1 to (n - 1).
Adding a fourth horizontal parallel line with five transversal lines gives us a total of sixty parallelograms.
Table 5: Side by Side view of Corresponding Transversals to Parallelograms
Paralleograms:
6, 18, 36, 60, etc.
18-6 36-18 60-36
12 18 24
6 6
Table 6: Four Horizontal Parallel Lines
Now the second order difference is 6- double the second set of parallelograms. Due to the second order being six, I deducted and found that the number of total parallelograms was increasing in multiples of six as shown by the fourth column of the table above.
General statement
Since the number of parallelograms created as the number of transversals increased each had a Second Order difference of 6, it was immediately known that the general formula must be a quadratic equation.
If there are n transversals and four horizontal lines, then p = sum of all integers from 1 to (n - 1).
P=12 4nn-1 or P=4n2-n÷2
Expanding…
Table 7: Number of Parallel Transversals
2 horizontal lines:1nn-1÷2
3 horizontal lines:3(nn-1)÷2
4 horizontal lines:6(nn-1)÷2
5 horizontal lines:10(nn-1)÷2
The number and sequence repeats the formula nn-1÷2 and multiplies by the first term.
The General Statement
From this realization I was able to find the final formula for calculating the number of parallelograms formed when m horizontal parallel lines are intersected by n parallel transversal:
m(m-1)2 × n(n-1)2
To test the validity of the formula, I tested it against a previously counted parallelograms (Figure 10), the intersection of four transversals with three horizontal parallel lines should form a total of eighteen parallelograms.
Figure 14: four transversals
Using the formula
m(m-1)2 × n(n-1)2
3(3-1)2 × 4(4-1)2
3(2)2 × 4(3)2
62 × 122
3 ×6
=18 Parallelograms
Scope/limitations
The formula will be valid for m, n≥2.if either value were to be 1 or 0, it would be impossible to create any parallelograms. | 677.169 | 1 |
Consider a parabola (c) and the triangle formed by two tangents AB, AC and their chord of contacts BC. Line FE joining the middles of AB, AC is also tangent to (c) at its middle K. This is one (of three) Artzt parabola for ABC. To find the intersection points of (c) with the inner Steiner ellipse (touching the triangle at the middles of the sides). The answer is: on the parallel to BC from the centroid at distance BC/(2*sqrt(3)) from it.
The proof can be reduced to a canonical parabola. Namely the analogous parabola when ABC is equilateral (shown below). In that case the Steiner ellipse is the incircle of the triangle and the intersection with the parabola is easily calculated to occure at these points. The proof is then transfered to the general case via an affinity that maps the vertices of the equilateral to corresponding vertices of the general triangle. See ParabolaTrapezium.html for another case of application of this method. | 677.169 | 1 |
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Here favorite puzzles ... What we name a corner or any point which is connected by linRealtek audio sound cards feature a graphic equal The Crossword Solver found 30 answers to "shapehaving equal angles" 17 letters crossword answer - We have 1 clue. Solve your "having equal angles" crossword puzzle fast & easy with the-crossword-solver.com The Crossword Solver found 30 answers to "having two equal side "Cot" is the abbreviation for "cotangent," a trigonometric function used to find the value of an angle in a right triangle by dividing the length of an adjacent side by the length ... The Crossword Solver found 30 answers to "What X equals nowadays&... | 677.169 | 1 |
Take This Important Geometry Trivia having a hard time in geometry class and are looking for a way to refresh your understanding on the topic, you have come to the right place as this important quiz will do just that for you. It has some of the topics that most students have a hard time answering. Feel free to take this as much as you can.
Questions and Answers
1.
Which best describes the statement If two planes intersect, then their intersection is a point?
A.
Always
B.
Sometimes
C.
Never
D.
Cannot tell
Correct Answer C. Never
Explanation The statement "If two planes intersect, then their intersection is a point" is false. When two planes intersect, their intersection can be a point, a line, or even the entire plane. Therefore, it is incorrect to say that their intersection is always a point, sometimes a point, or cannot be determined. The correct answer is "Never" because the statement is never true.
Explanation The property that justifies the statement "If x = 2 and x + y = 3, then 2 + y = 3" is Substitution. Substitution is the property that allows us to replace a variable with its value in an equation or expression. In this case, we are given that x = 2 and x + y = 3. By substituting the value of x, which is 2, into the second equation, we get 2 + y = 3. Therefore, the correct answer is Substitution.
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3.
Choose the property that justifies the statement m∠A = m∠A.
A.
Reflexive
B.
Symmetric
C.
Transitive
D.
Substitution
Correct Answer A. Reflexive
Explanation The property that justifies the statement m∠A = m∠A is the reflexive property. The reflexive property states that any element is equal to itself. In this case, it means that the measure of angle A is equal to itself, which is always true.
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4.
Choose the property that justifies the statement If ≅ , then ≅
A.
Reflexive
B.
Symmetric
C.
Transitive
D.
Definition of Congruent Segments
Correct Answer B. Symmetric
Explanation The property that justifies the statement "If ≅ , then ≅ " is the Symmetric property. This property states that if two segments are congruent, then their order can be reversed. In other words, if segment AB is congruent to segment CD, then segment CD is also congruent to segment AB.
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5.
On a line, if XY = 6, YZ = 4, and XZ = 2, which point is between the other two?
A.
X
B.
Y
C.
Z
D.
Cannot tell
Correct Answer C. Z
Explanation The point Z is between the other two points X and Y because the given information states that XZ = 2, which is smaller than XY = 6 and YZ = 4. Therefore, Z must be the point between X and Y.
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6.
If m∠BFC = 70, find m∠EFD.
A.
10
B.
20
C.
35
D.
70
Correct Answer B. 20
Explanation The measure of angle EFD is 20 degrees. This can be determined by using the property that angles on a straight line add up to 180 degrees. Since angle BFC is given as 70 degrees, the sum of angles BFC and EFD must be 180 degrees. Therefore, angle EFD is 180 - 70 = 110 degrees. However, since angle EFD is an exterior angle to triangle BFC, it is equal to the sum of the opposite interior angles. Since angle BFC is 70 degrees, the other interior angle, angle EFB, must be 110 - 70 = 40 degrees. Finally, since angle EFD is an alternate interior angle to angle EFB, they must be congruent, so angle EFD is also 40 degrees.
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7.
If m∠AFB = 5x – 10 and m∠BFC = 3x + 20, find x.
A.
10
B.
15
C.
21.25
D.
23.33
Correct Answer C. 21.25
Explanation To find x, we can set the two given angle measures equal to each other and solve for x. So, we have 5x - 10 = 3x + 20. By subtracting 3x from both sides and adding 10 to both sides, we get 2x = 30. Dividing both sides by 2, we find that x = 15. Therefore, the answer is 15. However, this contradicts the given correct answer of 21.25. Without further information or clarification, it is difficult to determine the correct value of x.
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0
1
8.
If ∠ABC ≅ ∠EFG, and m∠ABC = 72, find m∠GFH.
A.
18
B.
72
C.
90
D.
108
Correct Answer A. 18
Explanation If ∠ABC ≅ ∠EFG and m∠ABC = 72, then the measure of ∠EFG is also 72 degrees. Since ∠GFH is vertical to ∠EFG, it will also have a measure of 72 degrees. Therefore, the correct answer is 18.
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9.
If m∠ABJ = 28, ∠ABC ≅ ∠DBJ, find m∠JBC.
A.
90
B.
56
C.
45
D.
34
Correct Answer D. 34
Explanation Since ∠ABC ≅ ∠DBJ and m∠ABJ = 28, we can conclude that m∠ABC = m∠DBJ = 28. Since the sum of the angles in a triangle is 180 degrees, we can calculate m∠JBC by subtracting the known angles from 180: 180 - 90 - 28 = 62. However, we need to find m∠JBC, not m∠BCJ. Therefore, we subtract 62 from 90 to get the final answer of 28.
Explanation The missing information is Division. In step 4, the equation is divided by 2 on both sides to isolate the variable x. This step is necessary to solve for x and find its value.
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4
11.
If m∠1 = x + 50 and m∠2 = 3x – 20, find m∠1.
Correct Answer 85
12.
State the definition, property, postulate, or theorem that justifies each statement.If M is the midpoint of , then ≅ .
Correct Answer Definition of Midpoint Midpoint Theorem
Explanation State the definition, property, postulate, or theorem that justifies each statement.
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4
13.
State the definition, property, postulate, or theorem that justifies each statement.If ∠A ≅ ∠B and ∠B ≅ ∠C, then ∠A ≅ ∠C.
Correct Answer Transitive Substitution
Explanation The given statement "If ∠A ≅ ∠B and ∠B ≅ ∠C, then ∠A ≅ ∠C" can be justified by the transitive property of congruence. According to this property, if two angles are congruent to a third angle, then they are congruent to each other. In this case, ∠A ≅ ∠B and ∠B ≅ ∠C, so by the transitive property, ∠A ≅ ∠C. The term "substitution" is not applicable in this context and does not justify the statement.
Explanation The given statement is justified by the property of substitution. According to this property, if two angles have a sum of 90 degrees and one of the angles is known to be 20 degrees, then the other angle can be found by substituting the known value into the equation. In this case, m∠A + m∠B = 90, and m∠B = 20, so by substituting 20 for m∠B in the equation, we get m∠A + 20 = 90. Therefore, the answer is substitution.
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4
15.
State the definition, property, postulate, or theorem that justifies each statement.If ∠X and ∠Y are complementary, ∠Z and ∠Q are complementary, and ∠X ≅ ∠Z, then ∠Y ≅ ∠Q.
Correct Answer Complementary Angles Theorem
Explanation The Complementary Angles Theorem states that if two angles are complementary to the same angle or congruent angles, then they are congruent to each other. In this case, we are given that angle X and angle Z are congruent and angle X and angle Y are complementary, as well as angle Z and angle Q are complementary. Therefore, by applying the Complementary Angles Theorem, we can conclude that angle Y and angle Q are congruent.
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4
16.
State the definition, property, postulate, or theorem that justifies each statement.If ≅ then PR = QT.
Correct Answer Definition of Congruent Segments
Explanation The definition of congruent segments states that if two segments have the same length, then they are congruent. In this case, if segment PR is congruent to segment QT, it means that they have the same length. Therefore, PR must be equal to QT.
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4
17.
State the definition, property, postulate, or theorem that justifies each statement.AB + BC = AC
Correct Answer Segment Addition Postulate
Explanation The Segment Addition Postulate states that if point B is between points A and C on a line, then the sum of the lengths AB and BC is equal to the length of AC. In this case, AB and BC are added together to give the length of AC, which justifies the statement AB + BC = AC. | 677.169 | 1 |
Geometry – Diagonals of an n-Sided Polygon
In today's post, let us discuss sided polygons and the number of diagonals they have.
We will discuss the following:
1. How do we find the number of diagonals an sided polygon has?
2. How many diagonals are subtended by each vertex?
3. What happens when one or more vertices do not make diagonals?
4. How to handle cases when ignored vertices are adjacent/not adjacent?
First of all, a very basic question:
Given an sided polygon, how many diagonals will it have?
An sided polygon has vertices. If we join every distinct pair of vertices we will get lines. These lines include the sides of the polygon as well as its diagonals.
So the number of diagonals is given by .
Alternatively, think of it this way – every vertex makes a diagonal with vertices. It does not make a diagonal with itself, and the two vertices next to it on either side (since it forms sides with these two). So we get diagonals. But here, each diagonal is double counted, once for each of its two vertices. Hence we divide by 2 to get the actual number of diagonals. That is another way of arriving at
Let's take some examples to solidify this concept:
Example 1: How many diagonals does a polygon with 25 sides have?
No. of diagonals =
Example 2: How many diagonals does a polygon with 20 sides have, if one of its vertices does not send any diagonal?
The number of diagonals of a 20 sided figure
But one vertex does not send any diagonals. Each vertex makes a diagonal with other vertices – it makes no diagonal with 3 vertices: itself, the vertex immediately to its left, and the vertex immediately to its right. With all other vertices, it makes a diagonal.
So, we need to remove diagonals from the total.
Total number of diagonals if one vertex does not make any diagonals diagonals.
We hope everything done till now makes sense. Now let's we will give we a question with two solutions and two different answers. We have to find out the correct answer and explain why the other is wrong.
Question: How many diagonals does a polygon with 18 sides have if three of its vertices, which are adjacent to each other, do not send any diagonals?
Solution: We will use two different methods to solve this question:
Method 1: Using the formula discussed above
Number of diagonals in a polygon of 18 sides diagonals
Each vertex makes a diagonal with other vertices – as discussed before.
So, each vertex will make 15 diagonals.
Total number of diagonals if 3 vertices do not send any diagonals diagonals.
Method 2:
The polygon has a total of 18 vertices. 3 vertices do not participate so we need to make all diagonals that we can with 15 vertices.
Number of lines we can make with 15 vertices
But this 105 includes the sides as well. A polygon with 18 vertices has 18 sides. Since 3 adjacent vertices do not participate, 4 sides will not be formed. 15 vertices will have 14 sides which will be a part of the 105 we calculated before.
Total number of diagonals if 3 vertices do not send any diagonals
Note that the two answers do not match. Method 1 gives us 90 and method 2 gives us 91. Both methods look correct but only one is actually correct and the correct one is method 2. So then, what is the problem with method 1?
When we subtract 45 from 135 (for each of the 15 diagonals made by the 3 vertices which we need to ignore), we are double counting 1 diagonal in this figure of 45. We actually need to subtract only 44 diagonals. Which diagonal are we double counting? The one which connects 2 of the three ignored vertices. Try to make a polygon with a few vertices. Make a few diagonals. Remove 3 vertices next to each other. 2 of the three vertices which have a vertex between them will be joined by a diagonal. When we remove 15 diagonals for each vertex, we are removing that diagonal twice.
Hence, what we need to do is .
The correct answer is 91 diagonals.
Mind you, we assumed that the vertices which were removed were next to each other. If they are not, the answer would be different since there would be more double counting. Let's take an example of that.
Question: How many diagonals does a polygon with 18 sides have if three of its non-adjacent vertices, do not send any diagonals? (No two of the three vertices are next to each other)
Solution: Number of diagonals in a polygon of 18 sides diagonals
Each vertex makes a diagonal with other vertices – as discussed before.
So, each vertex will make 15 diagonals.
Since 3 vertices are not counted, they will not send out each of these 15 diagonals i.e. 45 diagonals. But in this figure of 45, we have double counted 3 diagonals. To understand this, say we number the vertices from 1 to 18. Say, we leave out vertices 1, 4 and 16.
In our total of 45, we have counted the diagonal of vertices 1 and 4 twice. We have also counted the diagonal of vertices 1 and 16 twice and diagonal of vertices 4 and 16 twice. So, we have double counted 3 diagonals in our figure of 45. We need to subtract only 42 diagonals from 135.
Total number of diagonals if 3 vertices do not send any diagonals diagonals.
Answer here is 93 diagonals.
Note that in the previous example, only 1 diagonal was double counted while here, 3 diagonals were double counted. Understand why – in the previous example, since the vertices were adjacent, they made sides of the polygon. So, assuming vertices 1, 2 and 3 were ignored, 1 and 2 joined to make a side of the polygon and 2 and 3 joined to make a side of the polygon. Only vertices 1 and 3 joined to make a diagonal and hence this diagonal was double counted. In our formula , we have already gotten rid of the sides so they don't come in the picture at all. The formula only gives us the number of diagonals. So, in the previous example, we had already ignored the two sides formed by adjacent vertices (when we used the formula ) and had to take care of double counting of only one diagonal. In this example, each ignored vertex made a diagonal with the other ignored vertex and hence we had to handle the double counting of 3 diagonals.
We hope all this is clear to you and you will be able to effortlessly handle any question regarding diagonals of polygons now. | 677.169 | 1 |
There are infinite many equilateral pentagons IJKLM, inscribed in a triangle ABC, as shown.
Essentially all these pentagons are parameterized by the angle (psi), determining the slope of side IJ to AB. Taking D arbitrary on AB and drawing DE parallel to NO (realizing the slope (psi)), we find E. Then from D and E we lay correspondingly segments equal to DE, finding points H and F. Then construct the isosceles with basis on HF and sides HG=GF=DE. This completes the construction of an equilateral pentagon DEFGH. The only shortcoming of the pentagon is that G is not on BC, in general. The clue is that all pentagons constructed that way, starting with an arbitrary point D on AB, are similar to each other and their vertices G lie on a fixed line [AL]. This determines point L on side BC, which, in turn, determines the pentagon IJKLM. Later e.g. by taking the similarity with center at A and ratio k = ML/HG and applying it to the pentagon DEFGH. The existence of G depends on the validity of HF < 2DE.
Info on animation: Points A, B, C are free movable (switch to the selection tool: ctrl+1). Moving them changes the shape of the triangle. Points N, O and D are modifiable through the tool [Select on contour] (ctrl+2). Moving N, O changes the slope (psi). Moving D changes DEFGH by similarity.
The subject handled here is a particular one of a more general problem, initiated at the DivisionProblem.html .
For the determination of all the equilateral pentagons inscribed in a triangle one can start by asking how are distributed the vertices on the sides of the triangle. The present case is the main one, for which the distribution is (2,2,1). There is also the case (1,1,3), which is degenerate. The pentagon has three of its vertices on one side of the triangle and degenerates to a quadrangle inscribed in the triangle and having three sides equal to t, say, and the other equal to 2t. This is handled in PentadivisionDegenerate.html . | 677.169 | 1 |
The equilateral triangle is a fascinating and significant geometric shape with a particular place in the geometry study. This specific triangle has three sides that are of the same length. The equilateral triangle differs from other kinds of triangles, which can have sides of varied lengths because of this special quality.
This article will thoroughly explore the unique attributes, characteristics, and mathematical formulas intrinsic to the equilateral triangle. Our investigation will encompass an in-depth analysis of its structural elements, angle measurements, and the interplay between its sides. Furthermore, we shall delve into precise methodologies for computing the equilateral triangle's area and perimeter. Additionally, we will shed light on this geometric shape's practical significance and applications across various domains.
Definition of the Equilateral Triangle
The term "Equilateral" is derived from the combination of "Equi" and "Lateral," both of which imply the notion of sides. Due to the same length of its sides an equilateral triangle is also referred to as a regular polygon or regular triangle, owing to the equal length of its sides. It is classified as an equilateral triangle when all three sides and angles are congruent—specifically, each angle within an equilateral triangle measures precisely 60 degrees.
Its angles are congruent, meaning they have the exact measurements, and all sides have identical lengths. Each angle in an equilateral triangle is precisely 60 degrees in length. Furthermore, the angle sum property of triangles holds, as the collective sum of these angles amounts to 180 degrees. The equilateral triangle has a sense of harmony, symmetry, and aesthetic appeal thanks to the equality of side lengths and angle measures.
The equilateral triangle is significant in many areas, including mathematics, architecture, engineering, and design, because of its unique properties and appealing features. It is regularly incorporated into structures and patterns to elicit a sense of harmony and visual attractiveness because of its symmetrical and proportionate nature in art and architecture.
Shape Description
An equilateral triangle is a straightforward yet beautiful geometric geometry that has three equal sides and three equal angles. It resembles a regular triangle with all sides having the same length. When the triangle's three corners meet, they form three straight lines. The equilateral triangle exhibits a balanced and symmetrical appearance, making it visually pleasing.
In terms of dimensions, an equilateral triangle possesses the following properties.
Sides
An equilateral triangle is characterized by having three sides of equal length. Consequently, the line segments connecting the triangle's three vertices are all the same length.
Angles
In an equilateral triangle, all three angles are congruent, meaning they have the same measure. Specifically, each angle within an equilateral triangle measures precisely 60 degrees. This uniform angle measure enhances the symmetry of the triangular shape, contributing to its visually pleasing appearance.
Vertex
The three corners or spots where the triangle's sides intersect are known as the vertices of an equilateral triangle. In an equilateral triangle, all three vertices are sharp and contribute to the triangular shape of the figure.
Lines and Segments
The sides of an equilateral triangle consist of straight lines that connect its vertices. These lines share the property of being congruent in length and collectively form the three edges of the triangle. Furthermore, the various segments within an equilateral triangle, such as the altitude (a perpendicular segment from a vertex to the opposite side) and the median (a segment connecting a vertex to the midpoint of the opposite side), exhibit specific relationships that arise from the triangle's inherent symmetry.
Symmetry
An equilateral triangle exhibits remarkable symmetry through multiple lines. The three medians of an equilateral triangle, which are segments connecting each vertex to the midpoint of the opposite side, intersect at a point known as the centroid. This centroid serves as the center of symmetry for the triangle.
Additionally, the equilateral triangle possesses rotational symmetry of order 3. This means it can be rotated by multiples of 120 degrees around its centroid and maintain its identical appearance. This rotational symmetry further highlights the balanced and harmonious nature of the equilateral triangle.
Regularity
The equilateral triangle belongs to the category of regular polygons. A regular polygon is characterized by having all sides and angles congruent. Specifically, in the case of the equilateral triangle, all three sides and angles are equal in measure, satisfying the conditions of a regular polygon.
Below is the geometrical diagram for an equilateral triangle.
Figure 1
Attributes
An equilateral triangle possesses specific attributes that distinguish it as a unique type. To identify an equilateral triangle, consider the following properties as indicators.
The sides of an equilateral triangle have equal measurements.
The angles of an equilateral triangle are congruent, and each measures 60 degrees.
In an equilateral triangle, if a perpendicular is drawn from any vertex to the opposite side, it will bisect the side into two equal lengths. Additionally, the perpendicular will divide the angle at the vertex into two halves, each measuring 30 degrees.
The orthocenter and centroid of an equilateral triangle coincide at the same point.
In an equilateral triangle, the medians, anglebisectors, and altitudes originating from each vertex are identical for all three sides.
The sum of all the angles in an equilateral triangle equals 180 degrees.
And many more.
Relevant Formulae
Side Length and Perimeter
The perimeter of an equilateral triangle can indeed be calculated by multiplying the length of any one side, denoted by "s," by three. So, the formula for the perimeter, denoted by "P," is P = 3s. This means the perimeter is three times the length of any one side of the equilateral triangle.
Interior Angles
In an equilateral triangle, each interior angle measures 60 degrees. This is so that the triangle's three angles, which are all congruent and each measure 60 degrees, must total up to 180 degrees. The equilateral triangle exhibits this consistent angle measurement, contributing to its symmetrical and balanced appearance.
Altitude and Height
In an equilateral triangle, the altitude is a line segment drawn from one vertex perpendicular to the opposite side. This altitude also serves as the height of the triangle. The height, denoted by "h," can be calculated using the formula:
$$h = \frac{\sqrt{3}}{2} \times s$$
Where s represents the length of the side of the equilateral triangle, by substituting the side length into the formula, one can determine the height of the equilateral triangle.
Area
The area of an equilateral triangle can be calculated using the formula:
$$A = \frac{s^2 \sqrt{3}}{4}$$
Where s represents the length of the side of the equilateral triangle. By substituting the side length into the formula, you can determine the area of the equilateral triangle. Thus, the area formula involves multiplying the side length squared by the square root of 3 and dividing the result by 4.
Centroid
The centroid of an equilateral triangle lies at the center of the triangle. Since all sides of the equilateral triangle are equal in length, finding the centroid is a straightforward process. To locate the centroid, perpendicular lines are drawn from each triangle vertex to the opposite side. These perpendicular lines, known as medians, are all equal in length and intersect at a single point, precisely the equilateral triangle's centroid. The centroid divides each median into two segments, with the segment connecting the vertex to the centroid being twice as long as the segment connecting the centroid to the midpoint of the opposite side. Below in Figure-2, the centroid for an equilateral triangle is depicted.
Figure-2: Equiletral triangle centroid.
Applications of Equilateral Triangle
Equilateral triangles find practical applications in various fields due to their unique properties and balanced structure. Here are a few notable applications.
Architecture and Engineering
Equilateral triangles play a crucial role in architecture and engineering, providing essential support for the structural integrity of buildings, bridges, and various structures. Their inherent equilibrium and balance allow them to withstand external forces, ensuring stability and durability effectively.
Computer Graphics and Design
Equilateral triangles serve as fundamental elements in the creation of three-dimensional objects and the generation of geometric patterns in the realm of computer graphics. Intricate and visually captivating designs can be meticulously crafted using these triangles, adding depth and complexity to digital imagery.
Trigonometry and Mathematics
Within trigonometry, equilateral triangles serve as foundational elements that enable the derivation of trigonometric functions and identities. These triangles play a crucial role in the study of ratios, angles, and the relationships between the sides of a triangle. By examining the properties and measurements of equilateral triangles, students and mathematicians can develop a deeper understanding of fundamental trigonometric concepts and expand their knowledge in this field of mathematics.
Robotics
Equilateral triangles find applications in robotics, especially in the design of robotic arms and mechanisms. The stability provided by equilateral triangles assists in precise movements, control, and efficient load distribution in robotic systems.
Surveying and Navigation
Equilateral triangles have practical applications in surveying and navigation. Land surveyors use the geometry of equilateral triangles to accurately measure angles and distances on the field. Similarly, navigational tools like compasses and astrolabes utilize equilateral triangles to determine direction and calculate bearings.
Traffic Signaling
Equilateral triangles are widely used in traffic signaling systems. Triangular road signs, such as yield signs and warning signs, often feature an equilateral shape. The distinctiveness of equilateral triangles helps drivers recognize important instructions and take appropriate actions on the road.
Crystallography
Equilateral triangles play a crucial role in crystallography, the study of crystal structures. The arrangement of atoms or molecules in certain crystal lattice structures can form equilateral triangles, aiding scientists in understanding the properties and behavior of crystals.
Chemistry
Equilateral triangles play a significant role in chemistry, particularly in examining crystal structures and determining atom arrangements within a lattice. By studying the symmetrical nature of equilateral triangles within the crystal lattice, scientists can gain insights into the properties and behavior of materials at the atomic level. Understanding the structure and composition of crystals, which in turn affects their physical and chemical properties, is crucial for improving our understanding of chemistry and related sciences. By utilizing equilateral triangles as a tool in crystallography, researchers can make significant contributions to various fields, including materials science, solid-state physics, and chemistry.
Exercise
Example 1
Find the area of the equilateral triangle given in Figure-3.
Figure-3
Solution
To find the area of an equilateral triangle, we can use the formula:
$$ A = \frac{s^2\sqrt{3}}{4} $$
Where A represents the area, and s represents the side length of the equilateral triangle.
In this case, the side length (s) of the equilateral triangle ABC is 4 cm.
Substituting the value of s into the formula, we get:
$$ A = \frac{4^2\sqrt{3}}{4} $$
Simplifying the expression, we have:
$$ A = \frac{16\sqrt{3}}{4} $$
$$ A = 4\sqrt{3}$$
Therefore, the area of an equilateral triangle ABC is $\boldsymbol{{4\sqrt{3}}}$square centimeters.
Example 2
What is the perimeter of an equilateral triangle with a side length of s = 8 units?
Solution
The perimeter of an equilateral triangle can be calculated by multiplying the length of any one side by three. In this case, the side length is s = 8 units.
P = 3 × s
P = 3 × 8
P = 24 units
Therefore, the perimeter of an equilateral triangle with a side length of 8 units is 24 units.
Example 3
What is the length (s) of each side of an equilateral triangle if its perimeter is 30 meters?
Solution
To find the length of each side of an equilateral triangle when the perimeter is known, we divide the perimeter by 3 since all sides are equal.
P = 30 meters
s = P/3
s = 30/3
s = 10 meters
Therefore, each side of the equilateral triangle is 10 meters long.
Example 4
Calculate the height(h) of an equilateral triangle with a side length of 16 inches.
Solution
To calculate the height of an equilateral triangle, we can use the formula:
$$h = \frac{\sqrt{3}}{2} \times s$$
Given that the side length is 16 inches, we can substitute this value into the formula:
$$h = \frac{\sqrt{3}}{2} \times 16 \ \textrm{inches}$$
$$h = \frac{1.732}{2} \times 16 \ \textrm{inches}$$
h = 0.0866 × 16 inches
h = 13.856 inches
Therefore, the height of the equilateral triangle with a side length of 16 inches is approximately 13.856 inches.
Example 5
If the area of an equilateral triangle is 36 square units, what is the length (s) of each side?
Solution
To find the length of each side of an equilateral triangle given its area, we can use the following formula:
$$s = \sqrt{\frac{4 \times A}{\sqrt{3}}}$$
Given that the area is 36 square units, we can substitute this value into the formula:
$$s = \sqrt{\frac{4 \times 36}{\sqrt{3}}}$$
$$s = \sqrt{\frac{144}{\sqrt{3}}}$$
$$s = \frac{\sqrt{144}}{\sqrt{\sqrt{3}}}$$
$$s = \frac{12}{\sqrt{\sqrt{3}}}$$
Simplifying the expression $\sqrt{\sqrt{3}}$, we get:
s = 121.732
s = 6.928
Therefore, the length of each side of the equilateral triangle with an area of 36 square units is approximately 6.928 units.
Example 6
For the equilateral triangle given in Figure-4, calculate the length (s) of each side of the triangle.
Figure-4
Solution
In an equilateral triangle, the centroid divides each median (segment connecting a vertex to the midpoint of the opposite side) in a 2:1 ratio. The length of each side of the triangle can be calculated as follows, given that the distance between each vertex and the centroid is 5 units:
Let's use x to symbolize each side's length.
From the centroid to each vertex, we have two segments: one segment of length 2x/3 and another segment of length x/3 (since the ratio of lengths is 2:1).
According to the given information, the length of each segment is 5 units. Therefore, we can set up the following equation:
2x/3 = 5
To solve for x, we can multiply both sides of the equation by 3/2:
2x/3 × 3/2 = 5 × 3/2
x = 7.5
Hence, each side of the equilateral triangle is 7.5 units.
Example 7
How many sides are there in total when an equilateral triangle is divided into smaller congruent equilateral triangles by connecting the midpoints of its sides, given that there are 16 small equilateral triangles?
Solution
Let's start by considering the original equilateral triangle. It has three sides.
When we connect the midpoints of its sides, we create four smaller equilateral triangles within it. Each of these smaller triangles also has three sides.
If we continue this process, we can see that the number of smaller triangles doubles with each iteration: 1, 4, 8, 16.
Since there are 16 small equilateral triangles, we can calculate the total number of sides in all the triangles combined as follows:
Number of sides = Number of triangles × Number of sides per triangle
Number of sides = 16 × 3
Number of sides = 48
Therefore, there are a total of 48 sides in all the triangles combined. | 677.169 | 1 |
56. How can you add eight 8's to get a final result of 1000? Answer: 888 + 88 + 8 + 8 + 8
57. What is the next number in the sequence 22, 21, 23, 22, 24, 23, _? Answer: 25
58. Does a concave shape curve inward or outward? Answer: Inward!
59. How would you write 0.81 as a percent? Answer: 81%
60. What is a polygon with eight sides called? Answer: Octagon!
61. What is the branch of mathematics that deals with the study of shapes called? Answer: Geometry!
62. What is the vertical axis in a graph called? Answer: Y-axis!
63. What is the perimeter of a rectangle? Answer: 2(L+B)
64. There are 10 pairs of white socks and 15 pairs of navy socks in a drawer. What is the ratio of white to navy socks? Answer: 2:3
65. Which is the longest side in a right angled triangle? Answer: The side opposite to the right angle is the longest and is called the hypotenuse!
66. How many sides does a rhombus have? Answer: Four!
67. What is the 3-D equivalent of a two dimensional rectangle called? Answer: A Cuboid!
Math Quiz Questions With Answers For High School
Gear up for high school math quiz with these exciting mathematics questions. Perfect for school competitions, this Math hard quiz for kids with answers is sure to test students' understanding of numbers, shapes, patterns and all things maths!
67. Can a right angled triangle have an obtuse angle? Answer: No! The sum of remaining angles must be equal to 90, making it impossible for a right angled triangle to have an obtuse angle.
68. How many zeros are there in a googol? Answer: 100 zeros!
69. What is the smallest perfect number? Answer: Six! A perfect number is a positive integer which is equal to the sum of its proper divisors.
70. What branch of mathematics deals with the study of numbers? Answer: Arithmetic
111. What is the name given to rectangles whose sides have a particular ratio of about 1.6 to 1? Answer: Golden rectangle!
112. What is the branch of mathematics that deals with the study of continuous change and motion called? Answer: Calculus!
113. What is the study of triangles and the relationship between their angles and sides called? Answer: Trigonometry!
114. What is a fraction with a numerator greater than the denominator called? Answer: Improper fraction!
115. What is the probability of rolling two dice and number five on both? Answer = 1/36
116. What does the sigma sign mean in maths? Answer: Sum of a number of similar terms.
117. What is the value of Pythagoras' Constant? Answer: Square root of 2!
118. What is the highest honour awarded in the field of mathematics to young mathematicians? Answer: The Fields Medal. It his awarded every four years to young mathematicians under the age of 40 years.
Math History Quiz For Kids
These quiz questions will test your knowledge of not math concepts but math history! From the founding fathers of mathematical concepts to great math brains, this math history quiz covers interesting history math facts!
119. Who was the first woman to win the Fields Medal? Answer: Maryam Mirzakhani! She won the Fields Medal in 2014 for her work on the dynamics and geometry of Riemann surfaces and their moduli spaces. She was widely known for her groundbreaking work on the theory of differential equations as well!
124. Who is known as the father of probability? Answer: Blaise Pascal and Pierre de Fermat in 1654!
125. Who is credited as the world's first computer programmer? Answer: Ada Lovelace, an English mathematician! She worked with Charles Babbage on his early mechanical general-purpose computer, the Analytical Engine, and wrote the first algorithm intended to be processed by a machine.
130. Who was the first woman to receive the prestigious Abel Prize? Answer: Karen Uhlenbeck. She received the Abel Prize in 2019 for her groundbreaking work in geometric partial differential equations, gauge theory, and integrable systems.
131. Who is considered the father of geometry? Answer: Euclid!
132. Which mathematician was nicknamed 'Human Computer'? Answer: Shakuntala Devi. The Indian woman mathematician won twice against a computer in a speed calculation race in 1977.
133. Who is the founder of Game Theory? Answer: John von Neumann
134. Who is the youngest recipient of the Fields Medal? Answer: French mathematician Jean-Pierre Serre who received the honour at the age of 27 years in 1954.
135. Who is known as the father of trigonometry? Answer: Ancient Greek mathematician Hipparchus!
Math Quiz Questions
FAQ
What are some fun Math quiz questions with answers for kids?
1. How many zeros are there in a googol? Answer: 100 zeros! 2. What is this Roman numeral 'MIX' equal to? Answer: 1009! 3. What is the value of Pythagoras' Constant? Answer: Square root of 2! 4. What is the probability of rolling two dice and getting the same number on both? Answer: 1/6 5. What is the sum of the exterior angles in a decagon? Answer: 360 degrees! 6. How many zeros are there in a quintillion? Answer: 18! 7. Who is considered the father of geometry? Answer: Euclid! 8. What is the square of 111,111,111? Answer: 12345678987654321! | 677.169 | 1 |
Hint: Use the property of the triangle that the sum of the angles of the triangle is $180^\circ $in the triangle AQM, it gives: \[\angle AQM{\text{ }} + \angle QAM + \angle AMQ{\text{ }} = {\text{ }}180^\circ \] After that use the property of the sum of the linear pairs of angles, which says that if we have a pair of linear angles then their sum is$180^\circ $. | 677.169 | 1 |
Consider triangle ABC a point D and the cevian triangle A1B1C1 of D (with vertices the traces of {DA,DB,DC} on opposite sides). Let also L be an arbitrary line in general position with respect to the side-lines of the triangle (i.e. no three lines out of the four pass through the same point). Let {A*, B*, C*} be correspondingly the intersection points of {B1C1, C1A1, A1B1} with L. Draw lines {AA*, BB*, CC*} and form through their intersections triangle A0B0C0. Then lines {AA0, BB0, C0} intersect at a common point D'.
The property of the previous section has an important application concerning cevian and circumscribed triangles formed by drawing parallels to the sides of a cevian triangle.
Let A'B'C' be the cevian triangle of point D with respect to triangle ABC. Draw from the vertices of ABC parallels to the sides of the triangle to form triangle A0B0C0 (B0C0 parallel to B'C', etc.). Then triangles ABC and A0B0C0 are perspective i.e. lines {AA0, BB0, CC0} concure at a point D'.
The proof follows from the property of the previous section by applying it to the case in which the line L is the line at infinity, hence the points {A*, B*, C*} appearing there are here the "common" points of the pairs of parallels {(B'C',B0C0), (C'A',C0A0), (A'B',A0B0)}.
Given two triangles ABC and A'B'C' we can in general inscribe and circumscribe in ABC triangles t1= A1B1C1 and t2=A2B2C2, both similar to the "prototype" A'B'C' so that their sides are parallel. Then rotate the system of {t1,t2} so that the triangles remain all the time inscribed/circumscribed and with sides corresondingly parallel. The result of previous section shows that when the system of three lines {AA1,BB1,CC1} becomes concurrent then the same happens with the system of three lines {AA2,BB2,CC2}. | 677.169 | 1 |
Prove that the tangents drawn at the end points of a chord of a circle make equal angles with the chord.
Video Solution
|
Answer
Step by step video & image solution for Prove that the tangents drawn at the end points of a chord of a circle make equal angles with the chord. by Maths experts to help you in doubts & scoring excellent marks in Class 10 exams. | 677.169 | 1 |
Proof
Since $OF\perp AB$ and $OG\perp AD$ the quadrilateral $ABOD$ is cyclic. Let $K$ be its circumcenter and define $L$ and $M$ as the intersection of its circumcircle $(K)$ with the given angles bisectors. $OK$ serves as a diameter of $(K).$
In the above diagram circle $(K)$ is divided into six arcs, with angle measurements $\alpha,$ $\beta,$ $\gamma,$ $\delta,$ $\epsilon,$ $\zeta$ that satisfy the following conditions: | 677.169 | 1 |
Elements of Plane and Spherical Trigonometry: With Practical Applications
61. In general, any one of the six trigonometric ratios of an angle being given, the relations expressed by the foregoing formulæ will enable us to find the value of all the rest. These are termed the elementary formulæ, and are collected in the following
62. To find the SINE and COSINE of the SUM of two angles by
means of their sines and cosines.
I
G
D
Let the two angles be COD, DOE. In the line OE take any point, E, draw EF perpendicular to OC, and ED perpendicular to O D. Draw D G perpendicular to EF, and DC perpendicular to OC. The triangles GED and COD have their sides perpendicular, hence they are similar (Geom., Prop. XXV. Bk. IV.), and the angles DEG and COD are equal. Let a = COD = DEG, and b— DO E.
F C
63. To find the SINE and COSINE of the DIFFERENCE of two
angles by means of their sines and cosines.
Let the two angles be FOD, D O E. In the line OE take any point, E, draw EF perpendicular to OF, and ED perpendicular to OD. Draw DG perpendicular to FE produced, and DC perpendicular to OF. The triangles GED and COD have their sides perpendicular, hence they are similar (Geom., Prop. XXV. Bk. IV.), and the angles DEG and COD are equal.
Let a COD = DE G, and b = = DOE.
D
C
F
G
E
64. The four formulæ last established apply to arcs as well as angles, and may be considered the fundamental formula of subsequent analysis. They are brought together in the following
read line 3
SIGNS OF THE TRIGONOMETRIC FUNCTIONS.
65. If on any line, straight or curved, different distances be measured from a fixed point of origin, the distances which have contrary situations may by convention be introduced into our calculations, by affecting the quantities representing their magnitudes by contrary signs.
A'
Let O be a fixed point in any line, A A", and suppose we have to determine the situations of other points in this line with respect to 0. The position of any point in the line will be known if we know the distance of the point from 0, and also know on which side of O the point lies. Now it is found convenient to adopt the following convention: distances measured in one direction from
A"
A
O along the line will be denoted by positive quantities, and distances measured in the contrary direction from O will be denoted by negative quantities.
Thus, for example, suppose that distances measured from O towards the right are denoted by positive numbers, and let A be a point, the distance of which from O is denoted by 2 or +2; then if A" be a point situated just as far to the left of O as A is to the right, the distance of A" from O will be denoted by 2.
In like manner, if distances originating in A A", and taken along O A', or only parallel to O A', when measured upwards be denoted by positive quantities, on being measured downwards they will be denoted by negative quantities.
66. A similar convention may conveniently be adopted with respect to angles.
Let any line, O B, revolve from the position OA, in one direction round 0, forming the angle BO A, and let this angle be denoted by a positive quantity; then, if the line OB revolve,
A
A'
B
A
B'
A"
from the position OA, round O in the contrary direction, forming the angle BOA, this angle may be denoted by a negative quantity.
Thus, for example, if each of the angles AOB and A OB is two ninths of a right angle, and we denote the former by 20° or +20°, the latter may be denoted by — 20°.
The direction of the positive distances is quite indifferent; but, being once fixed, the negative distances must lie in the contrary direction.
67. The representation of angles as the measure of the revolution of a line, turning in a plane about one of its own points, leads to the consideration of angles, not only greater than two right angles, but of all possible magnitudes.
Thus, when the line OB, starting from the initial position OA, has passed A", or made more than half a revolution, we have described an angular magnitude
of more than 180°; and when it has
passed on to A, we have an angular magnitude of 360°. If it now contin
ues to revolve in the same direction till A it arrives again at B, we have an an
gular magnitude of 360°+20°380°,
and thus we may conceive of angles
A'
B
A
'B'
A"
of all magnitudes. In like manner negative angles of all magnitudes may be formed by the describing line OB revolving from O A, but in a contrary direction. 68. The algebraic signs of the trigonometric functions can be readily fixed in the mind by being represented geometrically. Thus,
Let the extremity of a revolving line, starting from the initial position O A, describe the positive arc AB less than 90°, A B' between 90° and 180°, A A'B' between 180° and 270°, and A A'A"B" between 270° and 360°. Then, according to the definitions of Art. 54, BD, B'D', | 677.169 | 1 |
Python Example – Write a Python program to create a Pythagorean theorem calculator
(Python Example for Beginners)
Write a Python program to create a Pythagorean theorem calculator.
Note : In mathematics, the Pythagorean theorem, also known as Pythagoras' theorem, is a fundamental relation in Euclidean geometry among the three sides of a right triangle. It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
Pythagorean theorem calculator! Calculate your triangle sides.
Assume the sides are a, b, c and c is the hypotenuse (the side opposite the right angle
Which side (a, b, c) do you wish to calculate? side>a
Input the length of side b:10
Input the length of side c:15
The length of side a is
11.180339887498949
Python Example – Write a Python program to create a Pythagorean theorem calculator | 677.169 | 1 |
Points lines and planes answer key. 1.) Points One great way to start your points lines and planes in geom...
LessAnswerChapter 1 – Points, Lines, and Planes, Segments & Angles. Chapter 1 Test Review – Click ... Chapter 4 Test Review Answer Key – Click ... 90Plane HKP and plane RKP are two distinct planes. Name the intersection of plane HKP and plane RKP. <-->. KP. Study with Quizlet and memorize flashcards containing terms like What are the names of three collinear points?, Name the line and plane shown in the diagram., --> What is the name of the ray that is opposite BA and more.Use the diagram to answer the following questions. z 3. Use the diagram to answer the following questions. a) How many points appear in the figure? _____ b) How many lines appear in the figure? _____ c) How many planes appear in the figure? _____ d) Name three collinear points. _____ e) Name four non-coplanar points. _____ f) Give another name ...To summarize, some of the properties of planes include: Three points including at least one noncollinear point determine a plane. A line and a point not on the line determine a plane. The intersection of two distinct …Answer to Solved Name 1.1 Points, Lines, and Planes Naming Practice HW 2 rays that share the same endpoint and form a line. postulate or axium. an accepted statement of fact. postulate 1-1. through any 2 points there is exactly 1 line. postulate 1-2. if 2 distinct lines intersect, then they intersect in exactly 1 point. postulate 1-4. through any 3 noncollinear points, there is exactly 1 plane.II. Points, Lines, and Planes. The study of formal geometry begins with what we called the "undefined terms" of geometry: point line plane. Parallel lines are lines which lie in the same plane, but never meet, no matter how far they are extended. Parallel planes never meet, no matter how far they are extended. Answer1. Let's Practice! 2. Unit 1: Points, Lines, and Planes Homework 3. Lines and Angles - 4. GINA WILSON ALL THINGS ALGEBRA 2014 ANSWERS PDF 5. gina wilson all things algebra 2013 answers 6. Geometry Unit 10 Notes Circles 7. Geometry Unit Answer Key 8. 1.1 Identify Points, Lines, and Planes -.Perpendicular lines are those that form a right angle at the point at which they intersect. Parallel lines, though in the same plane, never intersect. Another fact about perpendicular lines is that their slopes are negative reciprocals of o...Terms in this set (23) collinear points. points that lie on the same line. coplanar points. points that line of the same plane. line segment. part of a line, doesn't have arrows. ray. a portion of a line which starts at a point and goes off in a particular direction to infinity.PlanesTitle Tutor-USA.com Worksheet Author: Tutor-USA.com Subject: Tutor-USA Worksheet Created Date: 10/3/2008 4:13:03 AMAnswer to Solved Name 1.1 Points, Lines, and Planes Naming Practice HWThis worksheet will help learners in deepening their knowledge about understanding points, lines, and planes. The activities in this worksheet will practice the learner's understanding and comprehension of points, lines, and planes. In addition to this, they can also practice their solving in the activities. Lastly, the answer key in the last ...Points Lines and Planes Worksheets. Points lines and planes worksheets can be used when a student is newly introduced to the topic of Geometry. Points, lines and planes form the basis of Geometry. The worksheets usually include problems based on collinear and coplanar concepts, descriptive charts, naming problems and the intersection of planes.A Line in three-dimensional geometry is defined as a set of points in 3D that extends infinitely in bothOne great way to start your points lines and planes in geometry lesson is to tell them to actually draw a point either on their paper or have one …Answer Key Observe the !gure and answer the followingone only one line perpendicular to the plane through that point Aug 26, 2013 · 1A. Points, Lines, and Planes A location in space, but has no size or shape A B Extends without end in one dimension (two directions) and always Called AB straight or line l A B C Extends without end in two dimensions (all directions), always flat, and has no thickness Called plane ABC or plane M Called point A M l A point is usually named by a capital letter. A line passes through two points. Lines consist of an infinite number of points. A line is often named by two points on the line or by a lowercase script letter. A plane is a flat surface that extends indefinitely in all directions. Planes are modeled by four-sided figures.Points, Lines, and Planes. 17. Draw a figure to fit each description. a. Through any two points there is exactly one line. b. Two distinct lines can ... Points, Lines & Planes Powerpoint knoxbaggett 17.3K views•8 slides. Point, Line and plane Kristine Joy Ramirez 1.8K views•40 slides. Basics Of Geometry 1 mpscils598s07 24.9K views•21 slides. Parallel lines and transversals Leslie Amoguis 8.5K views•34 slides. Polygons presentation Jessica 79.1K views•31 … Used16 jun 2015 ... For example, geometry plays a key role in construction, fashion design, architecture, and computer graphics. ... The answer is B. Drawing and ...Answer Key Name each gure using symbols. Part - A ... Points, Lines and Planes Printable Worksheets @ Name : Sheet 1. Created Date: Collinear points lie on the same line. Coplanar points lie on the same plane. plane CDE; Planes have two dimensions. Lines, line segments, and rays have one dimension. Sample answer: A, B, D, E DE, BC plane S plane T Q W, line g Sample answer: plane RST R, Q, S; Sample answer: T DB CA AC EB, EC, ED, EA EB and ED, EA and EC Sample answer: EC and ED LessThen click the add selected questions to a test button before moving to another page. Write the letter of the angle relationship next to each pair of angles based on the image below. Choose the correct missing reason from step 3. Given the parallel lines k k and m m, and transversal t t, prove that ∠1 ≅ ∠7 ∠ 1 ≅ ∠ 7. 1.point A line has one dimension. It extends without end in two directions. It is represented by a line with two arrowheads. line l or BC ^& *( l plane has two dimensions. It is represented by a shape that looks like a floor or wall. You have to imagine that it extends without end. plane M or plane DEF M D E FANSWMath. Algebra. Algebra questions and answers. Name 1-1 Points Lines and Planes PAP Day 2 1) Are N, P, Q, and coplanar? 2) Are N. l. P. and Q coplanar? 3) Are N, M, and copianar? 4) Name in two different ways the plane that contains F.C.D. and E 5) Are A, B, D, and Ecoplanar? 6) Are B and H collinear?These points, lines, and planes worksheets are ideal for 8th grade and high school students. Identifying and Naming Points, Lines, and Planes ... Members can download the answer key and cross-check the answers instantly! Members can share the worksheet with students instantly via WhatsApp, Email, or Google Classroom.Obj.: Use postulates involving points, lines, and planes. Key Vocabulary • Line perpendicular to a plane - A line is a line perpendicular to a plane if and only if the line intersects the plane in a point and is perpendicular to every line in the plane that intersects it at that point.This bundle includes 29 self-checking geometry mazes (each topic includes 2 or 3 mazes). Each maze is printable and includes an answer key.Topics included in this product:1. Interior and Exterior Angles of Polygons - 15 problems2. Midpoint and Distance - 17 problems3. Parallel Lines and Transversals. 14.We would like to show you a description here but the site won't allow us.One great way to start your points lines and planes in geometry lesson is to tell them to actually draw a point either on their paper or have one …Apr 19, 2021 · Lesson 1.1 Points, Lines, and Planes. Monitoring Progress. Question 1. Use the diagram in Example 1. Give two other names for . Name a point that is not coplanar with points Q. S, and T. Answer: The given plane is: The definition of a ray is: A ray has no starting and ending points So, The other names for \(\overline{S T}\) are: Line m and ...: Use the figure to name each of the following. a. a line containing point A The ...1 P LL ® ate ame Points, Lines & Planes • mathantics.com PLP 1. Instructions: Match each basic element of geometry with the correct picture. Basic Elements of GeometryJul 15, 2022 · Log In. The resource you requested requires you to enter a username and password below:901.1 POINTS, LINES, and PLANES DATE:_____ Use the figure on the right to answer 1-4. 1. Name a ray. 2. Name a line segment. 3. Name a pair of opposite rays. 4. Draw 𝑆𝑍⃖ ⃗ on the figure. Use the figure on the right to answer 5-12. NOTE: Line j intersects plane S at point T. Point M is not coplanar with plane S. 5. Name 3 collinear ...1 P LL ® ate ame Points, Lines & Planes • mathantics.com PLP 1. Instructions: Match each basic element of geometry with the correct picture. Basic Elements of GeometryCourse: Geometry (all content) > Unit 1 Lesson 4: Points, lines, & planes Specifying planes in three dimensions Points, lines, and planes Math > Geometry (all content) > Lines > Points, lines, & planes Points, lines, and planes Google Classroom What is another way to name line ℓ ? F B W I Y ℓ R Choose 1 answer: B F ↔ A B F ↔ I F ↔ B I F ↔ F B W ↔ C. Points, Lines, and Planes Points, Lines, and PlaneLog In. The resource you requested requires you to enter a use Planes Line Lesson 1-4 Segments, Rays, Parallel Lines Used to build the definitions of other figures. Point – in... | 677.169 | 1 |
This code calculates the distance between angles, particularly for n-tuples of angles. One example where this situation occurs is as follows:
I'm using a 2 arms, one 6 degree of freedom and the other 7 dof with revolute (rotating) joints. The position of the arm can be accurately represented as an n-d point representing the angle of each joint, with no need for orientation/pose on the surface of a torus, with 1 dimension for each joint.
1 Answer
1
This code is very hard to read. These are very simple geometric operations that you are programming (just things like subtracting vectors, or finding a vector's squared length) but their meaning is buried under a mountain of boilerplate. | 677.169 | 1 |
What is the exact value of sin 60. Find the Exact Value sin(120) Step 1. Apply the reference...
Find exact value of sin(105)Ans:= 4 6+ 2We know sin(A+B)=sinAcosB+cosAsinBHence sin105 ∘=sin(60 ∘+45 ∘)=sin60 ∘cos45 ∘+cos60 ∘sin45 ∘= 2 3× 21+ 21× 21= 2 23+1× 22= 4 6+ 2.Exact values of the ordered pairs of the unit circle is represented in pictorial form above in degrees and radians. ... What is the exact value of #sin 60 - cos 60#? What is the exact value of #cot 240#? What is the exact value of #sec 210#? If the terminal side of angle theta passes through the point ...Jun 8, 2015 · What Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. This video works to determine the exact values for the sin(30), cos(30), tan(30), sin(60), cos(60), and tan(60) using an equilateral triangle and the accompa...The exact value of sin 180 is zero. Sine is one of the primary trigonometric functions which helps in determining the angle or sides of a right-angled triangle. It is also called trigonometric ratio. ... 60: 90: 180: 270: 360: Angles (In Radians) 0: π/6: π/4: π/3: π/2: Note: Since, sine is an odd function, the value of sin(-240°) = -sin(240°). Methods to Find Value of Sin 240 Degrees. The sine function is negative in the 3rd quadrant. The value of sin 240° is given as -0.86602. . .. We can find the value of sin 240 degrees by: Using Trigonometric Functions; Using Unit Circle; Sin 240° in Terms of ... Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor.The problem reads as follows. An exact value for cos36° can be found using the following procedure. Begin by considering sin108°. Note that 108 = 72 + 36 and use the sine sum identity. Also note that 72 = 2 ⋅ 36 and use double angle identities.Find the Exact Value sin(45 degrees ) Step 1. The exact value of is . Step 2. The result can be shown in multiple forms. Exact Form: Decimal Form:This video works to determine the exact values for the sin(30), cos(30), tan(30), sin(60), cos(60), and tan(60) using an equilateral triangle and the accompa...It can be written in the form of sin(90 + 60) As it is in the 2nd quadrant and sin terms are positive in the second quadrant. so sin(150) = sin(90+60) = cos(60) =1/2. Example: What is cos(150) equal to? ... Find the exact value of sin 150 degrees If sin θ = 12/13, find the value of sin² θ- cos² θ/2 sin θ.cos θ x 1/tan² θ ...sqrt3 Use trig identity: tan 2a = (2tan a)/(1 - tan^2 a) In this case: tan (60) = (2tan (30))/( 1 - tan^2 30) Trig table gives --> tan 30 = sqrt3/3 --> tan^2 30 = 3/9 ...Trigonometry Find the Exact Value sin (105) sin(105) sin ( 105) Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. sin(75) sin ( 75) Split 75 75 into two angles where the values of the six trigonometric functions are known. sin(30+45) sin ( 30 + 45) Apply the sum of angles identity. Solve for ? sin (x)=1/2. sin(x) = 1 2 sin ( x) = 1 2. Take the inverse sine of both sides of the equation to extract x x from inside the sine. x = arcsin(1 2) x = arcsin ( 1 2) Simplify the right side. Tap for more steps... x = π 6 x = π 6. The sine function is positive in the first and second quadrants. To find the second solution, subtract ...Find the value of given trigonometric ratio. Given trigonometric ratio: sin 75 ∘. sin 75 ∘ can be expressed as, sin 75 ∘ = sin ( 45 ∘ + 30 ∘) We know that sin ( A + B) = sin A cos B + cos A sin B. So by applying the above formula we get, sin 75 ∘ = sin 45 ∘ cos 30 ∘ + cos 45 ∘ sin 30 ∘. As, sin 45 ∘ = 1 2, cos 30 ∘ = 3 2 ... (For cos 300 degrees, the angle 300° lies between 270° and 360° (Fourth Quadrant ). Since cosine function is positive in the fourth quadrant, thus cos 300° value = 1/2 or 0.5. Since the cosine function is a periodic function, we can represent cos 300° as, cos 300 degrees = cos (300° + n × 360°), n ∈ Z. ⇒ cos 300° = cos 660° = cos ...Method 2. Another method to derive the value of sineo 120 degrees is by using cosine functions. With the help of the trigonometry formula, sino (90 + a) = coso a ,we can determine sin 120 exact value. As we know, sino (90° + 30°) = sino 120 degrees. Hence, sin 120 degree = cos 30°.Revise trigonometric ratios of sine, ... 60° and 90 ° The trigonometric ... An equilateral triangle with side lengths of 2 cm can be used to find exact values for the trigonometric ratios of 30 ...Trigonometry. Find the Exact Value sin (pi/8) sin( π 8) sin ( π 8) Rewrite π 8 π 8 as an angle where the values of the six trigonometric functions are known divided by 2 2. sin( π 4 2) sin ( π 4 2) Apply the sine half - angle identity. ± ⎷ 1−cos(π 4) 2 ± 1 - cos ( π 4) 2. Change the ± ± to + + because sine is positive in the ...I have noticed that students cannot actually remember values of six trigonometric ratios (sin, cos, tan, cosec, sec and cot) for 0. , 30. , 45. , 60. and 90. . These values are used very often and it is recommended from my point of view that student should be able to tell the values instantly when asked. There is a proper method to memorize all ...How to Find the Value of Sin 600 Degrees? The value of sin 600 degrees can be calculated by constructing an angle of 600° with the x-axis, and then finding the coordinates of the corresponding point (-0.5, -0.866) on the unit circle. The value of sin 600° is equal to the y-coordinate (-0.866). ∴ sin 600° = -0.866.Hint: To find the exact value of the given function, at first, we will find the value of \[\sin {60^ \circ }\]and \[\cos {60^ \circ }\]separately. Then using these values we can …sin(60°) = sin( π 3) = √3 2. → sin(60°) −cos(60°) = √3 2 − 1 2 = √3 − 1 2. Answer link. sin (60°)-cos (60°)= (sqrt3-1)/2 The exact values of cos (60°) and sin (60°) …Trigonometry Find the Exact Value sin (60 degrees ) sin(60°) sin ( 60 °) The exact value of sin(60°) sin ( 60 °) is √3 2 3 2. √3 2 3 2 The result can be shown in multiple forms. Exact Form: √3 2 3 2 Decimal Form: 0.86602540… 0.86602540 … The exact value of sin(45) sin ( 45) is √2 2 2 2. − √2 2 - 2 2. The result can be shown in multiple forms. Exact Form: − √2 2 - 2 2. Decimal Form: −0.70710678… - 0.70710678 …. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just ...Therefore, the value of sin 180 degrees = 0. The value of sin pi can be derived from some other trigonometric angles and functions like sine and cosine functions from the trigonometry table. It is known that, 180° – 0° = 180° ———– (1) 270° – 90° = 180°———— (2) Sin 60 Values In trigonometry, there are three main functions (or ratios) that measure the angles and lengths of a right-angled triangle. The three ratios are sine, cosine, and tangent which are abbreviated as sin, cos, and tan, respectively. Trigonometric ratios have a lot of functions out of which solving a triangle is the most important purpose.However, as you might (and should!) remember, sine is a periodic function, so there are multiple numbers that have the same sine value. For example sin(0) = 0, but also sin(π) = 0, sin(2π) = 0, sin(-π) = 0 and sin(-326π) = 0. Therefore, if somebody wants to calculate arcsin(0), the answer can be 0, 2π (360°), or -π (-180°), to name a Find the Exact Value sec (60 degrees ) sec(60°) sec ( 60 °) The exact value of sec(60°) sec ( 60 °) is 2 2. 2 2. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor.Step 1: Create a table with the top row listing the angles such as 0°, 30°, 45°, 60°, 90°, 180°, 270° and 360°and write all trigonometric functions in the first column such as sin, cos, tan, cosec, sec, cot. Step 2: Determining the value of sin: Write the angles 0°, 30°, 45°, 60°, and 90° in ascending order. The values of sin forRevise trigonometric ratios of sine, ... 60° and 90 ° The trigonometric ... An equilateral triangle with side lengths of 2 cm can be used to find exact values for the trigonometric ratios of 30 ...Find the Exact Value sin (300) | MathwayThis video explains how to find the exact values of trig functions such as sine, cosine, and tangent using the 30-60-90 and 45-45-90 reference triangles and ...4. Write down the exact value of Sin 0°..... (1) 5. Write down the exact value of Cos 60°..... (1) 6. Write down the exact value of Sin 30°..... (1) 7. Write down the exact value of Tan 0°..... (1) 8. Write down the exact value of Tan 45°..... (1) 9. Write down the exact value of Cos 90°..... (1) 10. Write down the exact value of Sin 90 ...Find the Exact Value sin(pi/4) Step 1. The exact value of is . Step 2. The result can be shown in multiple forms. Exact Form: Decimal Form:: Get the latest SIN CARS INDUSTRY AD Registered Shs stock price and detailed information including news, historical charts and realtime prices. Indices Commodities Currencies StocksFind the Exact Value sin (600) sin(600) sin ( 600) Remove full rotations of 360 360 ° until the angle is between 0 0 ° and 360 360 °. sin(240) sin ( 240) Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the third quadrant. −sin(60 ...Find the Exact Value sin (225) sin(225) sin ( 225) Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the third quadrant. −sin(45) - sin ( 45) The exact value of sin(45) sin ( 45) is √2 2 2 2. − √2 2 - 2 2. The result can be shown in ...When it comes to determining used bicycle values, there are several venues that you can check. Before you get started, figure out the exact model and year of your bike to locate accurate information. This is more important when figuring how...The exact value of tan 30° is 0.57735. Tan 30° = 1/√3 = 0.57735. The value of tangent of angle 30 degrees can also be evaluated using the values of sin 30 degrees cos 30 degrees. Let us learn in detail in this article. …Find the Exact Value tan(60 degrees ) Step 1. The exact value of is . Step 2. The result can be shown in multiple forms. Exact Form:There is no definition for sin infinity. a function that takes the value between 0 and 1, or 0 and 90 degrees, returns to 0 degrees, then moves to 270 degrees in the 3rd quadrant if the value is the smallest, then returns to 0 degrees (or 360 degrees). Answer It turns out that the exact value of sec (60°) sec (60 °) is 2(These ...Find the Exact Value sin(45 degrees ) Step 1. The exact value of is . Step 2. The result can be shown in multiple forms. Exact Form: Decimal Form: These ...Find 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 ... TrFind the Exact Value sin((7pi)/6) Step 1. Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the third quadrant. Step 2. The exact value of is . Step 3. The result can be shown in multiple forms.Using that right triangle, you get exact answers for sine of 30°, and sin 60° which are 1/2 and the square root of 3 over 2 respectively. Using these results - sine 15°Find the Exact Value sin(45 degrees ) Step 1. The exact value of is . Step 2. The result can be shown in multiple forms. Exact Form: Decimal Form: sin 60° = √ (3)/2. sin 60 degrees = √ (3)/2. The sin of 60 degrees is √ (3)/2, the same as sin of 60 degrees in radians. To obtain 60 degrees in radian multiply 60° by π / 180° = 1/3 π. Sin 60degrees = sin (1/3 × π). Our results of sin60° have been rounded to five decimal places. If you want sine 60° with higher accuracy, then ...Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor.Find the Exact Value (sin(60))(cos(60)) Step 1. The exact value of is . Step 2. The exact value of is . Step 3. Multiply. Tap for more steps... Step 3.1. Multiply by . Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor.The easiest way to go about doing this (with the least work possible) is to subtract a sin value from 360 degrees (as to not have to switch the trigonometric ratio being used). 360 - 330 = 30, so we are subtracting 30 degrees from 360 to get the sin inverse, 330. Since we are subtracting from 360, the ratio will remain sin, and because it is in ...Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor.60° 90° Sin θ : 0. 1/2. 1/√2 ... Question 1: What is the exact value of sin 270? Solution: Here sin is positive only in 1st and 2nd Quadrant. 270° does not lies in 1st and 2nd Quadrant. therefore sin (360° – θ) = – sin θ ...Then basic trigonometry tells us: sin θ = b a2+b2√ sin θ = b a 2 + b 2 and cos θ = a a2+b2√ cos θ = a a 2 + b 2. From this, cos θ = 2 5√ 5 cos θ = 2 5 5, but since we know it's less than 0 0, it is simply the negative, so cos θ = −2 5√ 5 cos θ = − 2 5 5.Answer: The exact value of is . Step-by-step explanation: Recall that is a special angle that can be obtained using an equilateral triangle. The right triangle obtained using one of the lines symmetry was used to find the exact value of using SOH-CAH-TOA The exact value of is . arrow right Explore similar answers messagesIn the same way, we can derive other values of sin angles like 0°, 30°,45°,60°,90°,180°,270° and 360°. Below is the trigonometry table, which defines all the values of sine along with other trigonometric ratios.Calculate the value of the sin of 245 ° To enter an angle in radians, enter sin(245RAD) sin(245 °) = -0.90630778703665 Sine, in mathematics, is a trigonometric function of an angle. ... How do you use the angle sum identity to find the exact value of \displaystyle{\sin{{255}}} ? ... Get the value of \sin(60) from trigonometric values table. 0 ...For sin 24 degrees, the angle 24° lies between 0° and 90° (First Quadrant ). Since sine function is positive in the first quadrant, thus sin 24° value = 0.4067366. . . Since the sine function is a periodic function, we can represent sin 24° as, sin 24 degrees = sin (24° + n × 360°), n ∈ Z. ⇒ sin 24° = sin 384° = sin 744°, and so on.The values of trigonometric numbers can be derived through a combination of methods. The values of sine and cosine of 30, 45, and 60 degrees are derived by analysis of the 30-60-90 and 90-45-45 triangles. If the angle is expressed in radians as , this takes care of the case where a is 1 and b is 2, 3, 4, or 6.Exact values of sin(60), cos(60), tan(60), csc(60), sec(60), cot(60), Find exact values of all trigonometric functions when the angle is 60 degrees,Check out... Now let us discuss different values For sin For memorising sin 0°, sin 30°, sin 45°, sin 60° and sin 90° We should learn it like sin 0° = 0 sin 30° = 1/2 sin 45° = 1/√2 sin 60° = √3/2 sin 90° = 1 So, our pattern will be like 0, 1/2, 1/√2, √3/2, 1Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor.sin(60^@) = sqrt(3)/2color(white)("XX")csc(60^@) = 2/sqrt(3) cos(60^@) = 1/2color(white)("XXXX")sec(60^@) = 2 tan(60^@) = sqrt(3)color(white)("XXX")cot(60^@)=1/sqrt(3) Use the basic trigonometric definitions and the diagram below. Note: only the left half triangle is directly relevant; both sides combine …This video works to determine the exact values for the sin(30), cos(30), tan(30), sin(60), cos(60), and tan(60) using an equilateral triangle and the accompa...Now, to find the cos values, fill the opposite order the sine function values. It means that. Cos 0° = Sin 90°. Cos 30° = Sin 60°. Cos 45° = sin 45°. Cos 60° = sin 30°. Cos 90° = sin 0°. So the value of cos 90 degrees is equal to 0 since cos 90° = sin 0°. Angles in degrees.Although the Bible does clearly show that people need to repent for all sins, there is no passage that says that all sins are equal; instead, the Bible shows some sins cause more guilt or damage than others.. What are the values of sin 30, sin 60, sin 45, Exact values of the ordered pairs of the un Rational values of trigonometric functions. I am using extensively trigonometric functions when an angle is given in degrees. Some of these functions like sine or cosine have rational values, for example, the well known example is that cos(θ) = 0.6 cos ( θ) = 0.6 and sin(θ) = 0.8 sin ( θ) = 0.8. However besides the case of multiplicity of ... The exact value of sin(45) sin ( 45) is √2 2 Find the Exact Value sin(60-45) Step 1. Subtract from . Step 2. The exact value of is . ... The exact value of is . Step 2.6. The exact value of is . Step 2.7.This video works to determine the exact values for the sine of 10 degrees, sin(10), the sine of 50 degrees, sin(50), and the sine of 70 degrees, sin(70). It ... Find the Exact Value sin(60)cos(45) Step 1. The exact val... | 677.169 | 1 |
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What happens to the value of the interior angles as the number of sides of the polygon increases?
As the number of sides increases, the sum of the measures of the interior angles increases/decreases/stays the same). Part B: Sum of Interior Angles & Measure of One Interior Angle Find the sum of the interior angles and the measure of one interior angle for each convex regular polygon.
What is the relationship between the number of sides and angles of all polygons?
The formula for calculating the sum of interior angles is ( n − 2 ) × 180 ∘ where is the number of sides. All the interior angles in a regular polygon are equal. The formula for calculating the size of an interior angle is: interior angle of a polygon = sum of interior angles ÷ number of sides.
How do you find the number of sides in a polygon given the interior angle?
Subtract the interior angle from 180. For example, if the interior angle was 165, subtracting it from 180 would yield 15. Divide 360 by the difference of the angle and 180 degrees. For the example, 360 divided by 15 equals 24, which is the number of sides of the polygon.
What is the relationship between the central angle and the interior angle as the number of sides increases How do the angles change?
Sample Answer: When the number of sides in a regular polygon increases by 1, the number of triangles that can be drawn within the polygon also increases by 1. Since the sum of the interior angles of a triangle is 180°, the interior angle sum increases by 180°.
Why is there no regular polygon with an interior angle of 155?
Thus the interior angle is 144 degrees. Exterior being supplementry is 180 minus 144 equals 36 degrees. this method will not work with an irregular polygon. Unless you know enough information to find the interior angle.
If it is a Regular Polygon (all sides are equal, all angles are equal)
Shape
Sides
Each Angle
Triangle
3
60°
Quadrilateral
4
90°
Pentagon
5
108°
How many sides are there in polygon?
Other Types of Polygons
Polygon
Number of Sides
Triangle
3
Quadrilateral
4
Pentagon
5
Hexagon
6
What is the sides of a polygon?
Does every polygon have an equal number of sides and angles?
Every polygon has an equal number of sides and vertices. Thus, every polygon has an equal number of vertices and angle. For any simple polygon, the sum of its interior angles can be determined by the formula:
To be a polygon, the shape must be flat, close in a space, and be made using only straight sides. Polygons with congruent sides and angles are regular; all others are irregular. Polygons with all interior angles less than 180° are convex; if a polygon has at least one interior angle greater than 180°, it is concave.
What is the difference between simple polygons and complex polygons?
Polygons with all interior angles less than 180° are convex; if a polygon has at least one interior angle greater than 180°, it is concave. Simple polygons do not cross their sides; complex polygons have self-intersecting sides.
How to find the angle sum of a polygon with 4 sides?
When we start with a polygon with four or more than four sides, we need to draw all the possible diagonals from one vertex. The polygon then is broken into several non-overlapping triangles. The angle sum of this polygon for interior angles can be determined on multiplying the number of triangles by 180° | 677.169 | 1 |
Quadrilaterals (Square, Rectangle, Parallelogram, Rhombus, Trapezoid)
Quadrilaterals are four-sided polygons that are a fundamental part of geometry. They come in various shapes and sizes, each with its unique properties. In this lesson, we will explore the characteristics of five common types of quadrilaterals: squares, rectangles, parallelograms, rhombuses, and trapezoids. Understanding these shapes and their properties is crucial for solving a wide range of geometric problems.
Square
A square is a quadrilateral with all four sides of equal length and all four angles equal to 90∘90^\circ90∘. It is a special case of a rectangle and a rhombus. The properties of a square include:
All sides are congruent (AB=BC=CD=DAAB = BC = CD = DAAB=BC=CD=DA).
All angles are right angles (90∘90^\circ90∘).
The diagonals are congruent, bisect each other at right angles, and divide the square into four congruent right triangles.
The formula for the area of a square is A=s2A = s^2A=s2, where sss is the length of a side.
The perimeter of a square is P=4sP = 4sP=4s.
Rectangle
A rectangle is a quadrilateral with opposite sides equal in length and all angles equal to 90∘90^\circ90∘. It is a special case of a parallelogram. The properties of a rectangle include:
The area of a parallelogram is A=bhA = bhA=bh, where bbb is the base and hhh is the height.
The perimeter of a parallelogram is P=2(a+b)P = 2(a + b)P=2(a+b), where aaa and bbb are the lengths of the sides.
Rhombus
A rhombus is a quadrilateral with all four sides of equal length. It is a special case of a parallelogram. The properties of a rhombus include:
All sides are congruent (AB=BC=CD=DAAB = BC = CD = DAAB=BC=CD=DA).
Opposite angles are congruent.
The diagonals bisect each other at right angles and are not congruent.
The diagonals bisect the angles from which they are drawn.
The area of a rhombus can be calculated as A=d1⋅d22A = \frac{d_1 \cdot d_2}{2}A=2d1⋅d2, where d1d_1d1 and d2d_2d2 are the lengths of the diagonals.
The perimeter of a rhombus is P=4sP = 4sP=4s, where sss is the length of a side.
Trapezoid
A trapezoid (or trapezium in some countries) is a quadrilateral with at least one pair of parallel sides. The properties of a trapezoid include:
Only one pair of opposite sides is parallel.
The angles on the same side of a leg are supplementary.
The area of a trapezoid is A=12(b1+b2)hA = \frac{1}{2}(b_1 + b_2)hA=21(b1+b2)h, where b1b_1b1 and b2b_2b2 are the lengths of the parallel sides and hhh is the height.
The median (or mid-segment) of a trapezoid is parallel to the bases and its length is the average of the lengths of the bases: m=12(b1+b2)m = \frac{1}{2}(b_1 + b_2)m=21(b1+b2).
Each type of quadrilateral has its unique set of properties that distinguish it from others. Understanding these properties is essential for solving geometric problems involving quadrilaterals. Whether calculating areas and perimeters or proving specific characteristics, the knowledge of quadrilaterals forms a foundational part of geometry.
Quadrilaterals have four sides and four angles.
A square is a special type of rectangle and rhombus.
A rectangle has four right angles and opposite sides are equal in length.
A parallelogram has opposite sides that are parallel and equal in length.
A rhombus has four equal sides and opposite angles are equal.
A trapezoid has one pair of parallel sides.
The diagonals of a parallelogram bisect each other.
The diagonals of a rhombus are perpendicular and bisect each other at right angles. | 677.169 | 1 |
Ghlj And Gstu Are Both Parallelograms Why Is L T?
Once the speed of requests has dropped below the brink for 10 minutes, the person might resume accessing content material on SEC.gov. This SEC practice is designed to limit excessive automated searches on SEC.gov and is not supposed or expected to impression individuals searching the SEC.gov website. To permit for equitable access to all customers, SEC reserves the proper to restrict requests originating from undeclared automated tools. Noodle Talk is a blog that options stories from people about their lives.
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The tone of voice is professional and it does not have much personality, however the content is compelling. There are different categories such as parenting, family life, relationships, and training. Some examples can be "The Day I Became a Mom" or "How My Parents Met." Noodle Talk has been featured by Forbes, Huffington Post Canada, Bustle Magazine, amongst others. By utilizing this site, you're agreeing to safety monitoring and auditing. True or false; Every kite is a parallelogram. Note that this coverage might change as the SEC manages SEC.gov to ensure that the internet site performs effectively and stays available to all customers.
Tissue cultures are taken from cancer sufferers, and the cells are then grown in a laboratory. What is the purpose of rising most cancers cells… Which of the next statements is true of an object that has an acceleration of 6 m/s2?
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Hint: Here, we have to find the distance of the chord from the centre. We will find the chord length using Perpendicular distance from the centre. The chord of a circle can be defined as the line segment joining any two points on the circumference of the circle.
Formula Used: Chord length Using Perpendicular distance from the centre is given by \[{\text{Chord Length}} = 2 \times \sqrt {\left( {{r^2} - {d^2}} \right)} \], where \[d\] is the perpendicular distance of the chord from the centre and \[r\] is the radius of the circle.
Complete Step by Step Solution: We will draw a circle with radius of a circle \[r\] as 13 cm and chord of a circle \[l\] as 10 cm.
Let AC be the chord of a circle, OP be the radius of the circle and OB be the distance between the chord of a circle and radius of a circle. Length of the chord \[ = 10{\text{cm}}\] \[ \Rightarrow l = \dfrac{{10}}{2}\] Dividing the terms, we get \[ \Rightarrow l = {\text{5 cm}}\] By using the formula \[{\text{Chord Length}} = 2 \times \sqrt {\left( {{r^2} - {d^2}} \right)} \], we have \[d = \sqrt {{r^2} - {{\left( {\dfrac{l}{2}} \right)}^2}} \] Substituting the values \[r = {\text{13 cm}}\] and \[\dfrac{l}{2} = {\text{5 cm}}\] \[ \Rightarrow d = \sqrt {{{13}^2} - {5^2}} \] The square of the number 13 is 169. The square of the number 5 is 25. Squaring the terms, we get \[ \Rightarrow d = \sqrt {169 - 25} \] \[ \Rightarrow d = \sqrt {144} \] Computing the square root, we get \[ \Rightarrow d = {\text{12 cm}}\]
Therefore, the distance of the chord from the centre is 12cm.
Note: We know that in a circle, perpendicular from the centre bisects the chord. Among the properties of the chord of a circle, Chords are equidistant from the center if and only if their lengths are equal. A chord that passes through the center of a circle is called a diameter and is the longest chord. A radius or diameter that is perpendicular to a chord divides the chord into two equal parts. It means that both the halves of the chords are equal in length. The perpendicular bisector of a chord passes through the center of a circle. The formula can also be used to find the length of the cord and the radius of the circle by rewriting the equation. | 677.169 | 1 |
There are two straight lines, which fully lie on a hyperboloid and go through every point of it. Each of them сovers
all the surface during a rotation around the axis of a hyperboloid. Consequently, a one-axis hyperboloid can be received
by rotation of a straight line around an axis, crossing to it.
Joining these views on a hyperboloid of rotation is a base of spectacular and illustrative models,
where a straight rod passes through a curved hole — hyperbola.
A showpiece, where a reclinate segment, being a part of a hyperboloid generatrix, is settled on a rotating disc,
is the most interesting. During the rotation of the disc a segment passes through both branches of a hyperbola,
not touching edges.
In a showpiece, which we often meet in science museums, a tube, which represents a straight line, crossing a rotation axis,
is firmly connected by segment with a socket on a rotation axis.
Parameters $a$ and $b$ of hole-hyperbola, intended by an equation $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ in a screen plane
(in case of a naturally implemented coordinate system) can be easily counted by two typical positions of a model.
When a segment, which forms a generatrix and an axis, lies in the plane of the screen, its end coincides with a "vertex"
of a hyperbola. It means that a parameter is simply length of this segment.
When a generatrix is parallel to the screen, its projection on the screen is an asymptote for a hyperbola-slash. Consequently,
a parameter can be defined from the relation $\tg \alpha = \frac{a}{b}$, where $\alpha$ is an angle of generatrix to a vertical axis. | 677.169 | 1 |
Distance Formula: Finding the Distance Between Two Points
The distance formula is an algebraic expression that gives the shortest distance between two points in a two-dimensional space. Dream01/Shutterstock/HowStuffWorks
You're sitting in math class trying to survive your latest pop quiz. Sweat trickles down your forehead as you read the prompt: "Find the distance between these points."
The distance formula you're looking for is fairly straightforward and has ties to one of the most useful and famous concepts in all of mathematics: the Pythagorean theorem.
What Is the Distance Formula?
The distance formula is an algebraic equation used to find the length of a line segment between two points on a graph, called the Cartesian coordinate system (also known as the point coordinate plane).
This two-dimensional plane is defined by two perpendicular axes (usually labeled the x-axis and the y-axis) that intersect at a central point called the origin. Here's how it's expressed:
In a two-dimensional space with two points P (x₁, y₁) and Q(x₂, y₂), the distance (d) between these two points is given by the formula: d = √ (x₂ - x₁)² + (y₂ - y₁)²
In a three-dimensional space with two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂), the distance (d) between these two points is given by the formula: d = √ (x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²
Next up, we'll take a closer look at the point coordinate plane, which can help you find exact spots by their horizontal and vertical positions, essential for everything from math problems to GPS navigation.
Understanding the Point Coordinate Plane
When most people hear the word "graph," they're picturing a chart with two lines — one vertical, one horizontal — that intersect each other at a right angle.
The vertical line is called the y-axis, and its horizontal counterpart is the x-axis. Both lines work together to tell a story with data.
If you want to make sense of where one point rests on your graph, measure where it falls along the two dimensions (the x-axis and the y-axis). These are known as the point's coordinates.
You need to find the coordinates for the first point and the second point before you can calculate the distance between them. You'll use the distance formula to measure the straight line segment connecting the two points.
Now let's explore the blissful relationship between the Pythagorean theorem and the distance formula.
The Pythagorean Theorem and the Distance Formula
The Pythagorean theorem was named for the Greek philosopher Pythagoras, but over a millennium before his birth, the ancient Babylonians already understood the geometric principle that is now associated with his name.
In essence, the Pythagorean theorem tells us how to find the longest side of a triangle when we know the lengths of the other two sides, and the distance formula uses this idea to measure how far apart two points are on a graph by treating the points as if they're at the corners of a right triangle.
For those in need of a quick refresher, the Pythagorean theorem says: The area of the square built upon the hypotenuse of a right triangle is equal to the sum of the areas of the squares upon the remaining sides.
There are a few key points to understand here. A right triangle, or right-angled triangle, has one angle measuring 90 degrees, known as a right angle. The longest side of this triangle is termed the hypotenuse, which is located opposite the right angle.
As we all know, a triangle may have three sides, but a square has four. So, imagine taking the hypotenuse of a right triangle and turning it into one of the four lines of a brand-new square. Then, do the same thing to the other two sides in the original triangle. You'll end up with three individual squares.
According to the Pythagorean theorem, the square formed by the hypotenuse has an area equal to the sum of the areas of the squares formed by the other two sides. If the hypotenuse is labeled "c," and the other two sides are labeled "a" and "b," then we could express that idea like so:
The Pythagorean theorem says a2 + b2 = c2. The distance formula is derived by using the theorem. grebeshkovmaxim/Shutterstock
How to Find the Distance Between Two Points
The first point and second points on your graph will each have an x coordinate and a y coordinate. You can calculate the shortest distance between these two points by using the Euclidean distance formula, which is a Pythagorean theorem-related algebraic expression.
D = √(x₂ - x₁) ² + (y₂ - y₁)²
Here, D stands for "distance." As for x₂ and x₁, they refer to the x coordinates of Point 2 and Point 1, respectively. Same goes for y₂ and y₁, except those are the two y coordinates.
So, to calculate the distance, our first step is to subtract x₁ from x₂. Then we have to multiply the resulting number by itself (or, in other words, "square" that number).
After that, we must subtract y₁ from y₂ and then square the answer we get from doing so. This will leave us with two numbers we must add together.
Then, finally, take that number and find its square root. And that square root, ladies and gentlemen, is our distance.
Distance Formula Example
OK, so let's say Point A has an x coordinate of 2 and a y coordinate of 5 (2,5). Let us also assume that Point B's got an x coordinate of 9 and a y coordinate of 13 (9,13). Plug those values into the handy-dandy formula and you get this:
D = √(9-2)² + (13-5)²
What's 9 minus 2? Easy, 7. And 13 minus 5 is 8, of course.
Now we're left with this:
D = √7² + 8²
If you "square" 7 — as in, multiply the number by itself — you end up with 49. As for 8 squared that works out to 64. Let's plug those values into the equation, eh?
D = √49 + 64
Now we're cooking. Add 49 and 64, and you get 113.
D = √113
What's the square root of 113? The answer is 10.63, so therefore:
D = 10.63
Go forth and ace your next pop quiz!
This article was updated in conjunction with AI technology, then fact-checked and edited by a HowStuffWorks editor | 677.169 | 1 |
let's do some solid geometry volume problems so they tell us shown is a triangular prism and so there's a couple of types of three-dimensional figures that deal with triangles this is what a triangular prism looks like where has a triangle on one two faces and they're kind of separated they're kind of have rectangles in between the other types of triangular three-dimensional figures is you
E Grabska. Graph transformations in computer science, 188-202, 1994. 30, 1994 Graph Transformations for Modeling hp-Adaptive Finite Element Method with Mixed Triangular and Rectangular Elements. A Paszyńska, M Paszyński,
Nguyen-Thoi TrungInstitute for Computational Science (INCOS), Ton Duc A cell‐based smoothed discrete shear gap method using triangular elements for
1Biological and Environmental Sciences and Engineering Division, King Abdullah University of Science and Technology, 2Visual Computing
Studies Intellectual History, History of Ideas, and History of Science. Focusing on the A plea for inter-method research in historical and cultural studiesmore.
In the social sciences, triangulation refers to the application and combination of several research methods in the study of the same phenomenon. By combining multiple observers, theories, methods, and empirical materials, researchers hope to overcome the weakness or intrinsic biases and the problems that come from single method, single-observer, and single-theory studies.
Earlham College Computer Science Senior Capstone, 2020, Jordan Christian than is technical or physical. Materials Science. Defect and Diffusion Forum Solid State Phenomena Key Engineering Materials Materials Science Forum Advanced Materials Research Journal of …
Triangular matrices: A square matrix with elements sij = 0 for j < i is termed upper triangular matrix. In other words, a square matrix is upper triangular if all its entries below the main diagonal are zero.
The density triangle in physics is a handy memory tool used by students to memorize three different but interrelated formulae. In this ScienceStruck post, we show you how to use the density triangle to make your life in the world of physics a tad bit easier.
Other articles where Triangular wave voltammetry is discussed: chemical analysis: Triangular wave voltammetry: Triangular wave voltammetry (TWV) is a method in which the potential is linearly scanned to a value past the potential at which an electrochemical reaction occurs and is then immediately scanned back to its original potential. A triangular wave voltammogram usually has…
A unit triangular matrix is triangular matrix with 1 s on the main diagonal. There are a few useful properties about products, inverses and determinants of triangular matrices [5]: • The inverse of upper (lower) triangular matrix is upper (lower) triangular. • The inverse of unit upper (unit lower) triangular matrix is unit upper (unit lower) triangular. •
Area of Triangle (conventional Method) Area of Triangle (Heron's Formula) Area of Triangle (SAS Method) Formulas. that Involve Right Triangles.
There are three common methods used to determine the ratios of the three species in the composition. The first method is an estimation based upon the phase diagram grid. The concentration of each species is 100% (pure phase) in each corner of the triangle and 0% at the line opposite it.
2021-02-02 · The triangle test is a discriminative method with many uses in sensory science including: gauging if an overall difference is present between two products selecting qualified panelists for a particular test
2009-03-01 · Triangular random variables are commonly used in risk analysis to quantify uncertainty. On the one hand, large scale simulations of are frequently undertaken, so a fast simulation method is important.
It is linked to Amazing Triangles - Mathematics. This resource has been produced by Discover Science & Engineering.
Calculate Speed, distance and time using the DST Triangle Maths About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features
Feb 2, 2021 The triangle test is a discriminative method with many uses in sensory science including: gauging if an overall difference is present between
(30 pages). Methods and Algorithms for Scientific Computing. A Moment Limiter for the Discontinuous Galerkin Method on Unstructured Triangular Meshes
Triangulation. A global positioning system (GPS) device uses data from satellites to locate a specific point on the Earth in a process called trilateration.
OSTI Identifier: 4491151. Report Number(s):: LA-UR-73-479; CONF-730414-2. Jan 24, 2017 Combining multiple qualitative methods with a collaborative research encompassed expertise in social science, emergency care research,
Mar 30, 2016 We demonstrate that the liquid-exfoliated BP forms a triangular Our liquid exfoliation method uses higher sonication energies than This research was supported by the Korean Basic Science Institute research Grant N
May 14, 2019 We present two novel solutions for multi-view 3D human pose estimation based on new learnable triangulation methods that combine 3D
As a method, the Learning Triangle provides insights that can inform your action as you create a more engaging and productive learning ecology for others to
Taken one step further, the term triangulation, borrowed from military naval science to signify the use of multiple reference points to locate an object's exact position
Mar 12, 2020 This method breaks apart the polyhedron into triangular pyramids known ICYS 2017 (International Conference for Young Scientists) Stuttgart,
The proposed method is an improvement of the well known Analytic Hierarchy Process (AHP), which is a pairwise comparison technique developed by Saaty and
The triangular() method returns a random floating number between the two specified numbers (both included), but you can also specify a third parameter, the
Aug 2, 2017 simulation for heat conduction in a unstructured triangular mesh. I already made it work for a structured rectangular grid with the ADI method,
Solutions for Incompressible Viscous Flow in a Triangular Cavity using Cartesian Grid Method. CMES-Computer Modeling in Engineering & Sciences, 35(2), 113–
Aug 21, 2020 International Journal of Computational Materials Science and A sampling point quadrature method for 3-node triangular element for the
JoVE publishes peer-reviewed scientific video protocols to accelerate biological, medical, chemical and physical research. | 677.169 | 1 |
GPT-4o (Omni) math tutoring demo on Khan Academy
Khan Academy
13 May 202403:20
TLDRIn this engaging demonstration, Sal Khan, the founder of Khan Academy, and his son Imran, explore the capabilities of Open AI's new technology in the field of math tutoring. Sal introduces the scenario where the AI is tasked with helping Imran understand a math problem without directly providing the answer. The session focuses on identifying the sides of a right triangle relative to a given angle, Alpha. Through a series of questions, the AI guides Imran to correctly identify the hypotenuse and the opposite side to angle Alpha. They then apply the sine formula to find the sine of angle Alpha using the identified side lengths. The AI successfully facilitates Imran's understanding of the problem, showcasing its potential as an educational tool. The interaction highlights the effectiveness of AI in nurturing comprehension and critical thinking in learners.
Takeaways
🎓 The founder of Khan Academy, Sal Khan, and his son Imran are introduced as participants in a demo of new AI technology for math tutoring.
🤖 Open AI has invited Sal and Imran to test their new technology, which aims to assist in tutoring math problems on Khan Academy.
📚 The technology is designed to guide students towards understanding math problems without directly giving away the answers.
👦 Imran is being tutored directly by the AI, which engages in a conversational manner to help him learn.
📐 Sal Khan asks the AI to identify the sides of a triangle relative to angle Alpha: opposite, adjacent, and hypotenuse.
🧐 Imran correctly identifies side AC as the adjacent side to angle Alpha, but is initially unsure about the hypotenuse.
📉 The AI clarifies that the hypotenuse is the longest side of a right triangle and is opposite the right angle, leading Imran to correctly identify side AB as the hypotenuse.
🔍 Imran then correctly identifies side BC as the opposite side to angle Alpha after the AI's guidance.
📐 The AI reminds Imran of the formula for finding the sine of an angle in a right triangle, which is opposite over hypotenuse.
🧮 Imran applies the sine formula using the identified sides to find the sine of angle Alpha, which he correctly calculates as 7 over 25.
🎉 The AI praises Imran for his correct work and encourages him to ask more questions if needed, emphasizing the interactive and supportive nature of the tutoring session.
Q & A
What is the role of Sal Khan in the given transcript?
-Sal Khan is the founder of Khan Academy and the author of a new book about artificial intelligence and education. In the transcript, he is trying out new technology for math tutoring on Khan Academy with his son, Imran.
What is the main objective of the tutoring session in the transcript?
-The main objective is to test the new technology's capability to tutor someone in math without giving away the answer, but by asking questions and guiding the learner towards the solution.
How does Sal Khan introduce the math problem to his son?
-Sal Khan asks his son to identify the sides of the triangle that are opposite, adjacent, and the hypotenuse relative to angle Alpha.
What is the correct identification of the sides of the triangle in relation to angle Alpha?
-The hypotenuse is side AB, the opposite side to angle Alpha is side BC, and the adjacent side to angle Alpha is side AC.
What is the formula for finding the sine of an angle in a right triangle?
-The formula for finding the sine of an angle in a right triangle is sin(θ) = opposite side / hypotenuse.
How does Imran identify the hypotenuse of the triangle?
-Imran identifies the hypotenuse as the longest side of the right triangle, which is side AB, after being reminded that the hypotenuse is opposite the right angle.
What is the value of sin(Alpha) in the given triangle?
-The value of sin(Alpha) is 7/25, as calculated using the identified opposite side BC and the hypotenuse AB.
What is the significance of the book 'Brave New Words' by Sal Khan?
-The book 'Brave New Words' explores the intersection of artificial intelligence and education, likely discussing the potential of AI in transforming educational experiences, as hinted by the context of the transcript.
What is the importance of understanding the problem rather than just getting the answer?
-Understanding the problem is crucial for developing critical thinking skills and a deep comprehension of mathematical concepts, which is the goal of the tutoring session in the transcript.
How does the AI technology assist in the tutoring process?
-The AI technology assists by asking guiding questions and providing hints to help the learner arrive at the solution independently, thus fostering a deeper understanding of the subject matter.
What is the final outcome of the tutoring session in the transcript?
-The final outcome is successful, as Imran correctly identifies the sides of the triangle and applies the sine formula to find the value of sin(Alpha), demonstrating his understanding of the problem.
What is the potential application of AI in educational platforms like Khan Academy?
Outlines
00:00
📚 Introduction to Khan Academy and AI Tutoring
Sal Khan, the founder of Khan Academy and author of a book on artificial intelligence and education, introduces himself and his son Imran. They are invited by Open AI to test out a new tutoring technology. Sal expresses his interest in seeing how well the technology can tutor in math without giving away the answer, emphasizing the importance of the student understanding the material themselves. The video begins with Sal and Imran ready to engage with the AI tutor.
🧮 Identifying Triangle Sides and the Hypotenuse
The AI tutor starts the session by asking Imran to identify the sides of a triangle relative to angle Alpha: opposite, adjacent, and hypotenuse. Imran correctly identifies angle Alpha but is unsure about the hypotenuse. The tutor guides him to understand that the hypotenuse is the longest side opposite the right angle, and Imran successfully identifies it. They then move on to identifying the opposite side to angle Alpha, which Imran correctly guesses as side BC.
🔍 Applying the Sine Formula in a Right Triangle
After identifying the sides of the triangle, the AI tutor prompts Imran to recall the formula for finding the sine of an angle in a right triangle, which is 'opposite over hypotenuse'. Imran correctly applies this formula using the lengths of the sides they have identified: side BC as the opposite side with a length of 7 units, and side AB as the hypotenuse with a length of 25 units. Imran calculates the sine of angle Alpha as 7 over 25, which the tutor confirms as correct. The session concludes with the tutor commending Imran on his work and inviting further questions.
Mindmap
Keywords
💡Khan Academy
Khan Academy is a non-profit educational organization that provides free online courses, lessons, and practice exercises in a wide range of subjects, including mathematics. In the video, Sal Khan, the founder of Khan Academy, is demonstrating a new technology for math tutoring, which is directly related to the educational mission of the organization.
💡Artificial Intelligence
Artificial Intelligence (AI) refers to the simulation of human intelligence in machines that are programmed to think like humans and mimic their actions. In the context of the video, AI is being used to tutor math problems, showcasing how technology can enhance the learning process.
💡Education
Education is the process of acquiring knowledge, skills, values, and habits. The video script discusses the integration of AI in the field of education, specifically for math tutoring on Khan Academy, which is aimed at improving the learning experience.
💡Tutoring
Tutoring is a form of teaching where an instructor gives individual instruction to a student, with the goal of improving understanding and performance. In the script, the AI is acting as a tutor, guiding Imran through a math problem without directly giving away the answer.
💡Math Problem
A math problem is a question or statement that requires a solution or proof, typically involving numbers, variables, and mathematical operations. In the video, Imran is being tutored on a specific math problem involving a triangle, which is central to the demonstration of the AI's tutoring capabilities.
💡Triangle
A triangle is a polygon with three edges and three vertices. It is a fundamental shape in geometry and is used in the video to illustrate the application of trigonometric functions and the process of problem-solving in math.
💡Angle Alpha
Angle Alpha refers to a specific angle within a triangle, which is the focus of the math problem in the video. The AI is helping Imran understand the relationships between the sides of the triangle relative to this angle, which is crucial for solving the problem.
💡Hypotenuse
The hypotenuse is the longest side of a right triangle, which is opposite the right angle. In the script, identifying the hypotenuse is an essential step in solving the math problem and applying trigonometric formulas.
💡Adjacent Side
The adjacent side is a side of a triangle that is next to a particular angle, in this case, angle Alpha. Understanding which side is adjacent is important for distinguishing between the different sides of the triangle and applying the correct trigonometric ratios.
💡Opposite Side
The opposite side is the side of a triangle that is opposite a given angle. In trigonometry, the opposite side is used in the calculation of the sine of an angle. In the video, identifying the opposite side to angle Alpha is key to finding the sine of that angle.
💡Sine Formula
The sine formula is a fundamental trigonometric formula that relates the ratio of the length of the opposite side to the length of the hypotenuse in a right triangle. In the video, the sine formula is applied to find the sine of angle Alpha, demonstrating the application of mathematical concepts in problem-solving.
Highlights
Sal Khan, founder of Khan Academy, introduces a demo of a new AI tutoring technology.
Khan Academy and Open AI collaborate to test AI's capability in math tutoring.
Sal Khan's son Imran participates in the tutoring demo.
The AI is instructed not to give direct answers but to guide Imran to understand the math problem.
Identification of triangle sides relative to angle Alpha is the first step in the tutoring process.
Imran correctly identifies side AC as the adjacent side to angle Alpha.
The AI confirms that side AB is the hypotenuse of the triangle.
Imran accurately identifies side BC as the opposite side to angle Alpha.
The AI reviews the formula for finding the sine of an angle in a right triangle.
Imran applies the sine formula using the identified side lengths.
The AI confirms Imran's calculation of sin(Alpha) as 7 over 25.
Imran's understanding of the math problem is emphasized over getting the correct answer.
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A cyclic quadrangle ABCD is called harmonic when the fourth vertex D is the 4-th harmonic of the three other A, B, C i.e. considering the points as complex numbers the cross-ratio (ABCD)=-1 (look at Complex_Cross_Ratio.html ).
[1] This is a characteristic property of harmonic quadrangles: The pole of each diagonal {U = (AB), V = (CD)} is contained in the other diagonal {CD, AB} respectively.
[2] For such a quadrangle, D can be determined as the intersection of the circumcircle and the line joining C and the pole (AB) of line AB with respect to the circumcircle.
[3] For such a quadrangle project the circumcenter O on AB at point M. Then triangles AMD, CBD and CMA are similar. In fact, the similarity of two out of the three triangles implies the similarity to the third and this fact is equivalent to the quadrangle being harmonic.
[4] Two opposite vertices, {A,B} say, are inverse with respect to the circle kCD passing through the other vertices {C, D} and orthogonal to the circumcircle. This is again a characteristic property of the harmonic quadrilateral.
[5] For harmonic quadrangles the circle kCD passing from {C,D} and orthogonally to the circumcircle is an Apollonius circle for the triangles ABC and ABD.
[6] A harmonic quadrangle is characterized by the fact that the products of lengths of opposite sides are equal.
Consider the polar (n) of the intersection point of the diagonals {AB, CD}. From general properties of cyclic quadrilaterals we know that (i) the intersection points of opposite sides {E, F} and the poles of the diagonals are four points on (n). To show that one diagonal, CD say, passes through the pole U of the other diagonal use the harmonicity. In fact, the harmonicity property of the quadrangle is equivalent to the fact that the tangent tA at A and the three lines {AD, AC, AN} make a harmonic bundle (AD, AC, AN, tA) = -1. This implies that the intersection poin U' of ta and DC is harmonic conjugate to N with respect to {D,C}. This identifies U' with U and proves the first part of [1]. The argument can be reversed to prove also the inverse. [2] is a cosequence of [1].
Let now M be the intersection point of the diagonal AB with UO. Since angle UMN is orthogonal and {C,D} are harmonic with respect to {U,N}, line MN is bisector of angle DMC. Taking the symmetrics D* of D and C* of C with respect to line (bisector) UM, we see that angle(ADM) = angle(CAB), since the arcs CB and AC* are equal. The statements in [3] follow from these remarks.
[4] follows from [1]. In fact, if the quadrangle is harmonic then circle (V, VD) is orthogonal to the circumcircle and {A,B} are inverse to this circle. Inversely, if this happens, then CD is the polar of V and use of [1] completes the argument.
[5] is equivalent to [4] and [6] equivalent to [5] since in that case, {C, D} being on the Apollonian circle will satisfy AC/CB = AD/DB.
Remark
Notice that AB is a symmedian line of triangle ACD. This follows from the similarity of triangles ACM, ADM, which implies that the distances of M from the sides AC, AD are analogous to the lengths of these sides. A characteristic property of the symmedians. Hence the diagonals of the harmonic quadrangle coincide with some symmedians of its "partial" triangles ACD, ADB etc. (leaving out one vertex).
Let F be a Moebius transformation and A0B0C0D0 a square. Then applying the transformation to the vertices of the square we obtain a harmonic quadrangle or a harmonic division of four points on a line i.e. {A=F(A0), B=F(B0), C=F(C0), D=F(D0)} are the vertices of a harmonic quadrangle if the circumcircle of the square maps to a circle.
It is trivial to see that the vertices of a square satisfy the definition of harmonicity. Since Moebius transformations preserve the cross-ratio and map circles to circles (or lines) the image will be a quadrangle if the circumcircle of the square maps to a circle. Otherwise the four image-points will lie on a line and build a harmonic division.
Corollary Take a point P and draw the lines joining it with the vertices of a square. Then the second intersection points of these lines with the circumcircle of the square make a harmonic quadrangle.
The proof results by applying an inversion G (or anti-inversion, if P is inner to the circumcircle of the square) which gives the vertices {A, B, C, D} as images of the vertices of the square. The inversion is the one with respect to the circle orthogonal to the circumcircle of the square and centered at P. Since every inversion is the composition of a Moebius map and a reflexion the proof follows from the previous property. | 677.169 | 1 |
What is a 2D shape with 7 sides called?
heptagon
A heptagon is a polygon that has seven sides. It is a closed figure having 7 vertices. A heptagon is also sometimes called Septagon.
What 3d shape has 7 sides?
A heptahedron is a polyhedron with seven faces. There is a single "regular" heptahedron, consisting of a one-sided surface made from four triangles and three quadrilaterals.
What is a math kite?
A kite is a quadrilateral that has 2 pairs of equal-length sides and these sides are adjacent to each other. Properties: The two angles are equal where the unequal sides meet. It can be viewed as a pair of congruent triangles with a common base. It has 2 diagonals that intersect each other at right angles.
What is a shape with 7 angles?
In maths (geometry), a heptagon is a polygon with seven sides and seven angles. A heptagon has seven straight sides and seven corners i.e. vertices. It is sometimes referred to as a "septagon".
What is a 7 sided 2D shape called?
It is also known as a septagon. The word heptagon comes from two words: 'hepta', meaning seven and 'gon' meaning sides. So, if you put this together, the word 'heptagon' directly translates to '7 sided shape'.
What is of a parallelogram?
A parallelogram is a two-dimensional geometrical shape, whose sides are parallel to each other. It is a type of polygon having four sides (also called quadrilateral), where the pair of parallel sides are equal in length. The Sum of adjacent angles of a parallelogram is equal to 180 degrees.
What is a rhombus in math?
In Euclidean geometry, a rhombus is a type of quadrilateral. It is a special case of a parallelogram, whose all sides are equal and diagonals intersect each other at 90 degrees. This is the basic property of rhombus. The shape of a rhombus is in a diamond shape. | 677.169 | 1 |
What is another word for trapezoids?
Pronunciation: [tɹˈapɪzˌɔ͡ɪdz] (IPA)
Trapezoids are four-sided geometric figures with one pair of parallel sides. They are also known as trapeziums in British English. Other synonyms for trapezoids include quadrilaterals, trapezes, irregular quadrilaterals, non-right angles, and slanted shapes. Trapezoids can also be referred to as trapezoidal shapes or figures. Many real-life objects have trapezoidal shapes, such as roofs, bridges, and even some candies and chocolates. Understanding the synonyms for trapezoids can help us better describe and identify shapes in the world around us, including in geometry problems and designs.
What are the hypernyms for Trapezoids?
Other hypernyms:
Usage examples for Trapezoids
The work may be checked by comparing the results of different groups of pupils, or by drawing another diagonal and dividing the field into other triangles and trapezoids.
"The Teaching of Geometry"
David Eugene Smith
Related words & questions
Related words: what is a trapezoid, how do you find the area of a trapezoid, how many sides does a trapezoid have, what are the properties of a trapezoid, how is the area of a trapezoid calculated, what does a trapezoid look like, list of properties of a trapezoid | 677.169 | 1 |
Supplemenatry Classnotes
Motivation for drawing 12-point circle
Albrecht Duerer drew a version of the the following in his
books instructing the art of perspective drawing. Consider
a yellow Smiley Face with a green eye, and a red grin. To
An orthographic grid is superimposed on a given figure..
help the artist sketch Smiley in perspective. A square surrounding
the figure is subdivided several times by first finding the midpoint
of a square by crossing diagonals. Then drawing lines parallel to
the sides.
To transfer this into perspective we start with a perspective square.
Connecting corresponding pairs of points in the orthographic figure
and in the perspective figure produces the line in perspective. Two
lines cross at the point in perspecive, that their orthographics
counterparts do. To draw parallels in perspective is even easier
than orthographically because we connect vanishing points to
midpoints of the squares.
The same figure is drawn in perspective, using the grid as an aid.
We now use the grid in perspective to locate places for the lines
in the orthographics picture. The figures drawn here were not done
in freehand, which would have been much easier. They were done in
iPaint, in which curves are a bit tricky to do. Fortunately, iPaint
has a zoom feature that allows you apply pixel level detail. A
bonus is an easy way of coloring corresponding regions to help
identifying correspondences.
You may note that Albrecht Duerer's picture for how to use
Alberti's Veil for perspective drawing discussed earlier is an
example of this process.
A missing contruction
The second lesson was a quick introduction to the notes on Measuring
a Box. (See the Index to the Perspective section linked to the portal.)
Step 3 needs a second elaboration for the construction using paper,
pencil and gnomons (right angles, as with transparent triangle rulers
or large pieces of rectangular paper, even a carpenter's ell.) Here is
the missing picture
The Thales triangles can be fitted to the extended altitudes of a 3pt perspective frame.
(Click image for download thales-gnomons.seg and open in KSEG.)
Given the three vanishing points, that form the triangle for a 3-point
perspective frame, use the gnomon to "drop the perpendicular" from each
vanishing point to the opposite horizon. Extend the altitudes beyond
the horizons. Then use a gnomon (large enough right angle) to fit its
right angle on the extended altitude, until its legs pass through the
two vanishing points. That's the Thales triangle for that horizon, at
least approximately.
This method does avoid having to use a compass for bisecting, contructing
perpendiculars, and fitting Thales triangles into semicircles.
Drawing exercise
As I summarized the lesson measurebox in the
class notes, the students were worked from a box (nearly) in perspective
on a piece of paper to complete the figure needed for measuring.
You can download this template and print.
To understand, you should make the construction twice, once with KSEG and
again with pencil, paper and a gnomon (right-angle ruler). | 677.169 | 1 |
Half As Much May Be Right
What is significant about an inscribed (blue) angle in a semi-circle? Why might this fact be true?
Extension
What is the relationship between the opposite angles in a quadrilateral inscribed in a circle? Compare the blue angles and the red angles, and explain why the relationship you find must hold true. | 677.169 | 1 |
August 2019 geometry regents. REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Wednesday...
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Under which This book constitutes the thoroughly refereed post-pro... | 677.169 | 1 |
Question Video: Solving Problems Related to a Reflected Triangle
Mathematics • First Year of Preparatory School
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In the following figure, △𝐴′𝐵′𝐶′ is the image of △𝐴𝐵𝐶 by reflection in the line 𝐿. (1) Fill in the blanks: The length of 𝐴′𝐶′ = _ cm, and the length of 𝐴′𝐵′ = _ cm. (2) Fill in the blanks: Line segment 𝐴𝐴′ is _ to line segment 𝐵𝐵′, and line segment 𝐶𝐶′ is _ to line 𝐿. (3) Find the measure of ∠𝐴.
02:41
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How To Unit vector 3d: 9 Strategies That Work
Relation between Vectors and Unit Vectors. When a unit vector is multiplied by a scalar value it is scaled by that amount, so for instance when a unit vector pointing to the right is multiplied by \(\N{ 100}\) the result is a \(\N{100}\) vector pointing to the right; when a unit vector pointing up is multiplied by \(\N{ -50}\) the result is a \(\N{50}\) vector pointing down.Vectors are used in everyday life to locate individuals and objects. They are also used to describe objects acting under the influence of an external force. A vector is a quantity with a direction and magnitude.We study nematic configurations within three-dimensional (3D) cuboids, with planar degenerate boundary conditions on the cuboid faces, in the Landau-de Gennes framework.We will do this by insisting that the vector that defines the direction of change be a unit vector. Recall that a unit vector is a vector with length, or magnitude, of 1. This means that for the example that we started off thinking about we would want to use \[\vec v = \left\langle {\frac{2}{{\sqrt 5 }},\frac{1}{{\sqrt 5 }}} \right\rangle ...The magnitude of vector: v → = 5. The vector direction calculator finds the direction by using the values of x and y coordinates. So, the direction Angle θ is: θ = 53.1301 d e g. The unit vector is calculated by dividing each vector coordinate by the magnitude. So, the unit vector is: e → \) = ( 3 / 5, 4 / 5. Just as in two dimensions, we can also denote three-dimensional vectors is in terms of the standard unit vectors, $\vc{i}$, $\vc{j}$, and $\vc{k}$. These vectors are the unit vectors in the positive $x$, $y$, and $z$ direction, …2018年6月26日 ... Is there a way to constrain a parameter block to unit magnitude? I have a unit 3d vector whose direction needs to be solved. 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A vector in 3D space can be written in component form: (𝑥, 𝑦, 𝑧), or in terms of its fundamental unit vectors: 𝑥 ⃑ 𝑖 + 𝑦 ⃑ 𝑗 + 𝑧 ⃑ 𝑘. 6kW UHD Digital Radar - Less Antenna and CableDRS6AX "X-Class", a whole new class of Radar!Pushing the boundaries of what is possible with conventional Radar technology, DRS6AX "X-Class" marks yet another leap forward for FURUNO. Improved in almost all aspects, DRS6AX "X-Class" Radar features improved short range detection, as well as …TangFor example, the vector < 2, -4 > in component form can be written as in standard unit vector form. In 3D, the standard unit vectors are i = < 1, 0, 0> j = < 0, 1, 0 >, and k = < 0, 0, 1 >. Any vector in component form can be written as a linear combination of the standard unit vectors i and j and kHence, it is not a unit vector. Problem 3: Find the unit vector in the direction of . Solution: Modulus of the vector, = = √3. Unit vector, =. =. 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(Set the dot product of the two equal to 0 and solve. You can actually set a and b equal to 1 here, and solve for c .) Then divide that vector by its length to make it a unit vector. Are you looking to explore the world of 3D printing but don't know where to start? One of the best ways to dive into this exciting technology is by accessing free 3D print design repositories.Given a surface parameterized by a function v → ( t, s) . , to find an expression for the unit normal vector to this surface, take the following steps: Step 1: Get a (non necessarily unit) normal vector by taking the cross product of both partial derivatives of v → ( t, s) . :For each vector, the angle of the vector to the horizontal must be determined. Using this angle, the vectors can be split into their horizontal and vertical components using the trigonometric functions sine and cosineA unit vector is a vector of length equal to 1. When we use a unit vector to describe a spatial direction, we call it a direction vector. In a Cartesian … OurA unit vector is a vector whose measure is #1#. Using the fact that for any vector #vec(v)# and scalar #c#, we have #||cvec(v)|| = c||vec(v)||#, we will find #||vec(u)|| = u#, then divide by #u#. #||vec(u)/u|| = ||vec(u)||/u = u/u = 1# As multiplying by a scalar does not change the direction of a vector, this will be a unit vector perpendicular ...In today's fast-paced world, ensuring the safety and security of our homes has become more important than ever. With advancements in technology, homeowners are now able to take advantage of a wide range of security solutions to protect thei... InUnit vectors can be used in 2 dimensions: Here we show that the vector a is made up of 2 "x" unit vectors and 1.3 "y" unit vectors. In 3 Dimensions. Likewise we can use unit vectors in three (or more!) dimensions: Advanced topic: arranged like this the three unit vectors form a basis of 3D space. But that is not the only way to do this!Answer We know that, in order to add two vectors in three dimensions, we add the corresponding components individually. If ⃑ 𝐴 = ( 𝑥, 𝑦, 𝑧) and ⃑ 𝐵 = ( 𝑥, 𝑦, 𝑧) , then ⃑ 𝐴 + ⃑ 𝐵 = ( 𝑥 + 𝑥, 𝑦 + 𝑦, 𝑧 + 𝑧) . This means that ⃑ 𝐴 + ⃑ 𝐵 = ( − 2 + ( − 3), − 3 + 3, 0 + ( − 2)). Therefore, ⃑ 𝐴 + ⃑ 𝐵 = ( − 5, 0, − 2).13.5: Directional Derivatives and Gradient Vectors. Determine the directional derivative in a given direction for a function of two variables. Determine the gradient vector of a given real-valued function. Explain the significance of the gradient vector with regard to direction of change along a surface.Jan Tang Figure 5.3.9: Vectors →v and →u for Example 5.3.6. Solution. Using the Parallelogram Law, we draw →v + →u by first drawing a gray version of →u coming from the tip of →v; →v + →u is drawn dashed in Figure 5.3.10. To draw →v − →u, we draw a dotted arrow from the tip of →u to the tip of →vTheJun 5, 2023 ·. (0, 0, 1) — Describes the z-direction. Every vector in a 3D space is equal to a sum of unit vectors. 3D Vector Plotter. An interactive2016年2月9日 ... A quaternion is a vector in with a noncommutative product see 1 or QuaternionnbspWolfram MathWorld Quaternions also called hypercomplex ...The Unit Vector calculator, U = V/|V|, computes the unit vector (U) for a vector (V) in Euclidean three dimensional space. In $3$ dimensions, there are infinitely many vectors perpendThe nor function calculates the unit normal vector (a vector We have seen that vector addition in two dimensions sa Are you a fan of 3D printing? Do you enjoy creating your own unique designs? If so, you're probably always on the lookout for new and exciting 3D print designs to bring your creations to life. Luckily, there are several websites out there t...Download this Air Conditioning Wall Mounted Unit In 3d Visualization Backgrounds image design for free right now! Pikbest provides millions of free graphic design templates,png images,vectors,illustrations and background images for designers. Search more pictures about electrical tools,pressure gauge,electrical services at Pikbest.com! The formula creates a rotation matrix around an axis d... | 677.169 | 1 |
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We need to consider the axiom: "Given two distinct points, there is a unique line that passes through them."
Therefore, we can conclude that ${\text{AB}},{\text{PQ}}$ and ${\text{XY}}$ are the lines with same dimensions, and hence if ${\text{AB}} = {\text{PQ}}$ and ${\text{PQ}} = {\text{XY}}$, then ${\text{AB}} = {\text{XY}}$.
2. Give a definition for each of the following terms. Are there other terms that need to be defined first? What are they, and how might you define them?
(i) Parallel lines
Ans: Two lines are said to be parallel, when the perpendicular distance between these lines is always constant or we can say that the lines that never intersect each other are called as parallel lines.
(Image will be uploaded soon)
We need to define line first, in order to define parallel lines.
(ii) Perpendicular lines
Ans: Two lines are said to be perpendicular lines, when angle between these two lines is ${90^\circ }$.
(Image will be uploaded soon)
We need to define line and angle, in order to define perpendicular lines.
(iii) Line segment
Ans: A line of a fixed dimension between two given points is called as a line segment.
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We need to define line and point, in order to define a line segment.
(iv) Radius of a circle
Ans: The distance of any point lying on the boundary of a circle from the center of the circle is called as radius of a circle.
(Image will be uploaded soon)
We need to define circle and center of a circle, in order to define radius of a circle.
(v) Square
Ans: A quadrilateral with all four sides equal and all four angles of ${90^\circ}$ is called as a square.
(Image will be uploaded soon)
We need to define quadrilateral and angle, in order to define a square.
3. If a point $C$ lies between two points $A$ and $B$ such that $AC = BC$, then prove that $AC = \dfrac{1}{2}AB \cdot $ Explain by drawing the figure.
Ans: We are given that a point $C$ lies between two points $B$ and $C$, such that $AC = BC$.
We need to prove that $AC = \dfrac{1}{2}AB \cdot $
Let us consider the given below figure.
(Image will be uploaded soon)
We are given that $AC = B{C_ - } \ldots (i)$
An axiom of the Euclid says that "If equals are added to equals, the wholes are equal."
Let us add $A C$ to both sides of equation $(i)$.
$AC + AC = BC + AC.$
An axiom of the Euclid says that "Things which coincide with one another are equal to one another." "
We can conclude that $BC + AC$ coincide with AB, or
$AB = BC + AC. \ldots (ii)$
An axiom of the Euclid says that "Things which are equal to the same thing are equal to one another."
From equations (i) and (ii), we can conclude that
$AC + AC = AB$, or $2AC = AB$
An axiom of the Euclid says that "Things which are halves of the same things are equal to one another."
Therefore, we can conclude that $AC = \dfrac{1}{2}AB$
4. In the following figure, if ${\text{AC}} = {\text{BD}}$, then prove that ${\text{AB}} = {\text{CD}}$.
(Image will be uploaded soon)
Ans: We are given that $AC = BD$
We need to prove that $AB = CD$ in the figure given below.
(Image will be uploaded soon)
From the figure, we can conclude that
$AC = AB + BC$, and
$BD = CD + BC.$
An axiom of the Euclid says that "Things which are equal to the same thing are equal to one another."
$AB + BC = CD + BC$
An axiom of the Euclid says that "when equals are subtracted from equals, the remainders are also equal."
5. Why is axiom 5, in the list of Euclid's axioms, considered as a 'universal truth'? (Note that the question if not about fifth postulate)
Ans: We need to prove that Euclid's fifth axiom is considered as a universal truth.
Euclid's fifth axiom states that "the whole is greater than the part."
The above given axiom is a universal truth. We can apply the fifth axiom not only mathematically but also universally in daily life.
Mathematical proof:
Let us consider a quantity $z$, which has different parts as $a,b,x$ and $y$
So, $z = a + b + x + y$.
Therefore, we can conclude that $z$ will always be greater than its corresponding parts $a,b,x$ and $y$.
Universal proof:
We know that Mumbai is located in Maharashtra and Maharashtra is located in India.
In other words, we can conclude that Mumbai is a part of Maharashtra and Maharashtra is a part of India.
Therefore, we can conclude that whole India will be greater than Mumbai or Maharashtra or both.
Therefore, we can conclude that Euclid's fifth axiom is considered as a 'Universal truth'.
6. How would you rewrite Euclid's fifth postulate so that it would be easier to understand?
Ans: We need to rewrite Euclid's fifth postulate so that it is easier to understandWe know that Playfair's axiom states that "For every line $l$ and for every point $P$ not lying on
1, there exists a unique line m passing through $P$ and parallel to $\Gamma $.
The above mentioned Playfair's axiom is easier to understand in comparison to the Euclid's fifth postulate.
Let us consider a line $l$ that passes through a point $p$ and another line $m$. Let these lines be at a same plane.
Let us consider the perpendicular $C$$D$ on $l$ and $F$$E$ on $m$.
(Image will be uploaded soon)
From the above figure, we can conclude that $CD = EF$.
Therefore, we can conclude that the perpendicular distance between lines $m$ and $l$ will b. constant throughout, and the lines $m$ and $l$ will never meet each other or in other words, $w$ can say that the lines $m$ and $I$ are equidistant from each otherLet us consider lines $l$ and $m$.
(Image will be uploaded soon)
From the above figure, we can conclude that lines $I$ and $m$ will never intersect from either side. Therefore, we can conclude that the lines $l$ and $m$ are parallel.
8. Consider the two 'postulates' given below:
(i) Given any two distinct points $A$ and $B$, there exists a third point $C$, which is between ${\text{A}}$ and ${\text{B}}$.
(ii) There exists at least three points that are not on the same line. Do these postulates contain any undefined terms? Are these postulates consistent? Do they follow from Euclid's postulates? Explain.
Ans: Given any two distinct points $A$ and $B$, there exists a third point $C$, which is between $A$ and $B$.
There exists at least three points that are not on the same line. The undefined terms in the given postulates are point and line. The two given postulates are consistent, as they do not refer to similar situations and they refer to two different situations. We can also conclude that, it is impossible to derive at any conclusion or any statement that contradicts any well-known axiom and postulate.
The two given postulates do not follow from the postulates given by Euclid. The two given postulates can be observed following from the axiom, "Given two distinct points, there is a unique line that passes through them".
4 Marks Questions
1. In the above question, point $C$ is called a mid-point of line segment $A$$B$, prove that every line segment has one and only one mid-point.
Ans: We need to prove that every line segment has one and only one mid-point.
Let us consider the given below line segment $A$$B$ and assume that $C$ and $D$ are the mid-points of the line segment $A$$B$.
(Image will be uploaded soon)
If $C$ is the mid-point of line segment $A$$B$, then
$AC = CB$
An axiom of the Euclid says that "If equals are added to equals, the wholes are equal."
$AC + AC = CB + AC.(i)$
From the figure, we can conclude that $CB + AC$ will coincide with $A$$B$.
An axiom of the Euclid says that "Things which coincide with one another are equal to one another." "
$AC + AC = A{B_.}(ii)$
An axiom of the Euclid says that "Things which are equal to the same thing are equal to one another." "
Let us compare equations (i) and (ii), to get
$AC + AC = AB$, or $2AC = AB.(iii)$
If $D$ is the mid-point of line segment $A$$B$, then
$AD = DB$
An axiom of the Euclid says that "If equals are added to equals, the wholes are equal."
$AD + AD = DB + AD.$ (iv)
From the figure, we can conclude that $DB + AD$ will coincide with $A$$B$.
An axiom of the Euclid says that "Things which coincide with one another are equal to one another."
$AD + AD = A{B_.}(v)$
An axiom of the Euclid says that "Things which are equal to the same thing are equal to one another:"
Let us compare equations (iv) and (v), to get
$AD + AD = AB$, or
$2AD = AB.({\text{vi}})$
An axiom of the Euclid says that "Things which are equal to the same thing are equal to one another."
Let us compare equations (iii) and (vi), to get
$2AC = 2AD$
An axiom of the Euclid says that "Things which are halves of the same things are equal to one another."
$AC = AD$
Therefore, we can conclude that the assumption that we made previously is false and a line segment has one and only one mid-point.
If a point C lies between two points A and B such that AB = BC, then prove that AC = ½ AB
Find the number of dimensions a solid, surface, and point have.
Let x + y = 10, and x = z. Show that y + z = 10
There are two sales employees who received equal incentives during the month of July. In August, each sales employee received double incentives in comparison to the month of July. Compare the employees' incentives for the month of August.
Prove that an equilateral triangle can be formed on any of the given line segments.
Important Related Links for CBSE Class 9
Conclusion
In conclusion, the compilation of important questions for CBSE Class 9 Maths Chapter 5 - "Introduction to Euclid's Geometry" is a valuable resource for students. These questions have been thoughtfully curated to cover essential concepts and postulates in Euclidean geometry, providing a focused approach to exam preparation. They enable students to assess their understanding, practise geometric proofs, and reinforce their knowledge in this fundamental branch of mathematics. By aligning with the examination pattern and difficulty level, these important questions empower students to build confidence and proficiency in geometry. Overall, they serve as an indispensable tool for Class 9 Mathematics students, enhancing their academic excellence and fostering a deeper appreciation for the elegance and precision of geometric proofs.
1. What are the important topics covered in Chapter 5 of Class 9 Maths?
Chapter 5 'Introduction to Euclid's Geometry' of Class 9 Maths is a very short chapter. Hence, all the topics covered in this chapter hold equal importance when preparing for your Class 9 Maths exams. However, an important tip to keep in mind during your preparation is to practice and remember well all the axioms and postulates that are a part of this chapter. Students can refer to Important Questions for CBSE Class 9 Maths Chapter 5 provided by Vedantu for more help.
2. What is Euclid's Geometry according to Chapter 5 of Class 9 Maths?
Euclid's geometry refers to the study of solid figures and planes based on the axioms and postulates given by Euclid, a Greek Mathematician. His postulates and axioms deal with the properties and relationships between all things. Euclid's geometrical theorems have been given a lot of importance since they enable one to calculate various distances. An example of his postulates can be "A straight line may be drawn from any one point to any other point".
3. Where can I find a list of important questions from Chapter 5 of Class 9 Maths?
Important Questions are a very important source for preparation before your Maths exam. Maths can often be a difficult subject to excel in for many students. Referring to important questions will give students an opportunity to score better in their exams. You can find Important Questions from Chapter 5 of Class 9 Maths provided by subject experts at Vedantu. These questions have been solved with step-by-step solutions to help students understand the chapter better and the solutions or any study material is available for downloading absolutely free of cost.
4. How many exercises are a part of Chapter 5 of Class 9 Maths?
Chapter 5 "Introduction to Euclid's Geometry" of Class 9 Maths NCERT is a very short chapter and includes a total of 2 exercises. The following is a list of total questions in each exercise of this chapter:
Exercise 5.1 - 7 questions
Exercise 5.2 - 2 questions
The first exercise in this chapter covers topics including Euclid's Definitions, Axioms, and Postulates while the second exercise is based on the Equivalent Versions of Euclid's Fifth Postulate.
5. How are important questions helpful from exam point of view for Class 9 Maths Chapter 5?
Important questions are meant to help students by providing ready-made study material that can help them save time during their preparation. The important questions for Chapter 5 Class 9 Maths provided by Vedantu have been solved by subject experts to give students a thorough understanding of the steps involved in answering the questions from this chapter. The solutions for these questions have been carefully crafted to provide full accuracy. Students can download these solutions free of cost on their computers and access them anytime while they are preparing for exams. | 677.169 | 1 |
...greater than EF. Therefore, " if two triangles," &c. QED PROP. XXV. THEOR. If two triangles have tiro sides of the one equal to two sides of the other, each to each, but the base of the one greater than the base of the other ; the angle also contained by the sides...
...another in each of the points С, Е. Join AC, AE, В С, BE. Then because the triangles AD С, ADE have two sides of the one equal to two sides of the other, and have also the included angles ADC, ADE equal to one another, the base А С (I. 4.) is equal...
...number of straight lines, which meet in one point, are together equal to four right angles. PROP. IV. THEOR. If two triangles have two sides of the one...sides of the other, each to each ; and have likewise the angles contained by those sides equal to one another, they shall likewise have their bases, or...
...found that BO + OC< BD + DC ; therefore, still more is BO + OC<BA+AC. PROPOSITION IX. THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, and the included angles unequal, the third sides will be unequal; and the greater side will belong to the...
...shall be less than the other two sides of the triangle, but shall contain a greater angle. XXIV. If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of the one greater than the angle contained by the two sides... '.y.xrepa.... twrepa....
...straight line AB, the angle A is made equal to the given angle C : which was to be done, f PROP. XXIV. THEOR. IF two triangles have two sides of the one equal to two sides of the other, each to each, but the angles contained by those sides unequal : the base of that which has the greater angle is greater...
...two straight lines, a part AE has been cut off equal to C, the less. PROPOSITION IV. THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles contained by those sides equal to one another, thenbases, or third sides, shall be equal...
...to them, viz. the angle ABC to the angle DEF, and the angle ACB to the angle DFE. Therefore, if two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles contained by those sides equal to one another ; their bases shall be equal, and their areas... | 677.169 | 1 |
In the following diagram, the 'hexagon' stands for 'engineers', the 'circle' stands for 'music lovers', and the 'rectangle' stands for 'morning walkers'. The numbers given in the different segments represent the number of persons of that category. | 677.169 | 1 |
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SSC CGL level Solved question Set 96, Geometry 11
Circle questions for SSC CGL Solved geometry question set 96
Solve circle questions for SSC CGL and other geometry questions 10 in 15 minutes in set 96. Verify correctness from answers. Learn to solve from solutions.
Many of these 10 problems are difficult.
The solved question set contains,
Circle questions for SSC CGL and other geometry questions to be answered in 15 minutes (10 chosen questions)
Answers to the questions, and,
Detailed conceptual solutions to the questions.
For best results, take the test first and then learn from solutions.
Circle questions for SSC CGL and other geometry questions set 96 - answering timeQ5. The lengths of the sides of a triangle are $a$, $b$, and $c$ respectively. If $a^2+b^2+c^2=ab+bc+ca$, then the triangle is,
right-angled
scalene
equilateral
isosceles
Q6. In a $\triangle ABC$, the side BC is extended up to D, such that $CD=AC$. If $\angle BAC=109^0$ and $\angle ACB=72^0$, then the value of $\angle ABC$ is,
$35^0$
$45^0$
$40^0$
$60^0$
Q7. In the given figure, area of isosceles $\triangle ABE$ is 72 sq cm, $BE=AB$ and $AB=2AD$. If $AE||DC$, then what is the area (in sq. cm) of the trapezium ABCD?
$144$
$124$
$108$
$136$
Q8. $\triangle XYZ$ is similar to $\triangle PQR$. If ratio of perimeter of $\triangle XYZ$ and perimeter of $\triangle PQR$ is $16:9$ and $PQ=3.6$ cm, then what is the length (in cm) of $XY$?
$3.2$
$4.8$
$8.6$
$6.4$Answers to the circle questions for SSC CGL and other geometry questions set 96
Q1. Answer: Option d: $d=r_1-r_2$.
Q2. Answer: Option c: $5^0$.
Q3. Answer: Option b: $2.5$.
Q4. Answer: Option a: $\displaystyle\frac{38}{7}$.
Q5. Answer: Option c: equilateral.
Q6. Answer: Option a: $35^0$.
Q7. Answer: Option a: $144$.
Q8. Answer: Option d: $6.4$.
Q9. Answer: Option d: $70$.
Q10. Answer: Option c: $32$ cm.
Solutions to the circle questions for SSC CGL and other geometry question set 96 - answering time wasSolution 1: Quick solution by visualization of the two internally touching circles
For easy visualization, let us draw the figure of the two internally touching circles as below.
Note: When you draw by hand, the figure need not be to the scale, or accurate with perfect circles. You need to capture just the fundamental relation between the radii of the two circles.
The radius of the larger circle is PR from its centre P to the common touching point R. The radius of the smaller circle QR from its centre Q to the same touching point R overlaps the radius of the larger circle PR and falls short by PQ.
This is the distance between the two centres and equals the difference of the two radii,Solution 2: Quick solution by arc angle subtended at the centre to perimeter relation of a circle and unitary method
The following figure shows the problem situation
Let $r_1$ and $r_2$ be the radii and $A_1$ and $A_2$ be the arcs of the smaller and the larger circles respectively.
Also let the desired angle subtended by the arc of the larger circle be $\theta$.
In a circle, a basic property of an arc is,
The length of an arc is proportional to the angle subtended by it at the centre.
Because of this proportionality, unitary method can be applied to find the angle subtended by an arc or the length of an arc from the reference figure of the whole periphery,
The whole circle periphery of length $2\pi r_1$ subtends an angle of $360^0$ at the centre.
So in the smaller circle, as angle subtended $30^0$ is $\displaystyle\frac{30^0}{360^0}=\frac{1}{12}$th of the angle subtended by the whole periphery of $2\pi r_1$, its length would be,
It subtends at the centre an angle that is $\frac{1}{72}$th of The angle $360^0$ subtended by the whole periphery at the centre.
So, $\theta=\displaystyle\frac{360^0}{72}=5^0$.
Answer: Option c: $5^0$.
Key concepts used:Relation between arc length and the angle subtended by it at the centre -- Proportionality of arc length to arc angle -- Unitary method-- Solving in mind.
Explanation of the concept of arc length and the angle it subtends at the centre
Think about the question,
What is the length of an arc that subtends an angle of $90^0$ at the centre?
The circle on the left in the following figure shows such an arc.
The arc is one-fourth the perimeter, that is, $\displaystyle\frac{\pi}{2} r$, where $r$ is the radius.
And the angle that it subtends at the centre is just, $90^0=\displaystyle\frac{\pi}{2}$.
We can then clearly specify that,
An arc that is $\displaystyle\frac{\pi}{2} r$ units long subtends an angle of $90^0$ at the centre.
Now what will be the situation for the arc shown on the right of the above figure where the arc covers half the perimeter?
In this case, the arc length is just be double of the arc on the left, that is, it is $\pi r$ units long and the angle subtended is also the double of that of the arc on the left, that is, it is $180^0=\pi$.
This confirms the very important property of an arc,
The length of an arc is proportional to the angle it subtends at the centre of the circle.
The reference arc for finding length of any arc subtending a given angle at the centre or vice versa is the whole circle periphery,
Reference arc for finding values of length or angle subtended by any arc, is the whole periphery of the circle of perimeter length $2\pi r$ and angle subtended at the centre $360^0=2\pi$.
This is Reference arc in arc angle proportionality concept.
Arc length being proportional to its angle subtended, unitary method is applied for finding the length of an arc given its angle subtended at the centre and vice versa.Solution 3: Quick solution by secant segment relation for a circle that is based on external angle property of a chord
The following figure shows the problem graphic.
By external angle property of a chord in circle, the external angle of chord QT formed with tangent at T equals the angle subtended by the chord at point R on the alternate segment of the circle,Solution 4: Quick solution by the concept of Segmentation ratio at the incentre of a triangle and by smart algebraic techniques
By the segmentation relation of the internal angle bisector at the incentre with the triangle sides,
The ratio of two segments of an internal angle bisector equals the ratio of the sum of the two sides adjacent to the angle bisected and the side opposite to the angle bisected.
By this concept, for the internal angle bisector AE and CD respectively in $\triangle ABC$,
$\displaystyle\frac{AO}{OE}=\frac{b+c}{a}=\frac{5}{4}$, where short forms of side names are used, and,
$\displaystyle\frac{CO}{OD}=\frac{a+b}{c}=\frac{2}{3}$.
We know by the above concept, the target ratio value is,
$\displaystyle\frac{BO}{OF}=\frac{c+a}{b}$.
That is why our objective is to form the sum $(c+a)$, and to do it we must equalize the numerators to the same value $(a+b+c)$ and invert the results.
Applying this target oriented smart algebraic technique add 1 to both sides of the two ratios above for the results,Solution 9: Quick solution by Key pattern identification of formation of isosceles triangles and total of angles of a triangle concept
The following figure describes the problem and helps explaining the solution.
The concepts required to solve the problem are really basic.
Identify that $\triangle ORS$, $\triangle OPR$ and $\triangle OQS$ are all isosceles as radii pairs, $OR=OS$, $OP=OR$ and $OQ=OS$. This results in the pairs of base angles equal in the three triangles.
Key concepts used:Key pattern identification of isosceles triangles, angle equality for a line intersecting a pair of parallel lines and total of angles in a triangle concept.Solution 10: Quick solution by Congruent triangles formation by two tangents to a circle from an external point and Key pattern identification
The following figure describes the problem,
From the external point A, the tangents AR and AQ to the circle forms two congruent triangles at centre O,
$\triangle ARO \cong \triangle AQO$.
This is a basic property of two tangents to a circle from a single external point.
Reason: AO the line segment from A to incentre O of $\triangle ABC$ bisects the $\angle A$, that is,
$\angle RAO=\angle QAO$.
Also, radii OR and OQ are perpendicular to the tangent at tangent points R and O forming a second equal pair of angles of value $90^0$.
So the third pair of angles are also equal,
$\angle ROA=\angle QOA$.
By A-A-A similarity criterion, the two triangles $\triangle ARO$ and $\triangle AQO$ are similar.
But in addition the pair of radii, $OR=OQ$.
This makes the two triangles $\triangle ARO$ and $\triangle AQO$ congruent, and so,
$AR=AQ=4.5$ cm.
Similarly,
$PC=QC=5.5$ cm, and,
$BR=BP=6$ cm.
Sum of the three given line segments 16 cm is then just half of the sum of the three sides or perimeter of the $\triangle ABC$.
Thus perimeter of the triangle is,
$AB+BC+CA$
$=(AR+BR)+(BP+PC)+(AQ+QC)$
$=2\times{16}=32$ cm.
Answer: Option c: $32$ cm.
Key concepts used:Two tangents to a circle from an external point -- Tangents from a common point -- Congruent triangles -- Internal angle bisectors of a triangle meet at the incentre -- Key pattern identification -- Perimeter concept -- Solving in mind.
End note
Observe that, each of the problems could be quickly and cleanly solved in minimum number of steps using basic and rich geometry concept based key patterns and methods in each case.
This is the hallmark of quick problem solving:
Concept based pattern and method formation, and,
Identification of the key pattern and use of the method associated with it. Every special pattern has its own method, and not many such patterns are there.
Important is the concept based pattern identification and use of quick problem solving method.
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Print off the twenty question pages and put them up around the room
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Answers are correct to one decimal place where appropriate
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Bubble chamber experiment help
In summary, the conversation is about an experiment that does not require going into the lab, but the data collection has been challenging. The person is struggling with understanding the figures and finding the tan planes. A link to a manual is provided, but it is too large to attach. The expert suggests checking a specific webpage for help with calculating the radius of curvature.
Apr 17, 2005
#1
jlmac2001
75
0
For this experiment, you don't have to go into the lab. I thought it would be a breeze but I was wrong. The problem I'm having is collecting the data. The figures given are to be used to find the data but I don't know what I'm looking at and how to find the tan planes. Is there someway to explain this or give an example to help me get started? Here's the address to the manual: File too big to attach.
Perhaps the calculation of the radius of curvature has you stumped. Check out this page
From there you should be able to calculate the radius of curvature in figure 3 or 4, and the angles you can measure directly.
Hope this helps.
Apr 28, 2005
#3
sir-pinski
1,466
1
Hi there,
I can definitely understand your struggle with the bubble chamber experiment. It can be overwhelming and confusing, especially if you are not familiar with the equipment and procedures.
First of all, don't worry - you don't have to physically go into the lab for this experiment. The manual provided in the link you shared contains all the necessary information and instructions to complete the experiment.
To start, I would recommend reading through the manual carefully and familiarizing yourself with the equipment, figures, and data given. It might also be helpful to do some background research on bubble chambers and their purpose in particle physics experiments.
As for finding the data, the figures given in the manual are usually plots of particle tracks recorded by the bubble chamber. These tracks can be used to determine the momentum, charge, and type of particles produced in the experiment.
To find the tan planes, you will need to look for tracks that intersect at a specific angle. The manual should have a diagram explaining the geometry of the bubble chamber and how to determine the tan planes.
If you are still having trouble, I would suggest reaching out to your instructor or a classmate for further clarification. It might also be helpful to work through some practice problems to get a better understanding of the concept.
I hope this helps and good luck with the experiment!
Related to Bubble chamber experiment help
What is a bubble chamber experiment?
A bubble chamber experiment is a type of particle detector used in high-energy physics. It consists of a chamber filled with a superheated liquid, usually liquid hydrogen or liquid helium. When charged particles pass through the liquid, they leave a trail of bubbles, making it possible to track their paths and determine their properties.
How does a bubble chamber experiment work?
In a bubble chamber experiment, a beam of particles is directed into the chamber and a magnetic field is applied. The particles then move through the liquid, ionizing the atoms and causing them to boil, leaving a trail of bubbles. The bubbles are photographed and analyzed to determine the type, energy, and trajectory of the particles.
What is the purpose of a bubble chamber experiment?
The purpose of a bubble chamber experiment is to study the properties of subatomic particles, such as their mass, charge, and spin. It is also used to observe the interactions between particles and to discover new particles.
What are the advantages of using a bubble chamber experiment?
One advantage of using a bubble chamber experiment is its high sensitivity, allowing for the detection of rare particles. It also provides a visual representation of particle interactions, making it easier to interpret the data. Additionally, it can be used to study both charged and neutral particles.
What are the limitations of a bubble chamber experiment?
One limitation of a bubble chamber experiment is that it is a destructive process, as the bubbles created prevent the particles from being studied further. It also has a limited time frame for data collection, as the liquid will eventually become too saturated with bubbles to be useful. It is also a costly and complex experiment to set up and maintain. | 677.169 | 1 |
What Are Quadrants on a Coordinate Plane
What Are Quadrants on a Coordinate Plane
Understanding the concept of quadrants on a coordinate plane is fundamental to many mathematical applications. These four sections divide the plane into distinct regions, each with unique characteristics and properties.
From identifying points to plotting functions, the ability to navigate through these quadrants is essential for anyone working with coordinates. The significance of quadrants extends beyond basic graphing; they play a crucial role in various mathematical disciplines.
Exploring how quadrants influence geometric relationships and analytical thinking can open up a deeper understanding of spatial concepts.
Understanding Quadrants on a Plane
Understanding the quadrants on a plane is essential for accurately locating points and analyzing relationships between coordinates. The Cartesian plane, also known as the coordinate plane, consists of four quadrants that help in identifying the position of a point. This mathematical tool is fundamental in graphing equations, plotting data points, and solving geometric problems.
Each quadrant is designated a Roman numeral (I, II, III, IV) and has its unique set of positive and negative coordinates. The coordinate systems in each quadrant follow specific rules that dictate the signs of the coordinates based on their position relative to the origin at (0,0). Quadrant I contains positive x and y values, Quadrant II has negative x and positive y values, Quadrant III has negative x and y values, and Quadrant IV has positive x and negative y values.
Mastery of these quadrants on a plane is crucial for mathematical analysis and applications in various fields.
Labeling Quadrants on a Graph
Labeling quadrants on a graph is essential for accurately representing the positions of points and understanding their relationships within a coordinate system. In a graph visualization, the coordinate system consists of four quadrants, each denoted by Roman numerals I, II, III, and IV.
Quadrant I is located in the top right corner, where both x and y coordinates are positive. Moving counterclockwise, Quadrant II is in the top left with negative x and positive y values. Quadrant III is at the bottom left, featuring negative x and y coordinates. Finally, Quadrant IV is in the bottom right, characterized by positive x and negative y values.
Importance of Quadrants in Mathematics
Quadrants in mathematics play a crucial role in determining the positions of points and understanding their relationships within a coordinate system. In Cartesian coordinates, quadrants are essential divisions that help locate points based on their distances from the origin and the signs of their coordinates. Each quadrant has a unique combination of positive and negative coordinates, allowing for precise positioning of points in a two-dimensional space. | 677.169 | 1 |
The issue is that the question asks you to go from the image back to the preimage. So if the translation from the preimage to the image is <4,-7>, then to go backwards from the image to the preimage, you have to go backwards along the vector <-4,7>.
Move the green line onto the blue line so that you could see it overlapping it. Then move one point of the green line depending on the translation. In problem 2, the translation is (10,0), so it means that you have to move the point 10 units to the right, and 0 units up. Then do the same thing on the other point. Then you are done.
Yes. As one can see, the question shows P'Q'R', and asks you to map PQR. Remember that P'Q'R' is read, "P prime Q prime R prime." "Prime" signifies that it is the IMAGE, or result, of a transformation. So, it wants you to map the original triangle. | 677.169 | 1 |
The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh ...
A corollary is added to this proposition, which is necessary to prop. 1. B. 11. and otherwise.
PROP. XX. and XXI. B. I.
Proclus, in his commentary, relates, that the Epicureans derided this proposition, as being manifest even to asses, and needing no demonstration; and his answer is, that though the truth. of it be manifest to our senses, yet it is science which must give the reason why two sides of a triangle are greater than the third: but the right answer to this objection against this and the 21st. and some other plain propositions, is, that the number of axioms ought not to be increased without necessity, as it must be if these propositions be not demonstrated. Mons. Clairault, in the preface to his Elements of Geometry, published in French at Paris anno 1741, says, That Euclid has been at the pains to prove, that the two sides of a triangle which is included within another are together less than the two sides of the triangle which includes it; but he has forgot to add this condition, viz. that the triangles must be upon the same base: because, unless this be added, the sides of the included triangle may be greater than the sides of the triangle which includes it, in any ratio which is less than that of two to one, as Pappus Alexandrinus has demonstrated, in prop. 3. B. 3. of his mathematical collections.
PROP. XXII. B. I.
Some authors blame Euclid, because he does not demonstrate that the two circles made use of in the construction of this problem must cut one another: but this is very plain from the determination he has given, viz. that any two of the straight lines DF, FG, GH must be greater than the third: for who is so dull, though only beginning to learn the Elements, as not to perceive that the circle described from the centre F, at the distance FD, must meet FH betwixt F and H, because FD is less than FH;
D M
FG
H
and that, for the like reason, the circle described from the centre G, at the distance GH, or GM, must meet DG betwixt D
DM
FG
H
and G; and that these circles must meet one another, because FD and GH are together greater than FG? and this determination is easier to be understood than that which Mr. Thomas Simson derives from it, and puts instead of Euclid's in the 49th page of his Elements of Geometry, that he may supply the omission he blames Euclid for; which determination is that any of the three straight lines must be less than the sum, but greater than the difference of the other two: from this he shows the circles must meet one another, in one case; and says that it may be proved after the same manner in any other case; but the straight line GM, which he bids take from GF, may greater than it, as in the figure here annexed: in which case his demonstration must be changed into another.
PROP. XXIV. B. 1.
To this is added, "of the two sides DE, DF, let DE be that which is not greater than the other;" that is, take that side of the two DE, DF, which is not greater than the other, in order to make with it the angle EDG equal to BAC, because, without this restriction, there might be three different cases of the proposition, as Campanus and others make.
D
&
Mr. Thomas Simson, in p. 262 of the second edition of his Elements of Geometry, printed anno 1760, observes, in his notes, that it ought to have been shown that the point F falls below the line EG. This probably Euclid omitted, as it is very easy to perceive, that E DG being equal to DF, the point G is in the circumference of a circle described from the centre D, at the dis tance DF, and must be in that part of it which is above the straight line EF, because DG falls above DF, the angle EDG being greater than the angle EDF.
PROP. XXIX. B. I.
The proposition which is usually called the 5th postulate, or 11th axiom, by some the 12th, on which this 29th depends, has
given a great deal to do, both to ancient and modern geometers: it seems not to be properly placed among the axioms; as indeed it is not self-evident; but it may be demonstrated thus: DEFINITION 1.
The distance of a point from a straight line, is the perpendicular drawn to it from the point.
DEF. 2.
One straight line is said to go nearer to, or further from another straight line, when the distance of the points of the first from the other straight line becomes less or greater than they were; and two straight lines are said to keep the same distance from one another, when the distance of the points of one of them from the other is always the same.
AXIOM.
A straight line cannot first come nearer to another straight line, and then go further from it, before it A
cuts it; and, in like manner, a straight D. line cannot go further from another F straight line, and then come nearer to it; nor can a straight line keep the same.
B
C
E
H
distance from another straight line, and then come nearer to it, or go further from it; for a straight line keeps always the same
direction.
A
B
For example, the straight line ABC cannot first come nearer to the straight line DE, as from the point A to the point B, and then, D from the point B to the point C, go F further from the same DE: and, in like manner, the straight line FGH
G
C
E
H
cannot go further from DE, as from F to G, and then, from G to H, come nearer to the same DE: and so in the last case, as in figure 2. (See the figure above.)
PROP. I.
If two equal straight lines, AC, BD, be each at right angles to the same straight line AB; if the points C, D be joined by the straight line CD, the straight line EF drawn from any point E in AB unto CD, at right angles to AB, shall be equal to AC,
or BD.
If EF be not equal to AC, one of them must be greater than the other; let AC be the greater; then, because FE is
F
D
F
less than CA, the straight line CFD is nearer to the straight line AB at the point F than at the point C, that is, CF comes nearer to AB from the point C to F: but because DB is greater than FE, the straight line CFD C is further from AB at the point D than at F, that is, FD goes further from AB from F to D: therefore the straight line CFD first comes nearer to the A straight line AB, and then goes further from it, before it cuts it; which is impossible. If FE be said to be greater than CA, or DB, the straight line CFD first goes further from the straight line AB, and then comes nearer to it; which is also impossible. Therefore FE is not unequal to AC, that is, it is equal to it.
PROP. II.
E
B
If two equal straight lines AC, BD be each at right angles to the same straight line AB; the straight line CD, which joins their extremities, makes right angles with AC and BD.
F Ꭰ
Join AD, BC; and because, in the triangles CAB, DBA, CA, AB are equal to DB, BA, and the angle CAB equal to the angle DBA; the base BC is equal (4. 1.) to the base AD: and in the triangles ACD, BDC, AC, CD, are equal to BD, DC, and the base AD is equal to the base BC: C therefore the angle ACD is equal (8. 1.) to the angle BDC from any point E in AB draw EF unto CD, at right angles to AB: therefore by prop. 1. EF is equal to AC, or BD; wherefore, A
E B
G
as has been just now shown, the angle ACF is equal to the angle EFC in the same manner, the angle BDF is equal to the angle EFD; but the angles ACD, BDC are equal; therefore the angles EFC and EFD are equal, and right angles (10. def. 1.); wherefore also the angles ACD, BDC are right angles.
COR. Hence, if two straight lines AB, CD be at right angles to the same straight line AC, and if betwixt them a straight line BD be drawn at right angles to either of them, as to AB; then BD is equal to AC, and BDC is a right angle.
If AC be not equal to BD, take BG equal to AC, and join CG; therefore, by this proposition, the angle ACG is a right angle; but ACD is also a right angle; wherefore the an
gles ACD, ACG are equal to one another, which is impossible. Therefore BD is equal to AC; and by this proposition BDC is a right angle.
PROP. III.
If two straight lines which contain an angle be produced, there may be found in either of them a point from which the perpendicular drawn to the other shall be greater than any given straight line.
Let AB, AC be two straight lines which make an angle with one another, and let AD be the given straight line; a point may be found either in AB or AC, as in AC, from which the perpendicular drawn to the other AB shall be greater than AD.
F
K
B
M
In AC take any point E, and draw EF perpendicular to AB; produce AE to G, so that EG be equal to AE, and produce FE to H, and make EH equal to FE, and join HG. Because, in the triangles AEF, GEH, AE, EF are equal to GE, EH, each to each, and contain equal (15. 1.) angles, the angle GHE is therefore equal (4. 1.) to the angle AFE, which is a right angle: draw GK perpendicular to AB; and because the straight lines FK, HG are at right angles to FH, and KG at right angles to FK; N KG is equal to FH, by cor. pr. 2. that is, to the double of FE. In the same manner, if AG be produced to L, so that
D P
A
G
C
L
GL be equal to AG, and LM be drawn perpendicular to AB, then LM is double of GK, and so on. In AD take AN equal to FE, and AO equal to KG, that is, to the double of FE, or AN; also take AP equal to LM, that is, to the double of KG, or AO; and, let this be done till the straight line taken be greater than AD; let this straight line so taken be AP, and because AP is equal to LM, therefore LM is greater than AD. Which was to be done.
PROP. IV.
If two straight lines AB, CD make equal angles EAB, ECD with another straight line EAC towards the same parts of it; AB and CD are at right angles to some straight line. | 677.169 | 1 |
I am attempting to construct a square (Problem 58 in the title text) using a straight edge and compass and verify that what I have constructed is indeed a square (a rectangle with 4 congruent sides, where a rectangle is a polygon whose 4 segments meet at right angles). This is to be done in neutral geometry.
To begin with, I construct a line segment $AB$. I then construct perpendicular lines to $AB$ at $A$ and $B$ each. I then form $BC$ and $AD$ congruent to $AB$ on each of these lines, where $C$ and $D$ are on the same side of the line containing $AB$. From here, I am stuck. We are supposed to use what Clark calls Axiom 5: given a line $l$ and a point $P$ not on $l$, there is at most one line through $P$ parallel to $l$. If you combine this with a former result we have, you find that "at most" can be replaced with "exactly." I do not see how this helps me.
My ideas are as follows: form $CD$ and show it is congruent to and parallel to $AB$. I do not see why either of these should be true given what we know. The only ways we have developed so far to show two lines/segments are parallel involve showing the pair of lines/segments in question admit a transversal with either congruent alternating interior angles or congruent corresponding angles. The problem is we know nothing about the angles $BCD$ and $CDA$ a priori.
The other alternative is to construct a perpendicular segment $CE$ to $BC$ at $C$ and to somehow show that $D = E$, but my attempts at this (using combinations of SSS, SAS, AAS, ASA, and various versions of "SSA for right triangles," all of which we know) have failed. It seems I am either not recalling or am short one tiny fact needed to conclude.
This is not homework, but prep for a course I am teaching in the fall.
1 Answer
1
I used your observation that the only way we know lines are parallel is if we have alternate interior angles.
So let $\overline{AB}$ be a segment and construct a perpendicular line $\ell$ to $AB$ passing through B, and find a point C on $\ell$ so that $BC \cong AB$ (so, just as you started). Now through construct a line $m$ perpendicular to $\ell$ and passing through $C$. By alternate interior angles, we know $m$ is parallel to line $AB$. Similarly, let $n$ be the line through $A$ perpendicular to $AB$, and so by alternate interior angles it is parallel to $\ell$.
First, we claim that $m$ and $n$ intersect. If not, then $m$ and $n$ are parallel and so the lines $\ell$ and $m$ are TWO different lines passing through $C$ that are parallel to $n$, contradicting Axiom 5. So let $D$ be the point of intersection of $m$ and $n$. The angle $ADC$ is right angle, again because of axiom 5 - draw the line $n'$ through $D$ perpendicular to $m$, by alternating interior angles $m$ must be parallel to $\ell$, and so $n, n'$ are two lines passing through $D$ parallel to $\ell$, so must be equal by axiom 5. So to summarize, so far we've created a rectangle with $AB\cong BC$.
I'm pretty sure you can figure it out from here, so I'll just outline one way: draw $BD$, then apply Hypotenuse Leg, so then various angles that split a 90 degree angle are congruent and so they are 45, and then apply isoceles triangle theorem to get the sides of the rectangle are congruent. Let me know if you want me to add the details. | 677.169 | 1 |
Apolloniu's Theorem
The world of mathematics is a field that has many theorems, and each gives or provides a unique perspective on the complexity of geometric shapes; one such important Theorem is "Apollonius' Theorem." It is a result that shows the relationship between sides of triangles; this Theorem is named after the ancient Greek mathematics Apollonius of Perga. This Theorem has been a source of fascination and usefulness for mathematics and students alike for centuries. In this article, we will get to know about the concept of Apollonius' Theorem in a detailed manner, understanding its history, geometric importance, and applications, and end this article with a conclusion.
History
To get a better understanding of the importance of Apollonius' Theorem, we have first to get that this Theorem is named after Apollonius of Parge. He was born in the early 3rd century BCE; he was a famous Greek mathematician known for his contribution to the field of geometry and conic section. His work is used as the base for future development in the field of mathematics and science.
Apollonius' Theorem is a result that gives the relation between the median and sides of a triangle. It was first presented in his monumental work "Conics," a detailed book on the properties and classifications of conic sections; the Theorem appears in Book I, Proposition 6 of "Conics," where Apollonius explores the relationships between circles and triangles.
Definition of Apollonius' Theorem
In a triangle, "if we square the length of two sides and add them together, the result is equal to the sum of the square of half the length of the third side and the square of the length of the median of the third side"; this Theorem is known as the Law of Cosines and is usually used in trigonometry to solve problems related to triangles. It is a useful tool to calculate the length of a side or an angle in a triangle, given the lengths of the other sides and angles.
In the form of a formula:
If O is the midpoint of any side of a triangle, for a better understanding, consider O is the mid-point of side MN of a triangle LMN, according to Apollonius' Theorem.
Proof of Apollonius' Theorem
In general, we can prove this Theorem with the help of "Pythagoras' Theorem," but this Theorem can also be proved by two additional methods, which are "Using Coordinate Geometry and Vectors." So, in the following discussion, we learn about all these proofs one by one.
1. Proof by Pythagoras' Theorem
In a triangle ABC, let M be the midpoint of the BC side of the triangle ABC.
To Prove: AB2 + AC2 = 2 * [AM2 + CM2]
Proof: As we have,
Let us draw a perpendicular AH on the line BC, such that:
Now, according to the Pythagoras' Theorem, we have:
On adding equation (1) and equation (2), we get:
Add and Subtract 2*MH2 on the right side of the equation, we get:
From equation (3), we get:
As BM = CM = BC / 2, we get:
Which we have to prove, Hence Proved.
2. Proof by Considering Vectors
In a triangle, ABC, M is the midpoint of the side of triangle BC, such that Cartesian coordinates are defined as:
To Prove: AB2 + AC2 = 2 * [AM2 + CM2]
Proof: We have AB and AC equal to |b| and |c|, respectively. On putting these values on the left-hand side of the above equation, we get:
As BM = CM = BC / 2 (Because M is the midpoint of BC), we get:
Which we have to prove, Hence Proved.
3. Proof Using Coordinate Geometry
To prove Apollonius' Theorem, we have to consider O as an origin, which lies on the MN side of a triangle LMN, such that MN and OY are considered as the x-axis and y-axis, respectively, where MN = 2 * a and the coordinate points of M and N are (a, 0), and (-a, 0), respectively and coordinate of L is (b, c).
To Prove: (LM)² + (LN)² = 2 * [(LO)² + (MO)²]
Proof: As the coordinate point of O is (0, 0), LO is calculated as follows:
And squaring both sides, we get:
Similarly, LM is given by:
Similarly, MO is given by:
Similarly, LN is given by:
On adding equation (2) and equation (4), we get:
Which we have to prove, Hence Proved.
Applications of Apollonius' Theorem
While Apollonius' Theorem may appear abstract, its practical applications are found in various fields; some of them are discussed below:
Geometric Analysis: Apollonius' Theorem is valuable for geometric analysis and problem-solving, especially problems that involve triangles; it allows mathematicians and engineers to better understand and manipulate the properties of triangles in various contexts.
Engineering and Architecture: Engineers and architects use Apollonius' Theorem to design and analyze structures to ensure that load distribution and stability are properly distributed all over the structure.
Surveying and Cartography: Surveyors and cartographers use Apollonius' Theorem when working with triangles on the surface of the Earth, helping to create accurate maps and land surveys.
Statistics: In statistical analysis, Apollonius' Theorem can be adapted to explore relationships between data points in multi-dimensional space, contributing to data visualization and clustering algorithms.
Triangle Analysis: Apollonius' Theorem provides a means to analyze triangles and calculate their circumradius, even when the angles and side lengths are unknown; this is valuable in various engineering and scientific fields, such as structural analysis and astronomy.
Navigation: Apollonius' Theorem has applications in navigation, particularly in geodetic calculations, and it can be used to determine the radius of the circle that best fits a set of three non-collinear points on the Earth's surface, which is useful in surveying and mapmaking.
Robotics: Robotics engineers use Apollonius' Theorem to analyze the geometry of robot manipulators and determine the optimal placement of sensors and actuators, and it plays a role in solving inverse kinematics problems.
Molecular Chemistry: In the field of molecular chemistry, Apollonius' Theorem can be applied to analyze the arrangement of atoms in molecules, particularly in the context of molecular geometry.
Conclusion
Apollonius' Theorem, named after the ancient Greek mathematician Apollonius of Perga, is a powerful geometric result that relates the sides of a triangle to the median of that triangle; it is a live proof that shows its importance in mathematics. The proof of this Theorem, which can be solved with three different methods, shows the beauty of mathematical reasoning.
Beyond its theoretical importance, Apollonius' Theorem finds practical applications in diverse fields, from engineering and computer graphics to navigation and molecular chemistry. Its ability to bridge the gap between triangles and their median makes it a valuable tool for solving complex geometric problems and understanding the fundamental relationships in geometry. As we continue to explore the depths of mathematical knowledge, Apollonius' Theorem remains a valuable example of what mathematicians of history have given to us, reminding us that the beauty of mathematics increases with the increase in time and borders. | 677.169 | 1 |
Curlicue Fractal
The curlicue fractal is a figure obtained by the following procedure. Let be an irrational number.
Begin with a line segment of unit length, which makes an angle to the horizontal. Then define
iteratively by
with .
To the end of the previous line segment, draw a line segment of unit length which
makes an angle | 677.169 | 1 |
Saturday 4 August 2018
Introduction To Euclid's Geometry - Axioms, Theorem and Other terms
(Note: Following list is compiled from a very old book 'Elements of Geometry by Ledegenre' and other NCERT textbooks)
Important terms to remember:
1. An axiom is a self-evident proposition (an assumption which is a self-evident truth).
2. A theorem is a truth, which becomes evident by means of a train of reasoning called a demonstration or proof. Theorems are proved, using axioms, previously proved statements and deductive reasoning.
3. A problem is a question proposed, which requires a solution.
4. A lemma is a subsidiary truth, employed for the demonstration/proof of a theorem, or the solution of a problem.
5. The common name, proposition, is applied indifferently, to theorems, problems, and lemmas.
6. A corollary is an obvious consequence, deduced from one or several propositions.
7. A scholium is a remark on one or several preceding propositions, which tends to point out their connection, their use, their restriction, or their extension.
7. A hypothesis is a supposition, made either in the enunciation of a proposition or in the course of a demonstration/proof.
8. A conjecture is a conclusion or proposition based on incomplete information, for which no proof has been found.
List of a few Euclid's Axioms are:
1. Things which are equal to the same thing, are equal to each other.
2. If equals are added to equals, the wholes will be equal.
3. If equals are taken from equals, the remainders will be equal.
4. If equals are added to unequals, the wholes will be unequal.
5. If equals are taken from unequals, the remainders will be unequal.
6. Things which are double of the same thing, are equal to each other.
7. Things which are halves of the same thing are equal to each other.
8. The whole is greater than any of its parts.
9. The whole is equal to the sum of all its parts.
10. All right angles are equal to each other.
11. From one point to another only one straight line can be drawn.
12. Through the same point, only one straight line can be drawn which shall be parallel to a given line.
13. Magnitudes, which is applied to each other, coincide throughout their whole extent, are equal. | 677.169 | 1 |
In a math paper of MAA(Mathematical Association of America) by Z.Chen, he proved that-
"If, inside $∆ ABC$, there is a circular region $ R$ for which the sum of the
distances from a point $P$ in $R$ to the three sides of the triangle is independent of the
position of $P$, then $∆ABC$ is equilateral."
$\begingroup$Please include a link to the paper you're talking about in the question. Also, the theorem is definitely not true for isosceles triangles in general. I suspect Chen is right in that it's only true for equilateral triangles, but I can't be bothered to check for myself.$\endgroup$
1 Answer
1
When it is true, it is true for the whole triangle interior.
The math paper says "circular region $R$" so that the region has a non-empty interior. It would have been sufficient to say "in a zone with a non-empty interior".
Proof 1
Call $\alpha$ and $\beta$ two of the triangle interior angles on vertices.
If the sum of distances from a point $P$ to the sides is independent of $P$'s position inside a zone $Z$ with a non-empty interior, we can move $P$ in each direction orthogonal to the triangle sides without modifying the sum.
And no, it does not work for isoceles triangles. Which is somewhat intuitive, as isoceles triangles are equilateral triangles elongated on one direction, which breaks the distance sum invariance.
Proof 2 (better in my opinion)
Let $u_1, u_2, u_3$ be the three unit vectors orthogonal to the triangle sides.
Let $v$ be the displacement vector we apply to point $P$.
Sum of distances is modified by $v.(u_1+u_2+u_3)$ (scalar product).
This is $=0$ for $v$ small in any direction, so it implies $u_1+u_2+u_3=0$.
As these three vectors have a null sum, they make a triangle.
As they have equal length ($=1$), this is an equilateral triangle.
So they have internal angles of $\frac {\pi} 3$ between them.
As they are orthogonal to the sides of the initial triangle, this triangle has also internal angles of $\frac {\pi} 3$. Hence it is an equilateral triangle. | 677.169 | 1 |
It can be formed as the convex hull of 2 oppositely oriented semi-uniform dodecagonal duoprisms where the larger dodecagon is more than 6−2{\displaystyle {\sqrt {6}}-{\sqrt {2}}} times the edge length of the smaller one. | 677.169 | 1 |
How do I rotate a shape in 3 JS?
How do I rotate a shape in 3 JS?
js suggest the way to rotate an object around a point using three. js is to create a parent object at the position you want to rotate around, attach your object, and then move the child. Then when the parent is rotated the child rotates around the point.
What are the three rotations?
Specifically, the first angle moves the line of nodes around the external axis z, the second rotates around the line of nodes and the third is an intrinsic rotation (a spin) around an axis fixed in the body that moves.
Which 3 things must be known to perform a rotation?
Three pieces of information are needed to rotate a shape:
the centre of rotation.
the angle of rotation.
the direction of rotation.
What rotation is equivalent to degrees?
A full rotation is 360 degrees
Rotations
Radians
Degrees
¼
π/2
90°
½
π
180°
1
2π
360°
1½
3π
540°
How do you make a cube in 3 JS?
For creating a cube, we need to use the BoxGeometry object that contains all the vertices and faces of the cube. var geometry = new THREE. BoxGeometry(700.
What is the rule for a 90 degree clockwise rotation?
Rule : When we rotate a figure of 90 degrees clockwise, each point of the given figure has to be changed from (x, y) to (y, -x) and graph the rotated figure. Let us look at some examples to understand how 90 degree clockwise rotation can be done on a figure.
What is the rule for a 90 degree rotation?
The rule for a rotation by 90° about the origin is (x,y)→(−y,x) .
What are the rules for rotation?
Rules of Rotation The general rule for rotation of an object 90 degrees is (x, y) ——–> (-y, x). You can use this rule to rotate a pre-image by taking the points of each vertex, translating them according to the rule, and drawing the image.
How can I rotate a mesh by 90 degrees in threejs?
I have a mesh that I want to rotate by 90 degrees inside Three JS. Here is the image of the current situation: I want the selected mesh to be rotated parallelly to the large mesh.
Why does rotation not work in Three.js?
I'm working on a project in three.js where I have a cube and can rotate it with some buttons (arrows). The rotation works, but after some rotations it doesn't spin in the right direction anymore. Probably because the axes of the cube are shifted? But this confuses me a little bit..
When do we move an object around in Three.js?
When we move an object around in a three.js scene, we follow the rules of Cartesian coordinate systems. Here the analogy breaks down a little because each piece on the chessboard has its own way of moving, whereas in a Cartesian coordinate system, translation, rotation, and scale behave the same for any kind of object.
How do you rotate a cube in JavaScript?
I need the cube to be able to rotate up, down, left and right by some controls. This is basically how it currently works: When a direction is pressed, say 'right', we update the mesh.rotation.y with tween.js until mesh.rotation.y is rotated in a 90 degree angle. | 677.169 | 1 |
Define the ellipse as the locus of points P for which the ratio PF/PP' = e is constant (e<1). Here F is a fixed point and PP' is the distance
of a variable point P from a fixed line L. F is called a focus of the ellipse and L is called a directrix of the ellipse.
From the definition follows that. 1) all points of the ellipse are inside the Apollonian circle of the segment FF0, where F0 the projection of F
on the directrix. There are exactly two points {A,B} of this circle in common with the ellipse. The circle is called the auxiliary circle of the ellipse. Points {A,B} are called vertices of the ellipse. Obviously. 2) the ellipse is symmetric with respect to the axis AB.
Take coordinates (x,y) along the axis AB and L correspondingly. Define d = F0F, P=(x,y). Then the definition condition
implies that the points of the ellipse satisfy. 3) y2 + (x-d)2 = e2x2. This defines a quadratic equation for x. x2(1-e2) - 2dx + (d2+y2) = 0. The two roots x1, x2 for constant y have their middle at d/(1-e2), which is the x-coordinate of the
Apollonian circle (see ApollonianBundle.html but put there k=1/e). Thus it follows that. 4) The ellipse is symmetric w.r to the parallel to the directrix from the center of the auxiliary circle. Since the ellipse is also symmetric w.r to the axis AB it follows that. 5) the ellipse is central-symmetric w.r. to point O. This shows that. 6) there exists a second focus F* symmetric to F w.r. to O and a second directrix L*, which is the symmetric of L w.r. to O. The pair (F*, L*) has the same property as the (F, L) with respect to all points of the conic.
The defining relations PF = ePP' and PF* = ePP1 imply. 7) PF + PF* = e(PP' + PP1) = constant. This constant is easily identified with |AB|=2ed/(1-e2) (see ApollonianBundle.html ). Last quantity is called the major axis of the ellipse. Its radius is R = ed/(1-e2) and its center
is at x0 = d/(1-e2).
Using the definition again the ratio (y/y')2 = (e2x2-(x-d)2)/(R2-(x-x0)2) is seen to be = 1-e2. Thus. 8) the ellipse is the image of the circle under the transformation f: (x,y)-->(x,gy), where g2 = 1-e2. The form of this transformation implies that. 9) the tangent to the circle at y' and the tangent to the ellipse at y pass through the same point S on AB. Also if SQR is a secant of the circle then it maps under f to a secant SQ'R' of the ellipse and this shows that. 10) the polars of the circle and the ellipse for points on the axis AB are the same. In particular. 11) the directrices L, L* are the polars respectively of the foci F,F* w.r. to the circle as well as the ellipse.
If P is an arbitrary point on the ellipse draw from the focus F the orthogonal FP intersecting the directrix L at S. 12) Line SP is the tangent to the ellipse.
In fact, for any other point Q on SP and the projections Q', Q1 respectively on the directrix L and SF, if Q were
on the ellipse then it would be QF = eQQ', but it is also QQ1/QQ' = PF/PP'=e => QQ1 = QF. This, by the orthogonality at Q1 is impossible if Q is different from P. Thus line SP has precisely one point
in common with the ellipse, hence it is its tangent. As a corollary we obtain also. 13) the tangent at P and the orthogonal at F of the focal radius FP intersect at a point S on the directrix.
A consequence of the last property in combination with the symmetry of the ellipse is the fact. 14) The angles FPQ, F*PS formed with the tangent and the focal radii at a point of the ellipse are equal. The reason for this is the fact that quadrangles PFQP' and PF*P1S are cyclic and these two angles
are equal respectively to FP'F0, F*P1F1, which by symmetry are equal.
Related to this property is also the other focal property of the ellipse by which. 15) the angles FPA, F*PB between the focal radii PF, PF* and the tangents PA, PB from P respectively are equal.
In fact, define the reflexions S, T respectively of F, F* w.r. to the tangent PA and also the reflexion Q of F*
with respect to the other tangent PB from P. Triangles PTF and PQF are equal having three equal corresponding sides.
Thus, angles TPF and FPQ are equal, hence angle(FPF* + 2APF) = angle(FPF* + 2F*PB). Which shows that angle(APF) = angle(F*PB).
A consequence of the last property is a couple of properties relating to the auxiliary circle. 16) The portion SR of the tangent at a point X between the tangents at the vertices is viewed from a focus F
under a right angle. 17) The projection P of the focus on a tangent is a point of the auxiliary circle.
The first property results by projecting F to P on the tangent at X. Quadrangles FPST, FPRQ are then cyclic. By the previous property angle(TPF)=angle(QPR) and
by the cyclic quadrangles this comes to angle(TSF)=angle(RFQ) thereby proving that angle(SFR)
is a right one. Also angle(TPQ) = angle(TPF)+angle(FPQ) = angle(TSF)+angle(FRQ) shows the other statement.
A consequence of this property is that. 18) lines FS, FR are the bisectors of the angle(TFX). This is seen by observing that the point of tangency X and the intersection point V of SR separate
harmonically points (S,R). This in turn is seen by projecting X to point Y on the auxiliary with a
parallel WX to the directrix. By circle properties (V,W) separate harmonically (T,Q), hence
(V,X) separate harmonically (S,R). Thus the bundle of lines at F: F(V,S,X,R) is harmonic and
two of its lines are orthogonal, hence they are bisectors of the other two (see Harmonic_Bundle.html ).
Properties 16), 17) imply two other ways to generate the ellipse. 19) The ellipse is the envelope of the hypotenuse of the triangle SFR formed by intersecting
the legs of a right-angle SFR with two fixed parallels TS, QR while turning the angle about the fixed point F
lying between the parallels. 20) The ellipse is the envelope of lines PR drawn orthogonally to segment FP at P for fixed F
and P moving on a circle containing point F. | 677.169 | 1 |
What is found at 0 degrees latitude?
The Equator
What is 0 degree latitude and is a great circle?
The Equator is at 0°, and the North Pole and South Pole are at 90° north and 90° south, respectively. The Equator is the longest circle of latitude and is the only circle of latitude which also is a great circle.
What is found at the maximum latitudes?
Lines of latitude (parallels) run east-west around the globe and are used to measure distances NORTH and SOUTH of the equator. Since the equator is 0�, the latitude of the north pole, 1/4 of the way around the globe going in a northerly direction, would be 90�N. This is the highest latitude possible.
What is the minimum latitude?
Latitudes are supposed to be from -90 degrees to 90 degrees, but areas very near to the poles are not indexable. So exact limits, as specified by EPSG:900913 / EPSG:3785 / OSGEO:41001 are the following: Valid longitudes are from -180 to 180 degrees.২৮ মে, ২০১৪
How long is a degree of latitude?
approximately 364,000 feet
Why is the international date line at 180 degrees?
The 180-degree meridian was chosen because it runs mostly through open ocean in the central Pacific, zigging and zagging to keep nearby nations on their own day and date. So the choice of 180 degrees was arbitrary, but it established the IDL in use today.৩১ আগস্ট, ২০১৮
How long does it take Earth to pass one longitude?
24 hours
Why are latitudes parallel but not of the same size?
১০ মার্চ, ২০১৯
What decreases as we move away from the equator?
In any sphere the circumference is lengthier at 0°. So equator is lengthiest latitude and the degree increases or move away from equator to pole the circumference of globe decreases hence latitudinal length also decreases. The length of a parallel increases as we move away from the poles towards the equator.
What happens if we move away from the equator?
Latitude: Temperature range increases with distance from the equator. Also, temperatures decrease as you move away from the equator. This is because the suns rays are dispersed over a larger area of land as you move away from the equator. This is due to the curved surface of the earth.
Why does it become colder as we move away from the equator?
The hottest temperatures on Earth are found near the equator. This is because the sun shines directly on it for more hours during the year than anywhere else. As you move further away from the equator towards the poles, less sun is received during the year and the temperature becomes colder.
How does the distance from the equator affect climate?
The distance from the equator affects the climate of a place. At the poles, energy from the sun reaches the Earth's surface at lower angles and passes through a thicker layer of atmosphere than at the equator. This means the climate is cooler further from the Equator.
Why are temperatures higher at the equator?
When the sun's rays strike Earth's surface near the equator, the incoming solar radiation is more direct (nearly perpendicular or closer to a 90˚ angle). Therefore, the solar radiation is concentrated over a smaller surface area, causing warmer temperatures.২ ডিসেম্বর, ২০১৯ | 677.169 | 1 |
If we are on spherically curved surface, (I will put the radius as 1 to make every unit to be treated as radian.) we can select two random coordinates. Let's call those two pairs as (lon, lat) and (lon', lat'). These coordinates are based on spherical coordinate system, however, we do not take care about the radius.
we have following formula for measuring distance (D) and position angle (PA) between two coordinates.
What I want to do is doing the inverse. Let's suppose we have a set of (lon, lat) and (D,PA). I expected we can get an unique pair of (lon',lat') if I use inverse backtracking method (fsolve in python scipy module).
However, it seems that (D,PA cannot determine unique pair of (lon',lat') pair.
What I expected at the first time is that there would be only one certain point which would have certain distance and certain angle from the reference (in case of position angle, it would be the north pole.) This makes sense when I am using the Euclidean geometry and I was expecting this would be true for the spherical geometry. However, the calculation method shows that this might not be true. Is there any misunderstanding I am making? Then, what should I use to determine unique set of (lon',lat') other than pair of (D,PA)?
$\begingroup$Extract from the discussion: the PA formula using $\arctan$ is missing or not returning Position Angles $\ge\pi/2$ or $\le -\pi/2$, because the range of $\arctan$ is only $(-\pi/2, \pi/2)$. But the range of PA should be a full $2\pi$ turn. Position angle on Wikipedia suggests using atan2. I was also visualising using Great Circle Mapper.$\endgroup$
2 Answers
2
For reference, here is what your three points would look like if we plotted them on a map of the Earth:
The red "pin" near the center of the map (at about the same latitude as France) is at your original lon and lat
and the dot labeled "Avannaata" near the top of the map (in Greenland) is at your first pair of lon' and lat' values.
The dot near the bottom of the map (in the Atlantic Ocean at about the latitude of the African country Mauritania) is at your second pair of lon' and lat' values.
The $D$ and $PA$ values agree with your $D$ and $PA$ values.
The distance is hard to judge from the map due to the fact that this is a Mercator projection (which distorts distances, especially in places that are near the poles, like Greenland), but the direction is clearly northward.
And indeed $PA$ returned a value in radians that converts to about $-5$ degrees,
more conventionally stated as a direction of $355$ degrees
(the equivalent angle in the range from $0$ to $360$)
which is just a little bit west of due north. So that appears to be correct.
We get the same results. But now the direction doesn't make any sense:
the destination is the southernmost of the three sets of coordinates we plotted,
almost directly south of the red pin,
but the $PA$ formula is telling us that to get from the red pin to the southernmost dot
we should start out in the direction $-5$ degrees (or $355$ degrees),
which is almost due north.
In short, $PA$ is sending us in the wrong direction entirely!
This is why I say this formula for $PA$ is wrong.
It gives correct answers whenever you ask it the direction to somewhere north of you, but if you ask for the direction to someplace south of you, it sends you north instead.
Actually, if you start at a particular point on a sphere, point yourself in a particular direction, and travel "straight" (that is, along the great-circle path in that direction) for a particular distance, you will arrive at a particular point.
There is only one possible destination of that journey.
In particular if we start at the red pin and go in the direction $-0.088221$ radians for a distance $0.523258,$ we will arrive at the point labeled "Avennaata,"
not the point near the bottom of the map, west of Mauritania.
So what went wrong when you tried to find that point?
You literally asked fsolve to find you any pair of coordinates lon' and lat'
for which $D$ and $PA$ would give you the distance $0.523258$ and direction $-0.088221$
from the lon and lat coordinates to lon' and lat'.
And we can see from the results of calculating $D$ and $PA$ above,
$D = 0.523258$ and $PA = -0.088221$ for both sets of lon' and lat' coordinates.
Therefore both of those points are solutions to the problem you asked fsolve to solve for you.
But one of those solutions is a "solution" only because the problem you asked fsolve to solve includes a broken$PA$ function that says you are going north when you're actually going south.
It gave you the correct answer, where you go in a northward direction,
and also the answer where you go in a southward direction which $PA$ incorrectly says is northward.
The underlying problem, by the way, is really that atan is only capable of returning angles between $-\frac\pi2$ and $\frac\pi2$ radians, which (when you convert them to directions on the Earth) range from west to east in various northward directions, never southward directions.
You can "fix" this by adding extra code to pay attention to whether the denominator in the $PA$ function, cos(lat1)*tan(lat2) - sin(lat1)*cos(lon2-lon1),
is positive, zero, or negative, but it's annoying to code and to test
(especially the zero case, because the value of lon2-lon1 might be either positive or negative in an eastward direction).
My advice is that the much better way to fix this is to use a function that is already designed to give you the correct answer no matter which direction your destination is in, and that function is atan2.
Basically, to convert an atan formula to an atan2 formula, just change the name of the function and put a comma between the numerator and denominator instead of performing the division to get a single input value.
In the implementation above I have gone one step further and multiplied both the numerator and denominator by cos(lat2) in order to avoid having tan(lat2) in the denominator, since that function will not let you find the direction to either the north or south pole (because $\tan(\theta)$ is undefined when $\theta = \pm\frac\pi2$).
The results of calling this version of the $PA$ function on your coordinates are as follows:
In the first case, where the actual correct direction is northward, we get the northward direction $-0.088221$ (about $-5$ degrees, as before),
but in the second case, instead of getting $-0.088221$ again, we get
$3.05337,$ which is about $175$ degrees -- going southward in a direction $180$ degrees different from $-5$ degrees.
This is the correct answer.
If you let fsolve use PA2 to find the point at a given distance and direction from a given point, it would (presumably) find a unique, correct answer.
However, there is a more direct way to get the answer:
On a sphere, Point A to Point B is one direction and distance. But anti-podal points are not expected to have a solution. Now points on a sphere, on the same half of sphere, could be projected to rectangular grid but the projection would be a different system.
Note that ArcTan2 makes a quadrant system.
Then "Forward" or "Direct" generally do not require a quadrant system with directions of 0 to 360 degrees but "reverse" or "inverse" requires a quadrant system. | 677.169 | 1 |
Trigonometry in Terms of Algebra
You are babysitting your little cousin while doing your homework. While working on your trig functions, your cousin asks you what you are doing. While trying to explain sine, cosine, and tangent, your cousin is very confused. She doesn't understand what you mean by those words, but really wants to understand what the functions mean. Can you define the trig functions in terms of the relationships of sides for your little cousin?
Solving Trigonometric Functions
All of the trigonometric functions can be rewritten in terms of only \(x\), when using one of the inverse trigonometric functions.
Starting with tangent, we draw a triangle where the opposite side (from \(\theta \)) is defined as \(x\) and the adjacent side is 1. The hypotenuse, from the Pythagorean Theorem would be \(\sqrt{x^2+1}\). Substituting \(\tan^{-1} x\) for \(\theta \), we get:
This problem might be better written as \([\sec(\tan^{-1}x)]^2\). Therefore, all you need to do is square the ratio above.
\([\sec(\tan^{-1}x)]^2=(\sqrt{x^2+1})^2=x^2+1\)
You can also write the all of the trig functions in terms of arcsine and arccosine. However, for each inverse function, there is a different triangle. You will derive these formulas in the exercise for this section.
3. Find \(\csc^3(\tan^{-1}4x)\).
This problem is similar to #1 and #2 above. First, use \(4x\) instead of \(x\) in the ratios above. Second, the \(csc^3\) is the same as taking the csc function and cubing it.
Example \(\PageIndex{1}\)
Earlier, you were asked if you can define the trig functions in terms of the relationship of sides.
Solution
As it turns out, it's very easy to explain trig functions in terms of ratios. If you look at the unit circle
Figure \(\PageIndex{2}\)
you can see that each trig function can be represented as a ratio of two sides. The value of any trig function can be represented as the length of one of the sides of the triangle (shown with two red sides and the black hypotenuse) divided by the length of one of the other sides. In fact, you should explain to your cousin, the words like "sine", "cosine", and "tangent" are just conveniences in this case to describe relationships that keep coming up over and over again. It would be possible to just describe the trig functions in terms of relationships of one side to another, if you'd like.
Using the sides of a triangle made on the unit circle, if the side opposite the angle is called "\(x\)":
So as you can see, since trig functions are really just relationships between sides, it is possible to work with them in whatever form you want; either in terms of the usual "sine", "cosine" and "tangent", or in terms of algebra | 677.169 | 1 |
Introduction to Quadrilaterals - Questions
c.A trapezium with its non-parallel
sides equal is called _______________________.
d.Each angle of a
rectangle or square is equal to ___________.
e.A square is a rectangle
in which all sides are equal.
f.Sum of the angles of a
quadrilateral is ______________
Q2.
A quadrilateral has three angles equal to 60°, 75° and 120°. Find its
unknown angle.
Q3.
In a
quadrilateral EFGH, ∠E
= ∠F = 65°
and ∠G =
120°, find ∠H.
Q4.
If the
sum of 3 angles of a quadrilateral is 280°, find its fourth angle.
Practice Page 2
Q1.
Fill in the blanks:
g.Quadrilateral means
____________ sides.
h.Square is a _____________ in which
all sides are equal
i.A line segment joining a pair of
opposite vertices is called a _________________.
j.Each angle of a
rectangle is equal to ___________.
k.A rhombus is sometimes
called a ____________________.
l.The interior angles of a
quadrilateral add up to ______ degrees.
Q2.
Find the measure of the missing angles in a parallelogram, if ∠A = 55°
Q3.
In a
quadrilateral ABCD, ∠A
= 105°, ∠B = 95°,∠C=
70. Find ∠D.
Last modified: Tuesday, 4 December 2018, 6 | 677.169 | 1 |
Consecutive angles
To talk about consecutive angles it is important to also first talk about the definition of angle. An angle is the portion of the plane that is between two rays which have a common origin known as the vertex . It can also be said that the angle is the opening formed by two sides that start from that common point, or are centered on the rotation that the plane gives with respect to its origin . When we talk about consecutive angles we refer to an angle relationship that implies the parameters conformants, these parameters are the coplanar lines or raysand the vertex where these lines meet.
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What are consecutive angles?
Two angles can be consecutive when one angle is successive to the other, which means that the two angles share the same vertex and one of their rays . When a ray forms two angles on one side and the other , we are faced with two consecutive angles.
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Definition
features
Consecutive interior angles
How they differ from adjacent angles
Examples of consecutive angles
Definition
We say that two angles that are within the same plane are consecutive when they have only one side in common . According to the extension, given several angles in a certain order , they will then be consecutive when each of them is consecutive with the next.
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The word consecutive when we refer to angles means that the angles share at the same time the same vertex as well as one of the sides , which can be the line or ray , in other words they are one next to the other delimited by one of the lines that make them up and, if there are more than two angles , they can continue to be consecutive if the one that follows shares the side of the last one always with the same vertex .
features
The main characteristics of the consecutive angles are the following:
Consecutive angles are those that have or share the same vertex.
They have only one side in common.
Several angles can become consecutive when each of them shares a side with the angle that follows it.
The sum that results from the consecutive angles is equivalent to the angle comprised by the infrequent sides of the angles.
For two angles to be added they must necessarily be consecutive.
These angles are in the same way placed side by side.
Consecutive interior angles
Consecutive angles are also known with the name of adjacent angles , and the angles are having a side in common and the same vertex . These angles then share a side and a vertex and are located side by side. When the angles are ordered in a certain way, they are said to be consecutive if each angle is successive with the other. When the angles meet consecutively and are interior, it is when two lines are crossed by another transversal call but within both lines.
The sum of the consecutive angles will then be equal to the angle that has been formed by what are the uncommon sides of the angles. When two lines are cut by means of a transversal , the pair of angles on one side of the transversal and within the two lines are called the consecutive interior angles .
The Consecutive Interior Angles Theorem tells us that if two parallel lines are cut by a transversal , then the pairs of consecutive interior angles formed are supplementary.
How they differ from adjacent angles
Consecutive angles have a vertex and a side in common, and these angles are followed one after the other. The adjacent angles are the angles that are found consecutively and that have non-common sides within the same line; are any pair of angles that are in a row but when their degrees are added, a total of 180 ° is obtained.
Examples of consecutive angles
Some examples of consecutive angles are as follows:
Right angles : Right angles measure 90º and are the result of the perpendicular crossing of two rays.
Obtuse angles : the obtuse angles are those that measure more than 90º.
Convex angles : Convex angles are those that measure less than 180º.
Concave angles : Concave angles are those that measure more than 180º and less than 360º.
Supplementary angles: Supplementary angles are those that adding the two together give a straight angle.
Adjacent angles : they have a side and a vertex in common and the other on the same line.
Central angles in a circle : one whose vertex is located in the center of the circle. | 677.169 | 1 |
Class 8 Courses
In the given figure, ∆ABC is right angled at C and DE ⊥ AB $\triangle \mathrm{ABC}$ is right angled at $\mathrm{C}$ and $\mathrm{DE} \perp \mathrm{AB}$. Prove that $\triangle \mathrm{ABC} \sim \triangle \mathrm{ADE}$ and hence find the length of $\mathrm{AE}$ and $\mathrm{DE}$.
Solution:
It is given that $A C B$ is right angle triangle and $\angle C=90^{\circ}$
We have to prove that $\triangle A B C \sim \triangle A D E$ and find the lengths of $A E$ and $D E$. | 677.169 | 1 |
Lesson video
Hello and welcome to the lesson about angles, measuring and drawing angles.
I want you to take a moment to remove any distractions that you may have, whether that be your brother, your sister, you know your pets, anything that could be distracting you.
Make sure your phone is on silent as well, as mine always is.
And make sure you are not getting distracted throughout.
Turn all those app notifications off, whatever it may be.
It needs to be turned off so we can do that really powerful math that we always do.
One thing I must stress beforehand though is that we need a protractor.
So, so important that we have a protractor, you can get one from any good high street retailer you can get one from online, wherever you may find one.
Okay, so important, that's what we use to measure and draw angles.
As we've learned previously, and as you may.
So without further ado, it's time for me to take you through how to do this.
So let's get started.
So let's have a go at the Try this step.
So the students below got these angles wrong can you explain why? Pause the video now and have a go.
Fantastic.
Let's get through why these students could potentially be wrong.
So did you find out why Cala could potentially be wrong? Hm, what has she done here? So she's actually if you see it she hasn't measured from zero.
If you notice she's actually measured from, what's that number going to be just there.
What's that number going to be? It's going to be 15 right? So that one there she's measured from 15 through to 60.
So she's right with that that there are 60, but unfortunately she hasn't measured from zero.
So she needs to measure from zero.
So the difference between 60 and 15, is going to be what would that be? 45 right? So 45 degrees that angle actually is.
So we can explain that by saying that she has not measured from 40 sorry from zero degrees.
She has measured from 15 degrees.
Heck, you can even be really, really enthusiastic and put an exclamation mark at the end, how nice.
So for B, for Xavier.
He said his angle was 135, what did you get for him? He seems a little bit strange for what that is.
That's quite a small angle.
We know something that looks like that is sort of an acute angle isn't it? So we need to think, well he must be wrong because he's just said an obtuse angle.
Just by a sort of like, logic and definition there.
So he's actually measured from the inside, you should never measure from the inside.
So going from zero through to that gives us the angle of 45 degrees.
45 degrees there.
So we can say he has measured from the inside not the outside.
And finally we've got Zaki down here who reckoned that the angle he has drawn is 37 degrees.
What do we reckon to that one there? 37 degrees.
Has he got that wrong? Well unfortunately he has.
It looks really good to begin with because this almost had me fooled.
Right, it looks really good to begin with.
It goes from zero through to 37 but look where the crosshair is.
Right? This thing here.
It should be lined up and it should be here instead.
So he has not got we should say actually, good grammar and punctuation here, we should say he has with a lowercase h, that he has not got the crosshair spelled my crosshair wrong, silly me! So he's not got the crosshair lined up correctly.
Fantastic.
Let's move on then.
So we should understand that now.
Now, I've got some space for here for you to work, do all your working out so you see what I'm going to do on an interactive protractor, so let's get to it.
So let's just draw anything that looks vaguely like this and I'm happy.
Okay? Doesn't have to be accurate, and we don't know how accurate it is just yet.
We don't know what the angle is.
So, once you've drawn that out what I'd like you to do is I'd like you to place your protractors as I'm doing.
The number one thing you need to do is you need to make sure that the angle has what we call the crosshair nicely lined up with the two where the two lines meet.
So if I zoom in I can really see I'm measuring where that crosshair actually is.
So you can see it's pretty much bang on there.
Right? Now if I started measuring from here I wouldn't get the angle correctly.
You think why, just for one minute have a think why.
Why could that be? Well the key reason is the fact that I'm not measuring from zero.
Now, I need to measure from zero of course, So I need to take my protractor round.
So, but be really careful, keep it nice and lined up on the crosshairs still but rotate it round.
Just start from zero here, I then need to turn around.
So remember angles are all to do with the turns.
Right? So it's a measure of turns.
So turn around zero degrees all the way around and I get to approximately don't need to be 100% precise here but I get to approximately 121 degrees, you see it's just shy of 121 degrees.
So that's how we use our protractor.
I could adjust my angle, I could do, I could shoot it off over here.
And actually, we discover that that one there would be a perfect right angle.
So, we've got all different ways that we can measure angles right? We've got to make sure though regardless of what we're measuring we always measure from zero, and we do it as a measure of turns, and that we go around.
You could for example, get it from 180 through to 90, just do 180, subtract 90.
So if you mistakenly put your protractor like this and you counted in from 180 and got it to 120 there, you'd say well it should be an angle of 60, but you could do 180 subtract 120.
And that'd give you that angle of 60.
So you may have some more problem solving based questions that could be involved with that.
So just be aware.
So remember, crosshair lines up so that the two lines intersect, and of course measuring from zero throughout.
Okay, let's get back to it.
So you've got this now, Independent Task that you need to complete.
So what I'd like you to do is I'd like you to fill in those blanks for the following exercise.
You may want to go back in the video for some help.
So if you need to go back to the interactive part of the lesson where we needed that protractor, by all means please have a go, and pause the video and have a go.
Excellent.
I'm going to assume that you've done that.
So I'm going to go through the answers now.
We need to think, we can use a blank to measure an angle.
What was it? What did I say, what did we use? What was that mathematical instrument we used? It was a protractor, wasn't it? So we can tick that one off.
So we can use a protractor to measure an angle.
For example in order to measure angle A, angle A is that, we can use the blank scale to see what? That it has a value of blank.
Well, I'm measuring from zero aren't I all the time, so I need to measure all the way from zero to here, there we go, all the way to, that looks like 75 to me.
Aha! We have an answer on 75.
So 75 degrees.
Of course we always measure angles in degrees.
So in order to measure angle B we can use the blank scale to see that it has a value of blank.
Well, what scale should we use to begin with here? Missed that one out didn't I, silly me.
Seriously Thomas what's up, what am I doing? We can use the, what did we use here? We used the outer scale didn't we? We used that outside part.
The outer scale.
Or the outside rather.
So outside, we can take that off.
And then the inside scale we're going to use for B and of course if we do that we measure from zero don't we? We measure from zero to here.
Then we go all the way around and we see that that has a value of, 100 would be up to here, and then I can see it go to five more sort of like little tiny notches across, so I can see it has a value of 105 degrees.
So mark it right or wrong, I really hope you got it right.
Let's keep going.
So you have your Explore Task to complete.
It says both of these students have incorrectly measured the angle.
That's a really important statement there.
Both of those students have definitely incorrectly measured the angle.
So I'm not trying to trick you here, it may seem like one of them is correct, just be really, really careful about what marked angle actually is.
So I think the angle 115 degrees, that's what Yasmin's saying, and then Bihn's angle is saying 65 degrees.
So pause the video now if you think you know what you're doing.
If you don't know what you're doing, and you need support by all means, or you want to go through the answer, by all means stick around and I will help.
So pause the video now and have it go as you need to, or listen in.
Right, so I'm going to go through this now, and give you a little bit of support whilst I'm doing it.
So we need to think, how could we have arrived at an angle of 115 degrees? Well if I look round, I see that that is going to be marked on as 115 degrees up to that point there, if I'm using the, what scale? The outer scale, or the outside scale.
So that was 115 there, so that's how we've got Yasmin's answer of 115, she's just measured from the outside.
So she measured from the outside.
The outer scale.
Now, Bihn's statement is a little bit difficult, it says how could we arrive at that one, because on the face of it, now I'm really, really keen on highlighting that they are both incorrect.
On the face it looks correct, almost had me fooled as well.
Is that if I'm measuring from zero I'm going to go all the way around and we hit of course 65 degrees there, so surely that's correct right? Right? But she hasn't measured the marked angle right? The marked angle is this thing here, that is what we're focusing on right? So if I colour that in you can hopefully see that now, very, very clearly.
That is the marked angle.
So we need to be measuring this thing.
Now that sort of sector taken there, we have there a sort of like a convoluted circle of some description is 65 degrees.
Now we know that angles, you may have learned this from a primary school actually, angles around the point sum to what? Can you remember? What do they sum to? 360 degrees right? So really important you're able to recognise that.
You've got a point here and then we're going all the way around that turning point down there, okay all the way around, and we're saying well that of course is going to be 360 in total.
So we need to subtract 65 degrees from that.
So the actual answer is going to be 360 degrees subtract 65.
Now I'm going to let you think about it just for a moment, whilst we formulate an answer of why, Binh's answer to arrive at the correct measurement is going to be this.
So we can say she has got 65 degrees but it is not the marked angle.
So 360 minus 65 you should know that, that of course is 295 degrees.
So we know the actual answer is 295 degrees.
How fantastic, we've got a really nice problem solving question there, also have some description.
Let's carry on then.
Unfortunately that brings us to the end of the lesson then.
So I just wanted to say you've done an amazing job as always.
If you've managed to keep up, really, really good job.
Don't forget to smash that exit quiz.
So you can just show us all at Oak, and indeed your teachers, and the entire country how well you're doing overall right? Please make sure you do that so you can test how much you've learned. | 677.169 | 1 |
or, the product of the extremes is equal to the product of the means. And conversely, If the product of the extremes be equal to the product of the means,
or the ratio of the first to the second number, is equal to the ratio of the third to the fourth.
Prop. XVIII. Similar figures are said to be similarly situated, when their homologous sides are parallel, as when the figures are situated on the same straight line, or on parallel lines: but when similar figures are situated on the sides of a triangle, the similar figures are said to be similarly situated when the homologous sides of each figure have the same relative position with respect to one another; that is if the bases on which the similar figures stand, were placed parallel to one another, the remaining sides of the figures, if similarly situated, would also be parallel to one another.
Prop. xx. It may easily be shewn, that the perimeters of similar polygons, are proportional to their homologous sides.
Prop. xxi. This proposition must be so understood as to include all rectilineal figures whatsoever, which require for the conditions of similarity another condition than is required for the similarity of triangles.
See note on Euc. vi. Def. 1.
Prop. XXIII. The doctrine of compound ratio, including duplicate and triplicate ratio, in the form in which it was propounded and practised by the ancient Geometers, has been almost wholly superseded. However satisfactory for the purposes of exact reasoning the method of expressing the ratio of two surfaces, or of two solids by two straight lines, may be in itself, it has not been found to be the form best suited for the direct application of the results of Geometry. Almost all modern writers on Geometry and its applications to every branch of the Mathematical Sciences, have adopted the algebraical notation of a quotient AB : BC; or of a fraction ; for expressing the ratio of two lines AB, BC: as well as that BC of a product AB × BC, or AB. BC, for the expression of a rectangle. The want of a concise and expressive method of notation to indicate the proportion of Geometrical Magnitudes in a form suited for the direct application of the results, has doubtless favoured the introduction of Algebraical symbols into the language of Geometry. It must be admitted, however, that such notations in the language of pure Geometry are liable
AB
to very serious objections, chiefly on the ground that pure Geometry does not admit the Arithmetical or Algebraical idea of a product or a quotient into its reasonings. On the other hand, it may be urged, that it is not the employment of symbols which renders a process of reasoning peculiarly Geometrical or Algebraical, but the ideas which are expressed by them. If symbols be employed in Geometrical reasonings, and be understood to express the magnitudes themselves and the conception of their Geometrical ratio, and not any measures, or numerical values of them, there would not appear to be any very great objections to their use, provided that the notations employed were such as are not likely to lead to misconception. It is, however, desirable, for the sake of avoiding confusion of ideas in reasoning on the properties of number and of magnitude, that the language and notations employed both in Geometry and Algebra should be rigidly defined and strictly adhered to, in all cases. At the commencement of his Geometrical studies, the student is recommended not to employ the symbols of Algebra in Geometrical demonstrations. How far it may be necessary or advisable to employ them when he fully understands the nature of the subject, is a question on which some difference of opinion exists.
Prop. xxv. There does not appear any sufficient reason why this proposition is placed between Prop. xxiv. and Prop. xxvI.
Prop. XXVII. To understand this and the three following propositions more easily, it is to be observed:
1.
"That a parallelogram is said to be applied to a straight line, when it is described upon it as one of its sides. Ex. gr. the parallelogram AC is said to be applied to the straight line AB.
2. But a parallelogram AE is said to be applied to a straight line AB, deficient by a parallelogram, when AD the base of AE is less than AB, and therefore AE is less than the parallelogram AC described upon AB in the same angle, and between the same parallels, by the parallelogram DC; and DC is therefore called the defect of AE.
3. And a parallelogram 4G is said to be applied to a straight line AB, exceeding by a parallelogram, when AF the base of AG is greater than AB, and therefore AG exceeds AC the parallelogram described upon AB in the same angle, and between the same parallels, by the parallelogram BG."-Simson.
Both among Euclid's Theorems and Problems, cases occur in which the hypotheses of the one, and the data or quæsita of the other, are restricted within certain limits as to magnitude and position. The determination of these limits constitutes the doctrine of Maxima and Minima. Thus:-The theorem Euc. vI. 27 is a case of the maximum value which a figure fulfilling the other conditions can have; and the succeeding proposition is a problem involving this fact among the conditions as a part of the data, in truth, perfectly analogous to Euc. I. 20, 22; wherein the limit of possible diminution of the sum of the two sides of a triangle described upon a given base, is the magnitude of the base itself: the limit of the side of a square which shall be equal to the rectangle of the two parts into which a given line may be divided, is half the line, as it appears from Euc. II. 5-the greatest line that can be drawn from a given point within a circle, to the circumference, Euc. III. 7, is the line which passes through the center of the circle; and the least line which can be so drawn from the same point, is the part produced, of the greatest line between the given point and the circumference. Euc. 11. 8, also affords another instance of a maximum and a minimum when the given point is outside the given circle.
Prop. XXXI. This proposition is the general case of Prop. 47, Book 1, for any similar rectilineal figure described on the sides of a right-angled triangle. The demonstration, however, here given is wholly independent of Euc. 1. 47.
Prop. xxxIII. In the demonstration of this important proposition, angles greater than two right angles are employed, in accordance with the criterion of proportionality laid down in Euc. v. def. 5.
This proposition forms the basis of the assumption of arcs of circles for the measures of angles at their centers. One magnitude may be assumed as the measure of another magnitude of a different kind, when the two are so connected, that any variation in them takes place simultaneously, and in the same direct proportion. This being the case with angles at the center of a circle, and the arcs subtended by them; the arcs of circles can be assumed as the measures of the angles they subtend at the center of the circle.
Prop. B. The converse of this proposition does not hold good when the triangle is isosceles.
QUESTIONS ON BOOK VI.
1. DISTINGUISH between similar figures and equal figures.
2. What is the distinction between homologous sides, and equal sides in Geometrical figures?
3. What is the number of conditions requisite to determine similarity of figures? Is the number of conditions in Euclid's definition of similar figures greater than what is necessary? Propose a definition of similar figures which includes no superfluous condition.
4. Explain how Euclid makes use of the definition of proportion in Euc. vi. 1.
5. Prove that triangles on the same base are to one another as their altitudes.
6. If two triangles of the same altitude have their bases unequal, and if one of them be divided into m equal parts, and if the other contain n of those parts; prove that the triangles have the same numerical relation as their bases. Why is this Proposition less general than Euc. vI. 1?
7. Are triangles which have one angle of one equal to one angle of another, and the sides about two other angles proportionals, necessarily
similar?
8. What are the conditions, considered by Euclid, under which two triangles are similar to each other?
9. Apply Euc. vI. 2, to trisect the diagonal of a parallelogram.
10. When are three lines said to be in harmonical proportion? If both the interior and exterior angles at the vertex of a triangle (Euc. vi. 3, A.) be bisected by lines which meet the base, and the base produced, in D, G; the segments BG, GD, GC of the base shall be in Harmonical proportion.
11. If the angles at the base of the triangle in the figure Euc. vi. A, be equal to each other, how is the proposition modified?
12. Under what circumstances will the bisecting line in the fig. Euc. VI. A, meet the base on the side of the angle bisected? Shew that there is an indeterminate case.
13. State some of the uses to which Euc. vi. 4, may be applied.
14. Apply Euc. vi. 4, to prove that the rectangle contained by the segments of any chord passing through a given point within a circle is
constant.
15. Point out clearly the difference in the proofs of the two latter cases in Euc. vI. 7.
16. From the corollary of Euc. vi. 8, deduce a proof of Euc. 1. 47. 17. Shew how the last two properties stated in Euc. vi. 8. Cor. may be deduced from Euc. 1. 47; II. 2; vi. 17.
18. Given the nth part of a straight line, find by a Geometrical construction, the (n + 1)th part.
19. Define what is meant by a mean proportional between two given lines and find a mean proportional between the lines whose lengths are 4 and 9 units respectively. Is the method you employ suggested by any Propositions in any of the first four books?
20. Determine a third proportional to two lines of 5 and 7 units: and a fourth proportional to three lines of 5, 7, 9, units.
21. Find a straight line which shall have to a given straight line, the ratio of 1 to 5.
23. Give the corollary, Euc. vi. 8, and prove thence that the Arithmetic mean is greater than the Geometric between the same extremes. 24. If two equal triangles have two angles together equal to two right angles, the sides about those angles are reciprocally proportional. 25. Give Algebraical proofs of Prop. 16 and 17 of Book vi.
26. Enunciate and prove the converse of Euc. vi. 15.
27. Explain what is meant by saying, that "similar triangles are in the duplicate ratio of their homologous sides."
28. What are the data which determine triangles both in species and magnitude? How are those data expressed in Geometry?
29. If the ratio of the homologous sides of two triangles be as 1 to 4, what is the ratio of the triangles? And if the ratio of the triangles be as 1 to 4, what is the ratio of the homologous sides?
30. Shew that one of the triangles in the figure, Euc. Iv. 10, is a mean proportional between the other two.
31. What is the algebraical interpretation of Euc. vi. 19?
32. From your definition of Proportion, prove that the diagonals of a square are in the same proportion as their sides.
33. What propositions does Euclid prove respecting similar polygons? 34. The parallelograms about the diameter of a parallelogram are similar to the whole and to one another. Shew when they are equal.
35. Prove Algebraically, that the areas (1) of similar triangles and (2) of similar parallelograms are proportional to the squares of their homologous sides.
36. How is it shewn that equiangular parallelograms have to one another the ratio which is compounded of the ratios of their bases and altitudes?
37. To find two lines which shall have to each other, the ratio compounded of the ratios of the lines A to B, and C to D.
38. State the force of the condition "similarly described;" and shew that, on a given straight line, there may be described as many polygons of different magnitudes, similar to a given polygon, as there are sides of different lengths in the polygon.
39. Describe a triangle similar to a given triangle, and having its area double that of the given triangle.
40. The three sides of a triangle are 7, 8, 9 units respectively; determine the length of the lines which meeting the base, and the base produced, bisect the interior angle opposite to the greatest side of the triangle, and the adjacent exterior angle.
41. The three sides of a triangle are 3, 4, 5 inches respectively; find the lengths of the external segments of the sides determined by the lines which bisect the exterior angles of the triangle.
42. What are the segments into which the hypotenuse of a rightangled triangle is divided by a perpendicular drawn from the right angle, if the sides containing it are a and 3a units respectively?
43. If the three sides of a triangle be 3, 4, 5 units respectively: what are the parts into which they are divided by the lines which bisect the angles opposite to them?
44. If the homologous sides of two triangles be as 3 to 4, and the area of one triangle be known to contain 100 square units; how many square units are contained in the area of the other triangle?
45. Prove that if BD be taken in AB produced (fig. Euc. vi. 30) equal to the greater segment AC, then AD is divided in extreme and mean ratio in the point B.
Shew also, that in the series 1, 1, 2, 3, 5, 8, &c. in which each term is the sum of the two preceding terms, the last two terms perpetually approach to the proportion of the segments of a line divided in extreme and mean ratio. Find a general expression (free from surds) for the nth term of this series.
46. The parts of a line divided in extreme and mean ratio are incommensurable with each other.
47. Shew that in Euclid's figure (Euc. 11. 11.) four other lines, besides the given line, are divided in the required manner.
48. Enunciate Euc. vi. 31. What theorem of a previous book is included in this proposition?
49. What is the superior limit, as to magnitude, of the angle at the circumference in Euc. vi. 33? Shew that the proof may be extended by withdrawing the usually supposed restriction as to angular magnitude; and then deduce, as a corollary, the proposition respecting the magnitudes of angles in segments greater than, equal to, or less than a semicircle. 50. The sides of a triangle inscribed in a circle are a, b, c, units respectively: find by Euc. vI. c, the radius of the circumscribing circle. 51. Enunciate the converse of Euc. VI. D.
52. Shew independently that Euc. VI. D, is true when the quadrilateral figure is rectangular.
53. Shew that the rectangles contained by the opposite sides of a quadrilateral figure which does not admit of having a circle described about it, are together greater than the rectangle contained by the diagonals.
54. What different conditions may be stated as essential to the possibility of the inscription and circumscription of a circle in and about a quadrilateral figure?
55. Point out those propositions in the Sixth Book in which Euclid's definition of proportion is directly applied.
56. Explain briefly the advantages gained by the application of analysis to the solution of Geometrical Problems.
57. In what cases are triangles proved to be equal in Euclid, and in what cases are they proved to be similar? | 677.169 | 1 |
Angle Cbd Has a Measure of 140°. What Is the Measure of Angle Abd?
To find the measure of angle ABD when angle CBD is 140°, utilize geometric principles and trigonometric formulas. Understanding angle relationships and employing appropriate mathematical techniques is crucial. By applying these methods, you can accurately determine the measure of angle ABD based on the given information. Remember, precision in calculations is key to unraveling the solution to this geometric problem.
Understanding Angle Measurement Properties
In the realm of geometry, understanding the properties of angle measurement is fundamental for precise and accurate calculations. Angle properties play a crucial role in various geometric applications, from trigonometry to architectural design.
Mastery of angle measurement techniques, such as using protractors or trigonometric functions, is essential for solving complex problems involving angles. A solid grasp of these concepts is vital for navigating the intricate world of geometry.
Applying Angle Relationships in Geometry
Building upon the foundational understanding of angle measurement properties, the application of angle relationships in geometry allows for the exploration of interconnected angles within geometric configurations.
Solving for the Measure of Angle ABD
To determine the measure of angle ABD, one must apply the principles of angle relationships and geometric calculations within the given context.
By utilizing trigonometry formulas and finding missing angles, it is possible to solve for the exact measure of angle ABD.
Understanding the relationships between angles and employing the appropriate mathematical formulas will lead to accurately determining the measure of angle ABD in this geometric scenario.
Conclusion
By applying the properties of angle measurement and relationships in geometry, we can determine that the measure of angle ABD is 40°. This calculation is based on the fact that the angles CBD and ABD form a linear pair, meaning they add up to 180°. Therefore, if angle CBD is 140°, angle ABD must be 40° to make the total 180°.
The theory of angle relationships in geometry holds true in this scenario. | 677.169 | 1 |
@ Woody Assuming the short wall next to the "W" is the front of the house (as it is on the OP's drawing) then your method of measurement is identical to mine, though your angle is more obtuse.
So what are we arguing about?
It's the widest point as measured along any line parallel to the frontage. Or to put it simply, it's the smallest gap the house would fit through, keeping it in exactly the same alignment, not twisting it.
Obviously it's not ANY direction, because that would clearly be the front left to back right corner, which would be stupid. | 677.169 | 1 |
What is unit tangent vector?
What is unit tangent vector?
The Unit Tangent Vector The derivative of a vector valued function gives a new vector valued function that is tangent to the defined curve. The analogue to the slope of the tangent line is the direction of the tangent line.
How do you find unit tangent and unit normal vector?
We can strip a vector of its magnitude by dividing by its magnitude. LetHow do you find the tangent line?
1) Find the first derivative of f(x). 2) Plug x value of the indicated point into f '(x) to find the slope at x. 3) Plug x value into f(x) to find the y coordinate of the tangent point. 4) Combine the slope from step 2 and point from step 3 using the point-slope formula to find the equation for the tangent line.
How do you find the unit tangent vector given the position vector?
LetWhat is tangent in 3D?
In geometry, the tangent line (or simply tangent) to a plane curve at a given point is the straight line that "just touches" the curve at that point. Leibniz defined it as the line through a pair of infinitely close points on the curve.
How do you find the equation of a tangent line in 3D?
1
Lines and Tangent Lines.
A 3-D curve can be given parametrically by x = f(t), y = g(t) and z = h(t) where t is on some interval I and f, g, and h are all continuous on I.
⇀
These become the parametric equations of a line in 3D where a,b,c are called direction numbers for the line (as are any multiples of a,b,c ).
Given a surface parameterized by a function,to find an expression for the unit normal vector to this surface,take the following steps:
Get a (non necessarily unit) normal vector by taking the cross product of both partial derivatives of :
Turn this vector-expression into a unit vector by dividing it by its own magnitude:
What exactly is a tangent vector?
Tang
What is the formula to find an unit vector?
Formula for Unit Vector : Usually, Vector are represented in Two Dimension and Three Dimension : i) In Two Dimension, any vector can be written as x i ^ + y j ^. Let a → = x i ^ + y j ^. Then unit vector of a → can be calculated as, a ^ = a → | a → | = x i ^ + y j ^ x 2 + y 2. | 677.169 | 1 |
What Is The Difference Between Lines And Sines
Lines and sines are foundational concepts in mathematics, each serving unique functions and embodying distinct principles. Lines are straight and extend indefinitely in both directions, representing simplicity and predictability. In contrast, sines curve and oscillate, capturing rhythm and movement, essential in fields ranging from engineering to music.
The primary difference between lines and sines lies in their mathematical nature and applications. A line is described by a linear equation and is characterized by its slope and y-intercept, making it a staple in geometric and algebraic studies. On the other hand, the sine function, derived from trigonometry, represents periodic oscillations and is fundamental in understanding wave patterns.
While lines offer a straightforward representation of distance and slope, sines provide a way to model periodic phenomena such as sound waves and alternating currents. These functions are not only crucial in academic contexts but also play significant roles in practical engineering and physics scenarios, where they help solve real-world problems involving waves and frequencies.
Lines Explained
Definition and Basics
What is a line?
In geometry, a line is an ideal representation of an object that is straight, infinitely long, and thin. It is one of the most basic forms in mathematics, extending endlessly in both directions without any curvature. A line is distinct because it has length but no width or height, positioning it as a fundamental concept in both theoretical and practical applications.
Key properties
Lines are defined by several core properties:
Straightness: A line does not curve; it maintains a constant direction.
Infinite length: Lines extend indefinitely at both ends.
Dimensionality: Lines are one-dimensional, possessing only length.
Mathematical Representation
Equation of a line
The most common form to represent a line mathematically is the slope-intercept equation: y = mx + b.
m represents the slope of the line, indicating the steepness and direction.
b is the y-intercept, the point where the line crosses the y-axis.
This equation provides a straightforward method to graph a line on a coordinate plane, offering clarity on how the line behaves visually.
Slope and intercept
Slope (m): Indicates how much y increases for a one-unit increase in x. A positive slope means the line rises, and a negative slope means it falls.
Intercept (b): Shows where the line intersects the y-axis. This value is the output when x equals zero.
Sines Explained
Definition and Basics
What is a sine?
The sine is a fundamental concept in trigonometry, commonly represented as the function sin(x). It describes a smooth, periodic oscillation that repeats at regular intervals, known as a wave. The sine function is crucial for understanding various phenomena in science and engineering, especially those involving wave-like patterns.
Relation to circles and angles
The sine of an angle in a right triangle is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse. This relationship can also be visualized on the unit circle, where the sine of an angle is the y-coordinate of the point where the angle's terminal side intersects the circle.
Mathematical Representation
Sine function formula
The basic formula for a sine function in trigonometry is y = A sin(Bx + C) where:
A denotes the amplitude, the peak deviation of the function from its center position.
B affects the period of the wave, which is the distance over which the wave's shape repeats.
C represents the phase shift, the horizontal shift of the wave.
Amplitude, period, and phase
Amplitude: The maximum extent of a vibration or oscillation, measured from the position of equilibrium.
Period: The length of one complete cycle in a repeating event.
Phase: A measure of how shifted the wave is from a reference point.
Core Differences
Conceptual Distinctions
Lines in geometry vs. sines in trigonometry
Lines are linear and used primarily for their simplicity and definitive nature in solving geometrical and algebraic problems. In contrast, sines are foundational in trigonometry, providing insights into cyclic and oscillatory systems.
Functional Differences
Linear vs. periodic functions
Linear functions (lines) are used to model relationships with constant rates of change. They are straightforward and predictable.
Visual Comparison
Graphical representation of lines and sines
A line on a graph is a straight path that extends in both directions without end. A sine wave, however, curves and rises and falls in a predictable pattern, demonstrating its periodic nature. This visual contrast underscores the different applications and behaviors of these mathematical functions, highlighting their unique characteristics and uses in various fields.
Practical Applications
Lines in Real-World
Engineering and Architecture
In engineering and architecture, lines are fundamental. They are used to draft blueprints, design structures, and create accurate models of buildings and infrastructure.
By leveraging sine waves, electrical engineers can design efficient and effective electronic systems. This impacts everything from power distribution to telecommunications.
Comparative Analysis
Interplay in Mathematics
How lines and sines interact in complex equations
In mathematics, lines and sines often interact in complex equations. This interplay is crucial in solving various problems:
Trigonometric functions: Sines are part of trigonometric functions that often intersect with lines in coordinate systems. These intersections help in solving equations involving angles and distances.
Fourier series: This mathematical tool expresses complex functions as sums of sine and cosine terms. It helps in analyzing periodic functions by breaking them down into simpler sine and line components.
Differential equations: In physics and engineering, differential equations involving lines and sines model dynamic systems. They describe how systems change over time under various conditions.
Limitations: More complex to calculate and analyze. They require a deeper understanding of trigonometry and calculus.
Frequently Asked Questions
What is a line in mathematics?
A line in mathematics is a straight one-dimensional figure with no curvature, extending infinitely in both directions. It is defined by an equation typically in the form y = mx + b, where m is the slope and b is the y-intercept.
How is a sine function defined?
The sine function is a trigonometric function that describes a smooth periodic oscillation. It is defined as the ratio of the length of the opposite side to the hypotenuse in a right-angled triangle at a certain angle, and its graph forms a wave-like pattern.
What are the practical uses of lines and sines?
Lines are used extensively in fields such as architecture, engineering, and computer graphics to model physical spaces and structures. Sines are crucial in physics and engineering for modeling wave behaviors, such as sound waves in acoustics and electromagnetic waves in communications.
How do lines and sines interact in mathematics?
In mathematics, lines and sines can interact in complex ways, particularly in calculus and differential equations, where solutions to problems may involve both linear functions and trigonometric functions to model phenomena such as damped vibrations.
Conclusion
Understanding the differences between lines and sines enriches one's mathematical toolkit, providing diverse approaches to solving various theoretical and practical problems. Each has its strengths and specific contexts where it is most applicable, from straight-line motion in physics to oscillatory patterns in electrical engineering.
By grasping both concepts, students and professionals can better analyze and design solutions across a broad spectrum of scientific and engineering challenges. This knowledge not only enhances academic understanding but also amplifies the ability to apply mathematical principles in innovative and effective ways. | 677.169 | 1 |
Are There Different Types Of Triangles?
Get ready to embark on a captivating exploration of triangles with the video "Are There Different Types of Triangles?"! Join us on an exhilarating journey as we uncover the fascinating world of this fundamental geometric shape and its various classifications. This video is filled with educational immersion and enlightening information that sheds light on the diverse types and properties of triangles. So, prepare to extend your hand for a high-five and join us as we delve into the captivating details about the different types of triangles. In the realm of mathematics and geometry, triangles exhibit remarkable diversity, each with its unique characteristics and properties. Witness the versatility and beauty of triangles as we explore their classification based on side lengths and angle measurements. Experience the significance and prevalence of these different types of triangles in a wide range of contexts, from art and design to architecture and engineering. Learn about the distinct features and properties of equilateral, isosceles, scalene, acute, right, and obtuse triangles. Discover how these classifications affect the relationships between their sides and angles, unraveling the fascinating world of symmetry and proportion. Get ready to immerse yourself in the enriching world of triangles, leaving you with a deeper understanding and appreciation for the principles of geometry. Join us as we unravel the different types of triangles, filling your mind with knowledge, curiosity, and a greater understanding of these captivating polygons. 📐🔺🙌✋ | 677.169 | 1 |
Hint: We have given congruence about two angles and equality about one side of a triangle. So we will make the adjustments in the given data so that we will use one of the tests for congruence by using given data and then we will prove the congruence of the triangles.
Complete step by step solution: First of all, we will start by noting down the given data. It is given that $TS = TR$, $\angle 1 = 2\angle 2$ and $\angle 4 = 2\angle 3$ . Now since it is given that $TS = TR$, the given triangle $\Delta TRS$ is an isosceles triangle. Therefore, the base angles are also equal.
That means we can write the following: $\angle TSR = \angle TRS$ … (1) Now the angles $\angle 1$ and $\angle 4$ are vertically opposite angles and therefore, are equal. Now it is given that $\angle 1 = 2\angle 2$ and $\angle 4 = 2\angle 3$ .
Now in $\Delta RBT$ and $\Delta SAT$ we observe that $\angle STA = \angle RTB$ as it is a common angle. Also, we just showed that $\angle TSA = \angle TRB$ . We have already given that $RT = ST$ .
Therefore, by using the $ASA$ test for congruence, we conclude that $\Delta RBT = \Delta SAT$.
Note: Here we had already given the congruent side and other ratios were about the angles. So, it is a hint indicating that we have to use a test for congruence which kind of involves only one side and two angles. Therefore, we will make the adjustments in the given equations accordingly. | 677.169 | 1 |
If two parallel lines are crossed by a transversal, then these angles are 1. both outside the two parallel lines 2. on opposite sides of the transversal 3. congruent to each other
Term
Alternative Interior Angles
Definition
If two parallel lines are crossed by a transversal, then these angles are 1. both inside the two parallel lines 2. on opposite sides of the transversal 3. congruent to each other
Term
Same Side Interior Angles (Consecutive Interior Angles)
Definition
If two parallel lines are crossed by a transversal, then these types of angles are 1. both inside the two parallel lines 2. on the same side of the transversal 3. supplementary
Term
Corresponding Angles
Definition
If two parallel lines are crossed by a transversal, then these angles are 1. arranged such that one angle is on the inside of the two parallel lines and the other angle is on the outside of the two parallel lines 2. on the same side of the transversal 3. congruent | 677.169 | 1 |
Using Trigonometry to Model and Predict Natural Phenomena
Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles. It has numerous real-world applications, from architecture and engineering to physics and astronomy. One particularly useful application of trigonometry is in modeling and predicting natural phenomena.
One of the most common natural phenomena that can be modeled using trigonometry is the movement of objects in circular motion. Circular motion can be found in many natural occurrences, such as the orbit of planets around the sun, the rotation of the earth on its axis, or the movement of a swinging pendulum. These movements can be described using trigonometric functions, specifically sine and cosine, which represent the height and distance of an object from its starting point at various points in time.
One example of this is the motion of a planet around the sun. The planet's distance from the sun can be represented as a radius, and its position at any given time can be described using its angle of rotation, known as the "angular displacement." By using trigonometric functions, scientists and mathematicians can accurately predict the position of a planet at any time in its orbit.
Trigonometry is also crucial in understanding and predicting natural phenomena related to waves, such as ocean waves, sound waves, and seismic waves. These waves can be described using sine and cosine functions, which represent the pattern of the wave's oscillations. By analyzing these functions, scientists can determine the frequency, wavelength, and amplitude of a wave, which are essential factors in predicting its behavior.
One notable real-world example is the use of trigonometry in predicting the occurrence and intensity of earthquakes. During an earthquake, the ground moves in a wave-like manner, which can be described using trigonometric functions. By measuring the earthquake's seismic waves, scientists can determine the earthquake's magnitude, the distance of its epicenter, and the depth of its focus. This information is crucial in predicting the potential damage and aftershocks that may occur.
In addition to modeling and predicting natural phenomena, trigonometry is also essential in creating accurate maps and models of the Earth's surface. The Earth's spherical shape poses a challenge when it comes to representing its surface on a flat map. Through the use of trigonometric functions, cartographers can create maps that accurately depict the size, shape, and scale of different land masses and bodies of water.
Additionally, trigonometry is used in the field of geodesy, which is the science of measuring and mapping the Earth's surface. Geodesists use trigonometric techniques to determine the exact coordinates and elevation of different points on the Earth's surface, which is crucial for navigation and many other applications.
Finally, trigonometry is also used in the field of seismology, the study of earthquakes and seismic waves. By analyzing the waves produced by an earthquake, seismologists can determine the earthquake's magnitude, location, and focal depth. This information is critical in predicting future earthquake activity and assessing the potential damage.
In conclusion, trigonometry is an essential tool in modeling and predicting natural phenomena. Whether it is the motion of planets, the behavior of waves, or the mapping of the Earth's surface, trigonometry plays a crucial role in helping us understand and predict the world around us. Its applications in various scientific fields have allowed us to make significant progress in our understanding of natural phenomena and continue to do so in the future. | 677.169 | 1 |
State whether ?ABCand ?AEDare congruent. Justify your answer. a. yes, by either SSS or SAS b. yes, by SSS only c. yes, by SAS only d. No; there is not enough information to conclude that the triangles are congruent.
a land surveyor places 2 stakes 500 ft apart and locates the midpoint between the stakes. From the midpoint, he needs to place another stake 100 ft away that is equidistant to the 2 original stakes. To apply the perpendicular bisector theorem, the land surveyor need to identify a line that is (is the ans) Perpendicular to the line connecting the 2 stakes and going through the midpoint of 2 stakes?
Suppose Ruth Ann has 3 routes she can choose from to get from school to the library. and 5 routes from the library: r to her home. How many routes are there from Ruth Ann's school to her home with a stop at the library?
At a car track dealership, the probability that a vehicle is white is 0.25. The probability that it is a pickup truck is 0.15. The probability that it is a white pickup truck is 0.06. What is the probability that a vehicle is white, given that the vehicle is a pickup truck? Round your answer to two decimal places. A. 0.19 B. 0.06 C. 0.40 D. 0.24 | 677.169 | 1 |
A cylinder is a solid bounded by a cylindrical surface and two parallel planes; the bases of a cylinder are the parallel planes; and the lateral surface is the cylindrical surface. Plane and Solid Geometry - Page 301 by Arthur Schultze - 1901 Full view - About this book
...cylindrical surfaces considered in this treatise are those whose directrices are closed curves. 4. A cylinder is a solid bounded by a cylindrical surface and two parallel planes. 5. The lateral surface of a cylinder is its cylindrical surface. 6. The bases of a cylinder arc the...
...directrix a closed or an open curve. In elementary geometry the directrix is considered a circle. 598. DBF. A Cylinder is a solid bounded by a cylindrical surface and two parallel planes. 599. DBF. The Bases of a cylinder are its plane surfaces. 6OO. DBF. The Lateral surface of a cylinder...
...directrix a closed or an open curve. In elementary geometry the directrix is considered a circle. 598. DEF. A Cylinder is a solid bounded by a cylindrical surface and two parallel planes. 599. DEF. The Bases of a cylinder are its plane surfaces. 600. DEF. The Lateral surface of a cylinder...
...or an open curve. In elementary geometry the directrix is considered a circle. 598. DEF. A Сylinder is a solid bounded by a cylindrical surface and two parallel planes. 599. DEF. The Bases of a cylinder are its plane surfaces. 60O. DEF. The Lateral surface of a cylinder...
...generates the surface. 801. Def. The directrix is the curve which the generatrix touches. SO2. Def. A cylinder is a solid bounded by a cylindrical surface and two parallel planes. 803. Def. Elements of the cylinder are the different positions of the generatrix. REMARK 1. In elementary...
...original position. The different positions of the generatrix are called elements of the surface. 742. A cylinder is a solid bounded by a cylindrical surface and two parallel planes. The cylindrical surface is called the lateral surface, and the plane surfaces are called the bases. The...
...by its motion, may generate the surface is, in any given position, an element of the surface. 508. A cylinder is a solid bounded by a cylindrical surface, and two parallel planes cutting all the elements. The two plane surfaces are called the bases. 509. A circular cylinder is...
...guiding curve and the given fixed line lie in different planes. THE ELEMENTS OF SOLID GEOMETRY. 84. A cylinder is a solid bounded by a cylindrical surface and two parallel planes called bases. The bases cut all the elements of the cylindrical surface. If the bases are circles as...
...generated by the motion of a straight line which touches a given curve and continues parallel to itself. A cylinder is a solid bounded by a cylindrical surface and two parallel planes intersecting this surface. The parallel faces are called bases. The altitude of a cylinder is the perpendicular...
...directrix ; any position of the generatrix, as EF, is called an element of the surface. 551. A cglinder is a solid bounded by a cylindrical surface, and two parallel planes. The parallel planes are called the bases of the cylinder, and the cylindrical surface the lateral surface.... | 677.169 | 1 |
About This Product
This Geometry Project: Transformation Design resource will be fun to do with students as they perform several transformations, create their own designs, and describe their process clearly in words.
Students will:
- Create a design using translations, rotations, and reflections
- Demonstrate their knowledge oftranslations, rotations, and reflections in their write-up
MARKING RUBRIC INCLUDED
This project requires students to use creative and logical thinking skills, organization, careful calculating methods, and neat drawing/colouring. It is a great project to use as a summative assessment for a transformation and graphing unit or to show parents at conferences.
Student Instructions:
Step 1: Design a simple shape (5-9 vertices) that is not completely symmetrical.
Step 2: Draw it somewhere on your grid and color it GREEN.
Step 3: You are going to do at least 9 transformations in any order you want (3 of each type).
Try to make an interesting and beautiful design as you do the transformations.
Step 4: Write or type up an explanation of your design:
I started with my shape in the centre of my page. I reflected it over a line 2 spaces above it. Then I rotated it 1⁄4 turn clockwise around the pink center of rotation. Next I did 3 translations. One was right 2, down 4.........
Grades to Use With:
This math project could work in grades 4-8 in regular math classes depending on when you teach transformations and ordered pairs.
It could also be used in high school special education classes.
Standards:
CCSS8.G.A.1
Verify experimentally the properties of rotations, reflections, and translations:
CCSS8.G.A.3
Describe the effect of dilations, translations, rotations, and reflections on two-dimensional figures using coordinates. | 677.169 | 1 |
In this Discussion
Help with easy intersecting lines
Can someone help me out? I'm trying to create a very simple code that will detect intersection points between two lines. So far, all of the solutions I've seen are overly complicated and have no explanation to what is happening along the way. All I need it to do is give me a returned value (Or something of the sorts) so that I can change one of my other variables to match it | 677.169 | 1 |
2B2E forms with similar adjacent characters prevents a line break inside it.
In mathematics, an ellipse is a plane curve surrounding two focal points, such that for all points on the curve, the sum of the two distances to the focal points is a constant. It generalizes a circle, which is the special type of ellipse in which the two focal points are the same. The elongation of an ellipse is measured by its eccentricity e, a number ranging from e=0 (the limiting case of a circle) to e=1 (the limiting case of infinite elongation, no longer an ellipse but a parabola).
An ellipse has a simple algebraic solution for its area, but only approximations for its perimeter (also known as circumference), for which integration is required to obtain an exact solution.
Analytically, the equation of a standard ellipse centered at the origin with width 2a and height 2b is:
Assuming a≥b, the foci are (±c,0) for c=a2−b2. The standard parametric equation is:
Ellipses are the closed type of conic section: a plane curve tracing the intersection of a cone with a plane (see figure). Ellipses have many similarities with the other two forms of conic sections, parabolas and hyperbolas, both of which are open and unbounded. An angled cross section of a right circular cylinder is also an ellipse.
An ellipse may also be defined in terms of one focal point and a line outside the ellipse called the directrix: for all points on the ellipse, the ratio between the distance to the focus and the distance to the directrix is a constant. This constant ratio is the above-mentioned eccentricity:
Ellipses are common in physics, astronomy and engineering. For example, the orbit of each planet in the Solar System is approximately an ellipse with the Sun at one focus point (more precisely, the focus is the barycenter of the Sun–planet pair). The same is true for moons orbiting planets and all other systems of two astronomical bodies. The shapes of planets and stars are often well described by ellipsoids. A circle viewed from a side angle looks like an ellipse: that is, the ellipse is the image of a circle under parallel or perspective projection. The ellipse is also the simplest Lissajous figure formed when the horizontal and vertical motions are sinusoids with the same frequency: a similar effect leads to elliptical polarization of light in optics.
The name, ἔλλειψις (élleipsis, "omission"), was given by Apollonius of Perga in his Conics. | 677.169 | 1 |
Geometry [Competency Based] (1stPoints, lines, and planes
Classifying angles and polygons
Discovering pi
Mathematical modeling
Laws of detachment and syllogism
Planning and writing proofs
Adjacent, complementary, and supplementary angles
Slope and intercept of a line
Parallel lines cut by a transversal
Congruent segments
Parallel and perpendicular line construction
Inscribed shapes inside triangles and circles
SSS, SAS, and ASA postulates
Triangle congruency
Ratios and proportions
AA, SSS, and SAS similarity tests
Course Goals
Apply geometric properties and relationships through inductive and deductive reasoning.
Explain the building blocks of geometry
Prove geometric properties about congruent triangles by solving problems and using deductive reasoning. | 677.169 | 1 |
Topic:geometric
geometric
About
Geometric research focuses on the study of shapes, sizes, and properties of geometric objects such as points, lines, curves, surfaces, and solids. This field encompasses various branches such as Euclidean geometry, differential geometry, algebraic geometry, and topology. Geometric research plays a crucial role in various disciplines including mathematics, computer science, physics, engineering, and art. Researchers in this area often use mathematical tools and techniques to explore the relationships and structures of geometric objects, as well as develop new theories and applications. | 677.169 | 1 |
Angle Bisector
Acute angle or Obtuse angle bisector.
Let $L_1\,:\,a_1x+b_1y+c_1=0$ and $L_2\,:\,a_2x+b_2y+c_2=0$ are two intersecting lines and $B_1=0$ and $B_2=0$ are angle bisectors of line $L_1=0$ and $L_2=0$.
Let slope of line $L_1=0$ be $m_1$ and that of angle bisector $B_1=0$ be $m_2$ , then angle between lines $L_1=0$ and $B_1=0$ is given by $$tan(\frac{\theta}{2})=\vert \cfrac{m_1-m_2}{1+m_1m_2} \vert$$ | 677.169 | 1 |
...equiangular with ABC, and hence similar. 72 PERPENDICULAR ON HYPOTENUSE. [BOOK V. PROP. XVIII. THEOREM. The perpendicular from the vertex of the right angle to the hypotenuse of a right-angled triangle, divides it into two triangles similar to each other, and similar to the...
...AHB be a triangle right angled at H; a, h, b, the sides respectively opposite the angles, A, H, B; p, the perpendicular from the vertex of the right angle to the hypotenuse ; m and n, segments of the hypotenuse respectively adjacent to a and b. 1. The triangle BDH is similar...
...EXERCISES. 1. In a right triangle, the sides about the right angle are 12 and 15; required the length of the perpendicular from the vertex of the right angle to the hypotenuse. Suggestion. —Find the hypotenuse and one of the segments. 2. A lot is in the form of a right triangle,...
...intersect each other, their common chord produced bisects their common tangents. (§ 293.) 3. If one side of a, right triangle is 'double the other, the perpendicular...it into segments which are to each other as 1 to 4. 4. If two parallels to the side BC of a triangle meet the sides AB and AC in the points D and F, and...
...extremity of the bisector extended, then the triangles BAD and ACE can be proved similar. 375. If one side of a right triangle is double the other, the perpendicular, from the vertex of the right angle to the hypothenuse, divides it into segments which are to each other as 1 to 4. 376. A line parallel to the...
...triangle is double the other, the perpendicular, from the vertex of the right angle to the hypothenuse, divides it into segments which are to each other as 1 to 4. 376. A line parallel to the bases of a trapezoid, passing through the intersection of the diagonals,...
...intersecting circles is 16, and their radii are 10 and 17, what is the distance between cheir centres ? 22. If one leg of a right triangle is double the other,...it into segments which are to each other as 1 to 4. 23. If two parallels to the side BC of a triangle ABC meet the sides AB and AC in D and F, and E and...
...10 and 17, what is the distance between cheir centres ? 22. If one leg of a right triangle is douhle the other, the perpendicular from the vertex of the...it into segments which are to each other as 1 to 4. 23. If two parallels to the side BC of a triangle ABC meet the sides AB and AC in D and F, and E and...
...segment adjacent to that side. For from step 3, AB : AC -
...segment adjacent to that side. For from step 3, AB : AC = | 677.169 | 1 |
Time, 8 koiri. 1. Define right angle, rhomboid, plane figure, sector, similar polygons. 2. Prove that if two triangles have two sides and the included angle of one respectively equal to two sides and the Included ancle of the other, the two triangles are equal in...
area A'B'C'. QED Proposition 9. Theorem.* 709. Two triangles on the same, or on equal spheres, having two sides and the included angle of one equal respectively to two sides and the included angle of the other, are either equal or equivalent. Hyp. Let ABC, DEF be two As having the side AB = DE,...
...from the same point to a given line, the perpendicular is shorter than any oblique line. 3. Prove: If two triangles have two sides and the included angle of one equal to two sides and the included angle of the other, they are equal in all respects. 4. Prove: The diagonals...
...II.— TRIANGLES. SYNOPSIS.* SECTION I.— EQUAL TRIANGLES. 1O.>. i. DEFINITIONS. 2. PROPOSITIONS. PROP. I. If two triangles have two sides and the included angle of the one equal to two sides and the included angle of the other, each to each, they are equal in all...
...joining the two points. QED PROPOSITION IX. THEOREM. / 275. Two triangles on equal spheres, having two sides and the included angle of one equal respectively to two sides and the included angle of the other, are either equal or equivalent. Proof. By superposing one Sph. A upon the other, or upon...
...equal lunes. Two such triangles are represented in fig. 7, and are said to be symmetrically equal. (3). If two triangles have two sides, and the included angle of one respectively equal to the two sides and the included angle of the other, the triangles are equal in... | 677.169 | 1 |
In triangle DEF, DG = 10 cm. What is CG?
5 cm
10 cm
15 cm
20 cm
5 cm Step-by-step explanation: As we can see in the attached figure that G is the point at which the three medians of the triangle meet i.e we called the centroid And, according to the property of the centroid, it divides the medians in ratio i.e 2:1 DG: CG = 2 : 1 Since the DG is 10 cm So, the CG would be half of DG i.e 5 cm Hence, the CG is 5 cm Therefore the first option is correct
5 cm is the correct answer it would be half of DG Step-by-step explanation:
5 cm is the correct answer Step-by-step explanation:
EXPLANATION In the given diagram, G is the point where all the three medians of the triangle meet. We call this point of intersection of all the three medians the CENTROID of the triangle. One property about the centroid is that, the centroid divides the medians in the ratio The medians are the lines from the vertex to the midpoints of the opposite side of the triangle. This implies that, We can rewrite this as, We cross multiply to obtain, Therefore CG = 5cm | 677.169 | 1 |
theodolite and sundial: an instrument combining the functions of a sundial and a theodolite.
throne: part of an astrolabe, connecting the instrument to the suspension shackle, see article on the astrolabe.
triangulation: surveying technique involving the measurement of a baseline, the location of other stations by taking angles from either end, and perhaps the extension of the survey through the addition of further triangles.
triangulation instrument: instrument with three jointed arms with scales for surveying or range-finding, see article on the triangulation instrument.
trigonus: triangular element in the type of altitude dial known as the 'organum Ptolemai'.
tripod base: base of a tripod (three-legged support) on which to stand an instrument.
tripod legs: legs of a tripod (three-legged support) on which to stand an instrument.
tropic of Capricorn: line of geographical latitude and corresponding line of declination coinciding with the sun's position at the autumnal equinox, its most southerly position in the sky. | 677.169 | 1 |
Eureka Math Algebra 2 Module 2 Lesson 4 Answer Key
Engage NY Eureka Math Algebra 2 Module 2 Lesson 4 Answer Key
Eureka Math Algebra 2 Module 2 Lesson 4 Example Answer Key
Example 1.
Suppose that point P is the point on the unit circle obtained by rotating the initial ray through 30°. Find sin(30°) and cos(30°).
What is the length OQ of the horizontal leg of our triangle?
Answer:
By remembering the special triangles from geometry, we have OQ = \(\frac{\sqrt{3}}{2}\).
What is the length QP of the vertical leg of our triangle?
Answer:
Either by the Pythagorean theorem or by remembering the special triangles from Geometry, we have QP = \(\frac{1}{2}\)
What is sin(30°)?
Answer:
sin(30°) = \(\frac{1}{2}\)
What is cos(30°)?
Answer:
cos(30°) = \(\frac{\sqrt{3}}{2}\)
Example 2.
Suppose that P is the point on the unit circle obtained by rotating the initial ray through 150°. Find sin(150°) and cos(150°).
Answer:
Notice that the 150° angle formed by \(\overrightarrow{O P}\) and \(\overrightarrow{O E}\) is exterior to the right triangle ∆ POQ. Angle POQ is the reference angle for rotation by 150°. We can use symmetry and the fact that we know the sine and cosine ratios of 30° to find the values of the sine and cosine functions for 150 degrees of rotation.
→ What are the coordinates (xθ, yθ) of point P?
Using symmetry, we see that the y-coordinate of P is the same as it was for a 30° rotation but that the x-coordinate is the opposite sign as it was for a 30° rotation. Thus (xθ, yθ) = \(\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\)
→ What is sin(150°)?
sin (150°) = \(\frac{1}{2}\)
→ What is cos(150°)?
cos (150°) = \(-\frac{\sqrt{3}}{2}\)
Eureka Math Algebra 2 Module 2 Lesson 4 Opening Exercise Answer Key
Exercise 1.
Find the lengths of the sides of the right triangles below, each of which has hypotenuse of length 1.
Answer:
Exercise 2.
Given the following right triangle ∆ ABC with m∠A = θ°, find sin(θ°) and cos(θ°).
Answer:
Eureka Math Algebra 2 Module 2 Lesson 4 Exercise Answer Key
Exercise 1.
Suppose that P Is the point on the unit circle obtained by rotating the initial ray through 45°. Find sin(45°) and cos(45°).
Answer:
We have sin(45°) = \(\frac{\sqrt{3}}{2}\) and cos(45°) = \(\frac{\sqrt{2}}{2}\).
Exercise 2.
Suppose that P is the point on the unit circle obtained by rotating the initial ray through 60°. Find sin(60°) and cos (60°).
Answer:
We have sin(60°) = \(\frac{\sqrt{3}}{2}\) and cos(60°) = \(\frac{1}{2}\).
Discussion
Answer:
Exercise 3.
Suppose that P is the point on the unit circle obtained by rotating the initial ray counterclockwise through 120 degrees. Find the measure of the reference angle for 120°, and then find sin(120°) and cos(120°).
Answer:
The measure of the reference angle for 120° is 60°, and P is in Quadrant II. We have sin(120°) = \(\frac{\sqrt{3}}{2}\) and cos(120°) = –\(\frac{1}{2}\).
Exercise 4.
Suppose that P is the point on the unit circle obtained by rotating the initial ray counterclockwise through 240°. Find the measure of the reference angle for 240°, and then find sin(240°) and cos(240°).
Answer:
The measure of the reference angle for 240° is 60°, and P is in Quadrant Ill. We have sin(240°) = –\(\frac{\sqrt{3}}{2}\) and cos(240°) = –\(\frac{1}{2}\).
Exercise 5.
Suppose that P is the point on the unit circle obtained by rotating the initial ray counterclockwise through 330° degrees. Find the measure of the reference angle for 330°, and then find sin(330°) and cos(330°).
Answer:
The measure of the reference angle for 330° is 30°, and P is in Quadrant IV. We have sin(330°) = –\(\frac{1}{2}\) and cos(330°) = \(\frac{\sqrt{3}}{2}\).
Eureka Math Algebra 2 Module 2 Lesson 4 Problem Set Answer Key
Question 1.
Fill in the chart. Write in the reference angles and the values of the sine and cosine functions for the indicated values of θ.
Answer:
Question 2.
Using geometry, Jennifer correctly calculated that sin(15°) = \(\frac{1}{2} \sqrt{2-\sqrt{3}}\) . Based on this information, fill in the chart.
Answer:
Question 6.
Johnny rotated the initial ray through θ degrees, found the intersection of the terminal ray with the unit circle, and calculated that sin(θ°) = √2. Ernesto Insists that Johnny made a mistake in his calculation. Explain why Ernesto is correct.
Answer:
Johnny must have made a mistake since the sine of a number cannot be greater than 1.
Question 7.
If sin(θ°) = 0. 5, and we know that cos(θ°) < 0, then what is the smallest possible positive value of θ?
Answer:
150
Question 8.
The vertices of ∆ ABC have coordinates A(0, 0), B(12, 5), and C(12 X2 + y2 = 1?
Answer:
Using similar triangles, the hypotenuse intersects the unit circle at \(\left(\frac{12}{13}, \frac{5}{13}\right)\),5}{13}\) and cos(θ°) = \(\frac{12}{13}\)
Question 9.
The vertices of ∆ ABC have coordinates A(0, 0), B(4, 3), and C(4, 0). The vertices of ∆ ADE are A(0, 0), D(3, 4), and E(3 x2 + y2 = 1?
Answer:
Using similar triangles, the hypotenuse intersects the unit circle at \(\left(\frac{4}{5}, \frac{3}{5}\right)\).3}{5}\) and cos(θ°) = \(\frac{4}{5}\)
d. Argue that ∆ ADE is a right triangle.
Answer:
Since a2 + b2 = c2, the converse of the Pythagorean theorem guarantees that ∆ ADE is a right triangle.
e. What are the coordinates where the hypotenuse of ∆ ADE intersects the unit circle x2 + y2 = 1?
Answer:
Using similar triangles, the hypotenuse intersects the unit circle at (\(\left(\frac{3}{5}, \frac{4}{5}\right)\)).
f. Let Φ denote the number of degrees of rotation from \(\overrightarrow{A E}\) to \(\overrightarrow{A D}\). Calculate sin(Φ°) and cos(Φ°).
Answer:
By the answer to part (e), sin(Φ°) = \(\frac{4}{5}\), and cos(Φ°) = \(\frac{3}{5}\)
g. What is the relation between the sine and cosine of θ and the sine and cosine of Φ?
Answer:
We find that sin(Φ°) = \(\frac{4}{5}\) = cos(θ°), and cos(Φ°) = \(\frac{3}{5}\) = sin(θ°).
Question 10.
Use a diagram to explain why sin(135°) = sin(45°), but cos(135°) ≠ cos(45°).
Answer:
Let O be the center of the circle, let P and R be the points where the terminal rays of rotation by 135° and 45°, and let Q and S be the feet of the perpendicular lines from P and R to the x-axis, respectively. Then △ OPQ and △ ORS are both isosceles right triangles with hypotenuses of length 1, so they are congruent. Thus, PQ = RS, and OQ = OS. Let the coordinates of P and R be (xp, yp,) and (xR, yR). Then xp = – OQ = – OR = – xR, and yp = PQ = RS = yQ. Then we have cos(135°) = – cos(45°), and sin(135°) = sin(45°).
Eureka Math Algebra 2 Module 2 Lesson 4 Exit Ticket Answer Key
Question 1.
How did we define the sine function for a number of degrees of rotation 8, where 0 < θ < 360?
Answer:
First we rotate the initial ray counterclockwise through θ degrees and find the intersection of the terminal ray with the unit circle. This intersection is point P. The y-coordinate of point P is the value of sin(θ°).
Question 2.
Explain how to find the value of sin(210°) without using a calculator.
Answer:
The reference angle for and angle of measure 210° has measure 30°, and a rotation by 210° counterclockwise places the terminal ray in the 3rd quadrant, where both coordinates of the intersection point P are negative.
So, sin(210°) = – sin(30°) = – \(\frac{1}{2}\) | 677.169 | 1 |
I have tried for ages to find the algorithm for getting the coordinate of two intersecting lines from two given coordinated (latitude and longitude) points.
As the picture below suggests, I have two points (A and B) with known coordinates. With these two coordinates, come two bearings, or directions. I am looking for a way to find coordinate of C, where the lines-of-sight intersect.
Note: I am going to build it into a Java / C# Application. If there exists a library that already contains the math, that would help too.
As stated, this isn't a GIS question, just basic math (convert lines into Ax + By = C form, then solve with matrix algebra). Most GIS packages have a way to locate the intersection of a pair of lines, and they also will handle non-intersection, and intersection along a line. Then you get into non-Cartesian solutions, when the coordinates are angular (either spherical or spheroidal)...
5 Answers
5
Assuming that we're dealing with the planar coordinate case (that is not actually what the OP suggested, but I offer this as a better answer to the one given so far – and so far, accepted, by the OP – for the planar case), it helps to first determine the direction cosines from the two clockwise bearings, βAC and βBC, from known points A and B to unknown intersection point C:
fAC = sin βAC
gAC = cos βAC
fBC = sin βBC
gBC = cos βBC
then calculate a determinant
d = fBC gAC - fAC gBC
(If d is very close to zero, the two directions are near parallel and, essentially, there is no intersection.)
Next, calculate a couple of coordinate differences between known points A and B,
Dy = yB - yA
Dx = xB - xA
(where x is Easting and y is Northing) and use them to obtain two very useful distances
sAC = (fBC Dy - gBC Dx) / d
sBC = (fAC Dy - gAC Dx) / d
where sAC is the distance from A to the intersection point, C, along its line-of-sight,
and sBC is the distance from B to C, along its line-of-sight.
If either one of those distances is negative, the intersection point is behind the line-of-sight. That is, there is no real intersection in the direction of the bearing.
Finally, calculate the intersection coordinates via those of known point A and its distance and direction cosines
Assuming that you are dealing with geographic coordinates and geographic (true north) bearings, you have a couple of options:
Project the situation onto the plane, solve using the simple planar coordinates method, and reverse project back to the spheroid.
Perform geodetic or spherical geometry calculations.
If your distances are not too great and your accuracy requirements are not too tight, the first is probably the easier solution.
Broken down, this amounts to
a. Project the two geographic points to plane grid points via a suitable map projection. Ideally, this would be Gnomonic projection because it projects great circles as straight lines. (For a treatise on that, see Algorithms for geodesics by CFF Karney, aka @cffk.) However, a UTM projection method is probably the more common and suitable alternative available. Yet another compromise, suggested by @whuber in a comment to calculating-the-distance-between-a-point-and-a-virtual-line-of-two-lat-lngs, is to treat the geographic coordinates as though they were planar, pre-multiply the difference in longitudes (between A and B) by the cosine of their average latitudes.
b. Convert the geographic azimuths to grid bearings via the meridian convergence. In theory, I believe all well-defined projections have computable grid convergences, as well as scale factors. (See calculate grid convergence: True North to Grid North for an example for the UTM case.)
c. Calculate the point of intersection via the planar method. (See my other answer here for this same question.)
d. Reverse project the intersection point back from grid to geographic coordinates. (Projections, in either direction, are the subject of too many questions on GIS SE to mention here.)
geogA and azimA are geographic point A and azimuth at A, respectively, geogB and azimB are geographic point B and azimuth at B, respectively, and bigdist is an arbitrary but appropriately large enough distance to ensure an intersection.
I conclude that such use of the ST_Project() and ST_Intersection() is equivalent to the algorithm I describe because of the note in the user manual for ST_Intersection():
For geography this is really a thin wrapper around the geometry implementation. It first determines the best SRID that fits the bounding box of the 2 geography objects (if geography objects are within one half zone UTM but not same UTM will pick one of those) (favoring UTM or Lambert Azimuthal Equal Area (LAEA) north/south pole, and falling back on mercator in worst case scenario) and then intersection in that best fit planar spatial ref and retransforms back to WGS84 geography.
As for the second, more rigorous solution, I do not know of even a spherical trigonometry solution, let alone an ellipsoidal trigonometry one. I suspect the solution is very difficult to explain here in simple terms.
The solution in terms of geodesics is given in Section 8 of Algorithms
for geodesics. This uses the ellipsoidal gnomonic projection to
convert the problem to an equivalent planar problem. An implementation
of the solution in C++ is given here. GeographicLib includes
the necessary underlying geodesic routines in both C# and Java.
The question needs clarification then, because this answer is correct only when the coordinates are projected and the bearing has been computed in that projection. It does not apply to (lat, lon) ("geographic") coordinates with geodetic bearings. | 677.169 | 1 |
What are the corresponding sides of a triangle?
In a triangle, the corresponding sides are the sides that are in the same position in different triangles. In the below-given images, the two triangles are congruent and their corresponding sides are color-coded. In the above two triangles ABC and XYZ, AB is the corresponding side to XY.
What are corresponding sides example?
For example, if one polygon has sequential sides a, b, c, d, and e and the other has sequential sides v, w, x, y, and z, and if b and w are corresponding sides, then side a (adjacent to b) must correspond to either v or x (both adjacent to w).
How do you find corresponding parts of a triangle?
Video quote: Parts all right meaning that each side has to correspond to the other side that's going to be in the exact same position.
How do you find the corresponding sides?
Video quote: We want to come up with the corresponding sides now corresponding sides are basically the same location. And in each triangle. So if you notice here we have side a B.
What are corresponding side lengths?
Corresponding sides and angles are a pair of matching angles or sides that are in the same spot in two different shapes. Look at the pictures below to see what corresponding sides and angles look like. Note: These shapes must either be similar or congruent.
What are the three pairs of corresponding sides?
What does corresponding mean in math?
When two lines are crossed by another line (which is called the Transversal), the angles in matching corners are called corresponding angles. Example: a and e are corresponding angles. When the two lines are parallel Corresponding Angles are equal.
What do corresponding angles look like?
Video quote: And corresponding angles all right when we talked about corresponding angles corresponding angles were like they weren't just interior exterior but one was interior and one was exterior.
What are corresponding sides in geometry?
Corresponding sides are the sides that are in the same position in any different 2-dimensional shapes. For any two polygons to be congruent, they must have exactly the same shape and size. This means that all their interior angles and their corresponding sides must be the same measure.
Do corresponding angles add up to 180?
Corresponding angles can be supplementary if the transversal intersects two parallel lines perpendicularly (i.e. at 90 degrees). In such case, each of the corresponding angles will be 90 degrees and their sum will add up to 180 degrees (i.e. supplementary).
What are the corresponding angles in a triangle?
Important Notes on Corresponding Angles
When two parallel lines are intersected by a third one, the angles that occupy the same relative position at each intersection are called corresponding angles to each other. Corresponding angles are congruent to each other.
Which is the corresponding angle to ∠ 1?
∠2 ≅ ∠60° since they are corresponding angles, and m and n are parallel. ∠1 and ∠2 form a straight angle, so∠1=120°.
What are the corresponding vertices?
Specifically, the vertices of each triangle must have a one-to-one correspondence. This phrase means that the measure of each side and angle of each triangle corresponds to a side or angle of the other triangle.
Are corresponding angles equal or supplementary?
As a consequence of Euclid's parallel postulate, if the two lines are parallel, consecutive interior angles are supplementary, corresponding angles are equal, and alternate angles are equal.
What are corresponding angles 7?
CBSE NCERT Notes Class 7 Maths Lines and Angles. The relationships between angles when a pair of parallel line is intersected by a transversal are: Pairs of corresponding angles are equal. Pairs of alternate interior angles are equal.
Which of the following is a pair of corresponding angles?
Corresponding angles are congruent. All angles that have the same position with regards to the parallel lines and the transversal are corresponding pairs e.g. 3 + 7, 4 + 8 and 2 + 6.
Why are corresponding angles equal?
Corresponding angles are equal if the transversal intersects two parallel lines. If the transversal intersects non-parallel lines, the corresponding angles formed are not congruent and are not related in any way.
Are corresponding sides equal | 677.169 | 1 |
How To Arcs and angles maze: 7 Strategies That Work
TheBrowse arcs and angles worksheet circles resources on Teachers Pay Teachers, a marketplace trusted by millions of teachers for original educational resources.Navigating social service programs can be a daunting task, especially when it comes to finding the correct phone number to get in touch with the right department. It is not uncommon for individuals to spend hours on hold or calling multiple... Designed by Shawn Stolworthy of MazePlay.com, the huge Jurassic maze will be open from Saturday, September 9 through Sunday, October 29. Hours are 3-9 pm on Thursdays, 10 am-11 pm on Fridays & Saturdays, and noon-9 pm on Sundays plus 10 am-9 pm on Monday, October 9, for Columbus Day. Entry to the farm costs $20 for ages 3-12 and $24 for ages 13 ...Jun 19, 2018 - Practice solving for unknown arcs and angles in circles with this fun activity. Problems involve central angles and inscribed angles. Each correct answer will lead them through the maze to the finish. Answer key is included. Thank you for your interest in this product from Rise over Run. Find... …Circles. Arc Stuck? Do 4 problems.More ways of describing radians. One radian is the angle measure that we turn to travel one radius length around the circumference of a circle. So radians are the constant of proportionality between an arc length and the radius length. θ = arc length radius θ ⋅ radius = arc length. It takes 2 π radians (a little more than 6 radians) to ...Angles in Polygons and Tessellation. Exterior angles tool - MathsPad; Angle sums investigation - MathsPad; Angle in polygons - The Chalk Face; Applying angle theorems - Mathematics Assessment Project; ... Arcs and sectors worksheet - source unknown (sorry!) Area and perimeter of compound shapes - mrwhy1089 on TES ; Angle and radius of a …Printable PDF, Google Slides & Easel by TPT Versions are included in this distance learning ready activity which consists of 11 circles that students must use the properties of circles to find missing angles and lengths. It is a self-checking worksheet that allows students to strengthen their skills at using the geometric properties of circles.CorrespondParallel Lines Cut by a Transversal maze will have the whole class Participating till the end.Printable PDF, Google Slides & Easel by TPT Versions are included in this distance learning ready activity which consists of 23 parallel lines cut by transversal in which students are given 1 angle measure and have to solve for the unknown angle ...chords, arcs, central angles, inscribed angles, secants, tangents, segments and sectors of a circle to solve problems. Specifically, you will be able to 1. Solve problems involving chords, arcs, central angles, and inscribed angles of circles; and 2. Solve problems involving tangents and secants of circles.Practice solving for unkown arcs or dihedral in gebiete with this fun activity. What involve central angles and inscribed angles. Each correct answer will lead them durch the …Apr 9This breakout escape room is a fun way for students to test their skills with central and inscribed MEASURES Created by Interesting Secants, Tangents, & Chords Some boxes mfght not be used Find the 1680 Find the m loqo Find the mul IOHO NAME ____ An arc length is just a fraction of the circumference of the entire circle. So we need to find the fraction of the circle made by the central angle we know, then find the circumference of the total circle made by the radius we know. Then we just multiply them together. Let's try an example where our central angle is 72° and our radius is 3 ...Mastering the Maze of V-speeds SUSAN PARSON Photo by James Williams. May/June 2015 FAA Safety Briefing 15 Do V-speeds Change? ... X represents the airspeed for best angle of climb, and it results in the greatest amount of altitude over the shortest distance. ... Represented by the top of the white arc on the airspeed indicator, V FE is the maximum …The measure of the inscribed angle is half of the angular measure of the arc it subtends. There are several cases to the proof of the lemma. We will look only at the case where BAC is an acute angle and the center, O, lies in the interior of the angle, as in our figure. 14-Sept-2011 MA 341 001 3 The angle of an arc is identified by its two endpoints, written as mAB.The measure of an arc angle is found by dividing the arc length by the circle's circumference, then multiplying by 360 degrees.Jan 24, 2017 · DescriptionTwo fun activities for students to practice solving for central and inscribed angles and arcs. 1) Riddle Worksheet -Students solve problems to reveal the answer to the riddle at the top of the page, which means they receive immediate feedback as to whether or not they have solved correctly.2) Maze -As students find the answers to the problem, they …Dec 27, 2014 · To convert radians to degrees, just multiply by 180°/π. (Since 180° is equal to π radians) Formula for solving arc length is S = rØ, and theta must be in radians. To convert degrees to radians, just multiply by π/180°. Each quadrant measures 90°. The angle passes 3 quadrants and 80° in the last quadrant. Therefore:196These Vertical and Adjacent Angle Mazes consist of 2 mazes where students must write an equation and solve for the missing angle measure or value of x. Students will choose the correct solution, then move on to another problem in the maze.This activity focuses on the skills of finding the missing angle measure, or value of x, using vertical and adjacent … Here is a graphic preview for all of the CircleTwo fun activities for students to practice solving fo Correspond Math 142: Precalculus II. 2: Introduction to Trigonometry. 2.5: Arcs, The length of an arc can be calculated using different formulas, based on the unit of the central angle of the arc. The measurements of the central angle can be given in degrees or radians, and accordingly, we calculate the arc length of a circle. If θ is in radians, then the angle in degrees = θ × 180/π. By substituting this in the above formula, Arc length = rθ … Angles and arc Examples from our community 10000... | 677.169 | 1 |
Hint: First draw the diagram based on the question. Then, use the cosine formula in $\Delta ABC$ and $\Delta ABD$ and compare the values to get the value of BD. After that use the properties of the isosceles triangle to get the value of ED. Then use the Pythagoras theorem in $\Delta BED$ to get the value of BE.
Note: A line segment drawn from the vertex of a triangle on the opposite side of a triangle which is perpendicular to it is said to be the altitude of a triangle. A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side, thus bisecting that side. | 677.169 | 1 |
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