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How do you find Cos on a unit circle?
The process for determining the sine/cosine of any angle θ is as follows: Starting from (1,0)left parenthesis, 1, comma, 0, right parenthesis, move along the unit circle in the counterclockwise direction until the angle that is formed between your position, the origin, and the positive x-axis is equal to θ.
How do you find sine and cosine?
In any right angled triangle, for any angle:
The sine of the angle = the length of the opposite side. the length of the hypotenuse.
The cosine of the angle = the length of the adjacent side. the length of the hypotenuse.
The tangent of the angle = the length of the opposite side. the length of the adjacent side.
Where is sin positive on unit circle?
Quadrants and the "cast" Rule
In the first quadrant, the values for sin, cos and tan are positive. In the second quadrant, the values for sin are positive only. In the third quadrant, the values for tan are positive only. In the fourth quadrant, the values for cos are positive only.
How do you find sin 30 on unit circle?
Video quote: So now sine 150 is associated with sine 30. So let's focus on the 30 degree angle the reference angle. So sine is opposite over hypotenuse based on sohcahtoa. So opposite to the 30 is 1.
How do you find cos 30 with unit circle?
Video quote: Will give us both the cosine function value. And the sine. Function value the x-coordinate is equal to cosine theta and the y coordinate is equal to sine theta on the unit circle.
What is the Cos of 30 degrees on a unit circle?
0.866
The value of cos 30 degrees can be calculated by constructing an angle of 30° with the x-axis, and then finding the coordinates of the corresponding point (0.866, 0.5) on the unit circle. The value of cos 30° is equal to the x-coordinate (0.866). ∴ cos 30° = 0.866.
Is Cotangent Cos over sin?
The cotangent of x is defined to be the cosine of x divided by the sine of x: cot x = cos x sin x .
How do you find sin?
Video quote: So sine would simply be 5 thirteenth's.
How do you calculate sin?
Video quote: So sine is going to be the opposite over the hypotenuse cosine will be the adjacent. Over the hypotenuse. And tangent will be the opposite.
What is sine?
The inverse sine function or Sin–1takes the ratio, Opposite Side / Hypotenuse Side and produces angle θ. It is also written as arcsin. Let us see an example of inverse of sine function. Example: In a triangle, ABC, AB= 4.9m, BC=4.0 m, CA=2.8 m and angle B = 35°.
What is sin equal to?
The sine of an angle is equal to the ratio of the opposite side to the hypotenuse whereas the cosine of an angle is equal to the ratio of the adjacent side to the hypotenuse.
How do you find cosine theta cos divided by sin?
tan(θ) = Opposite / Adjacent
For a given angle θ each ratio stays the same. no matter how big or small the triangle is. When we divide Sine by Cosine we get: sin(θ)cos(θ) = Opposite/HypotenuseAdjacent/Hypotenuse = OppositeAdjacent = tan(θ)
How do you convert sine to cosine on a calculator?
Video quote: We need to do is you can see above the the buttons here we've got assigned to the minus 1 cos to the minus 1 and 10 4 minus 1 so to take to get into that mode all we need to do is press shift sign.
How do you use sin 2 on a calculator?
How can I reverse my sins?
Video quote: Do you remember what which function deals with opposite and hypotenuse. Sine right so therefore I can say the sine of theta equals. Go that way please equals the opposite over the hypotenuse.
How do you find trig functions on a calculator?
Video quote: So they want to find the sine of 30 degrees we simply press the sine key. 30 close the parentheses and press ENTER.
How do you find the six trig functions on the unit circle?
Video quote: So we're going to look at the six different trig functions how to find them using the unit circle and how to simplify. Them. So the first and easiest one to find is going to be cosine of theta.
How do you do Sohcahtoa?
Video quote: For car we have the cosine equals the adjacent over the hypotenuse. That's our C a H and for this part the Toa. We have the tan equals. The opposite over the adjacent. That's our Toa.
How do you know when to use SOH CAH and Toa?
The mnemonic SOHCAHTOA can be used to aid in remembering which function to use in what circumstance – SOH stands for Sine is opposite over hypotenuse; CAH stands for Cosine is adjacent over hypotenuse; and TOA stands for Tangent is opposite over adjacent. This will save confusion when working with these | 677.169 | 1 |
Geometry and Its Concepts
Concept Map
Understanding congruent and similar figures is fundamental in geometry. Congruent figures are identical in shape and size, while similar figures maintain the same shape but vary in size. This text delves into geometric transformations that preserve these properties, criteria for triangle congruence and similarity, and the proportional relationships of areas and volumes in similar figures. These concepts are not only crucial for mathematical comprehension but also have practical applications in various fields such as architecture and engineering.
Summary
Outline
Show More
Geometry and Its Concepts
Geometry
Properties and Relations
Geometry deals with the properties and relations of points, lines, angles, and surfaces
Congruence and Similarity
Congruent Figures
Congruent figures are identical in form and size, with each corresponding side and angle matching exactly
Similar Figures
Similar figures maintain the same shape but differ in size, with corresponding angles congruent and corresponding sides proportional
Geometric Transformations
Geometric transformations are operations that alter the position or size of a figure while preserving certain properties
Triangles
Congruence
Congruent triangles have all three corresponding sides and angles that are congruent
Similarity
Similar triangles have all three angles congruent and corresponding sides proportional
Criteria for Congruence and Similarity
There are specific postulates and theorems for establishing triangle congruence and similarity, including SSS, SAS, ASA, AAS, and HL
Proportional Relationships
Area
The ratio of corresponding sides of similar figures is directly related to the ratio of their areas, with a ratio of a:b resulting in a^2:b^2
Volume
The ratio of corresponding linear dimensions of similar three-dimensional figures is related to the ratio of their volumes, with a ratio of a:b resulting in a^3:b^3
Applications of Congruence and Similarity
Construction
Congruent figures are essential in ensuring structural integrity and uniformity in construction
Scale Models and Resizing
Similar figures are used in creating scale models and resizing images while maintaining correct proportions
Practical Applications
Congruence and similarity have numerous practical applications in various disciplines such as architecture, engineering, and art, congruence and similarity are crucial for comparing shapes.
mathematics
01
Congruent Figures Definition
Figures are congruent if one can be mapped to the other via rotation, reflection, or translation without altering size or shape.
02
Similar Figures and Dilation
Figures are similar if they have the same shape but different sizes; this can be achieved through dilation, which preserves shape but changes size.
Understanding Congruent and Similar Figures in Geometry
Geometry, a branch of mathematics, deals with the properties and relations of points, lines, angles, and surfaces. Within this field, congruence and similarity are key concepts used to compare figures. Congruent figures are identical in form and size, with each corresponding side and angle matching exactly. Similar figures, while maintaining the same shape, differ in size; their corresponding angles are congruent, and the lengths of corresponding sides are proportional. These concepts are integral to the study of geometric figures, allowing for a deeper comprehension of their attributes and relationships.
The Role of Geometric Transformations in Congruence and Similarity
Geometric transformations are operations that alter the position or size of a figure while preserving certain properties. They are essential in determining the congruence or similarity of figures. A figure is congruent to another if it can be mapped onto the other figure through transformations such as rotation, reflection, or translation, which do not change size or shape. Similarity, however, can involve dilation—a transformation that alters the size of a figure but keeps its shape. Understanding these transformations is crucial for identifying congruent or similar figures in geometry.
Criteria for Congruence and Similarity in Triangles
Triangles, being the simplest polygon, have specific criteria for establishing congruence and similarity. Congruent triangles have all three corresponding sides and angles that are congruent. There are several postulates to establish triangle congruence, including Side-Side-Side (SSS), Side-Angle-Side (SAS), Angle-Side-Angle (ASA), Angle-Angle-Side (AAS), and for right triangles, the Hypotenuse-Leg (HL) theorem. For similarity, triangles must have all three angles congruent, and the sides must be proportional. The criteria for triangle similarity include Angle-Angle (AA), Side-Angle-Side (SAS), and Side-Side-Side (SSS) theorems.
Proportional Relationships and Area in Similar Figures
In similar figures, the proportional relationship between corresponding sides is directly related to the ratio of their areas. If the corresponding sides of two similar figures are in the ratio \(a:b\), then the ratio of their areas is \(a^2:b^2\). This relationship arises from the area being a two-dimensional measure, and it is squared in proportion to the sides. This understanding is vital for calculating the area of similar figures when given a scale factor or the dimensions of one figure.
Volume Ratios in Similar Three-Dimensional Figures
When extending the concept of similarity to three-dimensional figures, the ratio of volumes is affected by the cube of the scale factor. If two similar three-dimensional figures have corresponding linear dimensions in the ratio \(a:b\), then the ratio of their volumes is \(a^3:b^3\). This cubic relationship is due to volume being a measure of three-dimensional space. This principle is fundamental for solving problems involving the volumes of similar solids, such as cylinders, spheres, and cones, where the scale factor or the volume of one solid can determine the volume of the other.
Practical Applications of Congruent and Similar Figures
Congruence and similarity have numerous practical applications across various disciplines such as architecture, engineering, art, and more. In construction, congruent figures ensure structural integrity and uniformity. Similar figures are utilized in the creation of scale models and resizing of images, maintaining the correct proportions while altering size. Proficiency in these geometric principles is essential for the precise design, replication, and analysis of objects and structures in real-world scenarios.
Key Takeaways on Congruent and Similar Figures
To summarize, congruent figures are identical in shape and size, while similar figures share the same shape but differ in size. Geometric transformations such as rotation, reflection, translation, and dilation are fundamental in identifying congruence and similarity. Triangles have specific postulates and theorems for congruence and similarity. The ratios of corresponding sides of similar figures are crucial in determining the ratios of their areas and volumes. These geometric concepts are foundational for mathematical understanding and have significant practical applications. | 677.169 | 1 |
Examination papers used at the examinations for admission to the Royal military college, Sandhurst
Fra bogen
Resultater 1-5 af 93
Side 2 ... equal to of 20 Find the value of of divided by 103 . 5. Reduce to a decimal that terminates . Express if in the form of a recurring decimal . Add together 13 57 , 004 , 31 . · Divide 625 by 00005 , prove the last result by vulgar ...
Side 4 ... equal and also into two unequal parts , the squares of the unequal parts are together double the square of half the line and of the square of the line between the points of section . ( 1. ) How many square yards of sheet iron would be ...
Side 18 ... equal to 24 inches , and making with each other an angle BAC equal to 45 ° . Bisect this angle without the use of a protractor . 2. Describe a circle of three inches diameter ; assume a point in its circumference , and through this ...
Side 7 ... equal to one another , and likewise those terminated in the other extremity . 2. If a straight line be divided into any two parts , the square of the whole line is equal to the squares of the two parts together with twice the rectangle | 677.169 | 1 |
Calculating the vector product
In summary, there is only one formula for finding the cross product of two vectors, which involves computing a determinant using the coordinates of the vectors. It does not matter if one vector has a negative sign, as the formula remains the same.
Feb 9, 2009
#1
intenzxboi
98
0
i have a question I'm trying to find the cross product of the vector but don't know which formula to use.
the first one is
i(...) + j(...) + k (...)
and the other one is the same as this one but has a negative sign
i(...) - j(...) + k (...)
where [itex]\hat{A}=\left<a_1,a_2,a_3\right>[/itex] and [itex]\hat{B}=\left<b_1,b_2,b_3\right>[/itex].
Related to Calculating the vector product
What is the vector product?
The vector product, also known as the cross product, is a mathematical operation that takes two vectors as inputs and produces a third vector that is perpendicular to both of the input vectors.
How do you calculate the vector product?
The vector product is calculated by taking the determinant of a 3x3 matrix composed of the unit vectors i, j, and k and the components of the two input vectors. The resulting vector is given by the coefficients of the i, j, and k unit vectors in the determinant.
What is the geometric interpretation of the vector product?
The vector product has several geometric interpretations, including determining the direction of a perpendicular line, finding the area of a parallelogram formed by the two input vectors, and determining the direction of a torque or moment of force.
What is the difference between the vector product and the scalar product?
The vector product and the scalar product are both mathematical operations on vectors, but they produce different types of results. The vector product produces a vector, while the scalar product produces a scalar (a single number). Additionally, the vector product is defined as the cross product, while the scalar product is defined as the dot product.
How is the vector product used in physics and engineering?
The vector product is an important tool in physics and engineering for calculating forces, moments, and other physical quantities. It is particularly useful in calculating torque, determining the direction of magnetic fields, and analyzing the motion of objects in three dimensions. | 677.169 | 1 |
...334), and two regular polygons of the same number of sides are similar. PROPOSITION IV. THEOREM 375. The perimeters of two regular polygons of the same number of sides are to each other as their radii, and also as their apothems. D' AMU A' W B' Given the regular polygons with perimeters...
...othems. D' AMB A'~ M' B' Given the regular polygons with perimeters...
...'ofhems. D r AMB A' M.' B' Given the regular polygons with perimeters...
...AO : A'O'. Hence, But .'. P: P' = OD : ' = AO:A'0'. (303) (Why?) QBD 417. COR. The areas of regular polygons of the same number of sides are to each other as the squares of their radii or apothems. Ex.- 1315. The lines joining the mid-points of the radii of a regular...
...P:P' = AB:A'B' = AD:A'D'. (Why?) .'. P: P'= OD : O'D'= AO:A'O'. QED 417. COR. The areas of regular polygons of the same number of sides are to each other as the squares of their radii or apothems. Ex. 1315. The lines joining the mid-points of the radii of a regular...
...polygon, and the sum of the areas of the triangles is the area of the polygon,number of sides are to each other as the squares of their sides. The proof is left to the student. 253. Theorem. — The perimeters of two regular polygons of the same number of sides are to each other as their radii, or as their apothems. Hypothesis. AB and CD are sides, and M and N the centers, respectively,... two regular polygons of the same number of sides are to...
...-— . = -— -^ = etc. Ax. V MN NO 9. .-. ABCD MNOP -. Def. sim. poly. 252. Corollary. — The areas of two regular polygons of the same number of sides are to each other as the squares of their sides. 253. Theorem. — The perimeters of two regular polygons of the same number... | 677.169 | 1 |
Detailed Description
This class converts Cartesian coordinates of point x to polar coordinates wrote into output field (output[0] = radius>= 0, output[1] = phi in [0, 2Pi).
Initial situation for the polar coordinate system is that angle phi lies in plane perpendicular to the _axisDirection and turns around the _cartesianOrigin. The radius is the distance of point x to the _axisDirection. | 677.169 | 1 |
Resource Type: Activity, Worksheet - Print
Practical outdoor activity on right-angled trigonometry. Students work in small groups to measure the height of a tree. They need a clinometer, measuring tape, scientific calculator, pen and this worksheet. | 677.169 | 1 |
The Armed Services Vocational Aptitude Battery (ASVAB) is a multiple-choice test used by the United States military to assess an individual's aptitude for various military occupati...Study with Quizlet and memorize flashcards containing terms like Which of the following is equal to (2x/3-7)+7, Which expression represents the perimeter of the rectangle above? (3x-3) (3x-1), Considering the one-time deposit and monthly rent, the equation y = 1200 + 400x can be used to represent y, the total cost of renting an apartment for x months. What is …Study with Quizlet and memorize flashcards containing terms like In the late 1800s, Mendel predicted the existence of units of hereditary information or factors. ... Biology 5.12: Unit Test Molecular Genetics - Part 1. 15 terms. majesticMillah. Preview. 4.10 Unit Test: Mendelian Genetics - Part 1. 14 terms. meowmeow1122. …Study Study with Quizlet and memorize flashcards containing terms like Triangle ABC is similar to triangle DEF. 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Profile Picture · csheridan6.View 1.10 Unit Test_ Triangle Similarity - Part 1.pdf from MATH 2200H at George Washington High School. 2/24/2021 Unit Test : Triangle Similarity - Part 1 Item 1 Calculator Triangle ABC is similar Students also viewed. SIMILARITY TRANFORMS UNIT TEST - ... What typ Paraprofessionals are an integral part of the educational system, providing support to teachers and students in a variety of ways. In order to become a paraprofessional, you must f... Divorce laws vary state by state in the United States. This means, for... | 677.169 | 1 |
Let's find 𝑥 first. Since 𝐷𝐵𝐶 is an isosceles
triangle, we know that angles 𝐷𝐶𝐵 and 𝐶𝐵𝐷 are equal. Therefore, angle 𝐵𝐷𝐶 has measure
180 minus 40 minus 40, which is 100 degrees. The question tells us that the
measure of angle 𝐵𝐷𝐶 is equal to eight 𝑥 plus 20 degrees. Substituting in our value of 100
for the measure of 𝐵𝐷𝐶 and simplifying, we find that 𝑥 equals 10.
Since angles 𝐷𝐵𝐸 and 𝐷𝐶𝐸 are
subtended by the same arc, they must be equal. We are told that the measure of
𝐷𝐶𝐸 is 15 degrees. Therefore, the measure of 𝐷𝐵𝐸 is
15 degrees too. Therefore, angle 𝐸𝐵𝐶, which is
the sum of angles 𝐶𝐵𝐷 and 𝐷𝐵𝐸, has measure 40 plus 15, which is 55. We are told that the measure of
𝐸𝐵𝐶 is equal to 11 times 𝑦 minus two degrees. We can now substitute in our value
of 55 for the measure of 𝐸𝐵𝐶 and simplify, giving us 𝑦 equals seven. The answer to the question is 𝑥
equals 10 and 𝑦 equals seven. | 677.169 | 1 |
This Manipulative is used for understanding the concept of Parabola, Hyperbola, Ellipse and Circle.
A student can able to determine the Standard Equation of Circle,
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Latus-Rectum Major And Minor Axis of Ellipse with help of Rubber Bands.
Set contains 4 X - Y Coordinate Geoboard, 1 Wooden Conic Section Model, 4 Set of cutout of Conic Section in Plastic, 400
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What is a rhombus in geometry?… a diamond a rhombus?A diamond is a quadrilateral, a 2-dimensional flat figure that has four closed, straight sides. But a diamond is also categorized as rhombus, because it has four equal sides and its opposite angles are equal.Likewise, is a square a rhombus? A rhombus is a quadrilateral with all sides equal in length. A square is a quadrilateral with all sides equal in length and all interior angles right angles. Thus a rhombus is not a square unless the angles are all right angles. A square however is a rhombus since all four of its sides are of the same length. One may also ask, what are the 4 properties of a rhombus? Summary of properties S.No. Property Rhombus 1 All sides are congruent ✓ 2 Opposite sides are parallel and congruent ✓ 3 All angles are congruent ✕ 4 Opposite angles are congruent ✓ What is an example of a rhombus?If a parallelogram is a rhombus, then its diagonals are perpendicular. For Example: If PQRS is a rhombus, then ¯PR⊥¯QS . If a parallelogram is a rhombus, then each diagonal bisects a pair of opposite angles. ∠1≅∠2, ∠3≅∠4, ∠5≅∠6, and ∠7≅∠8 .
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Jah Shaka also known as Zulu Warrior, was a Jamaican reggae or dub sound system operator who has been operating a south East London-based, roots reggae Jamaican Sound System since the early 1970s. Jah Shaka was 55 years old. Jah Shaka died on April 2023. Jah Shaka began out on the Freddie Cloudbrust Soun System…
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The First Six Books: Together with the Eleventh and Twelfth
Dentro del libro
Resultados 6-10 de 33
Página 102 ... of the circle . V. In like manner , a circle is faid to be infcri- 4 bed in a rectilineal figure , when the cir- cumference of the circle touches each fide of the figure . о VI . A circle is faid to be described about VI . 102 ...
Página 103 ... described about a rec- tilineal figure , when the circumference of the circle paffes through all the angu- lar points of the figure about which it is defcribed . VII . A ftraight line is faid to be placed in a circle , when the extre ...
Página 106 ... described from the centre D , at the diftance of any of them , fhall pass through the extremities of the other two , and touch the ftraight lines AB , BC , CA , because the angles at the points E , F , G are right angles , and the ...
Página 108 ... described about the circle ABCD . Which was to be done . 34. I. a 10. I. H PROP . VIII . PROB . To infcribe a circle in a given square . C K Let ABCD be the given fquare ; it is required to inscribe a circle in ABCD . Bifect each of the ...
Página 114 ... equal to one another : Wherefore the cif- cle described from the centre F , at the diftance of one of these five , fhall pass through the extremities of the other four , and touch touch the ftraight lines AB , BC , CD , 1.14 THE ELEMENTS.
Pasajes populares
Página 472 - The circumference of every circle is supposed to be divided into 360 equal parts, called degrees ; and each degree into 60 equal parts, called minutes ; and each minute into 60 equal parts, called seconds ; and these into thirds, &c.
Página 105 - DEF are likewise equal (13. i.) to two right angles ; therefore the angles AKB, AMB are equal to the angles DEG, DEF, of which AKB is equal to DEG ; wherefore the remaining angle AMB is equal to the remaining angle DEF.
Página 10 62 - AB be the given straight line ; it is required to divide it into two parts, so that the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part.
Página 112 - To describe an equilateral and equiangular pentagon about a given circle. • Let ABCDE be the given circle; it is required to describe an equilateral and equiangular pentagon about the circle ABCDE. Let the angles of a pentagon, inscribed in the circle...
Página 38 - F, which is the common vertex of the triangles ; that is, together with four right angles. Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides. | 677.169 | 1 |
Hint: To show that side AB = CD prove that$\Delta OAB \simeq \Delta {\text{OCD}}$ using the property of triangle Side-Angle-Side and also remember to join A, B, C and D to the center of the circles O to form the triangles, use this information to approach the solution.
Complete step-by-step answer: Before attempting this question let's join A to O, B to O, C to O and D to O so the figure will be
Note: Whenever we face such geometry problems the key concept is to prove congruent the two triangles in which the sides that we need to prove equal lies. Taking this approach will eventually help as if the triangles are proved congruent than using congruence property the sides can be proved equal. | 677.169 | 1 |
sum of angles on a straight line
In a previous article, we looked at how simple algebraic problems can be solved, the case in point being a rectangle. In this example, notice how similar steps have been followed, except that instead of a perimeter of a rectangle, we shall be considering the angles in a triangle. As before, the material has been | 677.169 | 1 |
Pythagorean Theorem Calculator
The Pythagorean theorem is one of the fundamental theorems of geometric theory, which establishes the ratio between the sides of the rectangular triangle: the square of the hypotenuse is equal to the sum of the squares of the catheters | 677.169 | 1 |
How do you calculate the reference angle?
So, if our given angle is 110°, then its reference angle is 180° – 110° = 70°. When the terminal side is in the third quadrant (angles from 180° to 270°), our reference angle is our given angle minus 180°. So, if our given angle is 214°, then its reference angle is 214° – 180° = 34°.
What is the reference angle for 2?
Finding the reference angle
Quadrant
Reference angle for θ
1
Same as θ
2
180 – θ
3
θ – 180
4
360 – θ
What is the reference angle for 127 degrees?
Trigonometry Examples Since the angle 127° is in the second quadrant, subtract 127° from 180° .
What is the reference angle for a angle?
Reference Angle Definition A reference angle is defined as the absolute difference between 180 degrees and the original angle.
How do I find my reference number?
Typically, it's at the end of an application form or provided in an email or letter from the company. Most reference numbers will be found at the top of the application submission form which shows up after submitting an application. It's also usually quoted at the top of a follow-up email or letter from the company.
What is the reference angle for 120?
How do you find the reference number?
What is the reference number of T 6?
Since 6° is in the first quadrant, the reference angle is 6° .
How to find the reference angle reference angle = 80°?
Input your angle data to find the reference angle reference angle = 80° Finding your reference angle in radians is similar to identifying it in degrees. 1. Find your angle. For this example, we'll use 28π/9 2. If your angle is larger than 2π, take away the multiples of 2π until you get a value that's smaller than the full angle. 10π9 3.
What is the reference angle of 332 degrees?
When the terminal side is in the fourth quadrant (angles from 270° to 360°), our reference angle is 360° minus our given angle. So, if our given angle is 332°, then its reference angle is 360° – 332° = 28° | 677.169 | 1 |
Exit quiz
5 Questions
When the coordinate (2,-3) is reflected by a line of reflection along the y-axis, what is the new reflected coordinate?
Correct answer: (-2,-3)
(-2,-3)
(-3,2)
(2,2)
(3,-2)
Q2.
When the coordinate (2,-3) is reflected by a line of reflection along the x-axis, what is the new reflected coordinate?
(-2,-3)
(-2,3)
Correct answer: (2,3)
(2,3)
(3,2)
Q3.
A triangle with coordinates (4,7), (4,1) and (6,1) is reflected by a line of reflection along the x-axis, what are the reflected coordinates?
(-4,-7) (-4,-1) (-6,-1)
(-4,7) (-4,1) (-6,1)
Correct answer: (4,-7) (4,-1) (6,-1)
(4,-7) (4,-1) (6,-1)
(4,8) (4,6) (6,3)
Q4.
(-5,-5) and (-6,-6) are 2 vertices of a square. When the shape is reflected by a line of reflection along the y-axis, what are the translated vertices of the missing 2 coordinates?
(-5,5) and (-6,6)
(-6,5) and (-5,6)
(5,6) and (6,5)
Correct answer: (6,-5) and (5,-6)
(6,-5) and (5,-6)
Q5.
(5,-4) and (10,-6) are two vertices of a rectangle, what are the coordinates of the missing two vertices after they have been reflected by a line of reflection along the x-axis and then reflected by a line of reflection along the y-axis? | 677.169 | 1 |
Given the triangle t = ABC, its circumcenter is the intersection point O of the medial lines OJ, OD, OF of its sides BC, CA and AB respectively. There are some interesting relations between the medial lines and the symmedians, leading to a couple of construction-methods of a symmedian of t. This is explained in the remarks below.
Given the triangle t = ABC, consider the intersection points G, H of the medial lines FO, DO with sides AC, AB respectively. 1) q = BCGH is a cyclic quadrilateral (angle(HCG) = angle(HBG) = angle(A)). 2) The circumcircle c of q passes through the circumcenter O of the triangle t. 3) The circumcircle d of AFD and c intersect at O and E, such that A, E, I are on a line. I being the intersection of the medial of BC and c, diametral to O (orthogonality of OE to line AE). 4) angle(A) = angle(BGI) = angle(BHI) = angle(BEI) = angle(IEC) = angle(IGC) and p = AHIG is a parallelogram. 5) HG is bisected by AI at M, hence AI is the symmedian w.r. to A (see Antiparallels.html ). 6) angle(FAE) = angle(FOE) = angle(GCE) and since, because of the symmedian AI, angle(JAC) = angle(FAE) the triangle ANC is isosceles and N is on the medial of AC. 7) Similarly APB is isosceles and P is on the medial line of AB. Hence E is also the intersection point of the sides BP and CN of the isosceli ABP, ACN, defined through the median AJ. 8) angle(IBC) = angle(ICB) = angle(A) leads to an easy construction of the symmedian AI. 9) angle(BAE) = angle(ECA), angle(ABE) = angle(EAC) shows that ABE and CAE are similar triangles.
For an application of these remarks in a case of determination of the focus of a parabola, look at the file ParabolaSkew.html . | 677.169 | 1 |
it's where the lines meet at the sharp end of an isoceles triangle........ isn't it ... ? & if you think this answer was written with just a trace of a smile on my face, you might just be right ! its where it cant move any more and it just stays in place ! | 677.169 | 1 |
Polar to cartesian calculator. You can transform this equation to polar form by substituting the po...
Sunglass Technologies - Polarization can occur either naturally or artificially and is what causes a glare. Learn about polarization and sunglass polarization. Advertisement Sunglasses use a variety of technologies to eliminate the problems...Explore math with our beautiful, free online graphing calculator. Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more.Get more lessons like this at how to work with rectangular and polar coordinates on the ti-84 calculator. Also learn how to...polar-cartesian-calculator. cartesian \left(-2, -6\right) en. Related Symbolab blog posts. My Notebook, the Symbolab way. Math notebooks have been around for hundreds ...Summary. To convert a point from the Polar coordinates to Cartesian coordinates the trigonometric functions sine and cosine are used to solve for the x and y component of the point. A point in the polar coordinate system is in the form of P = (r,θ) and a point in the cartesian coordinate system is in the form of P = (x,y). The radius ofLearn how to use the Pythagoras Theorem and the Tangent Function to find the long side and angle of a point in Polar Coordinates. See examples, formulas and tips for solving right triangles with two known sides Home | 18.022 | Tools · Tools Index Up Previous Next. Polar-Cartesian Coordinates Converterpolar-cartesian-calculator. cartesian \left(-2, -6\right) en. Related Symbolab blog posts. My Notebook, the Symbolab way. Math notebooks have been around for hundreds ...This Enter your data in the left hand box with each ...Conversion from cartesian coordinates to polar coordinates. x-coordinate. y-coordinate. Calculation precision. Digits after the decimal point: 2. Radial coordinate (radius) Angular coordinate (azimuth), radians. Angular coordinate (azimuth), degrees.Plug8 thg 9, 2017 ... There is no such an angle except using the inverse trigonometri functions. Otherwise, you can approximate it since it looks to be quite ...Converting Polar Coordinates to Cartesian. The polar coordinates are defined in terms of r r and \theta θ, where r r is the distance of the point from the origin and \theta θ is the angle made with the positive x x -axis. Clearly, using trigonometry, if the Cartesian coordinates are (x,y), (x,y), then. \begin {array} {c}&x = r \cos \theta, &yPolar To Rectangular Calculator: At times, we feel boring to do our calculations. In such cases, if you ever need help wih converting Polar Coordinates to Cartesian Coordinates …Polar Form Calculator + Online Solver With Free Easy Steps. The online Polar Form Calculator helps you easily convert a complex number into its polar form.. The Polar Form Calculator proves to be a powerful tool for mathematicians, allowing them to convert a complex number into its polar form instantly. This time-consuming conversion is done in …Rectangular-Polar Coordinate Conversion. Pol converts rectangular coordinates to polar coordinates, while Rec converts polar coordinates to rectangular coordinates. Specify the angle unit before performing calculations. Calculation result θ is displayed in the range of -180° < θ ≦ 180°. Example 1: To convert rectangular coordinates (√ 2 ...Apr 20, 2023 · Plug Polar to Rectangular Online Calculator. Below polar-calculator. cartesian \left(-2, 2\right) en. Related Symbolab blog posts. My Notebook, the Symbolab way. Math notebooks have been around for hundreds of years Cartesian to Polar. Save Copy. Log InorSign Up. This graph converts cartesian coordinates to polar coordinates for the interval [0,2π). Because there are multiple solutions to ...Hi to all, I have a problem with converting polar coordinates with cartesian points and back. The issue is to find cartesian coordinate in 0,0 origin based chart. Lets say that I have plot in 50,50 and I would like to find another cartesian plot from that point with angle and distance in polar coordinates for example (angle 180: distance: 10) Correct …polar-calculator. cartesian \left(-2, 2\right) en. Related Symbolab blog posts. My Notebook, the Symbolab way. Math notebooks have been around for hundreds of years ...Free Pre-Algebra, Algebra, Trigonometry, Calculus, Geometry, Statistics and Chemistry calculators step-by-step ... Complex Numbers Polar/Cartesian Functions ...Learn how to use the Pythagoras Theorem and the Tangent Function to find the long side and angle of a point in Polar Coordinates. See examples, formulas and tips for solving right triangles with two known sides.cartesian-calculator. polar \left(-3, 8\right) en. Related Symbolab blog posts. My Notebook, the Symbolab way. Math notebooks have been around for hundreds of years ... Get the free "cartesian to polar coordinates" widget for your website, blog, Wordpress, Blogger, or iGoogle. Find more Mathematics widgets in Wolfram|AlphaThe conversion can be seen as two consecutive Cartesian to Polar coordinates conversions, first one in the xy x y plane to convert (x,y) ( x , y ) to (R,φ) ...Polar - Cartesian Complex Numbers Calculator. Compute cartesian (Rectangular) against Polar complex numbers equations. U: P: Polar - Cartesian Home. Contact. Polar - Cartesian. Use this form for processing a polar number against a cartesian (rectangular) number. Similar forms are listed to the right.The calculator converts spherical coordinate value to cartesian or cylindrical one. ... • Cartesian and polar two-dimensional coordinate systems • Area of triangle by coordinates • Area of a rectangle by coordinates • Distance between two …Partial Derivatives: Changing to Polar Coordinates. A function say f f of x x, y y is away from the origin. This function can be written in polar coordinates as a function of r r and θ θ. Now, if we know what ∂f ∂x ∂ f ∂ x and ∂f ∂y ∂ f ∂ y, how can we find ∂f ∂r ∂ f ∂ r and ∂f ∂θ ∂ f ∂ θ and vice versa ...9 thg 5, 2022 ... The Cartesian equation is x2+y2=(3+2x)2. However, to graph it, especially using a graphing calculator or computer program, we want to isolate y Free polar/cartesian calculator - convert from polar to cartesian and vise verce step by step.ByStep 5: We could take the simplification on step further here, but that is not necessary. Both answers have been shown below. r2+3rcos (θ)=6. r (r+3cos (θ))=6. And there you have it! Follow our Five Step Process whenever converting Polar to Cartesian equations and soon enough it'll become second nature! Definition 6.2.1 6.2. 1: Polar Form of a Complex Number. Let z = a + bi z = a + b i be a complex number. Then the polar form of z z is written as. z = reiθ z = r e i θ. where r = a2 +b2− −−−−−√ r = a 2 + b 2 and θ θ is the argument of z z. When given z = reiθ z = r e i θ, the identity eiθ = cos θ + i sin θ e i θ = cos θWe have seen that when we convert 2D Cartesian coordinates to Polar coordinates, we use \[ dy\,dx = r\,dr\,d\theta \label{polar}\] with a geometrical argument, we showed why the "extra \(r\)" is included. Taking the analogy from the one variable case, the transformation to polar coordinates produces stretching and contracting This video shows how to convert from polar to Cartesian form. Show Step-by-step Solutions. Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.Get the free "Polar to cartesian coordinates" widget for your website, blog, Wordpress, Blogger, or iGoogle. Find more Mathematics widgets in Wolfram|Alpha.. Is 0 is a complex number? 0 is a complex number, CCl4 is a non-polar molecule. The four C-Cl bonds are polar, but they The modulus or magnitude of a complex number ( denoted by ∣z∣ ), is the distance between the origin and that If you choose cartesian, you need to enter Some of the real-life uses of polar coordinates include avoiding collisions between vessels and other ships or natural obstructions, guiding industrial robots in various production applications and calculating groundwater flow in radially s... polar-cartesian-calculator. polar \left(... | 677.169 | 1 |
How The Pythagorean Theorem Relates To The Distance Formula
The Pythagorean theorem and the distance formula are two fundamental concepts in mathematics that are closely related. The Pythagorean theorem, named after the ancient Greek mathematician Pythagoras, states that in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. On the other hand, the distance formula is a mathematical equation used to calculate the distance between two points in a coordinate plane. While these concepts may seem distinct, they are actually intricately connected and understanding one can help in comprehending the other.
**Answer:**
The Pythagorean theorem and the distance formula are closely related in the world of mathematics. The Pythagorean theorem, expressed as **c^2 = a^2 + b^2**, allows us to find the length of the hypotenuse of a right-angled triangle when the lengths of the other two sides are known. This theorem is fundamental in geometry and has numerous practical applications, from construction to navigation. On the other hand, the distance formula, given by **√((x2 – x1)^2 + (y2 – y1)^2)**, is used to calculate the distance between two points in a coordinate plane. By understanding how the Pythagorean theorem relates to the distance formula, we can gain a deeper understanding of the underlying principles and applications of these mathematical concepts.
How Is Pythagorean Theorem Related To Distance Formula?
The Pythagorean theorem and the distance formula are closely related concepts in mathematics This theorem is represented by the equation a^2 + b^2 = c^2, where a and b are the lengths of the two legs of the triangle, and c is the length of the hypotenuse.
The distance formula, on the other hand, is used to calculate the distance between two points in a coordinate plane. It is derived from the Pythagorean theorem and can be represented by the equation d = sqrt((x2 – x1)^2 + (y2 – y1)^2), where (x1, y1) and (x2, y2) are the coordinates of the two points and d is the distance between them.
The connection between the Pythagorean theorem and the distance formula can be understood by considering the coordinate plane as a right-angled triangle. The two points represent the endpoints of the hypotenuse, and the horizontal and vertical differences between the points represent the lengths of the legs of the triangle. By applying the Pythagorean theorem to this triangle, we can derive the distance formula and calculate the distance between the two points on the coordinate plane.
How Do You Use The Pythagorean Theorem To Find The Distance Notes?
Sure! Here's an example of how you can use the Pythagorean theorem to find the distance between two points:
Paragraph 1:
To use the Pythagorean theorem to find the distance between two points, we first need to understand what the theorem is. The Pythagorean theorem is a mathematical principle used in geometry that relates to right-angled triangles. 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. In the context of finding distance between two points, we can use this theorem to calculate the length of the straight line connecting them.
Paragraph 2:
Let's say we have two points on a Cartesian plane, (x1, y1) and (x2, y2), where x and y represent the coordinates of each point. To find the distance between these two points, we can use the formula: distance = √((x2 – x1)^2 + (y2 – y1)^2). This formula is derived from the Pythagorean theorem, where the differences in x and y coordinates are squared, added together, and then the square root is taken to find the magnitude of the distance.
Paragraph 3:
For example, let's consider two points, A(3, 4) and B(6, 8). Using the distance formula, we can calculate the distance between these points as follows: distance = √((6 – 3)^2 + (8 – 4)^2) = √(3^2 + 4^2) = √(9 + 16) = √25 = 5 units. Therefore, the distance between points A and B is 5 units. By applying the Pythagorean theorem and using the distance formula, we can easily find the distance between any two points on a plane.
Please note that the above example assumes a Cartesian plane and two-dimensional coordinates. The Pythagorean theorem can also be applied in three-dimensional space using similar principles.
How The Pythagorean Theorem Relates To The Distance Formula Class
The Pythagorean theorem is a fundamental concept in mathematics that relates to the distance formula. It states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. This theorem can be used to calculate distances in coordinate planes, which is where the distance formula comes into play.
The distance formula is a way to find the distance between two points in a coordinate plane. It is derived from the Pythagorean theorem and uses the concept of the distance between two points being the hypotenuse of a right triangle. The formula is as follows:
d = √((x2 – x1)^2 + (y2 – y1)^2)
Here, (x1, y1) and (x2, y2) represent the coordinates of the two points, and d represents the distance between them. By substituting the coordinates into the formula, we can calculate the distance accurately.
Overall, the Pythagorean theorem provides the foundation for the distance formula. It allows us to calculate distances between two points in a coordinate plane accurately. By understanding the relationship between these concepts, we can solve various mathematical problems involving distances and coordinate planes effectively.
How to use the Pythagorean theorem to calculate distances in a coordinate plane:
Identify the coordinates of the two points you want to find the distance between.
Simplify the equation by subtracting the x-values and y-values, then squaring them.
Add the squared differences together.
Take the square root of the sum to find the distance between the two points.
How Is The Pythagorean Theorem Related To The Distance Formula? Explain.
The Pythagorean theorem and the distance formula are two mathematical concepts that are closely related and commonly used in geometry On the other hand, the distance formula is used to find the distance between two points in a coordinate plane.
To understand how the Pythagorean theorem relates to the distance formula, consider a straight line connecting two points in a Cartesian coordinate system. Let's say we have two points, (x1, y1) and (x2, y2). The distance between these two points can be found using the distance formula:
d = √((x2 – x1)^2 + (y2 – y1)^2)
If we draw a right-angled triangle with the two points as the endpoints of the hypotenuse, the lengths of the horizontal and vertical sides of the triangle will be (x2 – x1) and (y2 – y1) respectively. By applying the Pythagorean theorem to this triangle, we can see that the square of the distance (d^2) is equal to the sum of the squares of the horizontal and vertical distances. This can be written as:
d^2 = (x2 – x1)^2 + (y2 – y1)^2
By taking the square root of both sides of the equation, we get the distance formula mentioned earlier. Therefore, the Pythagorean theorem is the underlying principle that connects the distance formula to the concept of finding the distance between two points in a coordinate plane.
In conclusion, the Pythagorean theorem and the distance formula are closely related in the context of finding the distance between two points in a coordinate plane. The Pythagorean theorem provides the foundation for the distance formula, allowing us to calculate the distance by considering the horizontal and vertical distances between the points. This relationship between the two concepts is fundamental in geometry and has practical applications in various fields such as physics, engineering, and computer science.
Explain How The Distance Formula And Pythagorean Theorem Are Similar And Different.
The Pythagorean theorem and the distance formula are both mathematical concepts that are used to calculate distances in different scenarios. While they serve similar purposes, they have distinct formulas and applications.
The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. This theorem is expressed as a^2 + b^2 = c^2, where c represents the length of the hypotenuse, and a and b represent the lengths of the other two sides. This theorem is primarily used to calculate distances in two-dimensional spaces, such as finding the length of a diagonal line in a square or rectangular shape.
On the other hand, the distance formula is used to calculate the distance between two points in a Cartesian coordinate system. It is expressed as √((x2 – x1)^2 + (y2 – y1)^2), where (x1, y1) and (x2, y2) represent the coordinates of the two points. The distance formula can be used in both two-dimensional and three-dimensional spaces to find the straight-line distance between any two given points.
To summarize, the Pythagorean theorem is primarily used to find distances within right triangles, while the distance formula is used to find distances between any two points in a Cartesian coordinate system. Both formulas are essential in various fields, including geometry, physics, and engineering, and they provide valuable tools for calculating distances accurately and efficiently.
Here's a step-by-step tutorial on how to use the distance formula:
1. Identify the coordinates of the two points you want to find the distance between.
2. Label the coordinates as (x1, y1) for the first point and (x2, y2) for the second point.
3. Plug the values into the distance formula: √((x2 – x1)^2 + (y2 – y1)^2).
4. Subtract x1 from x2 and square the result.
5. Subtract y1 from y2 and square the result.
6. Add the two squared values together.
7. Take the square root of the sum to find the distance between the two points.
The Distance Formula Is Derived From The Pythagorean Theorem
The Pythagorean theorem is a fundamental concept in geometry that relates to the relationship used to solve various mathematical problems, including finding the distance between two points on a coordinate plane.
The distance formula is derived from the Pythagorean theorem and is used to calculate the distance between two points in a coordinate plane. It is expressed as:
d = √((x2 – x1)^2 + (y2 – y1)^2)
Start by identifying the coordinates of the two points you want to find the distance between. Let's say the coordinates are (x1, y1) and (x2, y2).
Substitute the values into the distance formula.
Calculate the differences between the x-coordinates and the y-coordinates, and square each difference.
Add the squared differences together.
Take the square root of the sum to find the distance between the two points.
For example, if we have two points A(3, 4) and B(6, 8), we can use the distance formula to find the distance between them:
d = √((6 – 3)^2 + (8 – 4)^2) = √(3^2 + 4^2) = √(9 + 16) = √25 = 5
Therefore, the distance between points A(3, 4) and B(6, 8) is 5 units.
The Distance Formula Is Derived From The Pythagorean Theorem True Or False
The Pythagorean theorem is a fundamental concept in mathematics that relates to the relationships applied in various fields, including geometry, trigonometry, and physics.
The distance formula is derived from the Pythagorean theorem and is used to find the distance between two points in a coordinate plane. It is based on the idea that if we consider the two points as the endpoints of a right triangle, with the horizontal and vertical distances as the two legs, then the distance between the points is the length of the hypotenuse. By applying the Pythagorean theorem, we can derive the formula: d = √((x2 – x1)^2 + (y2 – y1)^2), where (x1, y1) and (x2, y2) are the coordinates of the two points.
So, to answer the question "Is the distance formula derived from the Pythagorean theorem true or false?", the answer is true. The distance formula is indeed derived from the Pythagorean theorem. By recognizing the geometrical relationship between the coordinates of two points and the lengths of the sides of a right triangle, we can use the Pythagorean theorem to determine the distance between the points.
Pythagorean Theorem And Distance Formula Worksheet
The Pythagorean theorem and the distance formula are two fundamental concepts in geometry that are closely related. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. The distance formula, on the other hand, is a formula used to calculate the distance between two points in a coordinate plane.
To understand how the Pythagorean theorem relates to the distance formula, consider a right triangle in a coordinate plane. Let's say we have two points, A(x1, y1) and B(x2, y2), and we want to find the distance between them. By drawing a line segment connecting A and B, we can create a right triangle. The length of the horizontal leg of the triangle is |x2 – x1|, and the length of the vertical leg is |y2 – y1|.
Using the Pythagorean theorem, we can find the length of the hypotenuse (which is the distance between A and B) by applying the formula: c^2 = a^2 + b^2. In this case, c represents the distance we want to find, a represents the length of the horizontal leg, and b represents the length of the vertical leg.
To summarize, the Pythagorean theorem provides a foundational principle for finding the length of the hypotenuse in a right triangle, which can then be used to calculate the distance between two points using the distance formula. These concepts are essential in various fields, such as physics, engineering, and navigation.
How to use the Pythagorean theorem and distance formula worksheet:
1. Start by reviewing the Pythagorean theorem and the distance formula.
2. Read the given problem or question carefully.
3. Identify the two points for which you need to find the distance.
4. Use the distance formula to calculate the horizontal and vertical legs of the right triangle.
5. Apply the Pythagorean theorem to find the length of the hypotenuse, which represents the distance between the two points.
6. Double-check your calculations and ensure that you have used the correct formulae.
7. Write down the final answer with appropriate units, such as inches, centimeters, or meters.
8. Repeat the steps for other problems on the worksheet.
9. If you encounter difficulties, consult your teacher or refer to your textbook for additional explanations and examples.
10. Once you have completed the worksheet, review your answers and make any necessary corrections.
Triangle Distance Formula Calculator
The Pythagorean theorem is a fundamental concept in geometry that relates to the distance formula. It establishes the relationship between the lengths of the sides of a right triangle. According to the theorem, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. This theorem is often used in various applications, including calculating distances in two-dimensional space.
To understand how the Pythagorean theorem relates to the distance formula, we can consider a two-dimensional coordinate plane. Let's say we have two points, A(x1, y1) and B(x2, y2). The distance between these two points can be calculated using the distance formula, which is derived from the Pythagorean theorem. The formula is:
Distance = sqrt((x2 – x1)^2 + (y2 – y1)^2)
This formula represents the square root of the sum of the squares of the differences in the x-coordinates and y-coordinates of the two points. By applying the Pythagorean theorem, we can determine the distance between any two points in a coordinate plane.
How to use a triangle distance formula calculator:
1. Input the coordinates of point A and point B.
2. Use the formula Distance = sqrt((x2 – x1)^2 + (y2 – y1)^2) to calculate the distance.
3. The calculator will provide you with the result, which represents the distance between the two points.
In summary, the Pythagorean theorem is a crucial concept that forms the basis for the distance formula. By using this formula, we can calculate the distance between any two points in a two-dimensional coordinate plane. Understanding this relationship is essential in various fields, such as navigation, engineering, and physics.
Pythagorean Theorem Distance Between Two Points Worksheet Pdf
The Pythagorean theorem is a fundamental concept in geometry that relates to the distance formula. It states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. This theorem is often used to find the distance between two points in a coordinate plane, which is where the distance formula comes into play.
The distance formula is derived from the Pythagorean theorem and allows us to calculate the distance between two points (x1, y1) and (x2, y2) in a two-dimensional plane. It is given by the equation:
distance = sqrt((x2 – x1)^2 + (y2 – y1)^2)
To understand how the Pythagorean theorem relates to the distance formula, we can consider a right-angled triangle formed by the two points and the vertical and horizontal lines connecting them. The length of the hypotenuse of this triangle is exactly the distance between the two points, which can be calculated using the distance formula.
How to use the Pythagorean theorem and distance formula to find the distance between two points:
1. Identify the coordinates of the two points: (x1, y1) and (x2, y2).
2. Plug the coordinates into the distance formula: distance = sqrt((x2 – x1)^2 + (y2 – y1)^2).
3. Simplify the equation by subtracting the x and y values, squaring them, and taking the square root of the sum.
4. Calculate the final result to find the distance between the two points.
By understanding the relationship between the Pythagorean theorem and the distance formula, you can efficiently calculate distances in a two-dimensional plane and apply this knowledge to various mathematical and real-world situations.
In conclusion, the Pythagorean theorem and the distance formula are deeply intertwined, revealing the underlying connection between geometry and algebra. The Pythagorean theorem, with its elegant equation (a^2 + b^2 = c^2), is a fundamental concept in geometry that allows us to calculate the length of the hypotenuse in a right triangle. On the other hand, the distance formula, derived from the Pythagorean theorem, enables us to find the distance between any two points in a Cartesian coordinate system.
By understanding the relationship between the Pythagorean theorem and the distance formula, we gain a powerful tool for solving a wide range of mathematical problems. It allows us to apply geometric principles to algebraic equations, making it possible to measure distances in a variety of contexts, from calculating the shortest route between two cities to determining the length of a diagonal in a rectangle. This connection between geometry and algebra not only enriches our understanding of mathematics, but also provides us with practical tools to analyze and solve real-world problems. So, whether we are exploring the depths of theoretical mathematics or applying mathematical concepts in everyday life, the Pythagorean theorem and the distance formula continue to be invaluable tools that bridge the gap between geometry and algebra | 677.169 | 1 |
Geometry Conditional Statements Worksheet
Geometry Conditional Statements Worksheet - When two statements are both true or both. Write the converse, the inverse, and the contrapositive of the following. If 2 divides evenly into x, then x is a positive number. If a ≠ 5, then a2 ≠ 25; Web showing 8 worksheets for geometry conditional statements. Web learn how to form and write converse statements of conditional statements in geometry and other topics.
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Conditional Statements Worksheet With Answers
If a ≠ 5, then a2 ≠ 25; Identify the hypothesis and conclusion of the conditional: Web.
14 Best Images of Chapter 1 Geometry Worksheets TShirts, Geometry
A conditional statement is a compound statement that comprises two. Web conditional statements in geometry worksheet. The worksheets provide examples, practice questions, and answers for. Worksheets are logic and conditional statements, geometry name for each conditional circle t. Web 2) a polygon is a triangle if and only if the sum of its interior angles is 180°. 3) if a.
Converse Meaning In Discrete Math.
Conditional Statements I Have, Who Has Game High school math lesson
Web learn how to form and write converse statements of conditional statements in geometry and other topics50 Conditional Statement Worksheet Geometry
Your students can practice variations on conditional statements with a guided example!. Web in today's geometry lesson, you're going to learn all about conditional statements! 6 write the converse of the conditional, and determine the truth value of each: • conditional statements • converse • truth value • counter example you will receive a worksheet as well as fill in.
Conditional Statement Worksheet Geometry
Use math goodies free lessons and conditional statement examples for understanding what makes something a conditional statementIf a2 = 25, then a = 5. The worksheets provide examples, practice questions, and answers for. If the weather is nice, then i will wash the car. If 2 divides evenly into x, then x is a positive number. Improve your math knowledge with free questions in conditionals.
• conditional statements • converse • truth value • counter example you will receive a worksheet as well as fill in the blank notes with the. That is, we can determine if they are true or false. (i) two points are collinear if they lie on. Web ðï ࡱ á> þÿ þÿÿÿ.
Improve your math knowledge with free questions in conditionals. The worksheets provide examples, practice questions, and answers for. Your students can practice variations on conditional statements with a guided example!.
When two statements are both true or both. The worksheets provide examples, practice questions, and answers for. Web ðï ࡱ á> þÿ þÿÿÿ.
That Is, We Can Determine If They Are True Or False.
Worksheets are logic and conditional statements, geometry name for each conditional circle t. Web showing 8 worksheets for geometry conditional statements. If a2 = 25, then a = 5. Web conditional statement a conditional statement is a logical statement that has two parts, a hypothesis p and a conclusion q.
If You Want To Be Fit, Then Get Plenty Of Exercise.
A conditional statement is a compound statement that comprises two. Improve your math knowledge with free questions in conditionals. • conditional statements • converse • truth value • counter example you will receive a worksheet as well as fill in the blank notes with the. Write the converse, the inverse, and the contrapositive of the following.
Identify The Hypothesis And Conclusion Of The Conditional: Click the card to flip 👆. Your students can practice variations on conditional statements with a guided example!.
From Prentice Hall Geometry Book, 4.1 Pg 185.
6 write the converse of the conditional, and determine the truth value of each: Use math goodies free lessons and conditional statement examples for understanding what makes something a conditional statement. Web a conditional statement or a conditional expression is often symbolized by variables, mostly p and q. The variable (p) is often used for indicating the hypothesis and the. | 677.169 | 1 |
Explore exterior angle property Worksheets by Grades
Explore Other Subject Worksheets for class 11
Explore printable exterior angle property worksheets for 11th Class
Exterior angle property worksheets for Class 11 are an essential resource for teachers looking to enhance their students' understanding of geometry concepts. These worksheets provide a wide range of problems and exercises that focus on the exterior angle theorem, which states that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles. By incorporating these worksheets into their lesson plans, teachers can help Class 11 students develop a strong foundation in geometry, particularly in the area of triangle properties and angle relationships. As students work through these exercises, they will gain valuable practice in applying the exterior angle theorem to various types of triangles, ultimately improving their problem-solving skills and overall confidence in tackling complex geometry problems. Exterior angle property worksheets for Class 11 are a must-have for any math teacher's toolkit.
Quizizz is an excellent platform for teachers to access a wide variety of resources, including exterior angle property worksheets for Class 11, as well as other math and geometry-related materials. This interactive platform allows educators to create engaging quizzes, assignments, and activities that cater to the specific needs of their students. With Quizizz, teachers can easily track student progress and identify areas where additional support may be needed. In addition to exterior angle property worksheets, Quizizz offers a vast library of resources covering various topics in mathematics, such as algebra, trigonometry, and calculus. By utilizing Quizizz in their classrooms, teachers can provide a comprehensive and interactive learning experience for their Class 11 students, ensuring they are well-prepared for future math courses and real-world applications of geometry concepts. | 677.169 | 1 |
Exploring the Geometry Spot-Investigating the Marvels of Mathematical Shapes
Welcome to the Geometry Spot, where points, lines, and shapes join to make a universe of numerical wonders. In this navigational excursion, we will dive into the perplexing scenes of math, uncovering its importance, applications, and magnificence. Whether you're a carefully prepared mathematician or an inquisitive voyager, go along with us as we explore through the entrancing domain of mathematical shapes.
Figuring out the Fundamentals of Geometry
Geometryspot is the part of science that arrangements with the properties and connections of focuses, lines, points, surfaces, and solids. At its center, calculation looks to depict and break down the major designs that make up our actual world. To explore the Calculation Spot actually, it's significant to embrace the essential ideas:
Euclidean Calculation:
Named after the antiquated Greek mathematician Euclid, Euclidean math shapes the premise of customary math. It includes ideas like focuses, lines, planes, points, and shapes like triangles, circles, and polygons.
Non-Euclidean Calculation:
Not at all like Euclidean math, non-Euclidean math investigates mathematical frameworks where Euclid's fifth hypothesize (the equal propose) doesn't turn out as expected. This branch incorporates exaggerated and elliptic calculation, offering elective systems for figuring out spatial connections.
Exploring geometry spot
Calculation is packed with a different cluster of shapes, each having remarkable properties and qualities. We should leave on an excursion through probably the most unmistakable mathematical shapes:
Triangles:
Triangles are polygons with three sides and three points. They act as crucial structure blocks in Geometry and come in different kinds, including symmetrical, isosceles, and scalene triangles. Grasping triangle properties, like the Pythagorean hypothesis and the triangle disparity, is fundamental for exploring mathematical scenes.
Circles:
Circles are wonderful bends comprising of all focuses equidistant from a main issue called the middle. They highlight conspicuously in calculation, offering rich answers for issues including evenness, region, periphery, and bend length.
Polygons:
Polygons are shut Geometry figures with straight sides. From straightforward shapes like quadrilaterals and pentagons to complex designs like dodecagons and polygons with endless sides, investigating polygons discloses a rich embroidery of mathematical potential outcomes.
Utilizations of Calculation
Calculation reaches out a long ways past the limits of the homeroom, finding applications in different fields going from design and designing to craftsmanship and plan. We should investigate a few genuine
Engineering:
Modelers depend on mathematical standards to configuration structures that are stylishly satisfying as well as primarily sound. From the exact points of a three-sided support to the smooth bends of a curve, math shapes the constructed climate around us.
Map making and Route:
Guides, route frameworks, and GPS depend on mathematical ideas to address geographic highlights precisely and ascertain distances, headings, and courses. Whether you're investigating another city or diagramming a course across the oceans, math directs your excursion.
Craftsmanship and Plan:
Specialists and creators influence calculation to make outwardly enthralling arrangements, examples, and figures. From the mathematical accuracy of Islamic tilework to the theoretical types of present day craftsmanship, calculation fills in as a wellspring of motivation and articulation.
FAQs
What is Euclidean Geometry?
Euclidean calculation, named after the old Greek Geometry Euclid, investigates the properties of focuses, lines, and shapes in a level plane.
Why is figuring out triangles significant in Geometry?
Triangles act as crucial structure blocks in math, offering experiences into spatial connections and empowering the use of mathematical standards in different settings.
How does calculation apply to certifiable situations?
Math finds applications in different fields like design, advanced mechanics, map making, and workmanship, forming the actual climate and driving development across ventures.
What are a few instances of Geometry shapes in day to day existence?
Instances of Geometry shapes flourish, from the roundabout wheels of vehicles to the three-sided tops of houses, showing the unavoidable impact of calculation in our environmental factors.
Could geometry spot be outwardly engaging?
Totally! Craftsmen and creators frequently consolidate Geometry examples, structures, and balances into their manifestations, exhibiting the stylish charm of mathematical standards in craftsmanship and plan.
Final Thoughts
As we close our navigational excursion through the Geometry Spot, we've acquired a more profound appreciation for the style and meaning of mathematical shapes. From the effortlessness of lines and points to the intricacy of three-layered solids, calculation pervades our reality, forming both the actual scenes and the scholarly scenes of numerical request. Whether you're graphing mathematical courses or investigating mathematical miracles, may your process be as improving as the Calculation Spot itself. | 677.169 | 1 |
Flat And Curved Surfaces Worksheet
Flat And Curved Surfaces Worksheet - Worksheets are the perception of length on curved and flat surfaces, three gradients and the percept. This cut and paste sort helps students practice identifying 3d shapes with curved, flat, or both. Use the shape to draw a picture of a nice object. Web are the 3d shapes made up of flat or curved surfaces? Web identify objects with flat and curved surfaces. Web showing 8 worksheets for flat surface and curved surface.
Web showing 8 worksheets for flat and curve surfaces. 131 views 1 year ago. Web displaying 8 worksheets for grade 2 flat surface and curved surface. Web 0:00 / 6:55. A surface that is smooth and even is called flat.
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Are the 3D shapes made up of flat or curved surfaces? Use the shape to
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Flat and Curved Surfaces Lesson 2 YouTube
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Lines and Surfaces Flat and Curved Surfaces YouTube
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Flat or Curved Surfaces? Labelled diagram
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What Is Flat Surface? Definition, Solved Examples, Facts
Observe the examples of flat and curved. 4.9 based on 257 votes. Web showing 8 worksheets for flat surface and curved surface. Worksheets are the perception of length on curved and flat surfaces, three gradients and the percept. Web identify objects with flat and curved surfaces.
Curved surfaces, shell structures and discrete element assemblies
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3D Shapes Anchor Chart (30 off) Shape anchor chart, Shape chart
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Flat And Curved Surfaces Worksheets
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4.9 based on 257 votes. A surface that is smooth and even is called flat. Curved and flat surfaces worksheet.
Web displaying 8 worksheets for grade 2 flat surface and curved surface. Web are the 3d shapes made up of flat or curved surfaces? Worksheets are the perception of length on curved and flat surfaces, three gradients and the percept.
Worksheets are the perception of length on curved and flat surfaces, three gradients and the percept. Observe the examples of flat and curved. Flat surfaces and curved surfaces. And the surface of balls, eggs, or lemons is curved. | 677.169 | 1 |
(i) Line segment: It is the part of a line whose both the ends are fixed(terminated).The given figure shows a line segment AB.
(ii) Collinear points: Three or more points lying on the same straight line are called collinear points. In the given figure, points A,B and C lie on the same straight line so these points are collinear.
(iii) Parallel lines: Two lines are said to be parallel to each other if they do not have common point, i.e. they do not intersect.
(iv) Intersecting lines: If two lines have a common point,the lines are said to be intersecting lines.In the given figure, line l and m have common point O, therefore these lines are intersecting lines.
(v) Concurrent lines: Three or more lines in a plane are said to be concurrent if all of them pass through the same point. In the given figure, four lines are passing through the same point O, therefore these lines are concurrent lines. The common point O is called the point of concurrency.
(vi) Ray: A straight line, generated by a point and moving in the same direction is called a ray. The given figure shows a ray AB. | 677.169 | 1 |
...(147) with (148).] Of PROPOSITION III. Two triangles, having an angle of the one equal to an (159) angle of the other, are to each other as the products of the sides about the equal angles. Let the equal apgles of the triangles A, B, be made vertical, and join...
...A'B'C'. Relation of Areas of Figures. THEOREM VI. Triangles which have one angle of the one equal to one angle of the other, are to each other as the products of the sides containing the equal angle. Let the triangles ABC, A'BC' have equal angles at B. Then shall ABC...
...similar when they are mutually equiangular. 4. Two triangles having an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. 5. What is the length of the side of a regular decagon inscribed...
...BOOK IV. THEOREMS. 219. Two triangles which have an angle of the one equal to the supplement of an angle of the other are to each other as the products of the sides including the supplementary angles. (IV. 22.) 220. Prove, geometrically, that the square described...
...PROPOSITION XX.—THEOREM. 57. Two tetraedrons which have a triedral angle of the one equal to a triedral angle of the other, are to each other as the products of the three edges of the equal triedral angles. Let AB CD, AB'C'D', be the given tetraedrons, placed with...
...area of any polygon 43 EXERCISES (4) 44 VIII. Two triangles which have an angle of the one equal to an angle of the other, are to each other as the products of the sides including the equal angles 47 IX. The areas of similar triangles are to each other as the squares...
...similar when they are mutually equiangular. 2. Two triangles having an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. 3. To inscribe A circle in a given triangle. 4. The side of a regular...
...as the products of their bases and altitudes. PROPOSITION XIX. THEOREM. 677. Two tetrahedrons having a trihedral angle of the one equal to a trihedral...the other are to each other as the products of the three edges of these trihedral angles. Let V and V denote the volumes of the two tetrahedrons D-ABС,...
...GEOMETRY. — BOOK VII. PROPOSITION XIX. THEOREM. 577. Two tetrahedrons having a trihedral angle ofth one equal to a trihedral angle of the other are to each other as the products of the three edges of these trihedral angles. Let V and V denote the vol-аmes of the two tetrahedrons D ABC,...
...VII. PROPOSITION XIX. THEOREM. 577. Tioo tetrahedrons having a trihedral angle of the one equal io a trihedral angle of the other are to each other as the products of the three edges of these trihedral angles. Let V and V denote the volumes of the two tetrahedrons DA BC,... | 677.169 | 1 |
2B22 forms with similar adjacent characters prevents a line break inside it.
In geometry, a hexagon (from Greek ἕξ, hex, meaning "six", and γωνία, gonía, meaning "corner, angle") is a six-sided polygon. The total of the internal angles of any simple (non-self-intersecting) hexagon is 720°. | 677.169 | 1 |
Consider a hyperbola c and its auxiliary circle c'. Draw two tangents {t, t'} parallel to asymptotes and intersecting at a point P. Then every tangent p to the auxiliary circle intersects lines {t, t'} correspondingly at points {Q,R} and the conic at two points {S,T} of which one, S say, is the middle of QR.
The proof starts by defining the polar FE of P with respect to the conic c. This is simultaneously the polar of P with respect to circle c' since (G,H,I,P) = -1 are harmonic with respect to either of the curves. Take then a tangent of c' at B as required and consider the intersection point A of CB with the polar EF. The polar pA of A with respect to the circle passes through P (by the reciprocity of relation pole-polar) and is parallel to tangent p. Besides (E,F,J,A) = -1 build a harmonic division, thus the bundle of lines at P: P(E,F,J,A) defines a harmonic division on every line it meets. Apply this to the tangent p. Since PJ is parallel to this tangent, S is the harmonic conjugate with respect to {R,Q} of the point at infinity of line p, hence it is the middle of RQ.
Remark This property has an inverse characterizing a certain geometric locus: Given a circle c' and two tangents {t,t'}, the middles S of segments RQ intercepted by tangents p of the circle on {t,t'} generate a hyperbola c having c' as auxiliary circle and asymptotes parallel to the given lines {t,t'}.
The proof of this inverse can be easily supplied by constructing the hyperbola from the given data and then using its proven property identify it with the required locus. A similar property defining hyperbolas is discussed in HyperbolaProperty.html . | 677.169 | 1 |
Brooklyn Anderson
Skidmore College
Trigonometry
Trigonometry teacher | Tutor for 9 years
I hold a degree in Trigonometry from Skidmore College, where my passion for mathematics flourished. With a knack for breaking down complex concepts, I thrive on helping students grasp the intricacies of trigonometry. Whether it's solving equations or understanding geometric principles, I'm here to guide you every step of the way. Let's unlock the mysteries of triangles and angles together!
Questions
How do you find the coordinates of the terminal points corresponding to the following arc length on the unit circle: 11π/6?
How do you find the coordinates of the terminal points corresponding to the following arc length on the unit circle: 22π/3?
The hypotenuse of a 30-60-90 right triangle measure 18cm. What are the lengths of the longer leg and shorter leg?
How do you find the coterminal angle for #(11pi) / 4#?
How do you find the exact values #sin (pi/6)# using the special triangles?
The given angle -307° is in standard position, how do you determine the quadrant in which the angle lies?
How do you evaluate tan 30° without using a calculator by using ratios in a reference triangle?
The Hyptoneus is 16, inside/above to the left of the 90 degrees angle is 56 degrees, how do you find the missing angle?
In a 45 degree-45 degree right triangle, the length of a hypotenuse is 22 ft. How long is a leg of the triangle?
How do you solve the right triangle given the Hypotenuse side is 7 and it is a 45-45-90 triangle?
How do you determine the quadrant in which #-2# radians lies?
How do you determine the quadrant in which #2.25# radians lies?
If #sin A= 1/(sqrt(2))#, what is #cot A/csc A#?
How do you solve #cos x = 3# ?
A telephone pole is 28 feet tall. A wire is stretched from the top of the pole to a point on the ground that is 5 feet from the bottom of the pole. How long is the wire?
How do you find the measures of the angles of an isoscles triangle whose sides are 6,6,and 8?
A helicopter is flying at an altitude of 1200°m. The angle of elevation from the tower on the ground to the plane measures 36°. How far is the building to the plane?
The hypotenuse of a right triangle measures 9 inches, and one of the acute angles measures 36 degrees. What is the area of the triangle?
How do you use #csctheta=5# to find #tantheta#?
How do you find the five remaining trigonometric function satisfying #costheta=-2/5#, #sintheta>0#? | 677.169 | 1 |
Outline:
Number Line: Distance
Open the simulation in HTML file format.
About the simulation interface.
View the simulation in different scenarios.
Use checkboxes to show and hide the details.
Represent integers on the number line using the variables.
Differentiate between absolute value and directed distance.
Swap the variables to change the directed distance.
Calculate the difference between two integers.
Calculate the distance between two integers.
Switch between horizontal and vertical number lines. | 677.169 | 1 |
Angles
Geometric Angles Lesson Plan: What's the Measure on that Angle?
Posted by allisyn on
June 5, 2012
In this lesson plan, which is adaptable for grades 3 through 5, students use BrainPOP resources to use and apply vocabulary associated with geometric angles. Students will identify examples of differe... | 677.169 | 1 |
A ray of light is inclined to one face of a prism at an angle of 60∘. If angle of prism is 60∘ and the ray deviated through an angle of 42∘ find the angle which the emergent ray makes with second face of the prism.
Video Solution
Text Solution
Verified by Experts
The correct Answer is:18∘
Here i1=90∘−60∘=30∘,A=60∘, δ=426∘,(90−i2)=? Use i1+i2=A+δ Remember that i1,i2 are angles with normal to the faces of the prism.
Knowledge Check
A ray of light is incident at an angle of 60∘ on one face of a prism of angle 30∘. The ray emerging out of the prism makes an angle of 30∘ with the incident ray. The emergent ray is
A0∘
B90∘
C30∘
D45∘
Question 2 - Select One
A ray of light is incident on one of the faces of the angle prism with an angle 40∘ with the surface. The angle of the prism is 60∘. The emergent ray is deviated through an angle 38∘. The angle of emergence is
A38∘
B48∘
C52∘
D58∘
Question 3 - Select One
A ray of light is incident on one face of a prism at an angle of 50∘ with the normal . The ray is deviated by the prism through 42∘ . What is the angle of the prism , if the angle of emergence is 51∘ ? | 677.169 | 1 |
Vectors in Component Form - Video Tutorials & Practice Problems
Position Vectors & Component Form
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Welcome back everyone. So up to this point, we spent a lot of time talking about vectors and what they look like. Now we're called that visually vectors are these arrows drawn in space. And we've talked about how these arrows can be stretched or shrink or can be combined with other arrows to create result in vectors. Well, what we're gonna be talking about in this video is position vectors and component form. And this might sound a bit random and confusing but don't sweat it because it turns out all we're really going to be learning about in this video is a simple way to write numbers that represent our vector. And I think you're going to find mathematically it's actually very intuitive. So without further ado let's get right into things because this is an important concept to understand. Now, let's say we have this vector which we'll call vector V as you can see this is an arrow or drawn in space. Now, what we can do is represent this as a position vector and a position vector is simply a vector where the initial point is drawn at the origin. So if we want to draw vector V as a position vector, I just need to relocate this. So the initial point is at the origin and that right there is the position vector and that's all there is to it, it's just moving your vector to the origin of the graph. Now, the question becomes, how can we write this vector with some kind of numbers? Well, we can see that we have some numbers here on the graph and we can use these numbers to figure out what our vector is because what we do is we represent these vectors using component form where we have an X component and a Y component. And all these components do is tell you the length of the vector in the X and Y directions. So we can see in the X direction we need to go three units to the right. So we'd have three and then in the Y direction, we need to go two units up. So we'd have two. So this vector is three comma two. And that's all there is to it. As you can see, it's really straightforward. Now it turns out that there are also ways that you can represent these vectors if you don't have a position vector. So say that you were given some initial point of the vector like a point right there. And then a terminal point over here, you could figure out what the vector is in component form by using this equation down here. And to really put this equation to use and understand it rather than just looking at it. Let's actually try an example where we have to do this. So in this example, we are told if a vector has initial 0.23 and a terminal 0.35 without drawing the vector, write the vector in component form. So we're not allowed to just graph this immediately and figure out what it looks like. What we need to do is use this equation to figure out what our vector is. But this equation is actually pretty simple to use. So all I need to do is recognize that our vector V is going to be the difference in the X components common, the difference in the Y components. So we can see that we have the final X minus the initial X and then we're going to have the final Y minus the initial Y. Now I can see what these values are based on the points above. So if I go ahead and go to this first point and see this first point is 23, I can see that the second point is 35. So what I'm going to have for the X points is the difference between three and two. So we're gonna have three minus two. And the reason that I put three first is because there, it's the final X minus the initial X, it's gonna be 0.2 minus 0.1. So we have three minus two and then we're going to subtract the Y values. So the final Y is five and then the initial Y is three. So this is what our vector is going to be. So we're gonna have three minus two, which is one and we're going to have five minus three, which is two. And that right there is the solution that is vector B. So this is how you can solve problems when you can't initially draw them or don't initially have some kind of graph of the vector. Now, if we want to know what this vector looks like, we actually can use this graph just for reference. Well, I can see that our vector is 12. And if we draw this as a position vector starting at the origin of our graph, we're going to go one to the right and we're going to go two up. And that right, there would be our vector V. So as you can see whenever you're dealing with these types of vectors, you're going to have the X component, which is how far we travel in the X direction and the Y component, which is how far we travel in the Y direction. And that's always going to be the case when using component form. So that is how you can represent vectors using numbers and how you can draw position vectors which are at the origin of your graph. So hope you found this video. Helpful. Thanks for watching.
2
Problem
Problem
True or false: If a⃗=⟨3,2⟩a ⃗=⟨3,2⟩a⃗=⟨3,2⟩ and b⃗b ⃗b⃗ has initial point (3,−1)(3,-1)(3,−1) & terminal point (6,1)(6,1)(6,1 given information
3
Problem
Problem
True or false: If a⃗=⟨3,−2⟩a ⃗=⟨3,-2⟩a⃗=⟨3,−2⟩ and b⃗b ⃗b⃗ has initial point (−10,5)(-10,5)(−10,5) & terminal point (−7,3)(-7,3)(−7,3 the given information
4
example
Position Vectors & Component Form Example 1
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Let's see if we can solve this example. So in this example, we're told if vector V has initial 0.43 and terminal 0.12 write vector V in component form and sketch its position vector. Now our first step is going to be to find vector V in component form. And we can do that using this equation. It's going to be X two minus X one comma Y two minus Y one. And this will give us the vector we're looking for. Now the X two is going to be the final X or basically the terminal X and I can see the terminal point here one is going to be that X value. So we're going to have one minus the initial X, which we can see is four. Now next we're going to have Y two minus Y one. Now Y two is going to be the Y value for the terminal point which is two and then we're going to subtract off the Y value from the initial point which is three. So this is the vector we end up getting now to solve this. What we can do is we can do the subtractions here. So we're going to have one minus four, which is negative three and we're going to have two minus three, which is negative one. So the vector is negative three, negative one. And that is the vector that we're looking for in component form. Now we're also asked to sketch the position vector and the position vector is just going to be this vector that starts at the origin of the graph. So if I start here at the origin, I can see we're at negative three so that we can go three units to the left. And then we're going to be at negative one, which is one unit down. And this right here would be our vector V. So notice how it goes backwards, it goes towards the negative side of the X axis instead of the positive side because our vector is negative three negative one. So this is how you can find the vector and sketch its corresponding position vector. I hope you found this video helpful. Thanks for watching.
5
concept
Finding Magnitude of a Vector
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4m
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Welcome back everyone. So in the last video, we talked about how to write vectors using component form, which ends up looking something like this where we have these brackets. Now, what we're gonna be talking about in this video is how you can use this component form. We learned about to find the magnitude of a vector. Now, the magnitude is something that we've talked about in previous videos, recall that it describes how much of something you have or basically when it comes to vectors, it describes the total length of the vector itself. And it turns out finding this length, there's actually a pretty straightforward way that you can calculate a number to represent it. So without further ado, let's get right into an example of this because this is an important concept to understand. Now, let's say that we have this vector down here and we want to find the magnitude. Well, our first step should be to figure out what this vector is in component form. And we can do that by just taking a look at our graph. So I can see that we need to find the X and Y components. Now, the X component is going to be four units to the right. So our X component is four and our Y component is going to be three units up. So our vector is four comma three and that's it, that's the vector in component form. But how could we use these components to figure out the length of this vector? Could you think of some sort of way that we could do that? Well, I want you to recall something we learned about called the Pythagorean theorem. The Pythagorean theorem is an equation that we use on right triangles. And what it does is relates all the sides of a triangle together. So we have a squared plus B squared equals C squared where C is the hypotenuse or alongside of the triangle, then you can rearrange this equation to get that C is equal to the square root of A squared plus B squared. And plugging in the two legs of the triangle will allow you to calculate the hypotenuse. It turns out we can actually use this exact same equation and the same strategy to find the magnitude of any vector because notice something these vectors here, they form a right triangle, these two vectors are perpendicular. And then this would just be the hypotenuse. So if we want to find the magnitude or length of this vector, all we need to do is take the square root of the X component squared and the Y component squared and add them together because that's the same thing as having these two legs of the triangle and treating them like the opposite and adjacent side. So let's go ahead and do that. So we have this vector magnitude that we're trying to calculate. And it's going to be the square root of our X component squared, which is four squared plus our Y component squared, which is three squared. Now four squared is 16 and three squared is 9, 16 plus nine is 25 and the square root of 25 is five. So this right here is the magnitude or length of our vector. It's five units long. And that is how you can calculate this using this pythagorean theorem that we've already learned about. So you can just use this equation whenever you have your vector in component form. Now to make sure we understand this. Well, let's actually try an example where there's not a nice graph given to us. So in this example, we're asked to calculate the magnitude of vector PQ if we're given the initial point or some point P 12 and the final 0.53. Now to figure this out, recall that we learned this equation in the previous video which allows us to calculate our vector in component form if we're given two points. Now this is going to be the final point which we'll call 0.2. And this is going to be the initial point point one So let's go ahead and use this equation. We have that our vector V is going to be equal to the final X minus the initial X comma, the final Y minus the initial Y. Now, what I'm going to do is plug the numbers in. So we're going to have five and one which are the ini initial X components. So we'll have five minus one and that's going to be comma and then we're going to have subtracting the Y component. So we're going to have three minus two. So this is what our vector is going to be. Now, five minus one is four and three minus two is one. So this right here is our vector in component form. Now, since we've calculated the vector in component form, recall that we can now use this equation to figure out what the magnitude of our vector is. So the magnitude of our vector is going to be the X component squared plus the Y component squared. And all of that is going to be underneath a square root. So let's go ahead and plug the values in. So we're going to have the square root of our X component squared, which is four plus our Y component squared, which is one now four squared that comes out to 16 and one squared is just 1, 16 plus one is 17. And we can't simplify the square root down any further. So the magnitude of our vector is 17. And that is the solution to this problem. So this is how you can find the magnitude of any vector. And sometimes you will be given the vector in component form. Other times you need to figure out what the vector is in component form. But you could always use this equation when you're trying to calculate it. So hope you found this video helpful. Thanks for watching.
If vector v⃗v ⃗ v⃗ has initial point (−1,2)(-1,2)(−1,2) and terminal point (9,5)(9,5)(9,5), calculate the magnitude ∣v⃗∣|v ⃗ |∣v⃗∣.
A
2√292√292√29
B
√109\surd109√109
C
√73\surd73√73
D
√113\surd113√113
8
example
Finding Magnitude of a Vector Example 1
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Let's give this problem a try. So in this problem, we're told if vector V has initial 0.12 and terminal 0.44 sketch V is a position vector and calculate its magnitude. Now to solve this problem, the first thing I'm going to do is figure out what vector V is in component form. This is always a good first step when you're dealing with these types of problems where you're given the initial and terminal point of a vector. Now to solve this, define this component form. What I can do is take the final X and subtract off the initial X and then take the final Y and subtract off the initial Y and that's going to give us our vector. Now, first off, we can see that for X two, this is gonna be the X value for the terminal point which is four. So what I can first do is plug in four to this equation. Now, next, I'm going to subtract off the initial X and the initial X is one. So we're going to have four minus one for the difference in the X's. Now, as for the Ys, I can see that we have Y two and Y two is going to be this Y value for the terminal point which is four. And then this is going to be minus the initial Y and the initial Y is two. So we're going to have four minus one, which is three comma four minus two, which is two. So this right here is the vector. So now that we have our vector in component form, what we can do is we can now sketch the position vector. And the position vector just means that our vector is going to start at the origin. So if I go here at the origin of our graph, I can see that our vector is three. So that means on the X axis, we're going to go over here 123 and I can see that our Y value is two. So that means we're going to go up two. So this is what our vector is going to look like. And that's vector V sketched as a position vector. So we've now found the component form and we've sketched our position vector. And our last step is going to be to calculate the magnitude of this vector and to calculate the magnitude, we can use this equation. The magnitude of V is equal to the square root of VX squared plus VY squared where VX is the X component and VY is the Y component. So we can see that we have these values that we calculated already. So we're going to have a VX which is three. So we have three squared plus two squared now three squared is nine. So we're gonna have the square root of nine plus two squared, which is four and nine plus four is 13. So the magnitude of our vector V is equal to the square root of 13. And this is the magnitude of our vector as well as our vector sketched as a position vector. And that is the solution to this problem. So hope you found this video helpful. Thanks for watching.
9
concept
Algebraic Operations on Vectors
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Welcome back everyone. So in previous videos, we've talked about how you can add or subtract vectors using the tip to tail method. So recall if we add one vector, we could add the tip of that vector to the tail of a second vector and then the resulted vector that connected the two points would give you the sum of those vectors. This should be reviewed from previous videos. Now it turns out there is also a way that you can add or subtract vectors if you are given numbers as opposed to just a graph. And that's what we're gonna be talking about in this video. This is a very important skill to have in this course as well as likely future math or science courses. So let's just go ahead and get right into things. Now, if you have two vectors and you're given them in this component form, the way that you can add or subtract these vectors is by adding or subtracting the individual X and Y components. So to understand this, let's say that I have vector V and I want to add or subtract it to vector U. Well, what I just need to do is add or subtract the individual components. So I can add or subtract the X components. So we'll have VX plus or minus UX and then likewise, I can add or subtract the individual Y components. And that's all there really is too adding or subtracting vectors. So let's go ahead and try it with this example down here. So here we have vector 23 and we wish to add it to vector three negative one. Well, to do this, I'm first going to add the X components which are two and three. And then I'm going to add the Y components which are three and negative 12 plus three is five and three plus negative one is positive two. So this right here is the vector V plus U and that's all there is to it, that's the solution. Now it turns out if you actually use the tip to tail method on these vectors, you're going to find that you actually will get this result. So let's say I have vector 23, which I can draw on this graph right here. And then let's also say I have this vector three negative one. So I can draw this right about there. So we have vector V and here we have vector U. And if I go ahead and use the tip to tail method where I take the initial point of this, of the first vector and connect it to the terminal point of the second vector, notice how this is the vector that we get for V plus U. But notice how this vector ended up being +12345 units to the right and it ended up being +12 units up. So we get the exact same result if we use this tip to tail method. So as you can see the math actually checks out here when we this visually. Now something else I do want to touch on is this idea of multiplying a vector by a scalar, which is something else we've discussed, you can do this with the numbers too. But I think the best way to understand this is going to be to try an example where we have to do this. So let's say we have this situation where we have two vectors V and U and we're asked to find the vector V minus three U. Now to solve this, what I'm first going to do is write out what we have. So we can see that we have vector V and that's given to us it's 85. Now, what I also see is that we have vector U, but what I need to do is figure out what vector three U is to define vector three U. Well, that means I need to take our vector U which is 24 and I need to multiply it by three. But how exactly do I do this? Well, it turns out if you want to multiply a vector by a scalar, all you need to do is distribute the scalar into each of the individual X and Y components. So let's say that we have some scalar constant K. All I need to do is multiply this by the X and the Y component. So we'll have K times the X component and then K times the Y component. So going down here, I can just distribute this three into each component. So we're going to end up with three times two, comma three times four. Well, three times two is six and three times four is 12. So this is the vector three U. So we have vector three U and we have vector V. Now, all I need to do from here is figure out what vector V minus three U is and to find V minus three U. Well, I just figured out what V and three U are. So I just subtract the two vectors. We said that V is 85 and we said that U is 6, 12 or three U I should say. So we end up with these two vector and I just need to subtract them. So what we're going to do is subtract the individual components. So we'll have eight minus six, which comes out to two and then we're going to have five minus 12 which comes out to negative seven. So this here is the solution to the problem. And this is how you can do these basic operations with vectors like adding or subtracting vectors or multiplying vectors by scalers when you're dealing with numbers rather than a graph. So hope you found this video helpful. Thanks for watching.
Algebraic Operations on Vectors Example 1
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Let's see if we can solve this example. So in this example, we're told if vector A equals 31 and vector B equals negative 49 and vector C is equal to five times vector A minus two times vector B, we're asked to calculate the magnitude of vector C. Now, what I can see here is that we're trying to ultimately find the magnitude of this vector. But this vector depends on the other two vectors we have in this problem. So we definitely have our work cut out for us here. But whenever solving these problems where we have multiple vector operations, I personally like to start small with the problem and to work my way to solving the whole thing. So let's actually do that. Now, the first thing I'm gonna do is figure out what five times vector A is because I'm gonna really be focusing on this vector C here and to find five times a well, five times vector A is just going to be five times this vector. So we won't have five times 31. Now five times three, we can multiply this scalar into each of the components. So we have five times three, which is 15 and then we'll have five times five, which is five. So this is the vector five A. Now next, I'm going to figure out what vector two B is. So vector two B is going to be two times vector B. Well, I can see here that vector B is negative 49. So that's gonna be our vector. And then what I need to do is distribute the scalar into this vector as well. So we're gonna have two times negative four, which is negative eight. And then we're going to have two times nine, which is 18. So this is vector five A, five times A and vector two times B. And what I need to do from here, they need to subtract these two vectors. So to find vector C, it's going to be five A minus two B. So we can see here that five A is 15 5. And then we can see that vector two B is negative 818. And then what I can do is I can subtract these two vectors. So if I subtract them, we're going to have 15 minus negative eight and 15 minus negative eight is the same thing as 15 plus eight, which is 23. And then what I'll have is I'll have five minus 18 which comes out to negative 13. So this right here is vector C and now that I've found vector C, our last step is to just find the magnitude of C, we can use the Pythagorean theorem. And the Pythagorean theorem for these vectors is going to be the square root of the X component of C square plus the Y component of C square. Now, what I can see here is that the X component is going to be 23. So we're going to have 23 squared plus the Y component squared, which is negative 13 squared. So this is what we get now 23 squared that turns out to be 529 and negative 13 squared. Where if you, well, if you square a negative number, you're gonna get a positive number and negative 13 squared comes out to positive 100 69. So we're going to have 500 29 plus 169 which comes out to 698. And it turns out this is actually not a radical that you can simplify down any farther. So the magnitude of C is the square root of 698. And that is the solution to this problem. So this is how you can handle multiple vector operations as well as finding the magnitude of a vector once you've already done operations on that vector. So I hope you found this video helpful. Thanks for watching. | 677.169 | 1 |
Geometry of 2D Shapes [Foundation]
In this course, we will be looking at 2D shapes and their interesting characteristics. We will investigate triangles and quadrilaterals, and learn how to recognise shapes through their properties. Each type of quadrilateral and triangle has rules that help us find the missing sizes of angles and lengths of sides. We will consolidate our knowledge at the end of the course by practising what we have learnt. | 677.169 | 1 |
Grafika Inżynierska D because wavy lines are used to represent breaks or interruptions in objects when they are drawn manually.
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4.
Grubości linii cienkiej, grubej i bardzo grubej to:
A.
A. 0.18 mm ; 0.36 mm ; 0.54 mm
B.
B. 0.18 mm ; 0.5 mm ; 1.0 mm
C.
C. 0.18 mm ; 0.7 mm ; 2.0 mm
D.
D. Żadna z odpowiedzi nie jest prawdziwa
Correct Answer B. B. 0.18 mm ; 0.5 mm ; 1.0 mm
Explanation The correct answer is B. 0.18 mm ; 0.5 mm ; 1.0 mm. This is because the question asks for the thickness of thin, thick, and very thick lines. The measurements given in option B match these criteria, with 0.18 mm being the thickness of a thin line, 0.5 mm being the thickness of a thick line, and 1.0 mm being the thickness of a very thick line.
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5.
Tabliczka rysunkowa powinna być umieszczona:
A.
A. W prawym dolnym narożniku rysunku
B.
B. W lewym dolnym narożniku rysunku
C.
C. Miejsce jest dowolne
D.
D. Żadna z odpowiedzi nie jest prawdziwa
Correct Answer A. A. W prawym dolnym narożniku rysunku
Explanation The correct answer is A. W prawym dolnym narożniku rysunku. The explanation for this is that the question is asking about the placement of a drawing table, and the correct answer states that it should be placed in the bottom right corner of the drawing. This suggests that there is a specific location for the drawing table and it is not arbitrary or irrelevant.
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6.
Pismo techniczne pochyłe rysujemy pod kątem:
A.
A. 45 stopni od poziomu
B.
B. 90 stopni od poziomu
C.
C. 25 stopni od poziomu
D.
D. 75 stopni od poziomu
Correct Answer D. D. 75 stopni od poziomu
Explanation In technical drawing, a "pismo techniczne pochyłe" refers to an oblique lettering style. The question asks at what angle this oblique lettering should be drawn from the horizontal level. The correct answer is D. 75 degrees from the horizontal level. This means that the oblique lettering should be slanted at a 75-degree angle from the horizontal line.
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7.
Podstawowym formatem rysunkowym w grafice inżynierskiej:
A.
A. Przedstawiony na żółto format A4
B.
B. Format A1 z którego można wyodrębnić wszystkie pozostałe formaty
C.
C. Wszystkie formaty są formatami podstawowymi
D.
D. Żadna z odpowiedzi nie jest prawdziwa
Correct Answer A. A. Przedstawiony na żółto format A4
Explanation The correct answer is A because the question asks for the "basic drawing format in engineering graphics" and the answer states that it is "the yellow format A4". This suggests that the yellow A4 format is commonly used and considered as the standard or foundational format in engineering graphics.
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8.
Podstawowymi skalami zwiększającymi w grafice inżynierskiej są skale
A.
A. 50:1, 20:1, 10:1, 5:1, 2:1
B.
B. 50:1, 25:1, 15:1, 5:1, 2,5:1
C.
C. 50:1, 25:1, 20:1, 15:1, 10:1, 5:1, 2:1
D.
D. 50:1, 40:1, 30:1, 20:1 10:1, 5:1 2.5:1
Correct Answer A. A. 50:1, 20:1, 10:1, 5:1, 2:1
Explanation The correct answer is A. 50:1, 20:1, 10:1, 5:1, 2:1. This answer is correct because it lists the scales in descending order, starting with the largest scale (50:1) and ending with the smallest scale (2:1). These scales are commonly used in engineering graphics to represent the ratio between the size of an object on paper and its actual size.
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9.
Czy na rysunku powyżej istnieją błędy wymiarowania?
A.
A. Tak, wszystkie groty strzałek wymiarowych powinny być zaczernione
B.
B. Tak, nie umieszcza się linii wymiarowych wewnątrz przedmiotu
C.
C. Nie, nie ma błędów
D.
D. Tak, zamknięto łańcuch wymiarowy
Correct Answer C. C. Nie, nie ma błędów
Explanation The correct answer is C, which states that there are no errors in the dimensioning. This means that there are no mistakes or inaccuracies in the way the dimensions are represented in the drawing.
Explanation The correct answer is D. This is because the given drawing is dimensioned correctly according to the acceptable standards and practices.
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12.
Na powyższym rysunku linią punktową oznaczono:
A.
A. Podwójną dźwignię wyłącznika krańcowego
B.
B. Niewidoczną część wyłącznika krańcowego
C.
C. Skrajne położenie wyłącznika krańcowego
D.
D. Żadna z odpowiedzi nie jest prawdziwa
Correct Answer C. C. Skrajne położenie wyłącznika krańcowego
Explanation The correct answer is C. Skrajne położenie wyłącznika krańcowego. This is indicated by the dotted line on the diagram, which represents the extreme position of the limit switch. The limit switch is a device that is used to detect the presence or absence of an object and is often used to control the movement of machinery or equipment. In this case, the dotted line indicates the maximum or minimum position that the limit switch can reach.
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13.
Podany wyżej przykład a) ; b) stanowią przykład rzutowania :
A.
A. Typu E (europejskiego)
B.
B. Typu A (amerykańskiego)
C.
C. Jest to przykład rzutowania aksonometrycznego
D.
D. Jest to jeszcze inny rodzaj rzutowania
Correct Answer A. A. Typu E (europejskiego)
Explanation The given example a) and b) are examples of projection type E (European).
Explanation The correct answer is C because there is an error visible in Projection III. The error is that the top surface of the object is not shown in the projection.
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15.
Powyższy sposób rzutowania nosi nazwę:
A.
A. Rzutowania prostokątnego
B.
B. Rzutowania aksonometrycznego
C.
C. Rzutów Kochańskiego
D.
D. Jest to jeszcze inny rodzaj rzutowania
Correct Answer A. A. Rzutowania prostokątnego
Explanation The correct answer is A. Rzutowania prostokątnego. This is because the question asks for the name of the given method of projection, and "rzutowanie prostokątne" translates to "orthographic projection" in English. Orthographic projection is a method of representing a three-dimensional object in two dimensions by projecting lines from the object onto a plane.
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16.
Powyższy typ rysowania nosi nazwę:
A.
A. Rysunków złożeniowych
B.
B. Rysunków przekrojów aksonometrycznych
C.
C. Przekrojów złożeniowych
D.
D. Żadna z odpowiedzi nie jest prawdziwa
Correct Answer B. B. Rysunków przekrojów aksonometrycznych
Explanation The correct answer is B. Rysunków przekrojów aksonometrycznych. This is because "przekrojów aksonometrycznych" translates to "axonometric sections" in English, which refers to a type of drawing technique that combines both oblique and orthographic projections. This type of drawing is often used in technical and architectural drawings to show the internal structure of an object or building.
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17.
Przelotowe otwory na poniższym widoku i przekroju zostały:
A.
A. Pokazane bez błędów
B.
B. Błąd jest tylko w przekroju
C.
C. Błąd jest tylko w widoku
D.
D. Błędy są zarówna w widoku jak i przekroju
Correct Answer A. A. Pokazane bez błędów
Explanation The given correct answer is A. Pokazane bez błędów (Shown without errors). This suggests that the holes shown in both the view and the section are accurate and do not contain any mistakes or discrepancies.
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18.
Zakreskowany element A –A na rysunku powyżej przedstawia:
A.
A. Przekrój
B.
B. Widok
C.
C. Kład
D.
D. Żadna z odpowiedzi nie jest prawdziwa
Correct Answer C. C. Kład
Explanation The correct answer is C. Kład. The element A in the highlighted area represents a floor or a platform, which is commonly referred to as a "kład" in Polish. It is a horizontal surface that people can walk or stand on.
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19.
Podaj właściwy sposób użycia przekrojów:
A.
A. Górne rysunki a) b) c) dobrze
B.
B. Dolne rysunki a) b) c) dobrze
C.
C. Dobre rysunki a) dolne b) górne c) dolne
D.
D. Żadna z odpowiedzi nie jest prawdziwa
Correct Answer C. C. Dobre rysunki a) dolne b) górne c) dolne
Explanation The correct answer states that the proper way to use cross-sections is to use the lower drawings for options a) and c), and the upper drawing for option b). This means that for options a) and c), the cross-sections should be taken from the lower part of the object, while for option b), the cross-section should be taken from the upper part of the object.
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20.
Przerywane linie na widoku rysunku powyżej przedstawiają:
A.
A. Miejsca na założenie uszczelek montażowych
B.
B. Powierzchnie podlegające obróbce powierzchniowej lub cieplnej
C.
C. Niewidoczne krawędzie przedmiotu
D.
D. Żadna z odpowiedzi nie jest prawdziwa
Correct Answer C. C. Niewidoczne krawędzie przedmiotu
Explanation The interrupted lines in the view drawing above represent invisible edges of the object. These lines indicate edges that are not visible from the current viewpoint but are present in the object. They are used to provide a complete representation of the object's shape and structure.
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21.
Rysunek poglądowy reduktora obrotów został wykonany w:
A.
A. Rzucie prostokątnym
B.
B. Rzucie aksonometrycznym prostokątnym
C.
C. Rzucie środkowym (geometrii wykreślnej)
D.
D. Żadna z odpowiedzi nie jest prawdziwa
Correct Answer B. B. Rzucie aksonometrycznym prostokątnym
Explanation The correct answer is B. Rzut aksonometryczny prostokątny. This is because an axonometric projection is a type of orthographic projection that shows an object in three dimensions. In this type of projection, the object is rotated to show all three axes (length, width, and height) equally foreshortened. A rectangular axonometric projection specifically shows the object in a rectangular frame, with the three axes forming equal angles with the plane of the frame.
Explanation The correct answer is C because it accurately describes the definition of "kład" as the outline of a flat figure located in the plane of the cross-section of an object, rotated along with that plane by 90º and positioned either within or outside the object's outline.
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23.
Podaj zalecany sposób wymiarowania widoków
A.
A. Rysunek a - dobrze
B.
B. Rysunek b - dobrze
C.
C. Obie formy dopuszczalne
D.
D. Obie formy nie są dopuszczalne
Correct Answer B. B. Rysunek b - dobrze
Explanation The recommended way of dimensioning views is shown in drawing b.
Explanation The correct answer is B. Rysunki dolne dobrze. This means that the lower drawings are correct. The question is asking for the correct way to use cross-sections in the above screw and coupling connections. Since the drawings in the lower part of the question are labeled as "dobrze," which means "good" or "correct" in Polish, it can be inferred that these drawings show the proper use of cross-sections in the given connections.
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25.
Oceń sposoby wymiarowania:
A.
A. Rysunek a) – źle zwymiarowany; rysunek b) dobrze
B.
B. Rysunek b) – źle zwymiarowany; rysunek a) dobrze
C.
C. W obydwu rysunkach są błędy
D.
D. Obydwa typy wymiarowania są dopuszczalne
Correct Answer A. A. Rysunek a) – źle zwymiarowany; rysunek b) dobrze
Explanation Drawing a) is poorly dimensioned, while drawing b) is well dimensioned.
Explanation The correct answer is B because it states that in drawing b, the most dimensions are given in the section. This implies that the section view provides the most detailed and accurate representation of the dimensions in the drawing.
Explanation The correct answer is C because it states that examples b) and e) contain threads with preferred pitches.
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30.
Otwór " a" na powyższym rysunku:
A.
A. Jest nagwintowany do połowy
B.
B. Posiada kołek wkręcony do połowy
C.
C. Stanowi fragment połączenia kielichowego
D.
D. Żadna z odpowiedzi nie jest prawdziwa
Correct Answer A. A. Jest nagwintowany do połowy
Explanation The correct answer is A. Jest nagwintowany do połowy (It is half-threaded).
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31.
Powyższy rysunek przedstawia w I stopniu uproszczenia :
A.
A. Połączenie śrubowe
B.
B. Układ śrubowy ze sprężyną
C.
C. Połączenie teleskopowo – sprężynowe
D.
D. Żadne z w/w połączeń
Correct Answer B. B. Układ śrubowy ze sprężyną
Explanation The above drawing represents a simplified level I:A. Screw connectionB. Screw system with a springC. Telescopic-spring connectionD. None of the above connectionsThe correct answer is B because the drawing shows a combination of screws and a spring, indicating that it is a screw system with a spring.
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32.
Powyższy rysunek przedstawia:
A.
A. Kołek do połączeń wciskowych
B.
B. Fragment tulei cienkościennej
C.
C. Ślimak (pojedynczy) z połączenia ślimak - ślimacznica
D.
D. Fragment sworznia nagwintowanego
Correct Answer D. D. Fragment sworznia nagwintowanego
Explanation The given drawing represents a fragment of a threaded pin, also known as a threaded shaft or bolt. The shape and features of the object in the drawing closely resemble those of a threaded pin, indicating that it is most likely a fragment of a threaded pin.
Explanation The correct answer is C. Łożysko kulkowe w przekroju i pierwszym stopniu uproszczenia. This is because the image shows a cross-section of a ball bearing, which is a type of rolling element bearing used to support rotational motion. The simplified representation in the image only shows the first stage of the bearing, where the balls are located between the inner and outer raceways.
Explanation The correct answer is C. Linia gwintu w połączeniu gwintowym (Thread line in threaded connection). The explanation for this answer is that the given description mentions a 3/4 circle drawn with a thin line inside the lower part of the construction element. This indicates the presence of a threaded connection, where the thread line is represented by the 3/4 circle. Therefore, option C is the correct answer.
Explanation In the given question, the correct answer is D. W przykładach a) oraz b) - prawe rysunki są dobrze wykonane. This means that the correct way of drawing the cross-sections of the given connections is shown in the right drawings of examples a) and b). The left drawings in both examples are not correctly executed.
Explanation The correct answer is C. Moduły obydwu kół zębatych muszą być takie same. This means that in order for two gears to work together without interference, the modules of both gears must be equal. The module of a gear is a measure of the size of its teeth and determines the spacing and size of the teeth on the gear. If the modules of the gears are not the same, the teeth will not mesh properly and will cause interference or incorrect operation. Therefore, for smooth and efficient operation, the modules of both gears must be the same.
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37.
Moduł koła zębatego jest:
A.
A. Iloczynem średnicy podziałowej i ilości zębów
B.
B. Sumą średnicy podziałowej i ilości zębów
C.
C. Ilorazem średnicy podziałowej i ilości zębów
D.
D. Różnicą średnicy podziałowej i ilości zębów
Correct Answer C. C. Ilorazem średnicy podziałowej i ilości zębów
Explanation The correct answer is C because the module of a gear is calculated by dividing the pitch diameter of the gear by the number of teeth.
Explanation The symbol (e) on the drawings indicates that there should be gaps in the welding of lengths "e".
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39.
Powyższy rysunek przedstawia przykład połączenia:
A.
A. Zszywanego ( korpus górny do korpusu dolnego)
B.
B. Zgrzewanego ( korpus górny do korpusu dolnego)
C.
C. Spawanego ( korpus górny do korpusu dolnego)
D.
D. Żadnego z wymienionych połączeń
Correct Answer C. C. Spawanego ( korpus górny do korpusu dolnego)
Explanation The above drawing shows an example of a welded connection, where the upper body is joined to the lower body.
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40.
Powyższe rysunki przedstawiają:
A.
A. Wały o zmiennym przekroju
B.
B. Sprzęgła
C.
C. Różne rodzaje siłowniki
D.
D. Żadna z odpowiedzi nie jest poprawna
Correct Answer B. B. Sprzęgła
Explanation The drawings above represent different types of couplings.
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41.
Do obliczenia modułu koła zębatego będzie potrzebna ilość zębów oraz:
A.
A. Średnica okręgu podstaw zęba – df
B.
B. Średnica podziałowa – d
C.
C. Podziałka – p
D.
D. Wysokość całkowita zęba ha + hf
Correct Answer B. B. Średnica podziałowa – d
Explanation To calculate the module of a gear, the number of teeth and the pitch diameter (diameter of the pitch circle) are needed. The pitch diameter is the diameter of the imaginary circle that passes through the center of the gear teeth and is used to determine the gear ratio. The pitch diameter can be calculated using the pitch diameter formula, which includes the pitch diameter (d), the number of teeth (z), and the module (m). Therefore, the correct answer is B. Średnica podziałowa – d (Pitch diameter - d).
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42.
Wymiar końcowy wałka zawiera informację φ180 e8 .
A.
A. Wartoś dopuszczalnego bicia promieniowego
B.
B. Wartos dopuszczalnego luzu wierzchołkowego
C.
C. Tolerancja i pasowanie wałka według - e8
D.
D. Żadna odpowiedź nie jest właściwa
Correct Answer C. C. Tolerancja i pasowanie wałka według - e8
Explanation The given correct answer, C. Tolerancja i pasowanie wałka według - e8, means that the final dimension of the shaft is specified with a tolerance and fit of - e8. This indicates that the shaft must be manufactured within certain limits to ensure proper fit and function with other components. The e8 tolerance and fit standard provides specific guidelines for the size and fit of the shaft, ensuring that it will work correctly in the intended application.
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43.
Znak w prawym górnym narożniku rysunku oznacza:
A.
A. Klasę chropowatości
B.
B. Odchyłkę górną w tolerancji i pasowaniu
C.
C. Nakaz skrawania (nakaz obróbki poprzez skrawanie)
D.
D. Żadna odpowiedź nie jest właściwa
Correct Answer A. A. Klasę chropowatości
Explanation The symbol in the upper right corner of the drawing represents the roughness class.
Explanation The symbols Ra1.25, Ra10, Ra5, etc. on the engineering drawing indicate the height of surface roughness. This means that they represent the average distance between the highest and lowest points on the surface of the pump cover. These symbols are used to specify the desired level of smoothness or roughness required for the surface finish.
Explanation The correct answer is D. Tolerancję równoległości dwóch płaszczyzn. This is because the image (b) shows two parallel planes, indicating the tolerance for parallelism between the two surfaces.
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46.
Przykład powyższy rysunku złożeniowego zawiera
A.
A. Łożyska ślizgowe
B.
B. Podwójne i pojedyncze uszczelnienia mechaniczne trzonu wału
C.
C. Łożyska ślizgowe wraz z uszczelnieniami
D.
D. Łożyska toczne
Correct Answer D. D. Łożyska toczne
Explanation The given correct answer is D. Łożyska toczne (rolling bearings). This suggests that the above diagram includes rolling bearings, which are used to reduce friction and support rotating parts in machinery. Rolling bearings consist of two rings with rolling elements (such as balls or rollers) between them, allowing for smooth and efficient rotation.
Explanation The answer D states that none of the drawings depict the flatness tolerance of machine parts. This means that all the drawings, including a), b), c), and d), do not represent the flatness tolerance.
Explanation The given correct answer is C. Bicia osiowego (wzdłużnego) części maszyn względem osi otworu mierzonej na średnicy 100 mm. This means that the tolerance shown in the diagram represents the axial (longitudinal) displacement of machine parts relative to the axis of the hole measured on a diameter of 100 mm.
Explanation The correct answer is B because the drawing shows two parallel planes with a tolerance for perpendicularity. The lines in the drawing indicate that the two planes should be perpendicular to each other within a certain tolerance. This means that the two planes should be at right angles to each other, with no significant deviation from perpendicularity. | 677.169 | 1 |
Introduction
Angles play a crucial role in geometry, and one important concept to understand is the converse of corresponding angles. In this article, we will review the converse of corresponding angles and its significance in geometry. Whether you are a student brushing up on your math skills or a teacher looking for a refresher, this article will provide you with a comprehensive review of the topic.
Understanding Angles
Before diving into the converse of corresponding angles, let's first understand what angles are. An angle is formed when two rays share a common endpoint, known as the vertex. Angles are typically measured in degrees, ranging from 0 to 360. Understanding how to measure and classify angles is crucial when dealing with geometric problems.
Corresponding Angles
When two lines are intersected by a third line, eight angles are formed. Corresponding angles are pairs of angles that lie on the same side of the transversal and are in corresponding positions. In other words, they are in the same relative position in relation to the transversal and the two lines being intersected.
The Converse of Corresponding Angles
The converse of corresponding angles states that if two lines are intersected by a transversal and the corresponding angles are congruent, then the lines are parallel. This concept is based on the idea that if the corresponding angles are equal, then the lines must be parallel, as the angles formed will be alternate interior angles.
Proof and Examples
To understand the converse of corresponding angles better, let's look at a simple proof and some examples.
Proof of the Converse of Corresponding Angles
Let's assume we have two lines, line AB and line CD, intersected by a transversal line EF. If angle 1 is congruent to angle 2, and angle 3 is congruent to angle 4, we can prove that line AB is parallel to line CD.
By the definition of corresponding angles, angle 1 and angle 3 are corresponding angles, as well as angle 2 and angle 4. Since angle 1 is congruent to angle 2, and angle 3 is congruent to angle 4, we can conclude that line AB is parallel to line CD.
Example 1: Parallel Lines
Consider two lines, line PQ and line RS, intersected by a transversal line TU. If angle 1 is congruent to angle 2, and angle 3 is congruent to angle 4, we can conclude that line PQ is parallel to line RS.
This example illustrates the application of the converse of corresponding angles. If the corresponding angles are congruent, the lines must be parallel.
Example 2: Non-Parallel Lines
Now, let's consider two lines, line WX and line YZ, intersected by a transversal line AB. If angle 1 is congruent to angle 2, but angle 3 is not congruent to angle 4, we can conclude that line WX is not parallel to line YZ.
This example demonstrates that if the corresponding angles are not congruent, the lines cannot be parallel.
Applications in Geometry
The converse of corresponding angles is a fundamental concept in geometry and has various applications. Some of the key applications include:
Proving Parallel Lines
One of the significant applications of the converse of corresponding angles is in proving whether two lines are parallel or not. By examining the congruence of corresponding angles, we can determine the nature of the lines.
Constructing Parallel Lines
The converse of corresponding angles can also be used to construct parallel lines. By ensuring the congruence of corresponding angles, we can construct lines that are parallel to a given line.
Solving Geometric Problems
Understanding the converse of corresponding angles helps in solving a wide range of geometric problems. Whether it's finding missing angles or proving theorems, this concept is an essential tool in geometry problem-solving.
FAQs
1. What are corresponding angles?
Corresponding angles are pairs of angles that lie on the same side of the transversal and are in corresponding positions. They are congruent if the lines are parallel.
2. What is the converse of corresponding angles?
The converse of corresponding angles states that if two lines are intersected by a transversal and the corresponding angles are congruent, then the lines are parallel.
3. How can I prove that two lines are parallel using corresponding angles?
If the corresponding angles are congruent, you can conclude that the lines are parallel.
4. Can corresponding angles be congruent if the lines are not parallel?
No, corresponding angles are congruent if and only if the lines are parallel. If the lines are not parallel, the corresponding angles will not be congruent.
5. What are some real-life applications of the converse of corresponding angles?
The converse of corresponding angles is used in various fields, such as architecture and engineering, where parallel lines play a significant role in designing structures. | 677.169 | 1 |
This problem feel more complicated to me than usual. I am currently stuck with finding some values in the following case.
Here the image illustrate the basic of the problem. What the rules are :
A and B are always on the orange axis
AB,DE and CE are of known length
ABF, ADB and BEC are 90 degree
BF is infinite in that direction
What i need to find is what angle will make BF coincide with C ?
By making BF falling down, A move up a bit while B goes toward the left so that make 2 unknowns that for some reason confuse me too much.
You can see the problem as ABF is a partial drawing of a book on angle in a shelf and the orange lines are the side and bottom of the shelf. The book as to touch the side and bottom and i need to know the angle it hold on the vertical toothpick CE
I am a developer so my current solution is to iterate increasing the length of AD and calculating the triangle of ADB and since ABF is also square i have the triangle BEF and F being infinite i get a point perfect inline with CE finally i can check if my F is higher than C if yes bring up A again a bit more and test until F is below C and get somewhere close to the real answer.
$\begingroup$The problem, I suppose, confuses you because the confuguration you are looking for is not constructible and uses more variables. By constructible I mean everything could be calculated recursively, or inductively. Here you have defined the solution by lots of equations that hold in the final configuration, and not recursively. Use similar triangles as Sai suggested and translate your constraints as some equations involving solutions of polynomials. You should be able to make it one polynomial. The root of it would be the tanget of the angle you want.$\endgroup$
1 Answer
1
You're correct that there's a better way. However, first note you could more simply, and I believe also easily, explain your goal by stating that with $AB$ being a fixed length, you're trying to determine the position of $B$ on $DE$ so that $AB$ and $BC$ are perpendicular to each other. Thus, this would be equivalent to $C$ and $F$ coinciding. The diagram below shows this, along with a few specified lengths and angles:
Next, as suggested by donaastor's comment, using the relations among the similar triangles, plus the Pythagorean theorem, results in a system of several equations, which can be reduced to a polynomial in one of the unknown values. First, the Pythagorean theorem with $\triangle ADB$ gives that
Since $a$, $b$ and $c$ are given, this is a quartic equation in the variable $d$. Although \eqref{eq3A} can be solved analytically, as shown in the Wikipedia article, it's usually solved numerically instead, e.g., such as by using the Durand–Kerner method that's also suggested in the article. However, I believe basically all programming languages have either built-in functionality or third-party libraries you can use instead to find the roots (e.g., in Python, there's numpy.roots).
Regardless of how you solve \eqref{eq3A}, with $d$ determined, then $y$ can be calculated in several ways, e.g., by
However, especially if you try to solve this with your own code, be careful since quartic equations always have $4$ roots (due to the Fundamental theorem of algebra). I believe there'll usually be $2$complex conjugates, with the other $2$ being real roots. You want the one with $0 \lt d \lt c$, with the other one being negative, i.e., with $B$ to the left of $D$ (and $A$ below $D$), as shown below:
From \eqref{eq2A}, we then also have that $e \lt 0$, as the diagram above indicates as well.
I assume you always want $B$ to be between $D$ and $E$. If so, especially if you're using your own root-finding method, you should try to ensure you get the correct root (e.g., by choosing an appropriate starting value) or, at the least, add appropriate checking and handling in case you get an out of bound root (i.e., $d \le 0$ or $d \ge c$) instead.
$\begingroup$I have been at trying to simplify the formula at point #3 for about 4 hours so far but no result yet. I'll still try to go at it for a little while then fall back to the trial and error iterative way as i know i can get a result. Not a good one but a close one and I'll have to do extra 2d collisions to adjust for the errors of the approximate.$\endgroup$
$\begingroup$@Franck Thank you for the feedback, but I'm not clear on a few things. By "point #3", I assume you mean my equation $(3)$. Also, if by "simplify" you mean solving it analytically, I recommend you don't try to do that, as it's fairly complicated and easy to make mistakes. Instead, you should try to solve it numerically, such as using a third-party library (my updated answer suggests numpy.roots if using Python). One last thing to note, as well, is that $\arcsin$ in many computer languages return angles in radians, so you may need to convert to degrees, in case that's the issue.$\endgroup$
$\begingroup$Yes i meant the formula at (3). I have finally managed to find a solution for d but it's long as hell and i double checked with Wolfram and it output 4 different solutions. I'll try finding a library in my langage that can solve equations for me to remove the chance of mistake in typing the formula.$\endgroup$
$\begingroup$After long hours of work neither mine or WolframAlpha solve for the formula for d = ? at (3) works. I'll have to fall back on iterative to get a close value and integrate the collision detection to adjust to a good enough value$\endgroup$
$\begingroup$@Franck I'm sorry you weren't able to get using $(3)$ working properly. However, I don't understand why. I've checked, and rechecked several times, my logic and algebra. After your initial comment, I also manually checked a simple case where $x = y = 45^{\circ}$, with $a = \sqrt{2}$, $b = d = 1$ and $c = 2$. I've just now checked a couple of additional cases. First, I generalized my earlier case with $a = \sqrt{2}d$ and $c = d + b$. Using this in the LHS of $(3)$ gives $d^4 - 2(d+b)d^3 + (b^2 + (d+b)^2)d^2 - 2b^2d^2$. Collecting the terms in the various powers of $d$ together gives for ...$\endgroup$ | 677.169 | 1 |
Pentagonal prism: characteristics, parts, vertices, edges, volume
A pentagonal prism It is a three-dimensional geometric figure whose identical bases are pentagon-shaped, and also has a total of 5 parallelogram-shaped faces.
If the faces are rectangular, it is said to be a right pentagonal prismwhile if the edges are inclined with respect to the bases, then it is a oblique pentagonal prism. In the following image there is an example of each one.
The base pentagon can be regular if its five sides have the same measure, as well as the internal angles, otherwise it is an irregular pentagon. If the base of the prism is regular, it is regular pentagonal prism. otherwise it is a prism irregular pentagonal.
The pentagonal prism is a harmonious structure used in architecture and object design, such as the modern building shown in the figure above. The windows in the shape of an irregular pentagon form the base of the prisms.
[toc]
Characteristics of the pentagonal prism
-It is a three-dimensional geometric figure, the surfaces that compose it enclose a certain volume.
-Their bases are pentagons and their lateral faces can be rectangles or parallelograms.
-It has vertices -the corners of the prism- and edges -edges or edges-.
-If the edges that join the bases are perpendicular to them, the prism is straight, and if they are inclined, the prism is oblique.
-When the base is a pentagon whose internal angles are less than 180º, the prism is convexbut if one or more interior angles is greater than 180º, it is a prism concave.
Elements of the pentagonal prism
–Bases: It has two pentagonal and congruent bases -their measurements are the same-, either regular or irregular.
–faces: A pentagonal prism has a total of 7 faces: the two pentagonal bases and the five parallelograms that make up the sides.
–Edge: segment that joins two bases, shown in red in figure 3 or the one that joins two sides.
–Height: distance between the faces. If the prism is straight, this distance is equal to the size of the edge.
–Vertex: point in common between a base and two lateral faces.
The lower figure shows a right pentagonal prism with a regular base, in which the segments that form the base have the same measure, called to.
This type of prism also has the following elements, typical of the regular pentagon:
–radius R: distance between the center of the pentagon and one of the vertices.
–Apothem LA: segment that joins the center with the midpoint of one of the sides of the pentagon.
How many vertices does a pentagonal prism have?
In a pentagon there are 5 vertices and since the pentagonal prism has two pentagons as bases, this body has a total of 10 vertices.
How many edges does a pentagonal prism have?
The number of edges for geometric solids with flat faces, such as prisms, can be calculated using the Euler's theorem for convex polyhedra. Leonhard Euler (1707-1783) is one of the greatest mathematicians and physicists in history.
The theorem establishes a relationship between the number of faces, which we will call C, the number of vertices V, and the total number of edges A as follows:
C+V = A+2
For the pentagonal prism we have: C = 7 and V = 10. Solving for A, the number of edges:
A = C+V-2
Substituting values:
A = 7 + 10 – 2 = 15
A pentagonal prism has 15 edges.
How to get the volume of a pentagonal prism?
The volume of the pentagonal prism measures the space enclosed by the sides and the bases. It is a positive quantity that is calculated by the following property:
Any plane that cuts the prism perpendicular to its edges generates an intersection with the same shape as the base, that is, a pentagon with the same dimensions.
Therefore, the volume of the pentagonal prism is the product of the area of the base and the height of the prism.
Be AB the area of the pentagonal base and h the height of the prism, then the volume V is:
V = AB xh
This formula is of a general nature, being valid for any prism, be it regular or irregular, straight or oblique.
The volume of a prism is always given in cubed units of length. If the length of the sides and the height of the prism are given in meters, then the volume is expressed in m3, which is read "cubic meters". Other units include cm3, km3, inches3, and more.
– Volume of regular pentagonal prism
In a regular pentagonal prism the bases are regular pentagons, which means that the side and internal angles are equal. Given the symmetry of the body, the area of the pentagon and therefore the volume are easily calculated in several ways:
Knowing the height and the measure of the side
Be to the measure of the side of the pentagonal base. In that case the area is calculated by:
Therefore the volume of the regular pentagonal prism of height h is:
V = 1.72048 a2⋅h
Knowing the height and the measure of the radius
When you know the radius R of the pentagonal base, this other equation can be used for the area of the base:
A = (5/2)R2⋅ sin 72º
In this way the volume of the pentagonal prism is given by:
V = (5/2)R2 ⋅ h ⋅ sin 72º
Where h is the height of the prism
Knowing the height, the measure of the apothem and the value of the perimeter
The area of the pentagonal base can be calculated if its perimeter P is known, which is simply the sum of the sides, as well as the measure of the apothem LA:
A = P. LA / 2
Multiplying this expression by the value of the height hwe have the volume of the prism:
V = P.LA .h / 2
– Volume of the irregular pentagonal prism
The formula given at the beginning is valid even when the base of the prism is an irregular pentagon:
V = AB xh
Various methods are used to calculate the area of the base, for example:
-Method of triangulation, which consists of dividing the pentagon into triangles and quadrilaterals, whose respective areas are easily calculated. The area of the pentagon will be the sum of the areas of these simpler figures.
-Method of Gauss's determinants, for which it is necessary to know the vertices of the figure.
Once the value of the area is determined, it is multiplied by the height of the prism to obtain the volume. | 677.169 | 1 |
However, if you have a box with rectangular ends (like a cereal box or a book), you would have to draw an oval (not a circle) inside it to touch all the edges of the rectangle. As the plane of the box with an oval drawn on it turns in space, it doesn't obey the rules of a circle turning in space. There won't be a minor axis cutting it in half. I think that's what you have going on.
It's hard to draw a perfect square in perspective, but I think the goal is to get close-ish here. Close enough that the ellipse is somewhat convincing. Maybe it's easier to start with the ellipses for a few, and draw the boxes around them until you get a feel for it?
2:48 AM, Wednesday March 22nd 2023
Mmh, that might be the case!! Gotta check Thanks!! | 677.169 | 1 |
In Fig. 2.47, O is the centre of the circle. Name a chord, which is not the diameter of the circle.
Video Solution
|
Answer
Step by step video & image solution for In Fig. 2.47, O is the centre of the circle. Name a chord, which is not the diameter of the circle. by Maths experts to help you in doubts & scoring excellent marks in Class 6 exams. | 677.169 | 1 |
The concepts of similarity and congruence apply only to triangles. A. True, B. False.
Answered question
Answer & Explanation
Frida King
Beginner2022-09-10Added 8 answers
The concepts of similarity and congruence applies to all figures with dimensions, not only polygons but closed curves such as circles also. Hence, the statement is false. So, the correct answer is choice B. Result: B | 677.169 | 1 |
Honors Geometry Companion Book, Volume 1
1.1 Review Worksheet (continued)
12. Sketch, draw, and construct a segment twice the length of AB .
A
B
13. D is between C and E , CE = 17.1, and DE = 8. Find CD.
14. Find MN.
M
N R x
2.5 x
5 x – 3
15. During a football game, a quarterback standing at the 9-yard line passes the ball to a receiver at the 24-yard line. The receiver then runs with the ball halfway to the 50-yard line. How many total yards (passing plus running) did the team gain on the play? | 677.169 | 1 |
Search
Revision history of "1950 AHSME Problems/Problem 4845, 18 January 2012 Mrdavid445(talk | contribs) . .(557 bytes)(+557) . .(Created page with "==Problem== A point is selected at random inside an equilateral triangle. From this point perpendiculars are dropped to each side. The sum of these perpendiculars is: <math>\te...") | 677.169 | 1 |
I have tried to solve this problem for very long all to no avail and none of my peers have been able to find a solution either. Can someone provide any clues?
"In the diagram B and D are points of contact of tangents drawn from A to
the circle. E is a point on the circle and C is the point on BE such that
AC ∥ DE."
"Prove ABCD is a cyclic quadrilateral".
What I know is that for it to be a cyclic quadrilateral, the opposite angles have to be supplementary and that the angle in the tangent is equal to the angle in the alternate segment. I already know that angleBED = angle BDA (alternate angle theorem) but still cannot make headway. But despite tons of angle labelling, I can't seem to get anything at all!
$\begingroup$@Blue I have edited my post to account for my knowing this equivalence already. I also marked the other pair of equal angles in alternate segments, and also, by two tangents, I know CBA and CDA are equal. That's all I've really got lol$\endgroup$
$\begingroup$$\angle ADB$ and $\angle DEB$ are congruent by the Inscribed Angle Theorem (the first angle being a special case of such a thing). More-or-less a converse of that Theorem will get you to the target result. (The directness of that argument depends upon things you've already learned about circle geometry.)$\endgroup$
$\begingroup$We can also show that $ABCOD$ is a cyclic pentagon, where $O$ is the centre of the circle. The circle with diameter $OA$ contains $B$ and $D$ (because $\triangle{ABO}$ and $\triangle{ADO}$ are right triangles). So, by virtue of the theorem in this question, it also contains $C$. So $ABCOD$ is a cyclic pentagon.$\endgroup$
4 Answers
4
The red lines are cut by a transversal BCE so we have $\alpha=\beta$ as equal alternate angles.
Angles in the alternate segment at D and E are equal so we have $\beta=\gamma$. Remove common angle $\beta$
So $ \alpha=\gamma$ are seen as equal angles on same side of AB, and by virtue of the converse theorem that when equal angles are subtended on same side of AB at C and D, we should have quadrilateral ABCD concyclic inside the green circle.
Hint: The converse of "angles in the same segment" holds. That is, if two angles subtended from a single line segment are equal and on the same side of the line segment, then the four points they define form a cyclic quadrilateral.
This reduces the problem to showing any one of the below pairs are equal (some are easier than others).
In figure $1$, radius $\overline{OB}$ and $\overline{OD}$ determine a cyclic quadrilateral $ABOD$ because they are perpendicular to tangents $\overline{AB}$ and $\overline{AD}$ respectively so we have $\angle{BAD}=180^{\circ}-2t$.
Besides, because of $\overline{ED}$ is parallel to $\overline{CA}$ we have $\angle{BCA}=t$. Consequently, in order that $ABCD$ be a cyclic quadrilateral it is necessary that $\angle{BCA}=\angle{ACD}=t$ because $\angle{BCD}$ should be suplementary of $\angle{BAD}$.
On the other hand the corresponding circumcircle is easily constructed in figure $2$ since it is the circumcircle of the triangle $\triangle{ABO}$ of radius $\dfrac{\overline{OA}}{2}$ and center the midpoint of $\overline{OA}$. This circle shows that in fact $\angle{BCA}=\angle{ACD}$ because $\overline{AB}=\overline{AD}$.
NOTE.- Not only points $A,B,C,D$ are on the same circle but also the center of the original circle of the problem. We could say that the irregular pentagon $ABCOD$ is cyclic and that if the first circle is centered at the origine the equation of the circle solution is $x^2-2rx+y^2=0$ where $2r=\overline{OA}$. Finally, on the red circle, point $C$ distinct of $A$ and $D$ can be any in the arc able to segment $\overline{BD}$ seen under the angle $2t$.
Since $\angle ECD$ is the supplement of $\angle DCB$, $ABCD$ is a cyclic quadrilateral if$$\angle ECD=\angle BAD$$
We start with a kite $HAGJ$, touching its inscribed circle at $B$, $D$, $E$, $F$, and complete the isosceles trapezoid $BDEF$.
Let kite diagonal $AJ$ meet trapezoid diagonal $BE$ at $C$. From the bilateral symmetry of the kite and isosceles trapezoid, it is clear that $C$ lies on the line of symmetry $HG$ and on the other trapezoid diagonal $DF$, and further that $AC\parallel DE$ since both are perpendicular to $HG$.
Therefore, by symmetry $\triangle CED$ is isosceles.
And since$$\angle CED=\angle ABD$$(tangent/alternate segment), then$$\triangle CED\sim\triangle ABD$$and$$\angle ECD=\angle BAD$$Conversely then: Given any triangle $BED$ inscribed in a circle, with tangents $AB$, $AD$, and with $AC$ drawn parallel to base $ED$, it is clear from the symmetry of the constructible kite and isosceles trapezoid that$$\angle ECD=\angle BAD$$ whence $\angle BAD$ is supplementary to $\angle BCD$, and $ABCD$ is a cyclic quadrilateral. | 677.169 | 1 |
Does a rhombus have equal adjacent angles?
Generally, no. A rhombus will have supplementary adjacent angles
(i.e. adding up to 180 degrees). The only time where the adjacent
angles will be equal is when they are 90 degrees which by the way
is a square. | 677.169 | 1 |
Pyramid shaped table
A pyramid is a three-dimensional shape. A pyramid has a polygonal base and flat triangular faces, which join at a common point called the apex. A pyramid is formed by connecting the bases to an apex. Each edge of the base is connected to the apex, and forms the triangular face, called the lateral face.
Surely a pyramid shaped table would be very impractical as a table. I suppose a dead ant could be supported on it but very little else. | 677.169 | 1 |
What is geometric solid in montessori?
What is geometric solid in montessori?
The Geometric Solids are a key part of the sensorial curriculum area, allowing children to understand 3D shapes by making them tangible objects. The Geometric Solids comprises of ten solid wooden shapes that are coloured in a bright blue. The shapes include: A Triangular prism. A Rectangular prism.
What is the purpose of geometry in montessori teaching?
The Primary Years Children trace a variety of geometric figures including squares, triangles, circles, curvilinear triangles, and quatrefoils, among others. The main objective of this work is to prepare the child's muscles for proper pencil grasp and handwriting.
How many Geometric Solids are there?
Plat
Who proposed the geometric solidsWhat is geometric cabinet?
The Geometry Cabinet is part of the Sensorial area of the Primary Montessori classroom. It is used to further develop the child's visual sense in the discrimination of shape/form. This material is a wooden cabinet with six drawers that contain plane, or two dimensional, geometric shapes.
What is the name of the geometric solid?
Regular polyhedrons: they are known as platonic solids and are characterized as having faces that are all equal. There are five: tetrahedron, hexahedron (cube), octahedron, dodecahedron, and icosahedron. | 677.169 | 1 |
Is Photomath app cheating?
Using Photomath app is not cheating if you use it for personal studies and improve your mathematics. It is meant to help the student have a clear and stepwise understanding of a problem. However, using Photomath to get answers during an exam translates to cheating because it gives you an undue advantage.
Photomath
Photomath. Photomath is the king of math solving apps. Not only is it the best at recognizing all kinds of math problems, it also explains the answer in detail, which is why it's the most popular of the bunch. It solves over 2.2 BILLION math problems every month.
Why do I struggle with trigonometry?
Trigonometry is hard because it deliberately makes difficult what is at heart easy. We know trig is about right triangles, and right triangles are about the Pythagorean Theorem. About the simplest math we can write is When this is the Pythagorean Theorem, we're referring to a right isosceles triangle.
Is it cheating to use Mathway?
Using Mathway is not cheating if you use it as a learning tool to help you excel in math. When used for studies only, it can improve your homework. However, like any other tool, it can be used for good or for the bad.
How does Snapchat do math?
To access the feature, users simply "press and hold" on the Snapchat camera when a math equation is in view, and Photomath's math solver engine will recognize the problem and immediately generate a solution on the Snapchat camera screen. Problem scanning works for both handwritten and printed math problems.
How would you solve this trigonometry word problem?
tan = sin/cos = 1/cot. sin = tan/ sec = 1/ cosec if you put it simply it implies that sin equals divided cosec. cosec = 1/sec = cot/cosec. As with any other subject, it's all about practice, so make sure to spend even as little as 30 minutes every day solving trigonometry problems if you want to master this subject.
How to solve trig word problems?
From a point on the ground 96 m from a tree,the angle to the top of the tree is 38 degrees.
The angle form the ground to the top of the Statue of Liberty is 7 degrees at a distance of 1220 ft from the building.
A ladder is leaning up against a house.
A surveyor measured BC to be 125 ft.
How do you solve trigonometry problems?
Solving basic trig equations proceeds by studying the various positions of the arc x on the trig circle, and by using trig conversion table (or calculator). To fully know how to solve these basic trig equations, and similar, see book titled :"Trigonometry: Solving trig equations and inequalities" (Amazon E-book 2010).
How to solve a geometry word problem?
Assign variables: Let x = length of the equal sides Sketch the figure
Write out the formula for perimeter of triangle. P = sum of the three sides | 677.169 | 1 |
What Shape Am I?
A shape is defined as the form or the outline of an object. It can also be defined as the boundary of an object and the kind of pattern it follows. It is a basic property of every object just like its other properties like color, texture, size etc.
Different objects follow different patterns of these boundaries. These patterns have been given different names. Shapes for kids is a very important lesson, so we have made this educational video for kids.
For example, a pattern which has four equal sides is referred to as a square. A pattern with round is referred to a circle. Similarly, there are other shapes like a rectangle, triangle, cone etc. Each object is based on its shape. These shapes have been made for the understanding of man and to make a sense out of things. Every object in this universe has a shape just like earth is a sphere and a ball is a circle. | 677.169 | 1 |
Symmetry
Symmetry - Sub Topics
In this chapter, you will find a more detailed definition of symmetry. Symmetry is when an object is evenly balanced with one side mirroring the other as if having a twin but in the world of shapes! Symmetry is a fundamental concept in geometry that plays a crucial role in understanding the balance and order within shapes and patterns. This introduction sets the stage for exploring the fascinating world of symmetry, where we'll cover its different uses and significance in various fields.
Symmetry
Symmetry is a method to balance an object when it can be divided into two perfectly identical halves. It is like a special kind of balance and sameness in a shape. This means that when you split it up into two pieces, one side of an object looks exactly like the other.
In order to create an imaginary line in the middle of it, we could have divided it into two parts. The figure shows a symmetric flower.
Symmetric and asymmetric pots are shown in the figure.
Line of Symmetry
The line of symmetry is an imaginary line drawn through a shape to achieve symmetry. There may be a single or several lines of symmetry in the shape.
Imagine a sheet of paper folded in half so that you can better understand symmetry. When the two sides are in perfect alignment it is possible to achieve symmetry. The "line of symmetry" is referred to as the fold itself.
Types of lines of symmetry are as follows:
a. Vertical Line of Symmetry
If a vertical line divides an object into two identical halves, it is called a vertical line of symmetry.
b. Horizontal Line of Symmetry
If a horizontal line divides an object into two identical halves, it is called a horizontal line of symmetry.
c. Diagonal Line of Symmetry
If a diagonal divides an object into two identical halves, it is called a diagonal line of symmetry.
Line of symmetry of alphabets:
Some alphabet letters have no lines of symmetry while others may have one or more lines of symmetry. The lines of symmetry of the alphabet are shown below:
The letters F, G, J, L, N, P, Q, R, S and Z have no line of symmetry as the letters cannot be divided into two or more equal halves.
The letters A, B, C, D, E, K and M have exactly one line of symmetry.
The letters H, I and X have two lines of symmetry.
Line of symmetry of digits:
Some digits have no lines of symmetry while others may have one or more lines of symmetry. The line of symmetry of digits is shown below:
Digits 1, 2, 4, 5, 6, 7 and 9 do not have any lines of symmetry.
Digit 3 has exactly one line of symmetry.
Digits 0 and 8 have two lines of symmetry.
The following diagram illustrates lines of symmetry for some figures:
Example: How many lines of symmetry does the given figure have?
a) No line of symmetry b) One line of symmetry c) Two lines of symmetry d) Four lines of symmetry
Answer: b) One line of symmetry
Explanation: There is only one line of symmetry, shown as:
Rotational Symmetry
Rotational symmetry is a geometric property where an object retains its symmetrical appearance when rotated around its vertical axis.
The given figure has rotational symmetry.
The given figure has no rotational symmetry.
Centre of Rotation
The centre of rotation is a point within an object where rotational symmetry takes place. A point where the plane figure will rotate is the centre of rotation. During the rotation, this point is not moving.
In a wheel, the centre of rotation is shown as follows:
The centre of rotation of the hexagon is shown as:
Order of Rotational Symmetry
Order of symmetry describes the number of times a figure or object appears identical when rotated through a complete angle of 360°. The extent of rotation symmetry acquired by a figure is defined by this concept.
Let's learn about the order of rotation.
→ The order of symmetry is the number of times a figure can be moved around and still look the same as it did before it was moved. The kite looks the same only once after 360° rotation. Therefore, the order of symmetry of a kite is one.
The rotation of a kite clockwise is shown as
The rotation of a kite anticlockwise is shown as
→ The hexagon looks the same after a 60° rotation. It looks the same six times to complete the rotation. Therefore, the order of symmetry is six.
The order of rotational symmetry of common polygons is shown as:
Example: Match the figures in Column I with their order of rotational symmetry in Column II | 677.169 | 1 |
2024-06-21T21:26:49Z A. L.19052650720 Monist 15, 463-466 (1905).A circular polygon.j | 677.169 | 1 |
What are the Practical Uses of Perimeter of Isosceles Triangles?
Introduction
A triangle is a closed 2D polygon with three straight sides and when two of the sides of such figure are equal these are called isosceles triangles. In an isosceles triangle the opposite angles of the equal sides are also equal. An isosceles triangle can also be a right-angle isosceles triangle. The perimeter of an isosceles triangle is the sum total of the length of its sides or the total length of the boundaries enclosing the triangular region.
Cuemath explains the topic in a very efficient and easy to understand manner.
In this article the perimeter of isosceles triangle is explained by giving the bullet points and quick to understand examples.
Some properties of an isosceles triangle
Two sides of an isosceles triangle are equal and congruent.
Angles opposite to the equal sides of an isosceles triangle are equal.
The equal angles form of base angles of the triangle.
The angle that is different from the other two angles is called the apex angle.
The Converse angle theorem of the isosceles triangle says that if two angles of any triangle are equal, we can conclude the sides opposite to them are also equal.
Perimeter of isosceles triangles
Two Greek words 'peri' meaning around and 'metron' meaning measure come together to form a single term perimeter which means the measure of boundaries.
The perimeter of an isosceles triangle can be measured by adding up all the sides of the triangle, i.e.
Perimeter of the isosceles triangle= sum of all sides
Right isosceles triangle
Right isosceles triangle is one in which one angle is equal to 90 degrees and the base and the height are always equal. The side opposite to the right angle that is 90° forms the hypotenuse.
So, the perimeter of a right-angled isosceles triangle can be found out by adding twice the equal side with length of the hypotenuse.
Perimeter of isosceles triangle = 2*s + b, where s represents the length of two congruent sides and b is the length of the base.
Perimeter of the right isosceles triangle =2*base + hypotenuse.
*Note: The perimeter of the isosceles triangle changes with the change in the value of congruent sides and base.
The practical uses of perimeter of the isosceles triangle are:
Fencing off a triangular area to plot a crop in a field.
Isosceles triangles appear in the architecture as the shapes of pediments and gables.
In graphic design and decorative arts isosceles triangles have been a frequent design element in cultures around the world.
Perimeter of an isosceles triangle can be used to cut a triangular plot of land.
Perimeter of isosceles triangle may help to calculate the distance travelled along a figure or an area of land.
Now, at last let us learn how to find perimeter of an isosceles triangle with an example:
Example1:
Find the perimeter of the isosceles triangle when the length of the two equal sides is 5 cm and third side is 6 cm.
Solution: Each equal side= 5cm.
Third side= 6cm.
We know that the formula to calculate the perimeter of an isosceles triangle is P = 2s + b units. Therefore, the perimeter of an isosceles triangle = 2*5+6= 16 cm.
Example 2:
Calculate the distance travelled by a man while walking along the fence of a triangular field whose two sides are equal to 3 m and the longest side is3√2m. (The two equal sides make an angle of 90 degree between them).
Solution:
As it is given that the angle weight between the two equal sides is 90 degrees so we will conclude that it is an isosceles right-angled triangle.
Hence two equal sides that make the angle of 90 degree between them make up the base and the height =3m each. | 677.169 | 1 |
Identify the similarities and differences between types of triangles by the lengths of their sides and internal angles to show equivalences of common fractions. Play and learn while solving basic operations with fractions of the same and different denominators. | 677.169 | 1 |
Elements of Geometry: Containing the First Six Books of Euclid, with a ...
Take mA and nB any multiples of A and B, by the numbers m and n; and first let mA7nB: to each of them add mB, then mA+mB7 mB+nB. But mA+mB=m(A+B) (Cor. 1. 5.), and mB+nB= (m+n)B (2. Cor. 2. 5.), therefore m(A+B7(m+n)B.
And because A+B: B :: C+D : D, if m(A+B)7(m+n) B, m(C+D)7(m+n)D, or mC+mD7mD÷nD, that is taking mD from both, mC7nD. Therefore, when mA is greater than nB, mC is greater than nD. In like manner,it is demonstrated,that if mA=nB, mC=nD, and if mAZnB, that mDnD; therefore A: B:: C: D (def. 5. 5.). Therefore, &c. Q. E. D.
PROP. XVIII. THEOR.
If magnitudes, taken separately, be proportionals, they will also be proportionals when taken jointly, that is, if the first be to the second as the third to the fourth, the first and second together will be to the second as the third and fourth together to the fourth.
If A: B::C: D, then, by composition, A+B : B ::C+D : D. Take m(A+B), and nB any multiples whatever of A+B and B: and first, let m be greater than n. Then, because A+B is also greater than B, m(A+B)7nB. For the same reason, m(C+D)7nD. In this case, therefore, that is, when m7n,m(A+B) is greater than nB, and m(C+D) is greater than nD. And in the same manner it may be proved, that when m≈n,m(A+B) is greater than nB, and m (C+D) greater than nD.
Next, let mn, or n7m, then m(A+B) may be greater than nB, or may be equal to it, or may be less; first, let m(A+B) be greater than nB; then also, mA+mB7nB; take mB, which is less than nB, from both, and mA7nB-mB, or mA7(nm)B(6. 5). But if mA7 (n- m) B, mC7 (n--m)D, because A: B::C:D Now, (n−m)D=nDmD (6. 5.), therefore, mC7nD-mD, and adding mD to both, mC+m D7D, that is,(1, 5.),m(C+D)7n.D. If therefore, m(A+B)7nB, m(C+D) 7nD.
In the same manner it will be proved, that if m(A+B)=nB,m(C+ D)=nD; and if m(A+B)≤nB,m(C+D) nD; therefore (def.5.5.), A+B: B::C+D: D. Therefore, &c. Q. E. D.
PROP. XIX. THEOR.
If a whole magnitude be to a whole, as a magnitude taken from the first is to a magnitude taken from the other; the remainder will be to the remainder as the whole to the whole.
If there be three magnitudes, and other three, which taken two and two, have the same ratio; if the first be greater than the third, the fourth is greater than the sixth; if equal, equal; and if less, less.
If there be three magnitudes, A, B, and C, and other three D, E, and F; and if A: B:: D: E; and also B: C
If there be three magnitudes, and other three, which have the same ratio taken two and two, but in a cross order; if the first magnitude be greater than the third, the fourth is greater than the sixth; if equal, equal; and if less, less.
If there be three magnitudes, A, B, C, and other three, D, E, and F, such that A: B:: E: F, and B: C:: D: E: if A7C, D7F; if A=C, D F, and if AC, D4F.
If there be any number of magnitudes, and as many others, which, taken two and two in order, have the same ratio; the first will have to the last of the first magnitudes, the same ratio which the first of the others has to the last.*
First, let there be three magnitudes, A, B, C, and other three, D, E, F, which, taken two and two, in order, have the same ratio, viz. A: BD: E, and B : C:: E: F; then A: C:: D: F.
A, B, C,
Take of A and D any equimultiples whatever, mA, mD; and of B and D any whatever, nB, nE: and of C and F any whatever, qC, qF. Because A: B :: D: E, mA : nB :: mD : nE (4.5); and for the same reason, nB: qC:: nE: F. Therefore (20. 5.), according as mA is greater than qC, equal to it, or less, mD is greater than qF, equal to it or less; but mA, mĎ are any equimultiples of A and D; and qC, qF are any and F ; therefore (def. 5. 5.), A: C:: D: F.
D, E, F, mA, nB, qC, mD, nE. qF. equimultiples of C
Again, let there be four magnitudes, and other four which, taken two and two in order, have the same ratio, viz. A: B :: E: F; B: CFG; C: D::G: H, then A: D:: E: H.
For since A, B, C are three magnitudes, and E, F, G other three, which, taken two and two have the same ratio, by the foregoA:CE: G. And because also
In the same
ing case,
CD::G: H, by that same case, A: DE: H.
manner is the demonstration extended to any number of magnitudes. Therefore, &c. Q. E. D.
*N. B. This proposition is usually cited by the words "ex æquali," or "ex æque:""
Q
PROP. XXIII. THEOR.
If there be any number of magnitudes, and as many others, which, taken two and two, in a cross order, have the same ratio; the first will have to the last of the first magnitudes the same ratio which the first of the others has to the last.*
First, let there be three magnitudes, A. B, C, and other three, D, E, and F, which, taken two and two in a cross order, have the same ratio, viz. A: B : : E ; F, and B : C : : D : E, then A : C : : D : F. Take of A, B. and D, any equimultiples mA, mB, mD; and of C, E, F any equimultiples nC, nE, nF.
ples of A and D, and nC, nF any equimultiples of C and F; therefore, AC: D: F (def. 5. 5.).
Next, Let there be four magnitudes, A, B, C, and D, and other four, E, F, G, and H, which taken two and two, in a cross order, have the same ratio, viz. A: B:: G: H; B
C::
A, B C, E, F, G, H.
F: G, and C: D:: E: F, then A: D:: E : H. For, since A, B, C, are three magnitudes, and F, G, H other three, which taken two and two, in a cross order, have the same ratio, by the first case, A: C:: F: H But C D E F, therefore, again, by the first case, A: DE: H. In the same manner, may the demonstration be extended to any number of magnitudes. Therefore, &c. Q. E. D.
PROP. XXIV. THEOR.
If the first has to the second the same ratio which the third has to the fourth; and the fifth to the second, the same ratio which the sixth has to the fourth; the first and fifth, together, shall have to the second, the same ratio which the third and sixth together, have to the fourth.
Ratios which are compounded of equal ratios, are equal to one another.
Let the ratios of A to B, and of B to C, which compound the ratio of A to C, be equal, each to each, to the ratios of D to E, and E to F, which compound the ratio of D to F, A: C:: D : F. For, first, if the ratio of A to B be equal to that of D to E, and the ratio of B to C equal to that of E to F, ex æquali (22. 5.), A : C : : D : F.
A, B, C,
D, E, F.
And next, if the ratio of A to B be equal to that of E to F, and the ratio of B to C equal to that of D to E, ex æquali inversely (23. 5.), A: C:: D: F. In the same manner may the proposition be demonstrated, whatever be the number of ratios. Therefore, &c. Q. E. D. | 677.169 | 1 |
Algebra 1: Common Core (15th Edition)
by
Charles, Randall I.
Answer
a=15
Work Step by Step
The hypotenuse is 39 and the leg is 36. To find the other leg plug in a and c into the Pythagorean formula $a^{2}$ + $b^{2}$= $c^{2}$ --> $a^{2}$ + $36^{2}$= $39^{2}$:
$a^{2}$ + $36^{2}$= $39^{2}$ -simplify the exponent-
$a^{2}$ + 1296 =1521 - subtract 1296 from both sides-
$a^{2}$=225 -find the square root of 225 to solve for a-
a=15
A is 15. | 677.169 | 1 |
Trig Word Problems Worksheet With Answers
Trig Word Problems Worksheet With Answers - The angle measured from the horizon or horizontal line, down. These trigonometry worksheets are a great resource for children in 5th, 6th grade, 7th grade, and 8th grade. Web the corbettmaths textbook exercise on trigonometry. From a point on the ground 47 feet from the foot of a tree, the angle of elevation of the top of the tree is 35º. Make all answers accurate to the nearest tenth. Web a trigonometry packet with notes, examples, and practice questions related to word problems.
To plan the tunnel's length, the queen first walks 7.0 km from yram to a point where she can see both cities. The answers can be found below. Web we begin with worksheets with the properties of sine, cosine, and tangent measurements and then move to trigonometric word problems (both with and without pictures). Click here to find more. What is the angle that the sun hits the building?
He turned and dug through the dirt diagonally for 80 meters until he was above ground. What is the angle of elevation? Plus each one comes with an answer key. Web in this engaging worksheet, the trig functions sine, cosine and tangent are reinforced as students discover the answer to the riddle.there are 8 questions including 2 word problems.this product can be used forskill practicegroup workindividual workhomeworkreviewand more!this product is part of a bundle! Carry the full calculator value until rounding the final answer.
50 Trigonometry Word Problems Worksheet
The answers can be found below. (use a calculator in degree mode to find that after rounding to two decimal places) (use a calculator in degree mode to find that after rounding to two decimal places) A 75 foot building casts an 82 foot shadow. Cazoom also has created graphing worksheets with ways for your child to interact with transformations.
50 Trigonometry Word Problems Worksheet Answers
Part 2 includes a variety of word problems. Find the height of the tree to the nearest foot. (use a calculator in degree mode to find that after rounding to two decimal places) (use a calculator in degree mode to find that after rounding to two decimal places) Web right triangle trigonometry word problems. Web a trigonometry packet with notes,.
Trig Word Problems Worksheet Answers
The trigonometric ratios multiple choice. Web general triangle word problems. Carry the full calculator value until rounding the final answer. Find the value of the indicated trigonometric function of the angle ό in the figure. Web this worksheet explains how to solve word problems by using trigonometry.
Trig Word Problems Worksheet
Web a trigonometry packet with notes, examples, and practice questions related to word problems. A practical application of the trigonometric functions is to find the measure of lengths that you cannot measure. Lastly, she walks 6.0 km to haras. From a point on the ground 47 feet from the foot of a tree, the angle of elevation of the top.
Right Triangle Trigonometry Word Problems Worksheet With Answers
27 33 38 67 2. Find the angle inside the triangle that is adjacent (next door) to the angle of depression. Web use what you know about right triangles to solve for the missing angle. Give an exact answer with a rational denominator. Each one has model problems worked out step by step, practice problems, as well as challenge questions.
Trigonometry Word Problems Worksheet with Answers
What is the angle that the sun hits the building? Find the height of the tree to the nearest foot. The trigonometric ratios multiple choice. Getting ready for right triangles and trigonometry. A 75 foot building casts an 82 foot shadow.
50 Trigonometry Word Problems Worksheet Answers
Our trigonometry worksheets are free to download, easy to use, and very flexible. A ladder placed against a wall such that it reaches the top of the wall of height 6. Side ratios in right triangles as a function of the angles. To plan the tunnel's length, the queen first walks 7.0 km from yram to a point where she.
50 Trig Word Problems Worksheet
This adjacent angle will always be the complement of our angle. Web this worksheet explains how to solve word problems by using trigonometry. Web the corbettmaths textbook exercise on trigonometry. Find the height of the building. Web word problems using right triangle trig draw pictures!
Trigonometry Word Problems Worksheet With Answers Pdf Uphandicrafts
Carry the full calculator value until rounding the final answer. Round all answers to the nearest tenth of a degree. Web there are two correct options: Part 1 reviews right triangle trig values, soh cah toa, and applications. Getting ready for right triangles and trigonometry.
Trig Word Problems Worksheet Key Escolagersonalvesgui
Find the height of the tree to the nearest foot. This adjacent angle will always be the complement of the angle of depression, since the horizontal line and the vertical line are perpendicular (90º). Each one has model problems worked out step by step, practice problems, as well as challenge questions at the sheets end. Using similarity to estimate ratio.
Trig Word Problems Worksheet With Answers - Web trigonometry word problems. Find the height of the building. The trigonometric ratios multiple choice. If needed, draw a picture. Cazoom also has created graphing worksheets with ways for your child to interact with transformations of common trigonometric graphs. Web the corbettmaths textbook exercise on trigonometry. Find the angle inside the triangle that is adjacent (next door) to the angle of depression. The answers can be found below. Find the height of the tree to the nearest foot. Find the angle adjacent (next door) to our angle.
A damsel is in distress and is being held captive in a tower. Plus each one comes with an answer key. 8) tan 30° 9) csc 60° 10) tan 45° 11) cot 60° Over 2 miles (horizontal), a road rises 300 feet (vertical). Using sine is probably the most common, but both options are detailed below.
The dwarf queen wants to build a tunnel between her cities yram and haras. This adjacent angle will always be the complement of the angle of depression, since the horizontal line and the vertical line are perpendicular (90º). Using right triangle ratios to approximate angle measure. 27 33 38 67 2.
Web use what you know about right triangles to solve for the missing angle. Lastly, she walks 6.0 km to haras. Web general triangle word problems.
Our trigonometry worksheets are free to download, easy to use, and very flexible. A practical application of the trigonometric functions is to find the measure of lengths that you cannot measure. From a point on the ground 47 feet from the foot of a tree, the angle of elevation of the top of the tree is 35º.
Use The Given Triangles To Evaluate The Expression.
In the diagram on the left, the adjacent angle is 55o. Web use what you know about right triangles to solve for the missing angle. Over 2 miles (horizontal), a road rises 300 feet (vertical). The answers can be found below.
What Is The Angle That The Sun Hits The Building?
Click here to find more. Lastly, she walks 6.0 km to haras. Web this worksheet explains how to solve word problems by using trigonometry. Calculate the height of a tree, the angle of elevation from a point a is 70 degrees.
What Is The Angle Of Elevation?
27 33 38 67 2. How high is the tree? When the knight stands 15 feet from the base of the tower and looks up at his precious damsel, the angle of. The dwarf queen wants to build a tunnel between her cities yram and haras.
Very Frequently, Angles Of Depression And Elevation Are Used In These Types Of Problems.
Web the corbettmaths textbook exercise on trigonometry. Web there are two correct options: Bugs bunny was 33 meters below ground, digging his way toward pismo beach, when he realized he wanted to be above ground. Web students find the final value of trigonometric word problems in assorted problems. | 677.169 | 1 |
Given triangle ABC and a number x, consider an orientation A- >B- >C and take on the sides AB, BC, CA respectively points D, E and F such that AD/AB = BE/BC = CA/CF = x. Show that:
(1) The locus of the middle H of DE is a line parallel to the side CA. Analogous statements hold for the other sides of the triangle DEF.
(2) More generally, given a number y, the locus of a point H on DE, such that DH/DE = y is a line.
(3) ABC and all the triangles DEF(x) share the same centroid I.
In the figure above varying y shows the position of the corresponding (blue) line, which is the geometric locus referred in (2) of the above proposition. This is discussed also in the file Analogon2.html . | 677.169 | 1 |
Here are some FAQs:
Which method is the most accurate?
The most accurate method for measuring angles without a protractor is to use a right triangle and a calculator. However, the ruler and compass method can also be valid if the line segment is accurately drawn and the ruler and compass are accurate within a millimeter.
What if I don't have a calculator?
If you don't have a calculator, you can still use the ruler and compass method to measure angles, but the results will not be as accurate. You can also use a smartphone app to measure angles.
What if I don't have a ruler or compass?
If you need a ruler or compass, you can still use the suitable triangle method to measure angles, but it will be more difficult. It would help if you used a straight edge and a piece of string to draw the right triangle.
How can I use a smartphone app to measure angles?
To use a smartphone app to measure angles, point your phone at the rise, and the app will display the angle in degrees. Most smartphone apps also allow you to save and share your measurements.
When would I need to measure angles without a protractor?
You may need to measure angles without a protractor if you do not have access to a protractor or if you need to measure an angle that is too large or too small to be measured accurately with a protractor | 677.169 | 1 |
MILOS 90° 3-way horizontal 3-way 90° Corner 90° down left, apex up down right, Corner 90° Triangle G Truss MILOS Lisää ostoskoriin Straight 1500 | 677.169 | 1 |
Hint: First we need to complete the triangle RTQ since the angles $\angle RQT$ and $\angle RTQ$ belong to it. Then, applying the cyclic quadrilateral property on the quadrilateral PQTS, which states that the sum of the opposite angles is equal to ${{180}^{\circ }}$, we can determine the angle $\angle TOS$. Then, using the linear pair property and the angle sum property, we can determine the remaining angles including the angles $\angle RQT$ and $\angle RTQ$ which are asked in the question.
Complete step by step solution: Since we have to determine the values of the angles $\angle RQT$ and $\angle RTQ$ which lie inside the triangle RTQ, we join the points T and Q to complete the triangle RTQ as shown in the below figure.
Now, since all the points P, Q, T, and S lie on the circumference of the circle, so the quadrilateral PQTS formed by these points is a cyclic quadrilateral. We know that the sum of the opposite angles in a cyclic quadrilateral is equal to ${{180}^{\circ }}$. So we can write $\Rightarrow \angle P+\angle QTS={{180}^{\circ }}$ From the above figure, we have $\angle P={{45}^{\circ }}$. Substituting this in the above equation, we get \[\begin{align} & \Rightarrow {{45}^{\circ }}+\angle QTS={{180}^{\circ }} \\ & \Rightarrow \angle QTS={{180}^{\circ }}-{{45}^{\circ }} \\ & \Rightarrow \angle QTS={{135}^{\circ }}........(i) \\ \end{align}\] Now, from the figure above, we can see that the angles \[\angle QTS\] and \[\angle RTQ\] are forming a linear pair. We know that the sum of the angles in a linear pair is equal to ${{180}^{\circ }}$. So we can write $\Rightarrow \angle QTS+\angle RTQ={{180}^{\circ }}$ Substituting (i) in the above equation, we get \[\begin{align} & \Rightarrow {{135}^{\circ }}+\angle RTQ={{180}^{\circ }} \\ & \Rightarrow \angle RTQ={{180}^{\circ }}-{{135}^{\circ }} \\ & \Rightarrow \angle RTQ={{45}^{\circ }}........(ii) \\ \end{align}\] Now, from the above figure we can see that the angles \[\angle POT\] and \[\angle TOQ\] are also forming a linear pair. So we have $\Rightarrow \angle POT+\angle TOQ={{180}^{\circ }}$ From the figure, $\angle POT={{140}^{\circ }}$. Substituting this in the above equation we get \[\begin{align} & \Rightarrow {{140}^{\circ }}+\angle TOQ={{180}^{\circ }} \\ & \Rightarrow \angle TOQ={{180}^{\circ }}-{{140}^{\circ }} \\ & \Rightarrow \angle TOQ={{40}^{\circ }}.........(iii) \\ \end{align}\] Now, we know that the distances of the centre from each point on a circle are equal. So we can write $\Rightarrow OT=OQ$ We know that the angles opposite to the equal sides are equal. Therefore from the triangle OTQ we have \[\Rightarrow \angle OQT=\angle OTQ........(iv)\] From angle sum property in the triangle OTQ we have \[\angle OQT+\angle OTQ+ \angle TOQ={{180}^{\circ }}\] Putting (ii) and (iv) we get \[\begin{align} & \Rightarrow \angle OTQ+\angle OTQ+\angle TOQ={{180}^{\circ }} \\ & \Rightarrow 2\angle OTQ+{{40}^{\circ }}={{180}^{\circ }} \\ & \Rightarrow 2\angle OTQ={{180}^{\circ }}-{{40}^{\circ }} \\ & \Rightarrow 2\angle OTQ={{140}^{\circ }} \\ & \Rightarrow \angle OTQ={{70}^{\circ }}........(v) \\ \end{align}\] Since \[\angle OTQ\] and \[\angle RTQ\] form a linear pair, we have \[\Rightarrow \angle RTQ+\angle QTQ={{180}^{\circ }}\] Putting (v) we get \[\begin{align} & \Rightarrow \angle RTQ+{{70}^{\circ }}={{180}^{\circ }} \\ & \Rightarrow \angle RTQ={{180}^{\circ }}-{{70}^{\circ }} \\ & \Rightarrow \angle RTQ={{110}^{\circ }} \\ \end{align}\] Hence, the angles $\angle RQT$ and $\angle RTQ$ are respectively equal to \[{{110}^{\circ }}\] and \[{{45}^{\circ }}\].
Note: We can also use the relation between the angle subtended by a chord on the centre of the circle and that subtended at some point on the circumference to solve this question. Also, the angle sum property may be applied in any of the triangles in the given figure. So there are many methods to solve this question and you can solve any of them. | 677.169 | 1 |
Kite test. Download the Kite Student Portal for use with an i...
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Two text versions survive: that printed in the Pennsylvania Gazette of October 19, 1752, reprinted below; and a copy in the hand of Peter Collinson, now in the Royal Society.7 Aside from unimportant variations in paragraphing, spelling, capitalization, and punctuation, the Collinson copya) The Q output is either SET or RESET as soon as the D input goes HIGH or LOW. b) The output complement follows the input when enabled. c) Only one of the inputs can be HIGH at a time. d) The output toggles if one of the inputs is held HIGH. View Answer. Take Digital Circuits Practice Tests - Chapterwise!. Sep 9, 2019 · KITE provides instant test resultYou can take the test , at any time, day or night. All you need i A Kite is a flat shape with straight sides. It has two pairs of equal-length adjacent (next to each other) sides. It often looks like. a kite! Two pairs of sides. Each pair is two equal-length sides that are adjacent (they meet) The angles are equal where the two pairs meet. Diagonals (dashed lines) cross at right angles,2/3 Test #2: kite-test .....***Failed 0.01 sec.. 67% tests passed, 1 tests failed out of 3 Total Test time (real) = 0.03 sec The following tests FAILED: 2 - kite-test (Failed) Errors while running CTest kite n (finance: cheque) شيك بدون رصيد : Dan wa ... Run unit tests python setup.py test. or. pytest-s tests/unit--cov …Page 117: Copying Individual Tests Using The Test Library Manager KITE project. See the How to display and manage test results, page 2-24 or the Reference manual, Keithley Interactive Test Environment (KITE), page 6-1 for further details on managing KITE tests and projects. Figure 3-8 Selecting multiple tests 4200-900-01 Rev. K / February 2017... The Kite ® Suite provides an online testing int... | 677.169 | 1 |
Angle Notation
Angles are typically denoted using the symbol ∠ followed by the vertex and two other points on the angle, with the vertex in the middle. For example, angle ABC can be denoted as ∠ABC.
Measuring Angles
Angles are measured using a protractor, which is a tool specifically designed for measuring angles. The measure starts from the initial side and rotates towards the terminal side in a counterclockwise direction. The measure is the amount of rotation between the two sides, typically in degrees.
Angle Relationships
Angles can have various relationships with each other, including complementary angles, supplementary angles, and vertical angles. Complementary angles add up to 90 degrees, supplementary angles add up to 180 degrees, and vertical angles are formed by two intersecting lines and are congruent.
Practice Problems
1. Find the complement of a 35-degree angle. Answer: The complement of a 35-degree angle is 55 degrees because 35 + 55 = 90 degrees.
2. Measure the angle XYZ using a protractor. Answer: Use a protractor to measure the angle and record the number of degrees.
3. Identify the type of angle formed by the hands of a clock at 3:00. Answer: The hands of a clock at 3:00 form a right angle, which measures 90 degrees.
Conclusion
Understanding angles and their properties is crucial for solving geometric problems and understanding spatial relationships. By practicing angle measurements and identifying angle types, you can develop a strong foundation in geometry. | 677.169 | 1 |
Vector Algebra
Vector Algebra
What is Vector Algebra?
Vector algebra is a branch of mathematics that deals with vectors, which are quantities having both magnitude and direction. It involves operations such as vector addition, subtraction, scalar multiplication, and the dot and cross products, used extensively in physics and engineering.
Types of Vectors – Vector Algebra
Zero Vector: A vector with zero magnitude and no specific direction.
Unit Vector: A vector with a magnitude of one, indicating direction.
Position Vector: A vector representing the position of a point relative to an origin.
Co-Initial Vectors: Vectors sharing the same initial point.
Collinear Vectors: Vectors that lie along the same line, regardless of magnitude and direction.
Equal Vectors: Vectors with the same magnitude and direction, regardless of their initial points.
Negative Vectors: Vectors with the same magnitude but opposite direction.
Parallel Vectors: Vectors that are parallel to each other, regardless of magnitude.
Vector Algebra Operations
In vector algebra, several basic operations can be performed geometrically without relying on a coordinate system. These operations include vector addition, subtraction, and scalar multiplication. Additionally, vectors can be multiplied in two distinct ways: the dot product and the cross product. These fundamental operations are essential for understanding and manipulating vectors in various applications.
Addition of Vectors
Subtraction of Vectors
Scalar Multiplication
Scalar Triple Product of Vectors
Multiplication of Vectors
Vector Addition:
Combining two vectors to form a resultant vector by adding their corresponding components.
Triangle Law of Addition of Vectors: The law, often referred to as the Triangle Law of Vector Addition, states that if two sides of a triangle represent two vectors in both magnitude and direction, acting simultaneously on a body and arranged sequentially, then the third side of the triangle, when drawn to close the triangle, represents the resultant vector. This principle is crucial for visually determining the combined effect of two vector quantities.
Parallelogram Law of Addition of Vectors: The Parallelogram Law of Vector Addition states that if two vectors, originating from the same point and acting simultaneously, are represented by the adjacent sides of a parallelogram, then the diagonal of the parallelogram, starting from the same point, represents the sum of these two vectors. This diagonal is the resultant vector that effectively combines the magnitude and direction of the original vectors
Vector Subtraction:
Subtracting one vector from another by subtracting their corresponding components.
3) Scalar Multiplication:
4) Dot Product (Scalar Product):
5) Cross Product (Vector Product):
Results in a vector perpendicular to both A and B in three-dimensional space.
6) Magnitude of a Vector:
Calculates the length or size of vector A.
7) Unit Vector:
A vector with a magnitude of one, indicating direction.
8) Projection of Vector A on B:
Projects vector A onto vector B.
9) Angle Between Two Vectors:
Finds the cosine of the angle θ between vectors A and B.
10) Distance Between Two Points:
Calculates the distance between points (x₁,y₁,z₁) and (x₂,y₂,z₂)
Applications of Vector Algebra
Vector algebra is a pivotal branch of mathematics with wide applications in Physics and Mathematics, particularly because it deals with quantities that have both magnitude and direction, such as velocity, acceleration, and force. These quantities are expressed mathematically as vectors. Here's an enhanced overview of the applications of vector algebra:
Differential Equations and Geometry: Vector algebra plays a critical role in the study of differential equations and differential geometry. It provides essential tools for describing and solving problems involving multi-dimensional spaces, helping to analyze curves, surfaces, and their properties.
Physics and Engineering Applications: In physics and engineering, vectors are indispensable for modeling and analyzing phenomena associated with electromagnetic fields, gravitational fields, and fluid dynamics. Vectors help in understanding how forces interact in space, offering a clear method for representing field strengths and directions.
Force Analysis: Vector algebra is crucial in breaking down forces into their component parts. This is particularly useful in mechanics and structural engineering, where it is essential to determine the effects of forces acting in various directions.
System Interactions: In physics, vector algebra is used extensively to explore the interactions between multiple vectors. This includes studying the resultant effects when forces, velocities, or other vector quantities combine.
Scalar Triple Product and Coplanarity: The scalar triple product—a calculation involving the dot product of one vector with the cross product of two others—has significant implications in spatial analysis. This product is zero if the vectors are coplanar, meaning they lie in the same plane. This property is used in vector calculus to determine the relationships between vectors in three-dimensional space.
Vector Fields and Flow Lines: Vector algebra also helps in the visualization and analysis of vector fields, which are essential in understanding flow lines in fluid dynamics, as well as field lines in electromagnetic and gravitational fields.
Navigational Techniques: Vectors are used in navigation to plot courses in three-dimensional space, combining direction and magnitude to chart paths effectively, whether in aviation, marine navigation, or space exploration.
Is Vector Algebra the Same as Linear Algebra?
Vector algebra is a subset of linear algebra focused on operations with vectors in two or three dimensions, whereas linear algebra encompasses a broader range of topics including matrices, vector spaces, and linear transformations, applicable in higher dimensions.
What Are the Three Rules of Vectors?
The three primary rules of vectors are vector addition (vectors are added tail-to-head), scalar multiplication (a vector's magnitude is multiplied by a scalar), and the parallelogram rule (vector addition can be represented as the diagonal of a parallelogram).
Why Is Vector Algebra Hard?
Vector algebra can be challenging due to its abstract nature and the need to visualize operations in three-dimensional space, which requires spatial reasoning and understanding of complex rules for operations like the cross product.
What Is the Use of Vector Algebra in Real Life?
Vector algebra is widely used in physics for describing velocities and forces, in engineering for stress and fluid dynamics analysis, and in computer graphics for rendering scenes and animations accurately.
What Is Vector Algebra Simple?
Vector algebra is the branch of mathematics that deals with vectors, which are quantities that have both magnitude and direction. It provides a way to perform calculations involving physical quantities like displacement, velocity, and acceleration.
What Math Is Vector In?
Vectors are primarily studied within the field of linear algebra, which deals with vector spaces and linear mappings between these spaces. They are also essential in calculus and physics for describing and analyzing multidimensional systems.
What's the Hardest Math Class?
The perception of the hardest math class varies by individual but often includes courses like Real Analysis, Abstract Algebra, and Topology, which involve rigorous proofs and abstract concepts that challenge even mathematically adept students.
What Is Harder, Calculus or Linear Algebra?
The difficulty between calculus and linear algebra can depend on one's strengths. Calculus, with its focus on rates of change and integrals, can seem more intuitive than linear algebra, which requires abstract thinking about spaces and transformations.
Why Are Vectors So Confusing?
Vectors can be confusing due to their dual nature: they are not just numbers but quantities with direction and magnitude. Grasping how vectors behave, especially in products like the cross product which results in a vector perpendicular to the original plane, requires robust spatial and abstract reasoning. | 677.169 | 1 |
Understanding the math.acos() Function in Pine Script
In Pine Script, the math.acos() function is a mathematical tool used to calculate the arc cosine (inverse cosine) of a given number. The function is crucial for trigonometric calculations, particularly when you need to find the angle whose cosine is a specific value. This operation is essential in various financial analysis tasks, such as computing angles and rotations in geometric pattern analyses.
Syntax
The math.acos() function can be applied in several ways, depending on the type of input and the desired output. It supports various types of input, such as constants, simple floats, series floats, and even user inputs, offering flexibility in its application within Pine Script:
math.acos(angle) → const float
math.acos(angle) → input float
math.acos(angle) → simple float
math.acos(angle) → series float
Arguments
angle (const int/float): The angle for which you wish to calculate the arc cosine, specified in radians. It's important to note that the value of the angle must be within the range [-1, 1] for the function to return a valid result.
Returns
The function returns the arc cosine of the given value, with the resulting angle in the range ([0, \pi]), or na (Pine Script's representation for "not available") if the input is outside the range ([-1, 1]).
Practical Example
To illustrate the use of math.acos() in Pine Script, consider a scenario where we need to calculate the angle in radians for a given cosine value within a script. This example demonstrates how to use the function with a constant value, modify it for uniqueness, and understand each step of the code.
Walkthrough
Indicator Declaration: The script begins with the declaration of the indicator using indicator(), specifying the name "Arc Cosine Example" and setting overlay to true to display the plot on the main chart.
Defining the Cosine Value: A cosine value of 0.5 is defined using cosineValue. This value is within the acceptable range for the math.acos() function.
Calculating Arc Cosine: The math.acos() function is used to calculate the arc cosine of cosineValue, storing the result in arcCosineResult.
Plotting the Result: The plot() function visualizes arcCosineResult on the chart with the label "Arc Cosine of 0.5" in blue color.
Key Features and Takeaways
Function Flexibility:math.acos() can handle inputs as constants, series, or user inputs, making it versatile for different scenarios.
Range Specificity: It operates within a specific range of input values ([-1, 1]), returning na for values outside this range.
Result Range: The output is always within the range ([0, \pi]) radians, suitable for trigonometric analyses in financial markets.
Use Case: Ideal for geometric and trigonometric calculations, especially in pattern recognition and angle computations in trading strategies.
By integrating the math.acos() function into your Pine Script arsenal, you enhance your ability to perform complex mathematical calculations, opening up new possibilities for strategy development and analysis. | 677.169 | 1 |
Breadcrumb
Section Formula in 3 Dimension
Section formula is applied in the three-dimensional geometry to find the coordinates of points dividing a line segment. This division of a line segment occurs internally and in a specific ratio. Section formula uses coordinate geometry to find the coordinates of these points in space.
The three-dimensional section formula uses a three-coordinate system to denote the coordinates of points. In the same way, the section formula is applicable in two-dimensional geometry as well, but in a two-coordinate system.
Section formula
Consider two points A (x1, y1, z1) and B (x2, y2, z2). Consider another point P on the line AB with coordinates (x, y, z) that divides the line in m:n ratio.
1. We need to draw perpendiculars AL, PN, and BM to the XY plane such that AL || PN || BM.
2. The points M, N and L lie on the straight line formed due to the intersection of AL, PN and BM on the XY plane.
3. A line ST is drawn from point P such that ST is parallel to LM.
4. The line ST intersects BM internally at T and AL at S.
5. Since ST is parallel to LM and PN || AL || BM ||, therefore, these lines form two parallelograms, LNPS and NMTP.
We can represent external section formula as:
Example: What are the coordinates of the line segment (1, -2, 3) and (3, 4, -5), which divide the line segment in the ratio of 2:3 internally and externally? | 677.169 | 1 |
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Equation of a Circle Example 1 YouTube
Learn how to sketch the graph of a circle.
How to draw a circle with equation. Web how to graph circles using the center and radius — krista king math | online math help given the equation of a circle, we can put the equation in standard form, find the center and radius of the circle from the standard form, and then use the center and radius to graph the circle. The symbols a and b represent the center of the circle as a point on an axis, with a as the horizontal displacement and b as the vertical displacement. What is the standard equation of a circle?
Web explore math with our beautiful, free online graphing calculator. The equation of a circle with (h, k) center and r radius is given by: Web after you draw the last side of a square, the turtle is sitting on the outer circle, facing 45 degrees off a radius to that point.
You walk around a circle which has a diameter of 100m, how far have you walked? Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more. Move it the appropriate distance to traverse that chord (this is where your math comes in). | 677.169 | 1 |
Which Of These Triangle Pairs Can Be Mapped To Each Other Using A Single Translation?
Triangle pairs are two separate triangles that can be mapped to each other using a single translation. This means that each triangle is moved in the same direction and the same distance. This article will explain how to determine which triangle pairs can be mapped to each other using a single translation.
Triangle Pairs
Triangle pairs are two triangles that are related to each other. To be mapped to each other using a single translation, the triangles must have the same number of sides and corresponding lengths of sides. In addition, the triangles must be similar, meaning that they must have the same angles and corresponding lengths of sides. This means that both triangles must have the same shape and size.
Mapping with a Translation
To determine whether two triangles can be mapped to each other using a single translation, the following steps should be taken:
Check that the two triangles have the same number of sides and corresponding lengths of sides.
Check that the angles of the two triangles are the same.
Check that the corresponding sides of the two triangles are the same length.
Calculate the translation vector, which is the distance and direction between the two triangles.
If the translation vector is the same for both triangles, then they can be mapped to each other using a single translation.
If all of these conditions are met, then the two triangles can be mapped to each other using a single translation.
By following the steps listed above, it is possible to determine whether two triangles can be mapped to each other using a single translation. This is a useful tool for understanding the relationship between two triangles and can be used to solve a variety of mathematical problems. | 677.169 | 1 |
In Exercises determine whether the statement is true or false. If it is false, explain why or
Question:
In Exercises determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false.
If C is the circumference of the ellipse
then 2πb ≤ C ≤ 2πα.
Transcribed Image Text:
b²
1, b < a
Fantastic news! We've Found the answer you've been seeking!
Step by Step Answer:
Answer rating: 77% (9 reviews) | 677.169 | 1 |
The center of a hyperbola is the midpoint of the line segment joining its foci. The transverse axis is the line segment that contains the center of the hyperbola and whose endpoints are the two vertices of the hyperbola.
What is the vertex of a hyperbola?
Definition of the vertex of the hyperbola: The vertex is the point of intersection of the line perpendicular to the directrix which passes through the focus cuts the hyperbola. The points A and A', where the hyperbola meets the line joining the foci S and S' are called the vertices of the hyperbola.
What is hyperbolic shape?
Hyperbola: A hyperbola is an open curve with two branches, the intersection of a plane with both halves of a double cone. The plane may or may not be parallel to the axis of the cone.
What is called to one axis of symmetry joining the vertices of the hyperbola?
The axis along the direction the hyperbola opens is called the transverse axis. The conjugate axis passes through the center of the hyperbola and is perpendicular to the transverse axis. The points of intersection of the hyperbola and the transverse axis are called the vertices (singular, vertex) of the hyperbola.
How do you find the center of a hyperbola?
Centre of the Hyperbola The mid-point of the line-segment joining the vertices of an hyperbola is called its centre. Suppose the equation of the hyperbola be x2a2 – y2b2 = 1 then, from the above figure we observe that C is the mid-point of the line-segment AA', where A and A' are the two vertices.
How do you find the center of a hyperbola from an equation?
Divide each side of the equation by 144, and you get the standard form. The hyperbola opens left and right, because the x term appears first in the standard form. The center of the hyperbola is (0, 0), the origin. To find the foci, solve for c with c2 = a2 + b2 = 9 + 16 = 25.
How do you find the center point of a hyperbola?
What connects the two vertices of a hyperbola?
The hyperbola crosses that axis at points called the vertices. The line segment connecting the vertices is called the transverse axis.
Is vertex and vertices the same?
What are vertices of a shape? Vertices is the plural of the word vertex, which is the point at which two or more lines/edges meet. Edges are straight lines that connect one vertex to another.
Is a parabola half of a hyperbola?
the pair of hyperbolas formed by the intersection of a plane with two equal cones on opposites of the same vertex. So this is suggesting that each half of what we'd normally consider a hyperbola is itself a hyperbola. They're saying a hyperbola is just one unbroken curve like a parab | 677.169 | 1 |
Find The Distance Between Two Points (Real World Problems) Worksheet
8 problems
Finding the distance between two points refers to the calculation of the straight-line distance between two points in a given space, such as a coordinate plane or a three-dimensional space. This distance is also known as the "Euclidean distance" and is represented by a line segment connecting the two points.
Grade 6
Negative Numbers
6.NS.C.6.A
Teaching Find The Distance Between Two Points (Real World Problems) Easily
Introduce the concept of distance and the distance formula: Explain the distance formula to the students, using the examples provided earlier in the answer. Show them how to use the formula to calculate the distance between two points in a two-dimensional coordinate plane.
Practice with worksheets: Give students practice worksheets to solve using the distance formula. Start with simpler problems and gradually increase the difficulty level.
Provide feedback: Provide feedback on the students' work and help them understand where they made mistakes. Point out the key concepts and formulas they used to solve the problem and how they can apply them to other problems given in the worksheet.
Why Should You Use Find The Distance Between Two Points (Real World Problems) Worksheet for your students?
Yours students will earn to use different formulas to find the distance between two pointsThe most common methods include:
Distance formula: The distance formula is a mathematical formula used to calculate the distance between two points in a two-dimensional coordinate plane. It is given by:
Distance = √((x2 - x1)^2 + (y2 - y1)^2), where (x1, y1) and (x2, y2) are the coordinates of the two points | 677.169 | 1 |
Introduction
We have already discussed about polygons in the
Polygon and its Types
chapter. There we learnt that a polygon is a closed curve, which is made up of only line segments.
Triangle is one type of polygon which has three sides or line segments. It is a polygon with least number
of sides, in other words, a polygon with less than three sides does not exist.
Triangle is a closed figure which is formed by joining three line segments. It has 3 sides, 3 vertices and 3
angles. Triangle is denoted by Δ, a delta symbol.
Lets learn about triangles with the following figure with a triangle whose name is written as ΔABC, where Δ is a symbol of triangle
and ABC is the name of a triangle, which always includes three vertices of a triangle.
Example of a triangle ΔABC
We can use any three vertices in any order to represent a triangle.
For example, ABC or BAC or CAB or ACB. Therefore, the triangle can also be written as ΔBAC or
ΔCAB or ΔACB.
Sides of triangle
As you know that triangle always has three sides. These sides are also called the
line segments.
So, in ΔABC, line segments AB, BC and CA are sides of the ΔABC.
These three sides of a triangle are written as
AB—,
BC—
and
CA—.
Vertex of triangle
Vertex of a triangle is that point where any two sides of a triangle meet out or intersect.
In the ΔABC in the above figure, sides AB and AC meet at point A. So, A is the vertex of ΔABC.
Similarly, sides BC and AB meet at point B. So B is a vertex of ΔABC.
Also, sides AC and BC meet at point C. So C is a vertex of ΔABC.
Therefore, A, B and C are the three vertices of ΔABC.
Angles of triangle
Angles
are formed at the vertex of a triangle. It is a measurement of how slanted two lines are to each other.
The angles are measured in degrees units.
Let's understand angles in a triangle ΔABC from the above figure.
As we already discussed above, ΔABC has three vertices A, B and C.
Therefore, we can say angles are formed at vertices A, B and C.
Angles are written as ∠ABC, ∠BAC and ∠ACB or in a short form as ∠B,
∠A and ∠C respectively.
In other words, we can say:
∠ABC or ∠B is formed at vertex B of ΔABC.
∠BAC or ∠A is formed at vertex A of ΔABC.
∠ACB or ∠C is formed at vertex C of ΔABC.
Types of triangle on the basis of length of sides
Triangle has many types depending upon the length of its sides.
Equilateral triangle
A triangle is said to be an equilateral triangle if all sides of a triangle are of equal length.
ΔABC is equilateral triangle because AB = BC = CA, where AB is length of side AB, BC is length of
side BC and CA is the length of side CA.
Equilateral triangle ΔABC
Isosceles triangle
A triangle is said to be an Isosceles triangle if any two sides of the triangle are of equal length.
ΔABC is Isosceles triangle because AB = AC.
Isosceles triangle ΔABC
Scalene triangle
A triangle is said to be a scalene triangle if all sides of the triangle are of unequal length.
ΔABC is Scalene triangle because AB ≠ BC ≠ AC.
Scalene triangle ΔABC
Types of triangle on the basis of size of angles
Triangle has many types depending upon the length of its angles.
Acute angled triangle
A triangle is said to be an acute angled triangle if each angle of a triangle is acute. Acute angle is that
angle
which is less than 90°.
So, the ΔABC in figure is an acute angled triangle because all angles are less than 90°.
That is, ∠A, ∠B and ∠C are all less than 90°.
Acute angled triangle ΔABC
Right angled triangle
A triangle is said to be a right angle triangle if one of its angles is right angle. Right angle is that angle
whose measure is 90°.
So, the ΔABC in figure is a right angled triangle because ∠C is equal to 90°.
Right angled triangle ΔABC
Obtuse angled triangle
A triangle is said to be an obtuse angled triangle if one of its angles is an obtuse angle. Obtuse angle is that
angle
which is greater than 90°.
So, the ΔABC in figure is an obtuse angled triangle because ∠C is greater than 90°.
Obtuse angled triangle ΔABC
List of types of triangles on the basis of sides
Name of triangle
Number of equal sides
Equilateral triangle
All 3 sides are equal
Isosceles triangle
Any 2 sides are equal
Scalene triangle
No sides are equal
List of types of triangles on the basis of angle
Name of triangle
Measure of angle
Acute angled triangle
All three angles are acute angles
Right angled triangle
One angle is 90°
Obtuse angled triangle
One angle is obtuse angle
Frequently Asked Questions
1) What is a triangle?
Triangle is a plane and closed figure which is formed by three line segments.
2) How many sides are there in a triangle?
There are 3 sides in a triangle.
3) How many angles are there in a triangle?
There are 3 angles in a triangle.
4) What are the types of a triangle according to its sides?
There are three types of a triangle according to its sides.
1. Equilateral triangle
2. Isosceles triangle
3. Scalene triangle
5) What are the types of a triangle according to its angles?
There are three types of a triangle according to its angles.
1. Right angled triangle
2. Obtuse angle triangle
3. Acute angle triangle
6) What is the sum of three angles of a triangle?
Sum of three angles of a triangle is equal to 1800.
Solved Examples
1) In isosceles triangle ΔABC, with AB=AC, if base angle = 300, find the other angles. | 677.169 | 1 |
Gina wilson all things algebra answer key 2012
Gina wilson 2012 algebra [PDF] ? drupal8.pvcc.edu drupal8.pvcc.edu gina wilson all things algebra 2012 2017 kiddy math Jan 08 2023 gina wilson all things algebra 2012 2017 displaying top 8 worksheets found for this concept some of the worksheets for this concept are ginaQuestions & Answers. This Similar Triangles Unit Bundle contains guided notes, homework assignments, two quizzes, a study guide and a unit test that cover the following topics:• Ratio and Proportion: Includes extended ratio problems.•. Similar Figures: Identifying scale factors, and solving problems with similar figures ... Convert
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TL;DR
Trigonometry is used to find heights and distances in real-world applications, such as measuring the height of buildings or the distance between two objects.
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Key Insights
🏑 Trigonometry has numerous practical applications in various fields, including architecture, engineering, and navigation.
🗯️ Forming a right triangle is an important step in solving trigonometry problems.
🔺 Trigonometric ratios, such as sine, cosine, and tangent, are used to find unknown angles or side lengths in a triangle.
😒 The example provided demonstrates how to use trigonometry to find the distance between the foot of a ladder and the bottom of a wall.
Transcript
in the previous video we saw basics of trigonometry in this video we will see some of the applications of trigonometry in real-life trigonometry has lots of applications in the real world and can be used in many practical situations around us they are mostly used in finding the heights of things such as towers buildings or they are also used to fin...
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Questions & Answers
Q: How is trigonometry used in real life?
Trigonometry is used in various real-life applications, such as architecture, engineering, and navigation. It helps determine distances, heights, angles, and more.
Q: What is the importance of forming a right triangle in trigonometry problems?
Forming a right triangle is crucial in trigonometry because it allows us to apply trigonometric ratios to find unknown angles or side lengths based on the given information.
Q: How are trigonometric ratios used in the example provided?
In the example, the trigonometric ratio of tangent is used to find the length of BC, which represents the distance between the foot of the ladder and the bottom of the wall.
Q: Can trigonometric ratios other than tangent be used in similar problems?
Yes, trigonometric ratios such as sine or cosine can also be used depending on the given information. In this specific example, tangent was used, but cosine or sine would yield the same result.
Summary & Key Takeaways
Trigonometry has practical applications in finding heights and distances in real-life situations.
Trigonometric ratios are used to form right triangles and solve problems.
In the example given, the distance between the foot of a ladder and the bottom of a wall is found using the trigonometric ratio of tangent. | 677.169 | 1 |
The hypotenuese of an isosceles right triangle is 5 inches. The midpoints of its sides are connected to form an inscribed triangle, and
this process is repeated. Find the sum of the areas of these triangles if this process is continued infinitely. | 677.169 | 1 |
If in a triangle ABC,∠B=π3andsinAsinC=λ. Find the set of all possible values of λ.
Video Solution
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Answer
Step by step video & image solution for If in a triangle ABC, /_ B= pi/3 and sin Asin C = lambda. Find the set of all possible values of lambda. by Maths experts to help you in doubts & scoring excellent marks in Class 12 exams. | 677.169 | 1 |
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Page 10 ... polygons , by more than four straight lines . 19. Of three sided figures , an equilateral triangle is that which has three equal sides . 20. An isosceles triangle is that which has only two sides equal . ΔΔΔ 21. A scalene triangle is ...
Page 31 ... right angle . SCHOLIUM . When ( Cor . 1. ) is applied to polygons , which have re - entrant angles , as ABC each re - entrant angle must be regarded as greater than two right angles . And , by joining BD , BE , BF , OF GEOMETRY . BOOK I.
Page 32 ... polygons with salient angles , which might otherwise be named convex polygons . Every convex polygon is such that a straight line , drawn at pleasure , cannot meet the contour of the polygon in more than two points . PROP . XXXIII ...
Page 45 ... polygon ABCDE will be equivalent to the polygon ABCF , which has one side less than the original polygon . For the triangles CDE , CFE , have the base ČE common , and they are between the same paral- lels ; since their vertices D , F ...
Page 46 ... polygon ; for , by suc- cessively diminishing the number of its sides , one being retrenched at each step of the process , the equivalent triangle will at length be found . COR . Since a triangle may be converted into an equivalent | 677.169 | 1 |
Hint:: To solve this problem, we have to know a basic rule that the sum of all angles on a straight line is ${180^\circ}$. First of all sum up the given angle values (in unknown term) which is equal to ${180^\circ}$and then find the value of $x$ which is also the value of $\angle QPB$.
Complete step by step solution: All the given angles on a straight line after summing up will be equal to ${180^\circ}$. So, $2x + 3x + x = {180^\circ}$ On simplifying the above $x$ terms, $\Rightarrow 6x = {180^\circ}$ Solving for the value of $x$, $\Rightarrow n = \dfrac{{{{180}^\circ}}}{6} = {30^\circ}$ As per given information in the question, $\angle QPB = {x^\circ} = {30^\circ}$.
Note: First of all, read the question attentively so that you don't get into any confusion about what is asked for as the ultimate answer. Solve for the unknown term in the linear equation obtained from the given information in the question. Do the calculations properly to ensure that you don't make a silly mistake even after knowing the procedure to solve it. The only basic thing to be kept in mind is that the sum of all angles on a straight line is ${180^\circ}$. | 677.169 | 1 |
From the Classroom: Winter Perspective Drawings
From the classroom!
Using parallel lines, angles, and vanishing points, students in Geometry Saxon learned how to draw things from a one-point or two-point perspective. This collection of drawings was inspired by what they saw over winter break last school year! | 677.169 | 1 |
Corresponding sides
Corresponding sides
In triangles that are corresponding, they are exactly the same in every way except location, while triangles that are similar are proportional, and can be of different size and orientation. The triangles are similar and congruent because of SSS. | 677.169 | 1 |
Hint: We need to know the formula for converting the radian measures into degree measures. We need to substitute the given radian value in that formula. By using arithmetic operations we can easily find the answer. We need to find the final answer in degree measures, so we need to know the degree value \[\pi \].
Note: Note that this problem can also be solved by substituting the value of\[\pi \]is equal to\[{180^ \circ }\]in the given problem. By using this method we can easily find the answer. This type of questions involves the operation of addition/ subtraction/ multiplication/ division. Note that each radian value must be in the form of\[\pi \] and in each degree value the degree symbol will be present on that term. To make an easy calculation first we would try to cancel the term\[\pi \]in the formula, next we can easily find the answer using normal multiplication and division. | 677.169 | 1 |
Here we will learn about Pythagoras theorem, including how to find sides of a right-angled triangle and how to use Pythagoras theorem to check if a triangle has a right angle or not.
There are also Pythagoras theorem worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you're still stuck.
Pythagoras theorem states that the square of the longest side of a right angled triangle (called the hypotenuse) is equal to the sum of the squares of the other two sides.
Pythagoras theorem is:
Side c is known as the hypotenuse, which is the longest side of a right-angled triangle and is opposite the right angle. Side a and side b are known as the adjacent sides because they are adjacent (next to) the right angle.
If we know any two sides of a right angled triangle, we can use Pythagoras theorem to work out the length of the third side.
We can only use Pythagoras theorem with right-angled triangles.
E.g. Let's look at this right-angled triangle:
We can see that three squares have been drawn next to each of the sides of the triangle. The area of the square with side length 3 = 3 \times 3=3^2=9 The area of the square with side length 4 = 4\times 4=4^2=16 The area of the square with side length 5 = 5\times5=5^2=25
We can see that when we add together the areas of the squares on the two shorter sides we get the area of the square on the longest side.
We can see that when we square the lengths of the two shorter sides of a right angled triangle and add them together, we get the square of the longest side.
3, 4, 5 is known as a Pythagorean triple. There are other Pythagorean triples such as 5, 12, 13 and 8, 15, 17 .
See also: 3D Pythagoras
In order to use Pythagoras theorem:
Get your free pythagoras theorem worksheet of 20+ questions and answers. Includes reasoning and applied questions.
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Get your free pythagoras theorem worksheet of 20+ questions and answers. Includes reasoning and applied questions.
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Find x and give your answer to 2 decimal places.
It is very important to label the hypotenuse (the longest side) correctly with c . The adjacent sides, next to the right angle can be labelled a and b either way around as they are interchangeable.
2Write down the formula and apply the numbers.
3Work out the answer.
Make sure you give your final answer in the correct form including units where appropriate.
The final answer is:
x = 8.54 cm to 2 decimal places
An alternative method is to rearrange the formula and put the entireMake sure you give your final answer in the correct form including units where appropriate.
The final answer is:
x = 11.4 cm to 3 significant figures
An alternative method is to rearrange the formula and put one calculation into a calculator.
Find x and give your answer to 2 6.24 cm to 2 significant figures
An alternative method is to rearrange the formula and put one 16.7 cm to 3 significant figures
An alternative method is to rearrange the formula and put one calculation into a calculator.
Is the triangle below a right-angled triangle?
Label the sides of the triangle.
If the triangle is a right-angled triangle, then Pythagoras theorem should work: ItPythagoras theorem only works with right-angled triangles.
So because,
The sides of the triangles do not fit with Pythagoras theorem. Therefore the triangle is NOT a right-angled triangle.
Is the triangle below a right-angled triangle?
Label the sides of the triangle.
If the triangle is a right-angled triangle, then Pythagoras theorem should work:This is correct.
Work out the answer.
Pythagoras theorem only works with right-angled triangles. So because,
The sides of the triangles fit with Pythagoras theorem. Therefore the triangle is a right-angled triangle.
It is very important to make sure that the hypotenuse is correctly identified and labelled c . You can find the hypotenuse by identifying the side which is opposite the right angle.
The triangles can be drawn in different orientations. E.g.
Lengths can be decimals, fractions or even irrational numbers such as surds,
E.g. \sqrt{2} .
Pythagoras theorem only works on right-angle triangles, however we can sometimes calculate the lengths of other triangles by splitting them into right angled triangles. E.g. An isosceles triangle can be made into 2 right-angled triangles by putting in the line of symmetry.
An application of Pythagoras theorem is to extend it to work on other shapes such as a trapezium.
If you need to use Pythagoras theorem in a question with multiple steps, do not round until the very end of the question or you will lose accuracy. For example, you may need to find the height of a triangle, and then use that height to find its area. | 677.169 | 1 |
The angle bisectors of $\measuredangle BAD$ and $\measuredangle ADC$ of the trapezoid $ABCD$$(AB\parallel CD)$ intersect at $O$. Find the lengths of $AD$ and $DC$ if $\cos\measuredangle BAD=\dfrac23,OC=\sqrt7,OB=3\sqrt{15}$ and $AB=5DC$.
The only thing I was able to gather is that $\measuredangle AOD=90^\circ$ as $$\measuredangle DAO+\measuredangle ADO=\dfrac12\measuredangle BAD+\dfrac12\measuredangle ADC=\dfrac12(\measuredangle BAD+\measuredangle ADC)=\dfrac12 180^\circ=90^\circ\\\Rightarrow \measuredangle AOD=90^\circ$$ I don't see how I can use the given lengths as for example in triangle $BOC$ we have only 2 elements. And of course we can say $DC=x\Rightarrow AB=5x$.
$\begingroup$Find cosines of $BAO$, $CDO$. Mark $AD=y$, $DC=x$, $AB=5x$. Express $DO$ in terms of $y$. Write cosine rule for $CO$ and $BO$. Then solve system of two equations for $x$ and $y$.$\endgroup$
Multiplying the first of these equations by $25$ and subtracting gives
$$40=\frac{10}{3}y^2\implies y=2\sqrt{3}$$
Substituting this back into the first equation gives
$$x^2-\frac23\sqrt{3}x-5=0\implies x=\frac53\sqrt{3}$$ | 677.169 | 1 |
Subject areas
For educators
V5 Projection of vectors
Open image
In Resolution of vectors we learned how to resolve vectors in two dimensions along horizontal and vertical axes.
It is also possible to resolve one vector along the line of another vector (instead of along the x-y axes).
Learn how to find the projection (resolution) of one vector in the direction of a second vector.
There are two types of vector projection:
Scalar projection
Vector projection
For scalar projection, we calculate the length (a scalar quantity) of a vector in a particular direction.
For vector projection we calculate the vector component of a vector in a given direction.
Often, in Physics, Engineering and Mathematics courses you are asked to resolve a vector into two component vectors that are perpendicular to one another. As an example, in the diagram below a vector \(\vec{a}\) is the projection of \(\overrightarrow{F}\) in the horizontal direction while \(\vec{b}\) is the projection of \(\overrightarrow{F}\) in the vertical direction.
You can project a vector in any direction, not only horizontally and vertically.
This module discusses both scalar and vector projections.
Scalar Projection
Consider the following diagram:
Let \(\overrightarrow{PQ}=\vec{a}\) and \(\overrightarrow{PS}=\vec{b}.\) The scalar projection of the vector \(\vec{a}\) in the direction of vector \(\vec{b}\) is the length of the straight line \(PR\) or \(\left|\overrightarrow{PR}\right|\)where 11 Remember that the magnitude (or length) of a vector \(\vec{v}=v_{1}\vec{i}+v_{2}\vec{j}+v_{3}\vec{k}\) is denoted by \[\begin{align*}
\left|\vec{v}\right| & =\sqrt{v_{1}^{2}+v_{2}^{2}+v_{3}^{2}}
\end{align*}\] For example, if \(\vec{v}=2\vec{i}-3\vec{j}+\vec{k}\) then \[\begin{align*}
\left|\vec{v}\right| & =\sqrt{2^{2}+\left(-3\right)^{2}+1^{2}}\\
& =\sqrt{4+9+1}\\
& =\sqrt{14}
\end{align*}\]
where \(\hat{b}=\frac{\vec{b}}{\left|\vec{b}\right|}\) is the unit vector in the direction of \(\vec{b}\).33 A unit vector for the vector \(\vec{b}\) is denoted by \(\hat{b}\) and is the vector \(\vec{b}\) divided by its length \(\left|\vec{b}\right|\). That is \[\begin{align*}
\hat{b} & =\frac{\vec{b}}{\left|\vec{b}\right|}.
\end{align*}\] A unit vector is a vector of length one in the direction of the original vector.
The scalar projection of a vector \(\vec{a}\) in the direction of vector \(\vec{b}\) is given by:
Vector Projection
The vector projection of a vector \(\vec{a}\) in the direction of vector \(\vec{b}\) is a vector in the direction of \(\vec{b}\) with magnitude equal to the length of the straight line \(PR\) or \(\left|\overrightarrow{PR}\right|\) as shown below.
Therefore the vector projection of \(\vec{a}\) in the direction of \(\vec{b}\) is the scalar projection multiplied by a unit vector in the direction of \(\vec{b}\).
The vector projection of vector \(\vec{a}\) in the direction of vector \(\vec{b}\) is:
so that the vector projection of \(\vec{a}\) in the direction of \(\vec{b}\) is \[\begin{alignat*}{1}
\left(\vec{a}\cdot\hat{b}\right)\hat{b} & =\left(\frac{13}{\sqrt{33}}\right)\frac{(5,-2,2)}{\sqrt{33}}\\
& =\frac{13(5,-2,2)}{33}\\
& =\frac{13}{33}(5,-2,2)\\
& =\frac{13}{33}\left(5\hat{i}-2\hat{j}+2\hat{k}\right)
\end{alignat*}\] The vector projection is \(\frac{13}{33}(5,-2,2)\) or \(\frac{13}{33}\left(5\hat{i}-2\hat{j}+2\hat{k}\right)\) | 677.169 | 1 |
Which of the following options represent the three undefined terms in geometry
Get an answer to your question ✅ "Which of the following options represent the three undefined terms in geometry ..." in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. | 677.169 | 1 |
Parallelograms: Quick Investigation sidesAny 2 interior angles of a parallelogram that are not opposite angles are called consecutive angles. In the animation above, the two different colored angles are consecutive angles. What can we conclude here?
ConsecutiveA diagonal of a parallelogram is a segment that connects one vertex (corner) to its opposite vertex. What can we conclude about the diagonals of a parallelogram? | 677.169 | 1 |
If you work in a business that requires the use of mathematics, for example then it would be very important that you are aware of t The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure. (In addition, the square is a special case or type of both the rectangle and the rhombus.) Synonyms: foursquare, quadrate, candid… Antonyms: biased, ex parte, inequitable… Find the right word. So, as you may have guessed, I'll start by consider your "shape" as a list of segments that completes a loop. 799k members in the shittyaskscience community. Opposite reciprocals are numbers with differing signs that are flipped fractions of each other. b. Compute the area of one The opposite (facing) sides are parallel. Properties A square is a special case of a rhombus (equal sides, opposite equal angles), a kite (two pairs of adjacent equal sides), a trapezoid (one pair of opposite sides parallel), a parallelogram (all opposite sides parallel), a quadrilateral or tetragon (four-sided polygon), and a rectangle (opposite sides equal, right-angles), and therefore has all the properties of all these shapes, namely: Once you insert shapes on a slide, you may realize that it's not the perfect size.Do you want it larger, or a wee bit smaller? The fundamental definition of a square is as follows: A square is both a rectangle and a rhombus and inherits the properties of both (except with both sides equal to each other). Square A square has four sides of equal length. Another word for shape. We come across different types of objects and materials that are fundamentally governed by specific geometric aspects, which make them appear unique in their own manner. A quadrilateral is a rhombus, if All the sides are of equal length-Specified 2 pairs of sides are parallel to each other. Find more ways to say shape, along with related words, antonyms and example phrases at Thesaurus.com, the world's most trusted free thesaurus. Square: having four equal sides and four right angles. 155 votes, 51 comments. For K-12 kids, teachers and parents. So, to get the properties of a Use the shape on … Right-angle A Square is a quadrilateral in which all its sides have equal length and all the four corners are right angles. Back to You can also drag the origin point at (0,0), or drag the square itself. For example, the square root of four is two, and two square… Although relatively simple and straightforward to deal with, squares have several interesting and notable properties. Opposite angles add up to 180 . The three-level hierarchy you see with in the above quadrilateral family tree works just like A dog is a […] Square The sides of a square are all equal (the same) length, making it a regular quadrilateral. I'm sure this is me. [1] [2] [3] This sense of the word " square " originated with the American jazz community in the 1940s, in reference to people out of touch with musical trends. The opposite of finding the square root of a number is squaring the number. But crop to shape is like one of those things few people in the world know about ... except likely the folks here. It has four right angles (90 ). noun — square adjective — square Phrases with "Square" We can talk about a square meal, meaning a proper meal that will kill your hunger and make you feel nice and happy and ready to take on the world.It may or There are six special quadrilaterals with different properties. So when is a square not a rectangle? It's just another type of quadrilateral. 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First principles of Euclid: an introduction to the study of the first book ...
want to speak of it, we might point to it and say, that straight line. But if we want to speak of it without pointing to it, we must give it a name. Just as we give names to things, that we may be able to talk about them; so we give names to points and lines, and circles and triangles, that we may be able to talk about them. Now we might call this line the line Fchn, or by any other such name. But we find it more convenient to call geometrical figures by the letters of the alphabet.
Thus we may call this line AB; placing the letter A at one end, and the letter B at the other. So that when we speak of the line AB, the letters A B are simply the name by which we distinguish that particular line from all other lines.
You know, I daresay, that a point marks position; and that the extremities (or ends) of lines are points. Now let us make a syllogism about the line AB. (Refer to general syllogism on page 10). Here is one:
(a.) Every line has two points.
(b.) The figure A B is a line.
(c.) Therefore the figure A B has two points. It is not always necessary to write out the whole of a syllogism in this way. Sometimes one of the premisses is so evident that it is left out, and only one premiss and the conclusion given. But, whether expressed or understood, two premisses are absolutely necessary to form a conclusion. To distinguish one premiss from the other, the first is called the major premiss, and the second the minor premiss.
In the above syllogism for instance, we might leave out the major premiss (a), and the argument would stand thus:
The figure A B is a line.
Therefore the figure A B has two points.
Here the major premiss (every line has two points) is left to be understood, but is none the less necessary to the conclusion.
EXERCISE.
Put in the Premiss which has been omitted in each of the following Syllogisms.
Conclusion... A B and CD if produced will never
meet.
F
D
Minor Premiss. A B and CD are E each equal to E F
Conclusion... AB is equal to CD. C
A PROPOSITION.
A proposition in geometry is something proposed to be done or proved. If something is to be done, the proposition is called a problem; if something is to be proved, the proposition is called a theorem. But in both cases a proof is required.
Thus in a problem, it is not enough to do what is required to be done. We must also prove, by syllogisms, that it is done.
Every proposition, whether problem or theorem, consists of two parts. That which is given, and that which is required.
Here is a proposition:
If two circles have equal radii, those circles are equal.
Here we are told that two circles have equal radii. That is said to be given. We are then asked to
prove that those circles are equal. This is said to be required.
Here is another proposition:
On a given straight line to describe an equilateral triangle.
NOTE.-To describe means to draw, to construct.
In this case we are given-a straight line; and we are required to describe an equilateral triangle on it. EXERCISE.-Rule your exercise paper in two columns, thus:
GIVEN.
REQUIRED.
and place each of the two parts of the following propositions in its proper column.
1. From a given point to draw a straight line equal to a given straight line.
2. From the greater of two given straight lines to cut off a part equal to the less.
3. If two triangles have two sides and the included angle of one, equal to two sides and the included angle of the other each to each, the bases shall be equal.
NOTE.-When two sides of a triangle have been mentioned, the third side is called the base.
4. The angles at the base of an isosceles triangle are equal.
5. If two straight lines cut one another, the vertical (or opposite) angles are equal.
AXIOMS.
An axiom is a self-evident truth. That is to say, an assertion so evidently true that it need not (or indeed cannot) be proved.
The following are the axioms of Euclid:
1. Things which are equal to the same thing are equal to one another.
2. If equals be added to equals, the wholes are equal.
2a. (Not given by Euclid, but assumed by him). If the same thing be added to equals the wholes are equal. 3. If equals be taken from equals, the remainders are equal.
3a. (Not given by Euclid, but assumed by him.) If the same thing be taken from equals, the remainders are equal.
4. If equals be added to unequals, the wholes are unequal.
4a. (Not given by Euclid, but assumed by him.) It the same thing be added to unequals the wholes are unequal.
5. If equals be taken from unequals the remainders are unequal.
6. Things which are doubles of the same thing are equal to one another.
6a. (Not given by Euclid, but assumed by him). Things which are doubles of equal things are equal to one another.
7. Things which are halves of the same thing are equal to one another.
8. Magnitudes which coincide with one another are equal to one another. | 677.169 | 1 |
Shape
Shapes are planes with edges. A plane is a flat surface extending infinitely in width and height. A plane can be parallel to the format containing space, or it can recede or protrude in space to create depth. Shapes have composition with boundaries which separate the space they enclose from what surrounds them.
Shapes have their own distinctive quality.
Geometric shapes are mathematically measured and possess straight edges and precise curves. Stylistically they have mechanical and somewhat unnatural appearance.
There are three basic geometric shapes: the square, triangle, and circle. These shapes have corresponding volumetric forms: the cube, pyramid, and sphere.
Geometric Shapes
Additionally, there are more complex geometric shapes such as parallelograms, trapezoids, octagons, equilateral, isosceles, and right triangles. Other volumetric forms include the cylinder, cone, or prism.
Organic shapes are derived from nature (plants and animals). Stylistically they are curvilinear, as straight lines are rare in nature, and suggest growth and movement.
Rectilinear shapes are bound by straight lines which are not mathematically related.
Curvilinear shapes appear to have no straight lines or edges.
Irregular shapes posses both straight and curved lines which are not mathematically related. This may include shapes accidentally or artistically created.
Works Cited
Brigham Young University of Idaho. "Design and Color". Notes from lectures and course materials, 2010. | 677.169 | 1 |
What is the cross section parallel to the base of a circular pyraimd?
A.) circle
B.) square
C.) parallelogram
D.) triangle
Get an answer to your question ✅ "What is the cross section parallel to the base of a circular pyraimd? A.) circle B.) square C.) parallelogram D.) triangle ..." in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. | 677.169 | 1 |
What do paths of comets, supersonic booms, ancient Grecian
pillars, and natural draft cooling towers have in common? They can
all be modeled by the same type of conic. For instance, when
something moves faster than the speed of sound, a shock wave in the
form of a cone is created. A portion of a conic is formed when the
wave intersects the ground, resulting in a sonic boom.
See Figure
1.
Figure 1 A shock
wave intersecting the ground forms a portion of a conic and results
in a sonic boom.
Most people are familiar with the sonic boom created by
supersonic aircraft, but humans were breaking the sound barrier
long before the first supersonic flight. The crack of a whip occurs
because the tip is exceeding the speed of sound. The bullets shot
from many firearms also break the sound barrier, although the bang
of the gun usually supersedes the sound of the sonic boom.
Locating the Vertices and Foci of a
Hyperbola
In analytic geometry, a hyperbola is
a conic section formed by intersecting a right circular cone with a
plane at an angle such that both halves of the cone are
intersected. This intersection produces two separate unbounded
curves that are mirror images of each other. See Figure
2.
Figure 2 A
hyperbola
Like the ellipse, the hyperbola can also be defined as a set of
points in the coordinate plane. A hyperbola is the set of all
points (x,y)(x,y) in a plane such that the difference of
the distances between (x,y)(x,y) and the foci is a
positive constant.
Notice that the definition of a hyperbola is very similar to
that of an ellipse. The distinction is that the hyperbola is
defined in terms of the difference of two distances, whereas the
ellipse is defined in terms of the sum of two distances.
As with the ellipse, every hyperbola has two axes of
symmetry. The transverse axis is a line
segment that passes through the center of the hyperbola and has
vertices as its endpoints. The foci lie on the line that contains
the transverse axis. The conjugate
axis is perpendicular to the transverse axis and has
the co-vertices as its endpoints. The center of a
hyperbola is the midpoint of both the transverse and
conjugate axes, where they intersect. Every hyperbola also has
two asymptotes that pass through its
center. As a hyperbola recedes from the center, its branches
approach these asymptotes. The central
rectangle of the hyperbola is centered at the origin
with sides that pass through each vertex and co-vertex; it is a
useful tool for graphing the hyperbola and its asymptotes. To
sketch the asymptotes of the hyperbola, simply sketch and extend
the diagonals of the central rectangle. See Figure
3.
Figure 3 Key
features of the hyperbola
In this section, we will limit our discussion to hyperbolas that
are positioned vertically or horizontally in the coordinate plane;
the axes will either lie on or be parallel to
the x-
and y-axes. We will consider
two cases: those that are centered at the origin, and those that
are centered at a point other than the origin.
Deriving the Equation of a Hyperbola
Centered at the Origin
Let (−c,0)(−c,0) and (c,0)(c,0) be
the foci of a hyperbola centered at the origin. The
hyperbola is the set of all points (x,y)(x,y) such that
the difference of the distances from (x,y)(x,y) to the
foci is constant. See Figure
4.
Figure 4
If (a,0)(a,0) is a vertex of the hyperbola, the
distance
from (−c,0)(−c,0) to (a,0)(a,0) is a−(−c)=a+c.a−(−c)=a+c. The
distance
from (c,0)(c,0) to (a,0)(a,0) is c−a.c−a. The
difference of the distances from the foci to the vertex is
(a+c)−(c−a)=2a(a+c)−(c−a)=2a
If (x,y)(x,y) is a point on the hyperbola, we can
define the following variables:
By definition of a hyperbola, d2−d1d2−d1 is constant
for any point (x,y)(x,y) on the hyperbola. We know that
the difference of these distances is 2a2a for the
vertex (a,0).(a,0). It follows
that d2−d1=2ad2−d1=2a for any point on the hyperbola. As
with the derivation of the equation of an ellipse, we will begin by
applying the distance formula. The rest of the derivation is
algebraic. Compare this derivation with the one from the previous
section for ellipses.
Note that the vertices, co-vertices, and foci are related by the
equation c2=a2+b2.c2=a2+b2. When we are given the
equation of a hyperbola, we can use this relationship to identify
its vertices and foci.
HOW
TO
Given the equation of a hyperbola in standard form,
locate its vertices and foci.
Determine whether the transverse axis lies on
the x- or y-axis. Notice that a2a2 is always under
the variable with the positive coefficient. So, if you set the
other variable equal to zero, you can easily find the intercepts.
In the case where the hyperbola is centered at the origin, the
intercepts coincide with the vertices.
If the equation has the
form x2a2−y2b2=1,x2a2−y2b2=1, then the transverse axis
lies on the x-axis. The
vertices are located at (±a,0),(±a,0), and the foci are
located at (±c,0).(±c,0).
If the equation has the
form y2a2−x2b2=1,y2a2−x2b2=1, then the transverse axis
lies on the y-axis. The
vertices are located at (0,±a),(0,±a), and the foci are
located at (0,±c).(0,±c).
Solve for aa using the
equation a=a2−−√.a=a2.
Solve for cc using the
equation c=a2+b2−−−−−−√.c=a2+b2.
EXAMPLE 1
Locating a Hyperbola's Vertices and
Foci
Identify the vertices and foci of the hyperbola with
equation y249−x232=1.y249−x232=1.
Answer
TRY
IT #1
Identify the vertices and foci of the hyperbola with
equation x29−y225=1.x29−y225=1.
Writing Equations of Hyperbolas in
Standard Form
Just as with ellipses, writing the equation for a hyperbola in
standard form allows us to calculate the key features: its center,
vertices, co-vertices, foci, asymptotes, and the lengths and
positions of the transverse and conjugate axes. Conversely, an
equation for a hyperbola can be found given its key features. We
begin by finding standard equations for hyperbolas centered at the
origin. Then we will turn our attention to finding standard
equations for hyperbolas centered at some point other than the
origin.
Hyperbolas Centered at the
Origin
Reviewing the standard forms given for hyperbolas centered
at (0,0),(0,0), we see that the vertices, co-vertices,
and foci are related by the
equation c2=a2+b2.c2=a2+b2. Note that this equation can
also be rewritten as b2=c2−a2.b2=c2−a2. This relationship
is used to write the equation for a hyperbola when given the
coordinates of its foci and vertices.
HOW
TO
Given the vertices and foci of a hyperbola centered
at (0,0),(0,0), write its equation in standard
form.
Determine whether the transverse axis lies on
the x- or y-axis.
If the given coordinates of the vertices and foci have the
form (±a,0)(±a,0) and (±c,0),(±c,0), respectively,
then the transverse axis is the x-axis. Use the standard
form x2a2−y2b2=1.x2a2−y2b2=1.
If the given coordinates of the vertices and foci have the
form (0,±a)(0,±a) and (0,±c),(0,±c), respectively,
then the transverse axis is the y-axis. Use the standard
form y2a2−x2b2=1.y2a2−x2b2=1.
Find b2b2 using the
equation b2=c2−a2.b2=c2−a2.
Substitute the values
for a2a2 and b2b2 into the standard form of the
equation determined in Step 1.
EXAMPLE 2
Finding the Equation of a Hyperbola
Centered at (0,0) Given its Foci and Vertices
What is the standard form equation of
the hyperbola that has
vertices (±6,0)(±6,0) and
foci (±210−−√,0)?(±210,0)?
Answer
TRY
IT #2
What is the standard form equation of the hyperbola that has
vertices (0,±2)(0,±2) and
foci (0,±25–√)?(0,±25)?
Hyperbolas Not Centered at the
Origin
Like the graphs for other equations, the graph of a hyperbola
can be translated. If a hyperbola is translated hh units
horizontally and kk units vertically, the center of
the hyperbola will be (h,k).(h,k). This
translation results in the standard form of the equation we saw
previously, with xx replaced
by (x−h)(x−h) and yy replaced
by (y−k).(y−k).
STANDARD FORMS OF THE EQUATION OF A
HYPERBOLA WITH CENTER (H, K)
The standard form of the equation of a hyperbola with
center (h,k)(h,k) and transverse axis parallel to
the x-axis is
(x−h)2a2−(y−k)2b2=1(x−h)2a2−(y−k)2b2=1
where
the length of the transverse axis is 2a2a
the coordinates of the vertices are (h±a,k)(h±a,k)
the length of the conjugate axis is 2b2b
the coordinates of the co-vertices are (h,k±b)(h,k±b)
the distance between the foci
is 2c,2c, where c2=a2+b2c2=a2+b2
the coordinates of the foci are (h±c,k)(h±c,k)
The asymptotes of the hyperbola coincide with the diagonals of
the central rectangle. The length of the rectangle
is 2a2a and its width is 2b.2b. The slopes of
the diagonals are ±ba,±ba, and each diagonal passes
through the center (h,k).(h,k). Using
the point-slope formula, it is simple to show
that the equations of the asymptotes
are y=±ba(x−h)+k.y=±ba(x−h)+k. See Figure
7a
The standard form of the equation of a hyperbola with
center (h,k)(h,k) and transverse axis parallel to
the y-axis is
(y−k)2a2−(x−h)2b2=1(y−k)2a2−(x−h)2b2=1
where
the length of the transverse axis is 2a2a
the coordinates of the vertices are (h,k±a)(h,k±a)
the length of the conjugate axis is 2b2b
the coordinates of the co-vertices are (h±b,k)(h±b,k)
the distance between the foci
is 2c,2c, where c2=a2+b2c2=a2+b2
the coordinates of the foci are (h,k±c)(h,k±c)
Using the reasoning above, the equations of the asymptotes
are y=±ab(x−h)+k.y=±ab(x−h)+k. See Figure
7b.
Like hyperbolas centered at the origin, hyperbolas centered at a
point (h,k)(h,k) have vertices, co-vertices, and foci
that are related by the equation c2=a2+b2.c2=a2+b2. We
can use this relationship along with the midpoint and distance
formulas to find the standard equation of a hyperbola when the
vertices and foci are given.
HOW
TO
Given the vertices and foci of a hyperbola centered
at (h,k),(h,k), write its equation in standard
form.
Determine whether the transverse axis is parallel to
the x- or y-axis.
If the y-coordinates of the
given vertices and foci are the same, then the transverse axis is
parallel to the x-axis. Use the
standard form (x−h)2a2−(y−k)2b2=1.(x−h)2a2−(y−k)2b2=1.
If the x-coordinates of the
given vertices and foci are the same, then the transverse axis is
parallel to the y-axis. Use the
standard form (y−k)2a2−(x−h)2b2=1.(y−k)2a2−(x−h)2b2=1.
Identify the center of the
hyperbola, (h,k),(h,k), using the midpoint formula and
the given coordinates for the vertices.
Find a2a2 by solving for the length of the transverse
axis, 2a2a , which is the distance between the given
vertices.
Find c2c2 using hh and kk found
in Step 2 along with the given coordinates for the foci.
Solve for b2b2 using the
equation b2=c2−a2.b2=c2−a2.
Substitute the values
for h,k,a2,h, k, a2, and b2b2 into
the standard form of the equation determined in Step 1.
EXAMPLE 3
Finding the Equation of a Hyperbola
Centered at (h, k) Given its
Foci and Vertices
What is the standard form equation of
the hyperbola that has vertices
at (0,−2)(0,−2) and (6,−2)(6,−2) and foci
at (−2,−2)(−2,−2) and (8,−2)?(8,−2)?
Answer
TRY
IT #3
What is the standard form equation of the hyperbola that has
vertices (1,−2)(1,−2) and (1,8)(1,8) and
foci (1,−10)(1,−10) and (1,16)?(1,16)?
Graphing Hyperbolas Centered at the
Origin
When we have an equation in standard form for a hyperbola
centered at the origin, we can interpret its parts to identify the
key features of its graph: the center, vertices, co-vertices,
asymptotes, foci, and lengths and positions of the transverse and
conjugate axes. To graph hyperbolas centered at the origin, we use
the standard form x2a2−y2b2=1x2a2−y2b2=1 for horizontal
hyperbolas and the standard
form y2a2−x2b2=1y2a2−x2b2=1 for vertical hyperbolas.
HOW
TO
Given a standard form equation for a hyperbola centered
at (0,0),(0,0), sketch the graph.
Determine which of the standard forms applies to the given
equation.
Use the standard form identified in Step 1 to determine the
position of the transverse axis; coordinates for the vertices,
co-vertices, and foci; and the equations for the asymptotes.
If the equation is in the
form x2a2−y2b2=1, x2a2−y2b2=1, then
the transverse axis is on the x-axis
the coordinates of the vertices are (±a,0)(±a,0)
the coordinates of the co-vertices are (0,±b)(0,±b)
the coordinates of the foci are (±c,0)(±c,0)
the equations of the asymptotes are y=±baxy=±bax
If the equation is in the
form y2a2−x2b2=1,y2a2−x2b2=1, then
the transverse axis is on the y-axis
the coordinates of the vertices are (0,±a)(0,±a)
the coordinates of the co-vertices are (±b,0)(±b,0)
the coordinates of the foci are (0,±c)(0,±c)
the equations of the asymptotes are y=±abxy=±abx
Solve for the coordinates of the foci using the
equation c=±a2+b2−−−−−−√.c=±a2+b2.
Plot the vertices, co-vertices, foci, and asymptotes in the
coordinate plane, and draw a smooth curve to form the
hyperbola.
EXAMPLE 4
Graphing a Hyperbola Centered at (0,
0) Given an Equation in Standard Form
Graph the hyperbola given by the
equation y264−x236=1. y264−x236=1. Identify and label
the vertices, co-vertices, foci, and asymptotes.
Answer
TRY
IT #4
Graph the hyperbola given by the
equation x2144−y281=1.x2144−y281=1. Identify and label
the vertices, co-vertices, foci, and asymptotes.
Graphing Hyperbolas Not Centered at
the Origin
Graphing hyperbolas centered at a
point (h,k)(h,k) other than the origin is similar to
graphing ellipses centered at a point other than the origin. We use
the standard
forms (x−h)2a2−(y−k)2b2=1(x−h)2a2−(y−k)2b2=1 for
horizontal hyperbolas,
and (y−k)2a2−(x−h)2b2=1(y−k)2a2−(x−h)2b2=1 for vertical
hyperbolas. From these standard form equations we can easily
calculate and plot key features of the graph: the coordinates of
its center, vertices, co-vertices, and foci; the equations of its
asymptotes; and the positions of the transverse and conjugate
axes.
HOW
TO
Given a general form for a hyperbola centered
at (h,k),(h, k), sketch the graph.
Convert the general form to that standard form. Determine which
of the standard forms applies to the given equation.
Use the standard form identified in Step 1 to determine the
position of the transverse axis; coordinates for the center,
vertices, co-vertices, foci; and equations for the asymptotes.
If the equation is in the
form (x−h)2a2−(y−k)2b2=1,(x−h)2a2−(y−k)2b2=1, then
the transverse axis is parallel to the x-axis
the center is (h,k)(h,k)
the coordinates of the vertices are (h±a,k)(h±a,k)
the coordinates of the co-vertices are (h,k±b)(h,k±b)
the coordinates of the foci are (h±c,k)(h±c,k)
the equations of the asymptotes
are y=±ba(x−h)+ky=±ba(x−h)+k
If the equation is in the
form (y−k)2a2−(x−h)2b2=1,(y−k)2a2−(x−h)2b2=1, then
the transverse axis is parallel to the y-axis
the center is (h,k)(h,k)
the coordinates of the vertices are (h,k±a)(h,k±a)
the coordinates of the co-vertices are (h±b,k)(h±b,k)
the coordinates of the foci are (h,k±c)(h,k±c)
the equations of the asymptotes
are y=±ab(x−h)+ky=±ab(x−h)+k
Solve for the coordinates of the foci using the
equation c=±a2+b2−−−−−−√.c=±a2+b2.
Plot the center, vertices, co-vertices, foci, and asymptotes in
the coordinate plane and draw a smooth curve to form the
hyperbola.
EXAMPLE 5
Graphing a Hyperbola Centered at
(h, k) Given an Equation in General Form
Graph the hyperbola given by the
equation 9x2−4y2−36x−40y−388=0.9x2−4y2−36x−40y−388=0. Identify
and label the center, vertices, co-vertices, foci, and
asymptotes.
Answer
TRY
IT #5
Graph the hyperbola given by the standard form of an
equation (y+4)2100−(x−3)264=1.(y+4)2100−(x−3)264=1. Identify
and label the center, vertices, co-vertices, foci, and
asymptotes.
Solving Applied Problems Involving
Hyperbolas
As we discussed at the beginning of this section, hyperbolas
have real-world applications in many fields, such as astronomy,
physics, engineering, and architecture. The design efficiency of
hyperbolic cooling towers is particularly interesting. Cooling
towers are used to transfer waste heat to the atmosphere and are
often touted for their ability to generate power efficiently.
Because of their hyperbolic form, these structures are able to
withstand extreme winds while requiring less material than any
other forms of their size and strength. See Figure
10. For example, a 500-foot tower can be made of a
reinforced concrete shell only 6 or 8 inches wide!
Figure 10 Cooling
towers at the Drax power station in North Yorkshire, United Kingdom
(credit: Les Haines, Flickr)
The first hyperbolic towers were designed in 1914 and were 35
meters high. Today, the tallest cooling towers are in France,
standing a remarkable 170 meters tall. In Example
6 we will use the design layout of a cooling tower to
find a hyperbolic equation that models its sides.
EXAMPLE 6
Solving Applied Problems Involving
Hyperbolas
The design layout of a cooling tower is shown
in Figure
11. The tower stands 179.6 meters tall. The diameter of the
top is 72 meters. At their closest, the sides of the tower are 60
meters apart.
Figure 11 Project
design for a natural draft cooling tower
Find the equation of the hyperbola that models the sides of the
cooling tower. Assume that the center of
the hyperbola—indicated by the intersection of dashed
perpendicular lines in the figure—is the origin of the coordinate
plane. Round final values to four decimal places.
Answer
TRY
IT #6
A design for a cooling tower project is shown
in Figure
12. Find the equation of the hyperbola that models the
sides of the cooling tower. Assume that the center of the
hyperbola—indicated by the intersection of dashed perpendicular
lines in the figure—is the origin of the coordinate plane. Round
final values to four decimal places.
Figure 12
MEDIA
Access these online resources for additional instruction and
practice with hyperbolas.
Extensions
For the following exercises, express the equation for the
hyperbola as two functions, with yy as a function
of x.x. Express as simply as possible. Use a graphing
calculator to sketch the graph of the two functions on the same
axes.
The hedge will follow the
asymptotes y=34xy=34x and y=−34x,y=−34x, and
its closest distance to the center fountain is 20 yards.
For the following exercises, assume an object enters our solar
system and we want to graph its path on a coordinate system with
the sun at the origin and the x-axis as the axis of symmetry for
the object's path. Give the equation of the flight path of each
object using the given information.
66.
The object enters along a path approximated by the
line y=x−2y=x−2 and passes within 1 au (astronomical
unit) of the sun at its closest approach, so that the sun is one
focus of the hyperbola. It then departs the solar system along a
path approximated by the line y=−x+2.y=−x+2.
The object enters along a path approximated by the
line y=2x−2y=2x−2 and passes within 0.5 au of the sun at
its closest approach, so the sun is one focus of the hyperbola. It
then departs the solar system along a path approximated by the
line y=−2x+2.y=−2x+2.
68.
The object enters along a path approximated by the
line y=0.5x+2y=0.5x+2 and passes within 1 au of the sun
at its closest approach, so the sun is one focus of the hyperbola.
It then departs the solar system along a path approximated by the
line y=−0.5x−2.y=−0.5x−2.
The object enters along a path approximated by the
line y=13x−1y=13x−1 and passes within 1 au of the sun at
its closest approach, so the sun is one focus of the hyperbola. It
then departs the solar system along a path approximated by the
line y=−13x+1.y=−13x+1.
70.
The object enters along a path approximated by the
line y=3x−9y=3x−9 and passes within 1 au of the sun at
its closest approach, so the sun is one focus of the hyperbola. It
then departs the solar system along a path approximated by the
line y=−3x+9.
This page titled 10.3: The Hyperbola | 677.169 | 1 |
If both circles had the same radius, X would be located in the center between Ea and Eb. In my case, I used $A = (350 ; 350), B = (150 ; 200), r_1 = 175, r_2 = 125$. X was about X = (226.326 ; 257.245). After thinking about it, I came up with the idea, that X has to be located at (r2/r1)*l.
But $(r_2/r_1) = (125/175) = 0.7142857142857143$, I measured a relation of $0.6896173016811695$.
$\begingroup$It'll have to wait a day or so. You can do it yourself if you want: I wrote $\ell$ for the distance between $A$ and $B$. In those terms, the distance between $E_A$ and $E_B$ (which is what you're calling $l$) is $$\ell - (\ell - r_1) - (\ell - r_2) = r_1 + r_2 - \ell.$$$\endgroup$ | 677.169 | 1 |
In this section I examine a kind of generation of a circumparabola connected with bundles of parallel lines.
I start with an exercise on some line-intersections:
Given a triangle ABC and two independent directions {OX, OY} draw from the vertices of the triangle lines parallel to the two directions. These parallels create three parallelograms having the sides of the given triangle as diagonals. The other three diagonals of these parallelograms pass trhough one point P.
The diagonals {A''C'',AC,A0C0} of the created parallelograms {A''B''C''B, CB''AB0, B0A0BC0} intersect at a point (see (2) of MenelausApp.html ). This implies that triangles {A''B''C'', A0B0C0} have pairs of sides intersecting along the line AC i.e. they are line-perspective. By Desargue's theorem they are also point-perspective. P is the center of this perspectivity.
In the previous exercise fix the direction OX and let OY describe all possible directions of the plane. Then the corresponding point P describes the parabola passing through the vertices of triangle A'B'C' and having its axis parallel to OX. Here {A',B',C'} are the middles of the sides of ABC.
A proof results by showing that P describes a conic and identifying the conic with the claimed parabola. This can be easily done using {A',B',C',D'} as a projective basis (D' being the centroid of A'B'C') and calculating the intersection point P of lines in this coordinate system. The resulting equation is a quadratic one.
Then it is also easy to see that for appropriate positions of OY point P obtains the positions {A',B',C',A'',B'',C''}, where the three last points are the intersections of lines parallel to OX from {A,B,C} with the sides of ABC.
Using the results of AnticomplementaryAndCircumparabola.html , the previous calculations show that the conic coincides with the parabola claimed.
Remark-1 Letting also direction OX obtain all possible values we get all parabolas circumscribed on triangle A'B'C'. This is discussed in AllParabolasCircumscribed.html . Remark-2 This method to generate a circumparabola can be very easily generalized to arbitrary circumconics which are also tangent to a given line. This is discussed in CircumconicsTangentGeneration.html . | 677.169 | 1 |
Geometry Unit 1: Basic Geometry Geometry Unit 1 ... - Weebly 1. What Is The Measure Of Each Interior Angle In A Regular Octagon? 1) 108º 2) 135º 3) 144º 4) 1080º 2. The Sum Of The Interior Angles Of A Regular Polygon Is 540°. Determine And State The Number Of Degrees In One Interior Angle Of The Polygon. 3. The Measure Of An Interior Angle Of A Regular Polygon May 4th, 2024
SAMPLE TEST ITEMS Geometry - BTW Geometry Class For Mathematics Will Be Assessed On The Geometry EOC Test. A Discussion Of Each Item Highlights The Knowledge And Skills The Item Is Intended To Measure. It Is Important To Remember That These Sample Items Represent Only A Portion Of The Knowledge And Skill Jun 13th, 2024
Chapter 8 Chapter Test A Answer Key Mcdougal Littell Geometry Across From The YMCA, Old Bridge. A Variety Of Shapes And Sizes Of Pumpkins Will Be For Sale, Community Bulletin Board: The Suburban (for Oct. 27) Memory And The Dissolution Of The Monasteries In Early Modern England - October 2021 Chapter 1 - 'no News But The Abbeys Shall Be Down' The Av May 2th, 2024
Chapter 5 Vector Geometry 5 VECTOR GEOMETRY Chapter 5 Vector Geometry Eliminating M And N, X =()1−z +()1−y ⇒ X +y +z =2 As Deduced In The Previous Example. Activity 3 Deduce The General Equation Of A Plane Passing Through The Point A, Where OA → =a, And Such That The Vectors S And T Are Parallel To The Plane. Example The Lines L1 And L2 Have Equations R=()3i+j−k +α()i+2j+3k Jan 14th, 2024
Geometry Notes – Chapter 1: Essentials Of Geometry 1.7 – Perimeter, Circumference And Area : Polygon: Formulas: A Polygon Is A Closed Plane Figure With At Least 3 Sides. None Of The Sides May Intersect At Any Other Point Than Their Endpoints. Square : Ps =4. As = 2. Rectangle . Pl W =22+ A =l May 7th, 2024
Chapter 1 Introducing Geometry And Geometry Proofs Chapter 1: Introducing Geometry And Geometry Proofs 13 5. Give Two Examples Of Theorems That Are Not Reversible And Explain Why The Reverse Of Each Is False. Hint: Flip Through This Book Or Your Geometry Textbook Looking At Various Theorems. Try Reversing Them And Ask Yourself Whether They Still Work. Solve It 6. Give Two Examples Of Theorems ... Jan 7th, 2024
High School Geometry Chapter 1: Essentials Of Geometry ... High School – Geometry Note: If A Chapter Section Is Not Listed, It Is Meant To Be Skipped. **This Is The First Time Students Are Seeing This Topic. Develop To An Appropriate Level For The Class. 1 Chapter 1: Essentials Of Geometry Approximately 10 Days Sections Covered: 1.1: Identify Points, Lines & Pl Mar 2th, 2024
Geometry Chapter : Tools Of Geometry Geometry Chapter : Tools Of Geometry This Assignment List Is Given For Your Convenience. Use This List To Keep Track Of The Upcoming Assignments. This List Is Subject To Change, I Reserve The Right To Add Or Alter Assignments. Remember: You Are Responsible Fo Apr 2th, 2024
Chapter 1 Basics Of Geometry Name Geometry 3. Are Coplanar ... 4. Two Intersecting Planes, P And Q, With Where G Is In Plane P And H Is In Plane Q. 5. Four Non-collinear Points I, J, K, And L, With Line Segments Connecting All Points To Each Other. 6. Label This Line In Five Different Ways: 7. Label The Geometric Figure Below In Two Different Ways: 8. Jan 13th, 2024 | 677.169 | 1 |
(a) Prove that the shear center of an equal-leg angle (Fig. 1) is at the corner of the angle, and prove that the resultant of the shear flow in the legs of the angle is equal to the total shear force on the section. The shear force acts normal to the axis of symmetry of the cross section. Assume that t \ll b. (b) Determine the maximum shear stress in the cross section in Fig. 1.
Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.
Plan the Solution There is a resultant shear force along each leg of the angle. Since the lines of action of these two forces pass through the corner O, this point must be the shear center. We could develop an expression for this using basic shear-flow concepts (i.e., Eq. 6.65), or we can make use of Eq. 6.95, setting h_w = 0 and Φ = 45°. The maximum shear stress should occur at the neutral axis.
(a) Determine shear-center location and prove that the resultant of flange shears is V. Since V_1 and V_2 both pass through point O, as indicated in Fig. 2, the resultant, V, must pass through this point.
Therefore, the shear center lies at the intersection of the legs of the angle.^{35} The flange shear is given by Eq. 6.95, with h_w = 0.
V_f = \frac{Vtb^3 \sin 45°}{3I} (1)
where the moment of inertia is obtained by specializing Eq. 6.92 to give
we have shown that combining Eqs. 6.92 and 6.95 leads to the correct resultant shear force.
(b) Determine the maximum shear stress. The shear flow distribution is given by Eq. 6.94, which specializes to
where I is given by Eq. (2). This shear-flow distribution is depicted in Fig. 3. Since τ_f = q_f /t,
(τ_f)_{max} = \frac{(q_f)_{s=b}}{t}
or
(τ_f)_{max} = \left(\frac{3\sqrt{2}}{4}\right) \frac{V}{bt} (5)
Review the Solution As a rough check on our answer for τ_{max}, we can compare the preceding result, Eq. (5), with τ_{max} for a rectangle. From Example Problem 6.14,
(τ_{max})_{\text{rectangle}} = \frac{3}{2}\frac{V}{A}
Since the area of the equal-leg angle is 2bt, Eq. (5) can be expressed as
(τ_{f max})_{\text{angle}} = \frac{3\sqrt{2}} {2} \frac{V}{A}
Since the shear flows in the angle are not parallel with the resultant V, it is reasonable that the shear stress in the angle section is greater than that in a rectangle by a factor of \sqrt{2}.
See Homework Problem 6.12–16 for a chance to use the basic shear flow formula, Eq. 6.65, to obtain the above results. | 677.169 | 1 |
Show that a triangle $T$ in ${\bf H}$ is an equilateral triangle
(that is, all sides have the same length) if and only if its
angles are all equal.
\medskip
\noindent Now, let $\alpha$ be the angle at a vertex of $T$ and
let $a$ be the hyperbolic length of a side of $T$. Show that
$2\cosh(\frac{1}{2} a) \sin(\frac{1}{2}\alpha) = 1$.
\medskip
{\bf Answer}
\begin{center}
\epsfig{file=407-10-3.eps, width=60mm}
\end{center}
\un{by lcI}: all angles are equal and have
$\cos(\alpha)=\ds\frac{\cosh^2(a)-\cosh(a)}{\sinh^2(a)}$ (and the
fact that an angle in the range $[0,\pi]$ is completely determined
by its cosine)
(the converse, that equal angles imply equal side lengths follows
immediately from lcII, with
$$\cosh(a)=\ds\frac{\cos(\alpha)-\cos^2(\alpha)}{\sin^2(\alpha)})$$
The bisector of the angles intersects the opposite side in a right
angle by a geometric argument, namely the triangle is taken to
itself by reflection in the bisecting line. The same argument
shows that the bisecting line intersects the opposite side in its
midpoint.
Now use \un{lcII}.
$\cos(\frac{1}{2}\alpha)=-\cos(\alpha)\cos(\frac{\pi}{2})+
\sin(\alpha)\sin(\frac{\pi}{2})\cosh(\frac{1}{2}a)$
$\cosh(\frac{1}{2}a)=\ds\frac{\cos(\frac{1}{2}\alpha)}{\sin(\alpha)}=
\ds\frac{1}{2\sin(\frac{1}{2}\alpha)}$
(since
$\sin(\alpha)=2\sin(\frac{1}{2}\alpha)\cos(\frac{1}{2}\alpha))$
and so $2\cosh(\frac{1}{2}a)\sin(\frac{1}{2}\alpha)=1$ as desired.
\end{document} | 677.169 | 1 |
Important Notes & Short Tricks on Height & Distance with Questions for SSC Exams 2023
By BYJU'S Exam Prep
Updated on: September 25th, 2023
Heights and Distance is a study of Trigonometry applications. It has various real-world uses, such as measuring the height of an object, the depth of an object, the distance between two celestial objects, and so on. Trigonometric formulas and ratios are useful in the study of heights and distances.
In this post, we will talk about heights and distances basics, short tricks along with some important questions.
Trigonometry is one of the important chapters for the upcoming SSC & Railways exams 2023. We will be covering a very important topic Height & Distance from this section. It is important to learn the basics of this topic before moving further.
Some of the important terms associated with Height & Distance are as follows:
Line of Sight
Angle of Elevation
Angle of Depression
Line of Sight
The line of sight is the imaginary line drawn from the observer's eye to an item. It indicates where an observer's gaze is directed. Furthermore, the line of sight indicates the type of inclination angle established with horizontal.
Angle of Elevation
The angle of elevation is the angle produced by the line of sight with the horizontal when the observer is looking up at an object. When the viewer raises his head over the horizontal line, this angle is generated.
Angle of Depression
It is the angle of inclination formed by the line of sight with the horizontal when the observer is viewing the object downwards. This angle is formed when the observer lowers his head downwards from the top to the ground.
Trigonometric Ratio
Trigonometric ratios are used to address height and distance problems. We know that the angle's trigonometric ratio gives the relationship between any two sides of a triangle. Let us go over all of the trigonometric ratios in depth.
Some specific Ratios are as follows:
Short Tricks on Height & Distance with Questions
Angle of Elevation: Let AB be a tower/pillar/shell/minar/pole etc.) standing at any point C on the level ground is viewing at A.
The angle, which the line AC makes with the horizontal line BC is called the angle of elevation .so angle ACB is the angle of elevation.
Angle of Depression: If an observer is at Q and is viewing an object R on the ground, then the angle between PQ and QR is the angle of depression .so angle PQR is the angle of depression.
Numerically angle of elevation is equal to the angle of depression.
Both angles are measured with the horizontal.
The thread of a kite is 120 m long and it is making 30° angular elevation with the ground. What is the height of the kite?
Solution:
Sin 30° = h/120
1/2 = h/120
h = 60m
A tree bent by the wind. The top of the tree meets the ground at an angle of 60°.If the distance between the top of the foot is 8 m then what was the height of the tree?
Solution:
tan 60° = x/8
√3 = x/8
x = 8 √3
y cos 60° = 8/y
1/2 = 8/y
y = 16
therefore the height of the tree = x+y
= 8√3+16
= 8(√3+2)
The angle of elevation of the top of a tower from a point on the ground is 30°. On walking 100m towards the tower the angle of elevation changes to 60°. Find the height of the tower.
Solution:
In right triangle ABD,
tan 60° = h/x
√3 x = h
x = h/√3
Again , in right triangle ABC ,
tan 30 = h/x+100
1/√3 = h/x+100
√3 h = x+100
√3 h = h/√3 + 100
√3 h – h/√3 =100
3 h – h/√3 =100
2 h = 100√3
h = 50√3
By short trick:
d = h (cot Ɵ1 – cot Ɵ2)
h = 100/(√3-1/√3) = 100*√3/2 = 50√3
Ɵ1 = small angle
Ɵ2 = large angle
d = distance between two places
h = height
From the top of a temple near a river the angles of depression of both the banks of river are 45° & 30°. If the height of the temple is 100 m then find out the width of the river.
Solution:
tan 45° = AB/BD
1 = 100/BD
BD = 100
tan 30 ° = AB/BC
1/√3 = 100/BC
BC = 100 √3
Width of the river , CD = BC – BD = 100 (√3-1)
When the height of the tower is 1 m then the width of the river is √3-1
Since the height of the tower is 100 m
Therefore,
Width of river is 100(√3-1)m
By short trick:
The same formula can be used in this question too i.e.
d= h (cot Ɵ1 – cot Ɵ2)
The angle of elevation of the top of a tower from a point is 30 °. On walking 40 m towards the tower the angle changes to 45°.Find the height of the tower?
Solution:
tan 45° = AB/BD
1 = AB/1
Therefore AB = 1
tan 30° = AB/BC =>1/√3 = 1/BC
therefore BC= √3
Now CD =√3-1 m and the height of the tower is 1 m
1 m = 1/√3-1
Therefore 40 m = 1/√3-1.40 = 40/√3-1
= 20 (√3+1)m
By trick:
40 = h(√3-1)
H = 40/(√3-1) = 20 (√3+1)m
Important Height and Distance Questions for Upcoming SSC Exams 2023
Some important Height and Distance questions for upcoming SSC exams 2023 are as follows:
1. The angle of elevation of the top of a tower from the top of a building whose height is 680 m is 45° and the angle of elevation of the top of same tower from the foot of the same building is 60°. What is the height (in m) of the tower?
A. 340(3 + √3) B. 310(3 – √3) C. 310(3 + √3) D. 340(3 – √3)
Answer ||| A
2. From the top of an upright pole 24 feet high, the angle of elevation of the top of an upright tower was 60°. If the foot of the pole was 60 feet away from the foot of the tower, what tall (in feet) was the tower?
A. B. C. D.
Answer ||| A
3. From the top of an upright pole 17.75 m high, the angle of elevation of the top of an upright tower was 60°. If the tower was 57.75 m tall, how far away (in m) from the foot of the pole was the foot of the tower?
A. B. C. D.
Answer ||| D
4. Two ships are on the opposite of a light house such that all three of them are collinear. The angles of depression of the two ships from the top of the light house are 30° and 60°. If the ships are 230 m apart, then find the height of the light house (in m).
A. 175.4 B. 165.2 C. 172.5 D. 180.5
Answer ||| C
5. An person 1.8 metre tall is metre away from a tower. If the angle of elevation from his eye to the top of the tower is 30 degree, then what is the height (in m) of the tower?
A. 32.5 B. 37.8 C. 30.5 D. 31.8
Answer ||| D
6. A ladder of length 3.5 m just reaches the top of a wall. If the ladder makes an angle of 60° with the wall, then what is the height of the wall (in m)?
A. 1.75 B. 3.5 C. D.
Answer ||| A
7. The length of the shadow of a vertical tower on level ground increases by 8.4 m when the altitude of the sun changes from 45° to 30°. What is the height of the tower (in m)?
A. 4.2 B. C. D.
Answer ||| D
8. Exactly midway between the foot of two towers P and Q, the angles of elevation of their tops are 45° and 60°, respectively. The ratio of the height of P and Q is:
A. B. 3 : 1 C. 1 : 3 D.
Answer ||| A
9. From the top of a 195 m high cliff, the angles of depression of the top and bottom of a tower are 30° and 60°, respectively. Find the height of the tower (in m).
A. B. 195 C. 130 D. 65
Answer ||| C
10. A kite is attached to a string. Find the length of the string (in m) when the height of the kite is 90m and the string makes an angle of 30° with the ground. | 677.169 | 1 |
Consider a triangle ABC and the reflexion of points on a bisector AD of angle A say. This transformation
is called isogonal with respect to the bisector AD. The transformation, being a particular reflexion
preserves distances and maps lines to lines. In particular a line AX through A is mapped to a line AX'
through A, which is fixed by the transformation. Also sides AB and AC are interchanged by the transformation. Some further properties of this transformation are: (1) The projections {Y,Z} of X and {Y',Z'} of its transform X' on the sides are on a circle c. (2) The center of this circle c is on the bisector AD. (3) The triangles XYZ and X'Y'Z' are equal. (4) If a line AX is the locus of points such that XY/XZ = k, then its transform AX' is the locus of points X'
such that X'Y'/X'Z' = 1/k.
Start from (3) showing the equality of triangles XYZ and X'Y'Z'. This is immediate, since by definition
the angles at X, X' are equal and complementary to A and the sides XY=X'Z', XZ=X'Y'. (4) is a consequence of (3). Properties (1) and (2) follow from the fact that YY'ZZ' is an equilateral
trapezium, since AY'=AZ etc..
Consider two isogonal lines AE, AE' of the triangle ABC and take an arbitrary point X on AE and
an arbitrary point X' on AE'. Then project X, X' on the sides of the triangle to points {Y,Y'} on AB
and {Z,Z'} on AC [Lalesco, p.40]. (1) Triangles XYZ and X'Y'Z' are similar. (2) The quadrilateral YY'Z'Z is cyclic on a circle c, and {Y'Z, YZ'} are antiparallels. (3) AE is orthogonal to Y'Z' and AE' is orthogonal to YZ. (4) The center of the circle c is the middle O of XX'. (5) The three lines {Y'Z, YZ', XX'} pass through the same point G, whose polar with respect to
the sides {AB, AC} is the orthogonal from A to XX'.
(1) is valid because quadrangles AYXZ and AY'X'Z' are cyclic, thus angle YAX = YZX, X'Y'Z' = X'AZ'
etc..(3) follows also from these angle equalities e.g. X'Y'Z' = YAX and since the sides {AY, X'Y'} are
orthogonal the other two sides {XA,Y'Z'} are also orthogonal. (2) follows also from the equality of the
angles YY'Z' = YY'X'+X'Y'Z' and YZZ'=YZX+XZZ' but X'Y'Z' = YXZ etc.. (4) follows from the fact
that O projects on the middle of YY' on AB and the middle of ZZ' on AC, hence it is the intersection of
the two medial lines of these segments of the same circle c. (5) is proved in section-2 of Pedal.html . Remark In the aforementioned reference it is also proved that the polar of G with respect to
the pair of lines AB, AC passes through the intersection point of lines {YZ, Y'Z'}.
Consider a triangle ABC and the three reflexions FA, FB, FC with respect to its bisectors at the
corresponding angles. (1) The result of applying the transformations successively to a point X, creating X'=FA(X), X''=FB(X'), X'''=FC(X''), defines a cyclic quadrangle XX'X''X''' centered at I, the incenter of
the triangle. (2) The composition F=FC*FB*FA is the reflexion on the line through I which is orthogonal to AC.
(1) is obvious since reflexions preserve distances, hence all IX, IX', IX'', etc. are equal. (2) follows from (1). In fact, angle X'XX''' is then complementary to X'X''X''', which in turn
has its legs orthogonal to the bisectors IB, IC. By measuring angle it follows that XX''' is parallel
to AC. Since then XIX''' is isosceles X and X''' are symmetric with respect to the line from I which
is orthogonal to AC. Remark-1 This property of the composition of reflexions is a special case of composition of
several reflexions on lines passing through a point. If the number of different lines is even then the
result-composition is a rotation if the number of lines is odd (as is here the case), then the
result is a reflexion. Remark-2 This composition of isogonal conjugation must be distinguished from the
isogonal conjugation with respect to a triangle which is another kind of important transformation, playing an essential role in the geometry of the triangle and being quadratic in nature and transforming
lines to conics passing through the vertices of the triangle (see Isogonal_Conjugation.html ). | 677.169 | 1 |
Law of cosines exercises pdf
Law of cosines exercises pdfThe advantage to using the Law of Cosines over the Law of Sines in cases like this is that unlike the sine function, the cosine function distinguishes between acute and obtuse angles. The cosine of an acute is positive, whereas the cosine of an obtuse angle is negative. Since the sine of both acute and obtuse angles are positive, the sine of an angle alone is not enough to determine if the
8 6 practice law of cosines form g Mon, 17 Dec 2018 03:10:00 GMT 8 6 practice law of pdf – Sec. 5. (a) The operation of any law repealed and replaced by
View practice law of cosines.pdf from IED 106 at Central Academy of Technology and Arts. Name: Period: Date: Practice Worksheet: Law of Cosines Using the Law of Cosines…
Essential Question What are the Law of Sines and the Law of Cosines? Go to BigIdeasMath.com for an interactive tool to investigate this exploration. Work with a partner.
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In Exercises 10 and 11, fill in the blanks to complete the theorems. 10. For any nABC with side lengths a, b, and c, sin ____A a 5 sin B _____ b 5
In this lesson, students will apply the laws of sines and cosines learned in the previous lesson to find missing side lengths of triangles. The goal of this lesson is to clarify when students can apply the law of sines and when they can apply the
Section 6.2 Law of Cosines 443 In Exercises 1–16, use the Law of Cosines to solve the triangle. Round your answers to two decimal places. 1. 2. 3.
Improve your math knowledge with free questions in "Law of Cosines" and thousands of other math skills.
The Law of Cosines states that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of the other …
Section 9.2 The Law of Cosines Note: A calculator is helpful on some exercises. Bring one to class for this lecture. OBJECTIVE 1: Determining If the Law of Sines or the Law of Cosines …
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Exercise Set 7.3: The Law of Sines and the Law of Cosines Math 1330, Precalculus The University of Houston Chapter 7: Trigonometric Applications to Triangles 4 cm 8 cm A T C 50o 14. ΔWIN w =4.9 ft D∠=I 42 D∠=N 100 n =? Find all possible measures for the indicated angle of the triangle. (There may be 0, 1, or 2 triangles with the given measures.) Round answers to the nearest hundredth. If
12/10/2015 · Get YouTube without the ads. Working… No thanks 3 months free. Find out why Close. Law of Sines and Cosines Word Problems ALLEN JONES. …
Section 6.2 Law of Cosines 443 In Exercises 1–16, use the Law of Cosines to solve the triangle. Round your answers to two decimal places. 1. 2. 3.
View Notes – Law_of_sines_and_cosines.pdf from MATH 101 at Cookeville High School.A2.A.73: Law of Cosines 6: Solve for an unknown side or angle, using the Law of Sines or the Law of Cosines 1 In a triangle, two sides that measure 6 cm and 10 cm form an angle that measures 80° . | 677.169 | 1 |
Start with an equilateral triangle ABC and the diametral point D of A on its circumcircle. For every line through D consider the intersection points {E, F, G} with sides {BC, AB, AC} correspondingly. Then their fourth harmonic point coincides with the second intersection point H of the line with the circumcircle. And inversely, joining every point H of the circumcircle to D and intersecting with line DH the sides of the triangle we find points such that (F,G,E,H) = -1.
The proof follows at once by observing that at point H the angles BHA and AHC are equal, each being 60 degrees. Thus lines {HA, HD} are bisectors of the angle BHC hence H(B,C,E,I) is a harmonic bundle consequently A(F,G,E,H) is also a harmonic bundle, proving the claim. The inverse
Given a triangle ABC and a circumconic (c) with perspector P. Let D be the other intersection point of AP with c. For every line through D let {B',C',A'} be the intersection points with sides {AB, AC, BC} respectively. Then the fourth harmonic A'' of these points i.e. point such that : (B',C',A',A'') = -1 is on the conic.
Inversely, given a triangle ABC and a point D not on the side-lines of the triangle, consider
all lines through D and the corresponding intersections with these lines {A',B',C'}. Let then
A'' be the fourth harmonic of these three points: A'' = A'(B',C') (i.e (B',C',A',A'') = -1).
Then point A'' describes a conic circumscribing the triangle ABC and passing through D.
Further the pivot P of the conic with respect to the triangle is on line AD.
The proof follows from the previous section by applying to the equilateral an appropriate projectivity transforming it to the general triangle ABC and the center of the equilateral to the perspector P of the conic c.
Given two parallel lines {b, c} and a third line (a) intersecting them, consider all lines through a fixed point D not lying on the given three lines {a, b, c}. For each line through D define the intersection points {A, B, C} with lines {a, b, c} correspondingly. The fourth harmonic E on each line i.e. point such that (B,C,A,E) = -1 describes a hyperbola.
This property follows from the previous section by letting point A of the triangle ABC shown there go to infinity. When A coincides with the intersection M of line (a) with the middle parallel (m) of lines {b,c} then E goes to infinity, from which it follows that line DM is parallel to an asymptote. The other asymptote is obviously parallel to lines {b,c}. When A obtains the position of M' (parallel to b projection of D on a) points {B,C,E} go to the same point at infinity. Besides the points {B0, C0, D} from which passes the hyperbola we know also point M', which is the projection of D on line (m) parallel to line (a). This position is obtained by E when A goes to infinity on a and DA becomes parallel to line (a).
An easily constructible point is point D' on the line through D which is orthogonal to line (a). Then the conic can be identified with the unique conic passing through the five points {D, C0, B0, M'', D'}.
[1] Find the hyperbola passing through the vertices of a trapezium and having known directions of asymptotes. (Could use the property shown in the previous paragraph, but this handles a special case of trapezia, such that the parallel to one asymptote from the middle of a side B0C0 passes through another vertex of the trapezium.)
[2] The same problem for general quadrangle instead of hyperbola.
[3] Find the structure of all hyperbolas passing through the vertices of a trapezium.
[4] Same problem for a general quadrangle. | 677.169 | 1 |
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