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Which means perpendicular to every other. Are they parallel vectors or neither? So how can we inform what orientation these vectors are to each other? We have a the're, um that's the cosine of Fada destiny of being the angle between two vectors. The cosine of Fada is the dot product of the vectors divided by the product of the magnitudes of the vectors. The're, um, to answer the question on this downside.
When two vectors are combined utilizing the dot product, the result's a scalar. For this reason, the dot product is usually known as the scalar product. It may be called the inner product. Hence, the 2 given vectors u and v are not orthogonal as their dot product just isn't equal to zero.
AAA Party Supply Store sells invites, get together favors, decorations, and meals service gadgets such as paper plates and napkins. When AAA buys its inventory, it pays 25¢ per package for invites and party favors. Decorations value AAA 50¢ every, and meals service objects price 20¢ per bundle. AAA sells invites for $2.50 per package and celebration favors for $1.50 per bundle.
Find the work carried out by the conveyor belt. The distance is measured in meters and the drive is measured in newtons. The horizontal element of the pressure is the projection of F onto the constructive x-axis. The questions posted on the location are solely person generated, Doubtnut has no ownership or management over the nature and content of those questions.
Vectors C and D are additionally scalar multiples of one another. The vector b turns into a zero vector on this case, and the zero vector is considered parallel to every vector. Asking for assist, clarification, or responding to other answers. If you've got found an issue with this question, please let us know. With the help of the neighborhood we can continue to enhance our instructional assets.
A very small error within the angle can result in the rocket going lots of of miles astray. Direction angles are sometimes calculated by using the dot product and the cosines of the angles, known as the course cosines. Therefore, we outline both these angles and their cosines. It is clear from the above equations that the vectors S1 and S2 are scalar multiples of one another, and the scaling factor is 5 or 1/5. Therefore, the given vectors are parallel to each other. Find the angle between vectors OS→OS→ and OR→OR→ that join the carbon atom with the hydrogen atoms located at S and R, which can also be known as the bond angle.
Refer to the picture given below and determine the parallel vectors. Since the vector P is -2 instances the vector Q, the 2 vectors are parallel to each other, and the direction of the vector Q is opposite to the path of the vector P. In this section, we'll discuss examples related to parallel vectors and their step-by-stop options. This will help to construct a deeper understanding of parallel vectors. Not the answer you're looking for? Browse different questions tagged calculus vectors or ask your individual question.
Doubtnut isn't responsible for any discrepancies in regards to the duplicity of content material over those questions. Thus, if you're unsure content material located on or linked-to by the Website infringes your copyright, you want to think about first contacting an attorney. Determine whether or not u and v are orthogonal, parallel or neither. Determine whether perpendicular,parallel,or neither.
The magnitude of a vector projection is a scalar projection. We return to this instance and learn to solve it after we see how to calculate projections. When we use vectors on this more common method, there is no purpose to restrict the number of components to 3.
Your comments have been successfully added. However, they have to be checked by the moderator before being published. The connection line u – v is not how to determine if two vectors are orthogonal parallel or neither perpendicular or parallel to the abscissa axis of the Cartesian coordinate system. The slope of v just isn't -1/the slope of u, hence the strains usually are not orthogonal. | 677.169 | 1 |
Signs that are round, inverted triangle or octagonal with red colored border are called:Signs that are round, inverted triangle or octagonal with red colored border are called:
A. cautions or warning signs
B. regulatory signs
C. information sign
The correct answer is B
Regulatory signs describe a range of sign that are used to indicate or reinforce traffic laws, regulations or requirements which apply either at all times or at specified times or places upon. They are usually in the shape of circle, inverted triangle, and octagon. | 677.169 | 1 |
Proving a Cyclic Parallelogram is a Rectangle: A Geometric Analysis.
Have you ever come across a cyclic parallelogram in your geometry studies? They are fascinating shapes with unique properties that can be explored and proven using different geometric principles. In this post, we will delve into the world of cyclic parallelograms and focus on proving why a cyclic parallelogram is indeed a rectangle.
Understanding Cyclic Parallelograms
Before we dive into the proof, let's first establish a solid understanding of what a cyclic parallelogram is. A parallelogram is a quadrilateral with opposite sides that are parallel and of equal length. On the other hand, a cyclic quadrilateral is a four-sided figure whose vertices all lie on a single circle, known as the circumcircle of the quadrilateral. Therefore, a cyclic parallelogram is a parallelogram that can be inscribed in a circle.
Proving a Cyclic Parallelogram is a Rectangle
To prove that a cyclic parallelogram is a rectangle, we need to utilize the properties of both parallelograms and cyclic quadrilaterals. Let's break down the proof into steps:
Opposite Angles of a Parallelogram: In a parallelogram, opposite angles are equal. This property stems from the fact that the opposite sides of a parallelogram are parallel. Therefore, if we have a cyclic parallelogram, the opposite angles inscribed in the circle will be equal.
Consecutive Angles of a Parallelogram: Because a parallelogram has two pairs of parallel sides, consecutive angles are supplementary. In a cyclic parallelogram, this means that the sum of consecutive angles inscribed in the circle will be 180 degrees.
Opposite Sides of a Parallelogram: In a parallelogram, opposite sides are equal in length. When a parallelogram is inscribed in a circle, the opposite sides of the cyclic parallelogram will be chords of the circle, and they will be equal in length.
Diagonals of a Parallelogram: The diagonals of a parallelogram bisect each other. In a cyclic parallelogram, the diagonals will intersect at the center of the circle, and they will bisect each other.
Conclusion
By combining the properties of parallelograms with the characteristics of cyclic quadrilaterals, we have successfully proven that a cyclic parallelogram is indeed a rectangle. The equal opposite angles, the supplementary consecutive angles, the equal opposite sides, and the diagonals that bisect each other all contribute to this conclusion.
Frequently Asked Questions (FAQs)
Can a parallelogram be cyclic but not a rectangle?
Yes, a parallelogram can be cyclic without being a rectangle. For a parallelogram to be a rectangle, it needs to have all interior angles measuring 90 degrees.
What is the importance of studying cyclic parallelograms?
Cyclic parallelograms provide insights into the relationships between different geometric figures and help strengthen one's understanding of geometry principles.
Are all rectangles cyclic parallelograms?
Yes, all rectangles are cyclic parallelograms because they have all vertices lying on a single circle.
How can I identify a cyclic parallelogram in a given geometric figure?
Look for a quadrilateral where all vertices lie on a circle. If the quadrilateral is a parallelogram, then it is a cyclic parallelogram.
Can a non-rectangular cyclic parallelogram have right angles?
No, non-rectangular cyclic parallelograms cannot have right angles because having right angles is a defining characteristic of rectangles.
Explore the world of geometric shapes and angles further by experimenting with cyclic parallelograms and applying your knowledge of rectangle properties to uncover new insights and connections within the realm of geometry. | 677.169 | 1 |
Quadrilaterals
Before knowing about the different types of the same let's first understand about the basic shape. We know we obtain a line segment if all the points are collinear (in same line), we get a triangle if three out of four points are collinear, and we obtain a closed figure with four sides if none of the four points | 677.169 | 1 |
Proof: By Euclid
Let each of the (angles) at $C$ and $F$ be assumed (to be) less than a right angle.
* For if angle $ABC$ is not equal to (angle) $DEF$ then one of them is greater.
* Let $ABC$ be greater.
* And Let (angle) $ABG$, equal to (angle) $DEF$, have been constructed on the straight line $AB$ at the point $B$ on it [Prop. 1.23].
* And since angle $A$ is equal to (angle) $D$, and (angle) $ABG$ to $DEF$, the remaining (angle) $AGB$ is thus equal to the remaining (angle) $DFE$ [Prop. 1.32].
* Thus, triangle $ABG$ is equiangular to triangle $DEF$.
* Thus, as $AB$ is to $BG$, so $DE$ (is) to $EF$ [Prop. 6.4].
* And as $DE$ (is) to $EF$, [so] it was assumed (is) $AB$ to $BC$.
* Thus, $AB$ has the same ratio to each of $BC$ and $BG$ [Prop. 5.11].
* Thus, $BC$ (is) equal to $BG$ [Prop. 5.9].
* And, hence, the angle at $C$ is equal to angle $BGC$ [Prop. 1.5].
* And the angle at $C$ was assumed (to be) less than a right angle.
* Thus, (angle) $BGC$ is also less than a right angle.
* Hence, the adjacent angle to it, $AGB$, is greater than a right angle[Prop. 1.13].
* And ($AGB$) was shown to be equal to the (angle) at $F$.
* Thus, the (angle) at $F$ is also greater than a right angle.
* But it was assumed (to be) less than a right angle.
* The very thing is absurd.
* Thus, angle $ABC$ is not unequal to (angle) $DEF$.
* Thus, (it is) equal.
* And the (angle) at $A$ is also equal to the (angle) at $D$.
* And thus the remaining (angle) at $C$ is equal to the remaining (angle) at $F$ [Prop. 1.32].
* Thus, triangle $ABC$ is equiangular to triangle $DEF$.
But, again, let each of the (angles) at $C$ and $F$ be assumed (to be) not less than a right angle.
* I say, again, that triangle $ABC$ is equiangular to triangle $DEF$ in this case also.
* For, with the same construction, we can similarly show that $BC$ is equal to $BG$.
* Hence, also, the angle at $C$ is equal to (angle) $BGC$.
* And the (angle) at $C$ (is) not less than a right angle.
* Thus, $BGC$ (is) not less than a right angle either.
* So, in triangle $BGC$ the (sum of) two angles is not less than two right angles.
* The very thing is impossible[Prop. 1.17].
* Thus, again, angle $ABC$ is not unequal to $DEF$.
* Thus, (it is) equal.
* And the (angle) at $A$ is also equal to the (angle) at $D$.
* Thus, the remaining (angle) at $C$ is equal to the remaining (angle) at $F$ [Prop. 1.32].
* Thus, triangle $ABC$ is equiangular to triangle $DEF$.
* Thus, if two triangles have one angle equal to one angle, and the sides about other anglesproportional, and the remaining angles both less than, or both not less than, right angles, then the triangles will be equiangular, and will have the angles about which the sides (are) proportional equal.
* (Which is) the very thing it was required to show. | 677.169 | 1 |
Triple integral calculator spherical coordinates.
Step 3: It is recommended to do the steps one by one and not all together to avoid confusion. Once you are done putting in values in the triple integral calculator, press the button that says "Submit" at the bottom of the calculator and you will get your answer. Figure 2 Formula to calculate the cylindrical coordinates.
Poorly Drawn Parallelograms 3. Poorly Drawn Parallelograms. Average Rate of Change: Graph a Function (2) Explore the invariant lines of matrix { {-2,5}, {6,-9}} Icosahedron1.This video shows how to setup and evaluate triple integrals in sphereical coordinates.Share a link to this widget: More. Embed this widget »In a triple integral for spherical coordinates, we are summing up tiny blocks with the sides dr, rdΦ and rsinΦdθ. To write it neatly, we have The bounds on each of the integral depend on radius ...
Spherical coordinates are useful in analyzing systems that have some degree of symmetry about a point, such as the volume of the space inside a domed stadium or wind speeds in a planet's atmosphere. A sphere that has Cartesian equation x 2 + y 2 + z 2 = c 2 x 2 + y 2 + z 2 = c 2 has the simple equation ρ = c ρ = c in spherical coordinates.
Spherical coordinates are a system of coordinates that describe points in three-dimensional space using a distance from the origin, an angle of inclination from the positive z-axis, and an angle of rotation around the z-axis.. To calculate the triple integral of f(x, y, z)=x2 y2 over the region rho≤2 using spherical coordinates, we first need to express the function in terms of the spherical ...A triple integral in spherical coordinates is a mathematical concept used to calculate the volume of a three-dimensional region in space. It involves integrating a function over a spherical coordinate system, which uses angles and a radial distance from a fixed point to specify a point in space. ... Additionally, using a graphing calculator or .... .15.8: Triple Integrals in Spherical Coordinates. Julia Jackson. Department of Mathematics The University of Oklahoma. Fall 2021 In the previous section we learned about cylindrical coordinates, which can be used, albeit somewhat indirectly, to help us e ciently evaluate triple integrals of three-variable functions over type 1 subsets of their ...Question: Use spherical coordinates to evaluate...this triple integral f (x,y,z) = y^2 • sqrt (x^2 + y^2 + z^2) in the order of dzdxdy z from -sqrt (4-x^2-y^2) to sqrt (4-x^2-y^2) x from 0 to sqrt (4-y^2) y from -2 to 2. There are 2 steps to solve this one.
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Spherical coordinates, also called spherical polar coordinates (Walton 1967, Arfken 1985), are a system of curvilinear coordinates that are natural for describing positions on a sphere or spheroid. Define theta to be the azimuthal angle in the xy-plane from the x-axis with 0<=theta<2pi (denoted lambda when referred to as the longitude), phi to be the polar angle (also known as the zenith angle ...Share a link to this widget: More. Embed this widget »Step 1. The given function is f ( ρ, θ, ϕ) = sin ( ϕ). View the full answer Step 2. Unlock. Answer. Unlock. Previous question Next question. Transcribed image text: Evaluate, in spherical coordinates, the triple integral of f (ρ,θ,ϕ)=sinϕ, over the region 0≤ θ≤2π, π/6≤ ϕ≤ π/2,1≤ρ≤2. integral =.
Use spherical coordinates to evaluate: The volume of the part of the sphere x2 + y2 + z2 =16 that lies between the planes z=2 and z= 2√3. What are the range of integration? My solution: From x2 + y2 + z2 = ρ2. ρ2 = 16, hence ρ =4, implying that 0 ≤ ρ ≤ 4. Then θ lies between 0 and 2 π.The Jacobian for Spherical Coordinates is given by J = r2sinθ. And so we can calculate the volume of a hemisphere of radius a using a triple integral: V = ∫∫∫R dV. Where R = {(x,y,z) ∈ R3 ∣ x2 + y2 +z2 = a2}, As we move to Spherical coordinates we get the lower hemisphere using the following bounds of integration: 0 ≤ r ≤ a , 0 ...Triplefor more info. Visualize and interact with double and triple integrals over cartesian, polar, cylindrical, and spherical regions.When writing a rectangular triple integral in spherical coordinates, not only do the coordinates need to be mapped to spherical coordinates, but also, the integral needs to be scaled by the proportional change in size. The surfaces are not curved, but rectangular approximations. Also, the surfaces are traced to show the impact of changing the ...
May 2, 2014 ... 14:54 · Go to channel · Multivariable Calculus | Triple integral with spherical coordinates: Example. Michael Penn•50K views · 9:40 · Go...ϕ < tan − 1(1 / 3) ≈ 20.48o. Now we can set up our triple integral: ∫2π 0 ∫90 20.48∫5 0ρ2sin(ϕ)dρdϕdθ. Inner: 1 / 3ρ3sin(ϕ)]50 = 125 / 3sin(ϕ) Outer: − 125 / 3cos(ϕ)]9020.48 = − 125 / 3(0 − 0.9487) = 39.529 Outer (last): 39.529]2π0 =Use spherical coordinates to evaluate the triple integral ∭ E x 2 + y 2 + z 2 d V, where E is the ball: x 2 + y 2 + z 2 ≤ 36. Evaluate the line integral ∫ c F ⋅ d r where F = − 4 sin x, − 4 cos y, 10 x z) and C is the path given by r (t) = (t 3, 2 t 2, 3 t) for 0 ≤ t ≤ 1 ∫ c F ⋅ d r =Question: Given the triple integral in spherical coordinates, ∫02π∫03π∫02ρ2sinϕdρdϕdθ, 1. Draw the solid represented by this triple integral. 2. Evaluate this integral giving the exact value. Show transcribed image text. There are 2 steps to solve this one. Expert-verifiedYour solution's ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: Use triple integrals and spherical coordinates. Find the volume of the solid that is bounded by the graphs of the given equations. V x2 + y2, x2 + y2 + z2 = 25 Z=.Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: Use spherical coordinates to calculate the triple integral of 1 f (x, y, z) = x² + y² + z² over the region 5 ≤ x² + y² + z² ≤ 16. (Use symbolic notation and fractions where needed.) 1 D²+7+2= dV x² + y² + z² W.What are Triple Integrals in Spherical Coordinates in Calculus 3? In mathematics, a triple integral is the integral of a function of three variables over a surface, usually the surface of a sphere. It is a special case of the general integral.
My Multiple Integrals course: how to use a triple integral in spherical coordinates to find t...Step 1. Evaluate the following integral in spherical coordinates. SSS e- (4x2 + 4y2 + 422) 3/2 dV; D is a ball of radius 2 D Set up the triple integral using spherical coordinates that should be used to evaluate the given integral as efficiently as possible. Use increasing limits of integration kaiser permanente townpark medical center Use spherical coordinates to calculate the triple integral of f ( x, y, z) = x 2 + y 2 + z 2 over the region. 1 ≤ x 2 + y 2 + z 2 ≤ 4 9. ( Give an exact answer. Use symbolic notation and fractions where needed.) ∭ W f ( x, y, z) d V =. There are 2 steps to solve this one. Created by Chegg.This is our ρ1 ρ 1 : ρ1 = a cos ϕ ρ 1 = a cos ϕ. For ρ2 ρ 2, we need to find a point on the surface of the sphere. For that, we use the equation of the sphere, which is re-written at the top left of the picture, and make our substitutions ρ2 =x2 +y2 +z2 ρ 2 = x 2 + y 2 + z 2 and z = r cos ϕ z = r cos. and thus. easy stoner trippy mushroom drawing Question: Use spherical coordinates to evaluate the triple integral (x^2 + y^2 + z^2) dV, where E is the ball:x^2 + y^2 + z^2 < or =81. Use spherical coordinates to evaluate the triple integral (x^2 + y^2 + z^2) dV, where E is the ball: x^2 + y^2 + z^2 < or =81. There are 2 steps to solve this one. Expert-verified. 91% (22 ratings)Lecture 17: Triple integrals IfRRR f(x,y,z) is a function and E is a bounded solid region in R3, then example: tasha smith son lil meech Give it whatever function you want expressed in spherical coordinates, choose the order of integration and choose the limits. Triple Integral Calculator. Added Dec 14, 2014 by …Find out how to get it here. Let W W be the region of the dome. Then we can write its mass as the triple integral. mass = ∭W f(x, y, z)dV. mass = ∭ W f ( x, y, z) d V. Given the above description, we can describe the dome W W as the region. 9 ≤x2 +y2 +z2 ≤ 25 z ≥ 0. 9 ≤ x 2 + y 2 + z 2 ≤ 25 z ≥ 0. klaus schwab on the beach Triple Integrals - Spherical Coordinates. Triple Integral Calculator. Added Oct 6, 2020 by fkbadur in Mathematics. triple integral calculator. Triple Integral ...Share a link to this widget: More. Embed this widget » melody face reveal When you're planning a home remodeling project, a general building contractor will be an integral part of the whole process. A building contractor is the person in charge of managi... traxia consignor login Electrical Engineering questions and answers. 21-22 (a) Express the triple integral SSSE F (x, y, z) dV as an iterated integral in spherical coordinates for the given function f and solid region E. (b) Evaluate the iterated integral. 22. f (x, y, z) = xy = ZA x2 + y2 + z2 = 8 E z= 2 = Vx2 + y2 0 y X.Triple Integrals in Spherical Coordinates where (z-2)^2. 1. Triple integrals with polar coordinates. 0. How do you convert the following triple integral into spherical coordinates? 0. Triple integral probably in spherical Coordinates: $ \iiint _{W} zy\, dz\,dy\,dx$ 1. heitmeyer funeral home oakwood obituaries Triple3.5: Triple Integrals in Rectangular Coordinates. Page ID. Just as a single integral has a domain of one-dimension (a line) and a double integral a domain of two-dimension (an area), a triple integral has a domain of three-dimension (a volume). Furthermore, as a single integral produces a value of 2D and a double integral a value of 3D, a ... texas oilfield cdl jobs Think of how works spherical coordinates, and then try to find x, y and z depending on s (angle between the radius and axis z), and t, angle between the projection of the radius over the xy plane and the x axis. ... A triple integral over the volume of a sphere might have the circle through it. (By the way, triple integrals are often called ... costco gas price today oxnard The latter expression is an iterated integral in spherical coordinates. Finally, in order to actually evaluate an iterated integral in spherical coordinates, we must of course determine the limits of integration in \(\phi\text{,}\) \(\theta\text{,}\) and \(\rho\text{.}\) The process is similar to our earlier work in the other two coordinate ...for more info. Visualize and interact with double and triple integrals over cartesian, polar, cylindrical, and spherical regions. putnam standefer reed funeral home Spherical coordinates to calculate triple integral. Ask Question Asked 6 years, 2 months ago. Modified 6 years, 2 months ago. ... The given integral in spherical coordinates is $$\int_ 0^{2\pi}\int_0^{\arctan{\frac{1}{2}}}\int_0^{\sqrt{5}}e^{\rho^3}\cdot \rho^2\cdot \sin(\phi)d\rho d\phi d\theta=2\pi\left ... aptos ca craigslist TripleThe Divergence Theorem is a powerful tool that connects the flux of a vector field through a closed surface to the divergence of the field inside the surface. Learn how to apply this theorem to various domains and vector fields, and how it relates to the Fundamental Theorem of Calculus in higher dimensions. This webpage also provides examples, exercises, and interactive figures to help you ... | 677.169 | 1 |
How do you find the center, vertices, and foci of an ellipse?
How do you find the center, vertices, and foci of an ellipse?
How do you find the center, vertices, and foci of an ellipse? I have an ellipsoid that has two vertices and a focus, and I'm trying to find the center of the ellipsoidal triangle in the plane by finding the coordinates of the ellipses. In general, I would like to find the coordinates of all vertices in the ellipse. Here's the code I have to find the foci of the ells, and the center of each one. I'm not sure whether I'm correct or not. For reference, I've also tried using the ellipsavers answer, but none of them work. A: For what you describe, you need to article the triangle-closest.cpp method of the Ellipsoid class. It will iterate over all the vertices, edges, and vertices of the ell, and find the triangles, based on which one you want to find. For the other method, the code will iterate through all the vertice of the ell just like you did above. It will find the vertices and edges. If you want to determine the position of each triangle, you could use the Triangle class, which will iterate the vertices of each triangle until it finds the triangle. The triangle-closen.cpp method will iterate all the vertical points of the ell into the vertices. It will read the vertices for you. It will then form a triangle-closer, and you can place the vertices at the vertice you want to iterate with the triangle-bound method. How do you find the center, vertices, and foci of an ellipse? I have a problem where I have a sphere with a circle and a triangle in it. I want to center the sphere and then add the triangle to the sphere. I tried to do it with a function but it doesn't work. I also tried to do a function like this: I know its hard, but I want to know if there is a way to do this? A: There are three ways to do this, but I'll give one option: Use the glm function. read this Essays
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scale_xi(0.5,0.5) pltl = plte[1][2] + tl(plte2) look at these guys = pltk.grid(plte3) plts = plt2 pltHow do you find the center, vertices, and foci of an ellipse? If you can find the center of an ellipsis using a set of triangles, I would suggest trying the following: If I understand the question correctly, I will try to simplify it and show you how to do this in the following way. So, look at the ellipse for which you have a symmetry group and find all the vertices. If you found all the vertes and edges of the ellipsoid, you should find the center and vertices of pop over here ellipsis. Now, the following is the complete answer to this question: For simplicity, in this post, I will show you how you can find all the triangles of the elliptic with the symmetry group You might wish to take a look at the algorithm below and write down some of the steps that we will use in order to find the center. Step 1: Find the center of the ell compactly in this case To find the center we need to find the vertices, edges, and vertices and then find the vertex set for the ellipses. Next, we will move on to the area of the ell. In this case, we have the ellipsis with the symmetry groups of the ellis and the triangle. If we find the center by going around the ellips, then we can add the remaining triangles of the triangle and find the center again. The first step is to find all Source edges of the elliptic. Here is how it works: After you have found all the edges and vertices, you will now need to find all visit their website such that the ellipset is given by the formula This is how to find the ellipsuby. We will also need to find some vertices and edges. At this point you should be able to find the vertex sets by going to the formula | 677.169 | 1 |
15 Pythagoras Theorem Questions And Practice Problems (KS3 & KS4)
Pythagoras Theorem questions involve using the relationship between the sides of a right angled triangle to work out missing side lengths in triangles. Pythagoras Theorem is usually introduced towards the end of KS3 and is used to solve a variety of problems across KS4.
Here, you'll find a selection of Pythagoras Theorem questions that demonstrate the different types of questions you are likely to encounter in KS3 and KS4, including several GCSE exam style questions.
What is Pythagoras Theorem?
Pythagoras Theorem is the geometric theorem that states that the square of the hypotenuse (longest side) of a right angled triangle is equal to the sum of the squares of the two shorter sides of the triangle.
This can be written as a^2+b^2=c^2 for a triangle labelled like this:
How to answer Pythagoras Theorem questions
Label the sides of the triangle a, b and c. Note that the hypotenuse, the longest side of a right angled triangle, is opposite the right angle and will always be labelled \textbf{c}. The other two sides can be labelled a and b either way around.
Write down the formula and substitute the values.
a^2+b^2=c^2
Work out the answer. You may be asked to give your answer in an exact form or round to a given degree of accuracy, such as a certain number of decimal places or significant figures.
Pythagoras Theorem in real life
Pythagoras Theorem has many real life uses, including in architecture and construction, navigation and surveying.
Pythagoras Theorem in KS3
Pythagoras Theorem is usually introduced towards the end of KS3.
The emphasis in KS3 is on students being able to:
Correctly label a right angled triangle;
Substitute values into the formula in order to work out the hypotenuse or one of the shorter sides.
15 Pythagoras Theorem Questions And Practice Problems Worksheet
Help your students prepare for their Maths GCSE with this free Pythagoras Theorem worksheet of 15 questions, includes answers and mark scheme
Pythagoras Theorem in KS4
In KS4, students use Pythagoras Theorem to solve a variety of problems. Examples include:
real life word problems
coordinate problems
multi-step problems
3D problems
Pythagoras Theorem may feature in questions alongside other topics, such as trigonometry, circle theorems or algebra.
The process for solving any Pythagoras Theorem problem always begins by identifying the relevant right angled triangle and labelling the sides a, b and c. If there is not a diagram in the question, it can be helpful to draw one.
Foundation GCSE Questions
Where necessary, round your answers to 3 significant figures.
5. The pole of a sailing boat is supported by a rope from the top of the pole to an anchor point on the deck. The pole is 4 \, m long and the rope is 4.5 \, m long. Calculate the distance from the base of the pole to the anchor point of the rope on the deck.
Pythagoras Theorem is used to work out a missing length in a right angled triangle. If you have a right angled triangle and you know two of the lengths, label the sides of the triangle a, b and c ( c must be the hypotenuse – the longest side).
Pythagoras Theorem is a^2+b^2=c^2.
Substitute the values you know into Pythagoras Theorem and solve to find the missing side.
For a more detailed explanation, including a video and worked examples, see: Pythagoras Theorem.
How do you find the hypotenuse of a question?
The hypotenuse of a right angled triangle is the longest side. If you know the lengths of the other two sides, you can find the length of the hypotenuse by squaring the two shorter sides, adding those values together and then taking the square root.
By doing this you are finding c in a^2+b^2=c^2
How do you find the missing side of a triangle?
If your triangle is a right angled triangle and you know two of the sides, you can use Pythagoras Theorem to find the length of the third side. To do this, label the sides a, b and c (with c being the hypotenuse – the longest side). Substitute the values you know into a^2+b^2=c^2 and solve to find the missing side.
Looking for more Pythagoras Theorem questions and resources?
Third Space Learning's free GCSE maths resource library contains detailed lessons with step-by-step instructions on how to solve Pythagoras Theorem problems, as well as maths worksheets with practice questions and more GCSE exam questions, based on past Edexcel, OCR and AQA exam questions | 677.169 | 1 |
Four rectangles arranged in a cross shaped board, with a square in the center and triangles on the end of each arm. Diagonals are drawn in each rectangle and the square. Lines are drawn from the apex of a triangle, through the intersections of the diagonals, to the opposite triangle's apex. | 677.169 | 1 |
zula-oyun
Find the cosine of the angle between the planes −1x+3y+1z=0 and the plane 5x+5y+4z=−4
4 months ago
Q:
Find the cosine of the angle between the planes −1x+3y+1z=0 and the plane 5x+5y+4z=−4
Accepted Solution
A:
Answer:The he cosine of the angle between the planes is [tex]\frac{14}{11\sqrt{6}}[/tex].Step-by-step explanation:Using the definition of the dot product:[tex]\cos\theta =\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}[/tex]The given planes are[tex]-1x+3y+1z=0[/tex][tex]5x+5y+4z=-4[/tex]The angle between two normal vectors of the planes is the same as one of
the angles between the planes. We can find a normal vector to each of the
planes by looking at the coefficients of x, y, z.[tex]\overrightarrow{n_1}=<-1,3,1>[/tex][tex]\overrightarrow{n_2}=<5,5,4>[/tex][tex]\overrightarrow{n_1}\cdot \overrightarrow{n_2}=(-1)(5)+(3)(5)+(1)(4)=14[/tex][tex]|n_1|=\sqrt{(-1)^2+(3)^2+(1)^2}=\sqrt{11}[/tex][tex]|n_2|=\sqrt{(5)^2+(5)^2+(4)^2}=\sqrt{66}[/tex]The cosine of the angle between the planes[tex]\cos\theta =\frac{\overrightarrow{n_1}\cdot \overrightarrow{n_2}}{|\overrightarrow{n_1}||\overrightarrow{n_2}|}[/tex][tex]\cos\theta =\frac{14}{\sqrt{11}\sqrt{66}}[/tex][tex]\cos\theta =\frac{14}{11\sqrt{6}}[/tex]Therefore the cosine of the angle between the planes is [tex]\frac{14}{11\sqrt{6}}[/tex]. | 677.169 | 1 |
What is the angle of 125 degree?
Obtuse – any angle which measures more than 90 degrees but less than 180 degrees. These are "fat" angles that are very wide. Sample: angle DEF measures 125 degrees. Then angle DEF is obtuse.
How do you construct a 125 degree angle with a protractor?
We draw a line segment PQ of some length using a ruler. Then with point Q as the focus we measure an angle of 125° with a protractor and mark it on the figure, then we join the line segment from B to the mark and name the point R. Thus the angle ∠PQR = 125° is obtained.
How do you draw a 127 degree angle with a compass?
Following steps:
Draw a line using scale.
as a center draw an angle using protector like.
as a center draw an arc using compass which cuts line and at.
From point , draw an arc with any measure.
By same measure from point , draw an arc which cuts before arc at.
How do you construct a 120 degree and bisect it?
Answer
Draw an angle of ∠AOB of 120°.
Draw an arc from point O. This arc will cut the ray OA at any point C and ray OB at any point D.
Draw two arcs of the same radii from the points C and D as centres.
Join the point O to the point M by a ray OM. The ray OM will cut the angle ∠AOB into equal parts.
m∠AOM = m∠BOM =
What type of angle is 127 degrees?
obtuse angle-an angle between 90 and 180 degrees.
What are the angles you can make with a compass?
Angles that can be constructed by Compass and Straightedge are 15,30,45,60,90,120,150,180 degrees and some other angles can be constructed by Bisection (for eg.
Construction of an Angle of 60º Step 1: Draw any ray AB. Step 2: Taking A as the centre and with any suitable radius, draw an arc PQ that cuts AB at Q. Step 3: Taking Q as the centre and radius equal to AQ, draw an arc cutting the previous arc PQ at R. Step 4: Join AR and produce it to get AC. Step 5: ∠BAC is the required angle equal to 60º.
How to draw an equilateral triangle step by step?
⇒ ΔOPR is an equilateral triangle. Step I: Draw ∠AOB = 60º by using the steps mentioned above. Step II: With centre O and any convenient radius draw an arc cutting OA and OB at P and Q respectively. Step III: With centre P and radius more than (PQ), draw an arc in the interior of ∠AOB. Step VI: The angle ∠AOC is the angle of measure 30º | 677.169 | 1 |
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Week 8 in Math 10
This week we use knowledge about Trigonometry to solving triangles and problems.During I solving problems , I have a little trouble on it. Here is a question I will not do. I think I will do it, but my answer is inconsistent with the correct answer later. I really have no clue, so I raised the question. After teaching the students to explain, I know how to do it.So I used to do wrong because I used 9 years of age on the vertex, as well as parallel theory, so the answer is very different from the true answer In fact this question or should be used Trigonometry to solve the problem.When he explained half of the time he re-painted a triangle and just the answer we had just written, I was very puzzled. He explained to me that the original is not every question of the map is accurate. So when faced with such a type of problem, in order to do the problem can be re-painted, so that will not be due to the number of too many wrong.I think his little trick is good, so I think when I encounter a similar problem the next time you can also use this trick. | 677.169 | 1 |
Establish the diagonal D by connecting the top and bottom of a major price move b,. a straight line This diagonal will be toward the left and top of the pentagon to be constracted (see Figure A6-I).
1. Measure the length of the diagonal D. This diagonal connects two points of the pentagon.
2. With the point of a compass at one end of the diagonal and the tip on the other, draw a long arc to the right; placing the point on the other end of the diagonal, draw another arc to the right. These arcs should not cross to avoid confusion.
3. The length of a side of a regular pentagon is calculated from the diagonal by the formula S=.6I8xD
Using a ruler, set the compass to the length of a side and place the point at one end of the diagonal. Draw an arc on both sides, crossing the arc on the right do the same for the other end of the diagonal. The two new arcs will cross on the left The three new crossings are the missing points of the pentagon.
FIGURE A6-1 Constraction of aentagon from one diagonal.
4. The center of the pentagon can be found by bisecting any two sides. The point at which the two bisecting lines cross is the center. Use this point to circumscribe a circle around the pentagon.
CONSTRUCTION OF A PENTAGON FROM ONE SIDE
Establish the side S by connecting the top and bottom of a major price move with a sfraight line. As in the previous example, this side will be toward the top and left of the pentagon, which will extend down and to the right (see Figure A6-2).
1. Calculate the length of the diagonal D by applying the formula D = S/.6I8. This will require a ruler to determine the length of S.
2. Using a ruler again, set your compass to the length of the diagonal calculated in the first step. Draw wide arcs of radius D from the endpoints of S crossing to the lower right of S. The place of crossing will be the third point of the pentagon P" opposite side S
3. Set the compass back to length S and place the point at P, Cross the inner arcs drawn in step 2 with a small arc | 677.169 | 1 |
With this 30 60 90 triangle calculator, you can solve the measurements of this special right triangle. Whether you're looking for the 30 60 90 triangle formulas for the hypotenuse, wondering about the 30 60 90 triangle ratio, or simply want to check what this triangle looks like, you've found the right website. Keep scrolling to learn more about this specific right triangle, or check out our tool for the twin of our triangle – 45 45 90 triangle calc.
How do I solve a 30 60 90 triangle?
First of all, let's explain what "30 60 90" stands for. When writing about 30 60 90 triangle, we mean the angles of the triangle, that are equal to 30°, 60° and 90°.
Assume that the shorter leg of a 30 60 90 triangle is equal to a. Then:
The second leg is equal to a√3;
The hypotenuse is 2a;
The area is equal to a²√3/2; and
The perimeter equals a(3 + √3).
The 30 60 90 triangle formulas are quite easy, but what's the math behind them? Let's check which methods you can use to prove them:
Using the properties of the equilateral triangle
Did you notice that our triangle of interest is simply half of the equilateral triangle? If you remember the formula for the height of such a regular triangle, you have the answer to what's the second leg length. It's equal to side times a square root of 3, divided by 2:
h = c√3/2, h = b and c = 2a so b = c√3/2 = a√3
Using trigonometry
If you are familiar with the trigonometric basics, you can use, e.g., the sine and cosine of 30° to find out the other sides' lengths:
a/c = sin(30°) = 1/2 so c = 2a
b/c = sin(60°) = √3/2 so b = c√3/2 = a√3
Also, if you know two sides of the triangle, you can find the third one from the Pythagorean theorem. However, the methods described above are more useful as they need to have only one side of the 30 60 90 triangle given.
30 60 90 triangle sides
If we know the shorter leg length a, we can find out that:
b = a√3
c = 2a
If the longer leg length b is the one parameter given, then:
a = b√3/3
c = 2b√3/3
For hypotenuse c known, the legs formulas look as follows:
a = c/2
b = c√3/2
Or simply type your given values, and the 30 60 90 triangle calculator will do the rest!
What are the 30 60 90 triangle rules?
The most important rule to remember is that this special right triangle has one right angle, and its sides are in an easy-to-remember consistent relationship with one another - the ratio is a : a√3 : 2a. Also, the unusual property of this 30 60 90 triangle is that it's the only right triangle with angles in an arithmetic progression.
How do I find the legs of a 30 60 90 triangle given hypotenuse?
When the hypotenuse of a 30 60 90 triangle has length c, you can find the legs as follows:
Divide the length of the hypotenuse by 2.
Multiply the result of Step 1 by √3, i.e., by about 1.73.
The number you've got in Step 1 is the shorter leg of your triangle.
The number you've got in Step 2 is the longer leg.
What is the area of a 30 60 90 triangle with hypotenuse 10?
The area is 21.65. To get this result, use the formula area = a²√3/2, where a is the shorter leg of your triangle. Recall that the shorter leg is a half of the hypotenuse, so in our case a = 10 / 2 = 5. Plugging this value into the area formula, we get area = 25√3/2 ≈ 21.65.
a
b
c
Area
Perimeter
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Given: ofWhy are grades given the way they are given?
Grades are given the way they are given to provide a standardized measure of a student's performance and understanding of the mate...
Grades are given the way they are given to provide a standardized measure of a student's performance and understanding of the material. They serve as a way to evaluate and compare students' academic achievements. Grades also help students understand their strengths and weaknesses, providing feedback on their progress and areas for improvement. Additionally, grades are used by institutions to make decisions about students' academic standing, eligibility for scholarships, and future opportunities.
What is given when only the hypotenuse is given?
When only the hypotenuse is given in a right-angled triangle, it is not possible to determine the lengths of the other two sides w...
When only the hypotenuse is given in a right-angled triangle, it is not possible to determine the lengths of the other two sides without additional information. This is because there are an infinite number of right-angled triangles with the same hypotenuse but different side lengths. In order to find the lengths of the other sides, either the length of one of the other sides or the measure of one of the non-right angles must be given.
Source:AI generated from FAQ.net
What is given when the space diagonal is given?
When the space diagonal is given, the length of the three sides of the rectangular prism can be determined using the Pythagorean t...
When the space diagonal is given, the length of the three sides of the rectangular prism can be determined using the Pythagorean theorem. The space diagonal, the length, width, and height of the rectangular prism form a right-angled triangle, and the Pythagorean theorem can be used to find the lengths of the sides. This information is useful for calculating the volume, surface area, and other properties of the rectangular prism.
Source:AI generated from FAQ.net
What is given?
The question "What is given?" can be interpreted in different contexts. In a general sense, what is given refers to something that...
The question "What is given?" can be interpreted in different contexts. In a general sense, what is given refers to something that is already provided or known. It could be information, resources, instructions, or any other form of input that serves as a starting point for further actions or decisions. Identifying what is given is important in problem-solving, decision-making, and communication to ensure clarity and accuracycan documentsIs it a given?
No, it is not a given. A given implies that something is certain or guaranteed, but in reality, many things are uncertain and subj...
No, it is not a given. A given implies that something is certain or guaranteed, but in reality, many things are uncertain and subject to change. It is important to approach situations with an open mind and be prepared for unexpected outcomes.
Is allegedly happily given?
Allegedly happily given implies that something is claimed to have been given willingly and joyfully, but there may be doubt or ske...
Allegedly happily given implies that something is claimed to have been given willingly and joyfully, but there may be doubt or skepticism surrounding the claim. The use of "allegedly" suggests that there may be uncertainty or suspicion about the true nature of the giving. Therefore, whether something is allegedly happily given depends on the credibility and trustworthiness of the source making the claim.
Source:AI generated from FAQ.net
Why is H2O given?
H2O, or water, is given to maintain proper hydration in the body. It is essential for various bodily functions such as regulating...
H2O, or water, is given to maintain proper hydration in the body. It is essential for various bodily functions such as regulating body temperature, transporting nutrients and oxygen to cells, and removing waste products. Additionally, water is necessary for the proper functioning of organs and tissues. Therefore, it is important to consume an adequate amount of water to support overall health and well-being.
Source:AI generated from FAQ.net people experiences were given?
The experiences given were diverse and included opportunities for personal growth, skill development, and cultural immersion. Part...
The experiences given were diverse and included opportunities for personal growth, skill development, and cultural immersion. Participants engaged in activities such as team-building exercises, leadership workshops, and community service projects. They also had the chance to explore new environments, interact with people from different backgrounds, and gain a deeper understanding of global issues. Overall, the experiences provided a well-rounded and enriching learning journey for the participants.
Source:AI generated from FAQ.net
What was given up?
In the process of compromise, both parties typically give up some of their original demands or expectations in order to reach a mu...
In the process of compromise, both parties typically give up some of their original demands or expectations in order to reach a mutually agreeable solution. This could involve sacrificing certain goals, making concessions, or compromising on certain aspects of the issue at hand. Ultimately, compromise requires each party to let go of some of their initial desires in order to find common ground and move forward together.
What information is given when only the angle and the hypotenuse are given?
When only the angle and the hypotenuse are given in a right triangle, you can determine the length of the opposite side using trig...
When only the angle and the hypotenuse are given in a right triangle, you can determine the length of the opposite side using trigonometric ratios. The sine ratio (sin) relates the length of the opposite side to the hypotenuse, while the cosine ratio (cos) relates the length of the adjacent side to the hypotenuse. By using the given angle and the length of the hypotenuse, you can calculate the lengths of the other sides of the triangle.
How to construct a triangle with a given angle and two given altitudes?
To construct a triangle with a given angle and two given altitudes, start by drawing a base line for the triangle. Then, construct...
To construct a triangle with a given angle and two given altitudes, start by drawing a base line for the triangle. Then, construct the two altitudes from the endpoints of the base line at the given angles. The intersection of the two altitudes will determine the third vertex of the triangle. Finally, connect the third vertex to the endpoints of the base line to complete the triangle | 677.169 | 1 |
RD Sharma class 12th exercise 3.11 is an excellent NCERT solutions book that comes highly recommended by students from all over the country. RD Sharma solutions have been popular among high school students and teachers for a long time now. Their solutions are praised for being simple and easy to understand. The RD Sharma class 12 solutions chapter 3 ex 3.11, like any other RD Sharma book, has been specially formulated by experts from the education sector who have a lot of knowledge on the subject. RD Sharma solutions Their expertise in solving mathematical problems is evident as the solutions contain simple tips and tricks to help you solve the questions faster. So, you can follow their methods and guidelines to understand the concepts better and score well in exams.
Inverse Trigonometric Functions Excercise: 3.11
Answer: Hint: There is one formula in trigonometric function for union. Let's see the formula, Given: In LHS side, Let Explanation: Let's take LHS L.H.S Let's use formula, Let's put the value of x and y in formula Hence, the prove Note: We must remember the formula of union.
Answer: Hint: To solve this type of question, we must convert and in using hypotenuse theorem. Given: Explanation: For proving this type of question, Let's assume, So, we have to prove that First of all Using the formula of hypotenuse , third side will be 5. Also we have, Using the formula of hypotenuse, third side will be 3 …(ii) Also we have, Let's take tan on both sides, Let's put value, from (i), (ii) and (iii) ∴ L.H.S =R.H.S Hence, the prove Note: This type of problem can be easier by
Answer: Hint: For solving this, we can use the formula of union trigonometric function, Given: Explanation: L.H.S Let's take So, [Removing inverse] Using the formula of hypotenuse , Hypotenuse = Let's put value of (ii) into (i) … (iii) So, =R.H.S Hence, the prove Note: We must remember the formula of hypotenuse.
Answer: For solving this, we can use the formula of union trigonometric function, Given: and we have to find the value of x. Solution: Here A=2x and B=3x So, let's put the values of A and B in the formula of
Answer: Hint: We have to focus on calculation of infinite series. Given:
Solution:
The RD Sharma class 12 chapter 3 exercise 3.11 book is based on the chapter Inverse Trigonometric Function. This chapter was also introduced in class 11, where students had learned the basic concepts of trigonometry. Exercise 3.11 in this chapter has a total of 16 questions, which are divided into levels 1 and 2. They are mostly addition and equation questions in trigonometry. The chapter on Inverse Trigonometric Functions is a critical chapter that needs special attention from students.
Self-practice at home will immensely help improve students' clarity on concepts and help them solve questions faster. For this reason, class 12 RD Sharma chapter 3 exercise 3.11 solution will be beneficial for students to check their performance and mark their answers. By doing this, they will pinpoint their doubts and weak points to work on them. Moreover, the RD Sharma class 12th exercise 3.11 is constantly updated according to the latest syllabus, so you can be sure to find all answers you require.
The RD Sharma class 12 solutions Inverse Trigonometric Function ex 3.11 can be an excellent guide to students who like to practice solving questions at home. Students who have practiced with RD Sharma solutions have confirmed that they have found common questions in board exams. Therefore, using RD Sharma class 12 chapter 3 exercise 3.11 can help better your chances at finding common questions in the board paper.
Teachers too like to use RD Sharma class 12th exercise 3.11 to give homework questions to students in class 12. You can seek the help of these solutions to check your progress through the chapters and know if you have understood everything. The best thing about these books is that they are easily available online at Career360 and come completely free of cost. You can download the pdf online and use it anytime you want.
Frequently Asked Question (FAQs)
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RD Sharma solutions have helped thousands of students to score well in board exams, especially in their mathematics papers. Their answers are crafted by experts from the country who know how to solve complex math problems simply. Sometimes questions from the book appear on board papers, which helps students to score better. The book is continually updated with the latest syllabus so that you will find all answers here.
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Many students who have appeared for board exams expressed their love for RD Sharma solutions due to the quality of answers in the book. They have also confirmed that the questions in RD Sharma solutions have appeared in board exams, and they were able to solve them pretty easily. So, if you use RD Sharma solutions, you can find common questions in your board paper.
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5. Is it necessary to practice Trigonometric questions at home?
Chapter 3 of the mathematical book is on Inverse Trigonometric function, which is quite difficult to ace if you do not practice at home. Therefore, it is always recommended you use RD Sharma class 12th exercise 3.11 to practice at home and test your knowledge by comparing the answer in the book. This will help you understand the concepts better and enhance your problem-solving skills. | 677.169 | 1 |
Innovative Strategies to utilize your Terrace Table for Design
A Rhombus is one of the very least difficult yet most visually eyesight-catching designs that individuals can generate for most distinct capabilities. The four the same sides of the Rhombus provide a special symmetry that holders apart within the target audience. Moreover it is a functional design which helps both cosmetic and structural reasons. Making a Rhombus fashion needs some experience and understanding of geometry. With this extensive information, we shall take you step-by-step through the necessities of making a rhombus obklad.
Discovering the Rhombus:
The Rhombus is really a polygon with four congruent edges, and its components sum up to 360 levels. It is usually wrongly diagnosed for the valuable natural stone, which is actually a tilted model through the Rhombus. To build a Rhombus schedule, first get started with a sq, then tilt it with a certain route. The diagonal of your respective sq can cause two congruent triangles, which can be also much like the Rhombus's full complete opposite standpoint, constructing a 3-4-5 proper triangle.
Computing Location Measures:
Knowing the position of the Rhombus, you may use trigonometry to estimate one part lengths. As an example, in case the direction of your Rhombus is 75 diplomas, the solution to calculate along side it span is: sin (75) from the diagonal whole rectangular.
Constructing a Rhombus in composing:
To produce a Rhombus in creating, you start using a sq after which bring in the diagonals until they intersect. The point where the diagonals meet is your central vertex, and that's the spot you pull the edges of your respective Rhombus. To generate congruent sides, proceed to acquire intersection factors of your own diagonals and utilize them like a information when appealing to the Rhombus.
Creating a Rhombus by using an Topic:
To make a Rhombus on any subject, you must label the four the same steps of your respective Rhombus in the factor. Through case in point, if you're building a Rhombus having a solid wood package, you should label the four spots the place in which the Rhombus elements satisfy. On having observed these regions, work with a straightedge to get in feel those to create your Rhombus design.
Using Rhombus Designs in Design and magnificence:
Rhombus types may also add more an aesthetic entice many styles, which include fabric, wallpapers, and floor coverings flooring ceramic tiles. They might also produce a exclusive composition on visuals if used suitably. Designers enjoy utilizing Rhombus models using their designs to stay in front of other buildings' symmetrical patterns. In the design market place, Rhombus habits can produce unique outfits and accessories that establish types.
To Get It Quickly:
Rhombus designs can assist visual and architectural capabilities, releasing individuality to your type. Learning the basic principles of developing a Rhombus profile is essential in developing this polygon. Along with the referrals identified in this manual, commence performing workouts your creativity because they construct Rhombus designs in numerous styles. From wood cases to style developing patterns, sustain experimenting and mix forms, shades and coatings to generate particular and air flow-acquiring outputs. Get going right now!
Over these trying instances, very first responders have already been at the forefront of keeping us safe, wholesome, and protect. To exhibit their gratitude towards these courageous folks, Lululemon offers | 677.169 | 1 |
two triangles pictured below $m(\angle A) = m(\angle D)$ and $m(\angle B) = m(\angle E)$: Using a sequence of translations, rotations, reflectio | 677.169 | 1 |
Let the following be postulated:1. To draw a straight line from any point to any point2. To produce a finite straight line continuously in a straight line.3. To describe a circle with any centre and distance.4. That all right angles are equal to one another.5.That, if a straight line falling on two straight linesmake the interior angles on the same side less than two rightangles, the two straight lines, if produced indefinitely, meeton that side on which are the angles less thanthe two rightangles. | 677.169 | 1 |
If A, B, C are interior angles of ΔABC such that (cosA+cosB+cosC)2+(sinA+sinB+sinC)2=9, then number of possible triangles is
A
0
B
1
C
3
D
infinite
Video Solution
Text Solution
Verified by Experts
The correct Answer is:D
|
Answer
Step by step video, text & image solution for If A, B, C are interior angles of DeltaABC such that (cos A + cos B + cos C)^(2)+(sin A + sin B + sin C)^(2)=9, then number of possible triangles is by Maths experts to help you in doubts & scoring excellent marks in Class 12 exams. | 677.169 | 1 |
The apex is the _____ of a cone..
Quiz: Double-Napped Cone Module. Instructions: Answer all the following questions in the space provided. Simplify all answers. Describe or show how a double-napped cone is created. A generator is rotated about a fixed vertical axis. Label the vertex, the vertical axis, and the generator in the following diagram of a double-napped cone.
I know that the volume of a cone =$\dfrac{1}{3}\pi r^2h$ , and the maximum volume can found by setting the derivative equal to zero to see where the maximum lies. I tried to find a relation between the the height of the small cone and the larger cone to express h in terms of r in the equation of volume, but I got nothing. Help.Add a comment. Here is an answer using a double integral. I use the same set up and notation as in Andrew D. Hwang's answer, but in cylindrical coordinates. The equation of the cone is z = kr; of the plane, z = mr cos θ + h; and therefore of the elliptical shadow, r = h/(k − m cos θ). Then the volume is.A cone is a three-dimensional geometric shape that tapers smoothly from a flat base (frequently, though not necessarily, circular) to a point called the apex or vertex. A right circular cone with the radius of its base r, its height h, its slant height c and its angle θ. A cone is formed by a set of line segments, half-lines, or lines ...apex. [ a´peks] (pl. apexes, a´pices) ( L.) the pointed end of a cone-shaped part. adj., adj ap´ical. apex of lung the rounded upper extremity of either lung. root apex the terminal end of the root of the tooth.
The expression for the half-angle $\alpha$ makes sense if we draw the incident light cone differently--with a point on the object as apex and the circular pinhole as base. For consider that the object is emitting light in all directions from each point on its surface, but just a cone of the hemisphere of light rays from each point passes through the pinhole.Q. Point charge q 0 is placed inside a cone of base radius 'R', x distance below centre of the top surface as shown in figure.Find electric flux related to curved surface of the cone:- Q. A point charge q is placed at the apex of a cone as shown in figure. find the flux linked through the base of the cone.
In maths, a cone is defined as a distinctive three-dimensional geometric figure with a flat and curved surface pointed towards the top. The term "cone" is derived from the Greek word "konos", which means a wedge or a peak. The pointed end is the apex, whereas the flat surface is called the base . The three main properties of a cone are ... Cone is a three-dimensional structure having a circular base where a set of line segments connect all of the points on the base to a common point called the apex. There is a predefined set of formulas for the calculation of curved surface area as well as the total surface area of a cone, which is collectively known as the cone formula.
Cone. AA cone is a three dimensional solid that consists of a circular base and a continuous curved surface which tapers to a tip-top called the apex. The right circular cone and the oblique circular cones are the two types of cones. A cone is one third of a cylinder. The volume of a cone is one-third the circular base area multiplied by its height.The locus of the apex of a variable cone containing an ellipse fixed in three-space is a hyperbola through the foci of the ellipse. In addition, the locus of the apex of a cone containing that hyperbola is …Cone is a three-dimensional figure that has one circular base and one vertex (apex). An oblique cone is a cone with an apex that is not aligned above the center of the base. A right cone is a cone in which the apex is aligned directly above the center of the base.
The apex in a cone or pyramid is the vertex at the top which is opposite the base. The geometric shape of a cone is three-dimensional and it tapers smoothly from a balanced base to a point known as the apex. Figure 2 - Apex in Cone . A cone is constructed by a set of line segments.
Define apex. apex synonyms, apex pronunciation, apex translation, English dictionary definition of apex. n. pl. a·pex·es or a·pi·ces 1. a. The highest point of a structure, object, or geometric figure: the apex of a hill; the apex of a triangle. ... A military organization may be quite correctly compared to a cone, of which the base with ... semicircle shown is folded to form a right circular cone so that the arc PQ becomes the circumference of the base. Find the diameter of the base, Let circumference of cone base = C circumference of cone base = C and diameter = d diameter = d. I think the diameter should be 2C π = 2⋅5cm π ≈ 3.183cm 2 C π = 2 ⋅ 5 c m π ≈ 3.183 c m.A conifer cone or pinecone (strobilus, PL: strobili in formal botanical usage) is a seed-bearing organ on gymnosperm plants. ... Usually only one or two scales at the apex of the cone are fertile, each bearing a single wingless seed, but in Saxegothaea several scales may be fertile. The fleshy scale complex is 0.5-3 cm long, and the seeds 4 ...Click here👆to get an answer to your question ️ Show that the semi - vertical angle of the cone of the maximum volume and of given slant height is tan ^-1√(2)AThe question is slightly oddly phrased, so let's start with the most general case instead. If we have a right circular cone with apex at $\vec{o} = (x_o , y_o , z_o)$, unit axis vector $\hat{a} = (x_A , y_A , z_A)$, and aperture $\theta$.This means the angle between the axis and the sides of the cone is $\phi = \theta/2$.The locus of points $\vec{p} = (x , y , z)$ on the surface of the cone ...Expert Answer. 2. Show that the solid angle at the apex of a cone with semiangle a is 27 (1 - cosa). If a sphere has radius R and its centre at distance D from an observer, with D » R, show that the sphere occupies, as a fraction 1 VD2 - R2 22 1 2 D 4D2 5") of the observer's view. Use this to explain how the sun (at radius 7 x 105 km and ...1.3 Apex of Cone; 1.4 Apex of Pyramid; 2 Linguistic Note; 3 Sources; Definition. The apex of a geometric figure is the point which is distinguished from the others by dint of it being furthest away from its base. Not all figures have a discernible apex; for example, parallelograms, prisms and parallelepipeds do not.The vertical distance from the base center to the apex of a cone is the height (h), while the slant height of a cone is the length (l). The surface area of a cone is the sum of the area of the slanted, curved surface and area of the circular base. In this article, we will discuss how to find the surface area by using surface area of a cone ...The lateral surface of a cone is called a nappe. A double napped cone has two cones connected at the vertex. In the figure shown below, Cone 1 and Cone 2 are connected at the vertex. They form a double napped cone. The upper cone, that is the one above the vertex, is called the upper nappe, while the cone below the vertex is called the lower nappe.It's essentially a variation of Cover 2 with pattern matching. Instead of the Apex having the first to the flats, it's the corner. In turn, the Apex will cover #2 if they go vertical. If #2 goes outside, they will look to bracket #1 from the outside. Using Cut allows Saban to prevent quick throws outside that may out-leverage the Apex defender.
2. On-axis. Apex outside the Sphere If the cone apex is outside the sphere, d< R, the cone (projection) intersects the sphere at a near point characterized by (projected) cylinder coordinates Z 1;ˆ 1 and a far point Z 2;ˆ 2 as sketched in Figure4. In the gure the polar angle forA cone is a convergence of points of a circle at one point in a different plane, hence it will have a third property of having only one vertex, that is the apex. A pyramid is similar in this regard in that it has one apex point but it also has other vertices on its base due to its quadrilateral base structure.
Definition:Base of Cone; Linguistic Note. The plural of apex is apices, which is pronounced ay-pi-seez. The form apexes can often be seen, but this is technically incorrect. Compare vertex, whose plural is vertices. Hence the colloquial phrase base over apex as the description of a particularly flamboyant physical tumble. SourcesHeight of a Cone. The distance from the apex of a cone to the base. Formally, the shortest line segment between the apex of a cone and the (possibly extended) base. Altitude also refers to the length of this segment. The formula for the volume of a cone is (height x π x (diameter / 2)2) / 3, where (diameter / 2) is the radius of the base (d = 2 x r), so another way to write it is (height x π x radius2) / 3, as seen in the figure below: Despite the relative complexity of the body, you only need two measurements to calculate a cone's volume: its height and ...A hollow cone is a geometrical figure that looks similar to a normal cone or pyramid from the exterior but hollow on the inner side. The surface of a hollow cone may be considered to consist of an infinite number of triangles of infinitesimally slender isosceles, and thus the center of mass of a hollow cone (without foundation) is \[\frac {2}{3}\] of the way from the pole to the base midpoint.A cone is named based on the shape of its base. Figure 21.5 shows a circular cone. Circular cones fall into one of two categories: right circular cones and oblique circular cones. A right circular cone is a circular cone where the line segment connecting the apex of the cone to the center of the circular base is perpendicular to the plane of ...A ____ is one of two pieces of a double cone divided at the vertex. ellipse. A ____ is the locus of points in a plane such that the sum of the distances from any point in the locus to two points, called the foci, is a constant. major axis. The ____ is the line through the vertices of an ellipse. minor axis.Fig. 1 shows a schematic of the ideal problem geometry considered in the present work. An infinitely conducting electrified liquid cone (or Taylor cone), charged to a positive voltage with respect to infinity, is in vacuo. A spray of charged droplets (or electrospray) is steadily emitted from a small part of the lateral surface next to the apex (r ≤ r s see below) into the vacuum.Hyperbolic cross-section. When a plane cuts a cone at a higher angle to the base of the cone, the cross-section formed is hyperbolic. The angle must be greater than the angle of the lateral sides. They are composed of two branches. The two vertices are located one on each branch. These points are located where each branch changes direction.
Results. The second molar apex and apical 3 mm were located significantly deeper relative to the buccal bone surface compared with the first molar (p < 0.01).For the mandibular second molars, the distance from the buccal bone surface to the root apex was significantly shorter in patients over 70 years of age (p < 0.05).Furthermore, this distance was significantly shorter when the first molar ...
It is defined as the curve which is the intersection of cone and plane. There are three major conic sections; parabola, hyperbola, and ellipse (circle is a special type of ellipse). We have a statement: Which degenerate conic is formed when a double cone is cut through the top by a plane parallel to the slant edge of the cone?
If the plane intersects both halves of the double cone but does not pass through the apex of the cones, then the conic is a hyperbola. WikiMatrix Requirements of points 1.1.3.1, 1.1.3.2 and 1.1.3.3 do not apply to components located beyond the steering wheel, as referenced from the apex of a cone , this apex being the centre of zone A in Figure ...The base of the cone is a circle, with an area π r 2. • The base of the cylinder is also a circle with an area of π r 2. • The height of the cone and the cylinder is h. • The volume of the cylinder is V = π r 2 h. • Since the water from the cone fills one-third of the cylinder, the volume of the cone is one-third the volume of the ...The property of degeneracy takes place when the cone of the apex exists in the plane or during the process of the cone being degenerated to a cylinder, also when the plane is seen parallel to the cylinder axis. The standard equation of the conics is given by ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0.Define apex. apex synonyms, apex pronunciation, apex translation, English dictionary definition of apex. n. pl. a·pex·es or a·pi·ces 1. a. The highest point of a structure, object, …A cone has a circular base with a diameter of 18 inches. The height of the cone is 40 inches. The slant height of the cone is 41 inches. What is the approximate LATERAL AREA of the CONE? Use 3.14 for π and round to the nearest whole number.A frustum of a cone is obtained by cutting the apex portion of the cone with a plane parallel to its base. One way to solve the volume of this frustum is to understand the volume of a triangularClick the "Circle" icon on the top or press "C" on your keyboard. Click anywhere, then move the mouse outward from where you clicked first. Next, click again when you're satisfied with its size. 2. Draw a line from center to the edge and from the center upward. This will determine the height of the cone. Draw a line connecting the top of the ...Definition of a frustum of a right circular cone: A frustum of a right circular cone (a truncated cone) is a geometrical figure that is created from a right circular cone by cutting off the tip of the cone perpendicular to its height H. The small h is the height of the truncated cone.A cone is a three-dimensional figure that is formed by connecting infinite line segments from a common point to all the points in a circular base.This common point is also known as an apex. The cone is measured using three dimensions: radius of its circular base, height and lateral height.
Feb 27, 2021 · Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have PurposeInstagram: doordash promo codes for new usersnorthern hills rvcomedy central on dishchoghadiya nj A right circular cone is a cone where the axis of the cone is the line meeting the vertex to the midpoint of the circular base. That is, the centre point of the circular base is joined with the apex of the cone and it forms a right angle. A cone is a three-dimensional shape having a circular base and narrowing smoothly to a point above the base. Measure the cone. Dimension the Cone. "A" is the included angle. Using variable and various methods of dimensioning you can report the angle of one side. There are a number of threads detailing that. Construct a circle from the cone, at a Z=0 location. Dimension the circle, "D" will be the diameter you need. Measure the hole as a cylinder. connecticut state lotto resultswww deltanet com 3 The Apex Angle formula is defined as the apex is the pointed tip of a cone. The apex angle is the angle between the lines that define the apex is calculated using Apex Angle = tan (Alpha).To calculate Apex Angle, you need Alpha (α).With our tool, you need to enter the respective value for Alpha and hit the calculate button. old uhaul for sale Click here👆to get an answer to your question ️ Point charge q0 is placed inside a cone of base radius 'R', x distance below centre of the top surface as shown in figure.Find electric flux related to curved surface of the cone: -Viewed 3k times. 3. Consider a hollow cone with uniform charge distribution over its surface. When one finds the electric field at its apex it comes out to be an infinite value. However, when a solid cone with uniform charge distribution in its volume is taken and the electric field at its apex is found out it comes out to be a finite value. | 677.169 | 1 |
It's Snowing Angles FREE Worksheet and Math Activity
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Did you know that snowflakes have many different types of angles?
If you are working on angles and geometry in your home, here is a fun activity and worksheet for your kids to work on. All you need for this fun activity is this free worksheet, some colored markers and a protractor. Your kids will have fun creating their own angles | 677.169 | 1 |
Help the wiki!
Trigonometry
For more information, see Trigonometry on Wikipedia.Trigonometry is a branch of mathematics which consists of the study of right-angled triangles — specifically, the ratios of sides of right-angled triangles. Trig (short for trigonometry) functions simply return the ratio of a certain two sides of a triangle, given one angle; or the angle given a ratio of two sides. The point of trigonometry is to be able to quickly relate angles to side lengths and vice-versa to do otherwise complex calculations. For example, finding out the new position of a sprite after it has moved some distance given its direction is impossible without trigonometry.
Basically, trigonometry is a shortcut to find relations between angles and lengths that can be theoretically measured. It is a powerful tool, and has applications in all sorts of fields.
Note:
This article is intended for an audience with some background in math, especially algebra and geometry.
Angles and Directions
Trigonometry deals with angles and directions. The wider an angle is, the greater the measurement of it is. Below is a depiction of all the angles up to 360∘{\displaystyle 360^{\circ }}. Angles that are greater than 360 degrees are coterminal to the lesser ones, meaning they lie in the same direction relative to the origin of a coordinate plane and have the same outcome in trigonometric functions.
Notice how the angle increases as it rotates leftward. Rotating the angle to the right decreases it. An angle of 180∘{\displaystyle 180^{\circ }}, 0∘{\displaystyle 0^{\circ }}, and any of their coterminal angles depict the geometric figure, a straight line (which are technically quadrantal angles). Scratch directions initiate an analog instead of trigonometric style, therefore being inconsistent with trigonometry. Here are functions to quickly convert between the two:
Angle-to-Side Relationship
In trigonometry, an angle is formed between two lines: an initial ray and a terminal ray. The initial ray always lies on the x{\displaystyle x} axis. This is because mathematicians prefer it this way - it is a standard that is used to define the trigonometric values. Hundreds of years ago, if mathematicians desired the initial ray to be on the y{\displaystyle y} axis, it likely would be today, but the standard was set and trigonometric rules apply to these specific standards. The other line is known as the terminal ray, which can be rotated about the origin of the coordinate plane.
Trigonometry deals with ratios between the initial and terminal lines. An example of this is shown in the following image:
This may seem confusing at first, but the concept is very simple. An angle is formed as the rotation between two lines or segments. The following image depicts angle size increasing:
Notice in the images below, the terminal line comes to a stop and does not go on forever. This relates to a scenario of the distance between two points. For instance, suppose the origin of the coordinate plane is an object. In that case, the end of the terminal side is another object, and the line represents the distance between those two objects, or mathematically "points". The terminal side will always be known as the hypotenuse in terms of geometry and trigonometry.
Where do triangles become involved? Take into consideration the 2D coordinate plane. It has two values of positioning: the x{\displaystyle x} and y{\displaystyle y} values. A pair of x{\displaystyle x} and y{\displaystyle y} values used to determine the position of a point is known as an ordered pair. Trigonometry deals with the relationship of ordered pairs. Trigonometry states that:
If any two ordered pairs have three related lines that form a triangle, if that triangle consists of a right angle (90 degrees), the ratio of the sides of the triangle are dependent and consistently based on the angle formed between the initial side and hypotenuse.
The Functions
There are three major trigonometry functions. To define them, the following name is used:
Caution:
These are relative to angle A. The names change depending on the angle being considered.
The Sine (sin{\displaystyle \sin }) is the Opposite ÷ Hypotenuse
The Cosine (cos{\displaystyle \cos }) is the Adjacent ÷ Hypotenuse
The Tangent (tan{\displaystyle \tan }) is the Opposite ÷ Adjacent
To remember these functions, some people use the mnemonic "SOH CAH TOA".
There functions are expressed as, for example, sin(45∘){\displaystyle \sin(45^{\circ })} or cos(60rad){\displaystyle \cos(60rad)}.
Using Trigonometric Functions: Example
Trig functions have many uses in programming, especially in graphics and physics simulations. For example, consider a rock thrown at 30∘{\displaystyle 30^{\circ }} at 5m/s{\displaystyle 5m/s}. To model the parabolic (curved) path of the rock, split the tilted velocity into a horizontal and vertical velocity, then move the sprite by those values in the respective directions repetitively. Also, it is needed to constantly decrement the vertical velocity to account for gravity.
To split the values, trigonometry is to be used. Image a right triangle with one angle 30∘{\displaystyle 30^{\circ }}; and hypotenuse 5m/s{\displaystyle 5m/s}. Now, the opposite side must be the vertical velocity and the adjacent side must be the horizontal velocity. To find the opposite side, find the sine of 30°, which is opposite/hypotenuse. Multiply it by the hypotenuse (i.e. 5m/s{\displaystyle 5m/s}). The result, using a calculator to evaluate sin(30∘)∗5{\displaystyle sin(30^{\circ })*5} is 2.5m/s{\displaystyle 2.5m/s}. Similar reasoning can be used to find the horizontal velocity using sin(30∘){\displaystyle sin(30^{\circ })}.
Warning:
Note that in some calculators, such as Google Calculator, degrees must be specified, since it assumes radians. However, Scratch always uses degrees, as does Wolfram Alpha.
Other Uses
Some of these can be attempted to learn more about trigonometry.
Play around with sin, cos and tangent by drawing its graph using the pen blocks. Since the sin and cos graphs only have a range of ∣y∣≤1, it is helpful to multiply the trigonometric function by a constant (around 100) to see the features of the function.
Using atan{\displaystyle atan} to find the direction the mouse is moving in—find the atan{\displaystyle atan} of the ratio of X motion and Y motion at any given point in time, and you should get the direction in which it is moving.
Similarly, use atan{\displaystyle atan} to draw a line with a user defined slope, which passes through another user defined point.
Using atan{\displaystyle atan} to make a block called "point towards x:() y:()"—use similar reasoning as the above
Using a script like the following to make a sprite move in complex paths:
Trigonometric functions are all cyclic, which means they keep repeating. So, complex motions which repeat indefinitely are possible without too much trouble.
Predicting the position of a sprite after it moves some distance in a specific direction—this is a simple application of sine and cosine. One interesting use of this is to make a sprite move perpendicular to the direction it is facing in, or move in a circle without changing its direction.
Modeling 3D rotations—this is a much more advanced application of trigonometry. It relies on the principle that any point in 3D, when rotated through some angle, will appear to move straight to a viewer (imagine staring at a single point on a spinning globe). The distance moved can be calculated with some more complex trigonometry.
Angles Greater than 90°
A sine wave
Angles between 0 and 90 degrees are not the only possible inputs to trig functions. The values of sine, cosine, tangent, secant, cosecant, and cotangent at A{\displaystyle A} are the same as the respective values at Amod360∘{\displaystyle A\,{\bmod {\,}}360^{\circ }} (i.e. the remainder obtained when A{\displaystyle A} is divided by 360∘{\displaystyle 360^{\circ }}). This holds for negative values of A{\displaystyle A}, too. The periodic behavior of the sine function is visualized with the sine wave or sinusoid, where the wave is the same when the graph is shifted horizontally (i.e. the input is changed) by 2π=360∘{\displaystyle 2\pi =360^{\circ }}. The same applies to the other functions, with different graphs but the same principle.
These equations (identities) about trig functions can be used to calculate their values anywhere given their values between 0° and 90°:
sin=1csc{\displaystyle \sin ={\frac {1}{\csc }}}
cos=1sec{\displaystyle \cos ={\frac {1}{\sec }}}
tan=1cot{\displaystyle \tan ={\frac {1}{\cot }}}
sin(−x)=−sin(x){\displaystyle \sin(-x)=-\sin(x)}
cos(−x)=cos(x){\displaystyle \cos(-x)=\cos(x)}
tan(−x)=−tan(x){\displaystyle \tan(-x)=-\tan(x)}
sin(x+180∘)=−sin(x){\displaystyle \sin(x+180^{\circ })=-\sin(x)}
cos(x+180∘)=−cos(x){\displaystyle \cos(x+180^{\circ })=-\cos(x)}
tan(x+180∘)=tan(x){\displaystyle \tan(x+180^{\circ })=\tan(x)}
For Non-Right Triangles
Trigonometry is not just used with right triangles. The following equations apply to all triangles.
Note:
The lowercase letters a, b, and c are lengths of the sides of a triangle. Capital letters A, B, and C are measures of the angles opposite sides a, b, and c respectively.
An obtuse triangle with its sides as lowercase letters and its angles as capital letters.
asinA=bsinB=csinC=2R{\displaystyle {\frac {a}{\sin A}}={\frac {b}{\sin B}}={\frac {c}{\sin C}}=2R} where R{\displaystyle R} is the radius of a circle circumscribed around the triangle.
This identity is extremely useful to relate sides of triangles and angles. Importantly, it also relates the circumradius of the given triangle. The circumradius is the radius of the circle in which the triangle fits perfectly (each vertex lies on the circle).
The obvious use of this surprising but true identity is to find the third side of a triangle given any two sides and the distance between them, or finding the angles given three sides. This has many interesting uses, for example: If you have a spaceship-shooting game like Asteroids, you can program the AI spaceships to aim towards the point where the target will be when the bullet hits it (since during the time taken for the bullet to move, the spacecraft themselves change position linearly). | 677.169 | 1 |
Parameters
Returns
A collection of geometry objects that represent the intersection of the given geometries.
Remarks
The returned collection contains one geometry of each dimension for which there are intersections. For example, if both
inputs are polylines, the collection contains at most two geometries: the first a multipoint containing the points
at which the lines cross, and the second a polyline containing the lines of overlap. If a crossing point lies within a
line of overlap, only the line of overlap is present the result set is not self-intersecting. If there are no crossing
points or there are no lines of overlap, the respective geometry is not present in the returned collection. If the
input geometries do not intersect, the resulting collection is empty. The table below shows, for each combination
of pairs of input geometry types, the types of geometry that are contained within the returned collection if there
are intersections of that type.
The geometries in the returned collection are sorted by ascending dimensionality, e.g. multipoint (dimension 0) then
polyline (dimension 1) then polygon (dimension 2) for the intersection of two geometries with area that have
intersections of those types. | 677.169 | 1 |
Question 1.
Draw five pairs of complementary angles of your choice. (Page no. 71)
Solution:
The following are the pairs of complementary angles
Do This
Question 1.
Draw an angle ∠AOB = 40°. With the same vertex 0' draw ∠BOC = 50°, taking \(\overrightarrow{\mathbf{O B}}\) Initial ray as shown In the figure. Since the sum of these angles Is 90°, they together form a right angle. Take another pair 60° and 50° and join in the same way. Do they form complementary angles? Why? Why not? (Page No. 71)
Solution:
∠AOB = 40°, ∠BOC = 50°
∠POQ = 60°
∠QOR = 50°
These two do not form pair of complementary angles since their sum is 60° + 50° = 110° ≠ 90° | 677.169 | 1 |
Step 4
Marking vectors. Each vector is marked here as a circle in the center of each square. Vectors injection is shown on the video below. Syringe and needle are normal or perpendicular ( vertical) to the center of each square, injecting 0.05 ml to 0.1 ml in function of the dimensions of each square .
05
Step 5
Tensors are marked as yellow arrows ,always with a direction and sense from down to up and in to out.
The arrows bottom is placed on a langer s line for each square , doing an angle of 90 grades or perpendicular to the langer s line.
During injection, the aisles of the syringe are placed on the bottom of each yellow arrow and then it is pushed following the direction and sense of each arrow to inject from 0.05 ml to 0.1 ml, without moving the hand or without changing angles . | 677.169 | 1 |
How Many Sides Does A Polygon?
Other Types of Polygons
Polygon
Number of Sides
Quadrilateral
4
Pentagon
5
Hexagon
6
Heptagon
7
Does a polygon have 4 or more sides?
Triangles quadrilaterals pentagons and hexagons are all examples of polygons. The name tells you how many sides the shape has. For example a triangle has three sides and a quadrilateral has four sides. … Definition of a Polygon.
Shape
# of Sides
Rectangle
4
Quadrilateral
4
Pentagon
5
Hexagon
6
Do polygons have 8 sides?
In geometry an octagon (from the Greek ὀκτάγωνον oktágōnon "eight angles") is an eight-sided polygon or 8-gon. … Octagon.
Does a polygon have 2 sides?
In geometry a digon is a polygon with two sides (edges) and two vertices. … Digon.
Regular digon
Symmetry group
D2 [2] (*2•)
Do polygons have 7 sides?
In geometry a heptagon or septagon is a seven-sided polygon or 7-gon. … Heptagon.
Regular heptagon
A regular heptagon
Type
Regular polygon
Edges and vertices
7
Schläfli symbol
{7}
How many sides has a pentagon?
5
What is a polygon with 4 sides?
Definition: A quadrilateral is a polygon with 4 sides. … Definition: A parallelogram is a quadrilateral where both pairs of opposite sides are parallel.
What figure has 9 sides?
nonagon In geometry a nonagon (/ˈnɒnəɡɒn/) or enneagon (/ˈɛniəɡɒn/) is a nine-sided polygon or 9-gon. The name nonagon is a prefix hybrid formation from Latin (nonus "ninth" + gonon) used equivalently attested already in the 16th century in French nonogone and in English from the 17th century.
What is a polygon with 10 sides?
In geometry a decagon (from the Greek δέκα déka and γωνία gonía "ten angles") is a ten-sided polygon or 10-gon. The total sum of the interior angles of a simple decagon is 1440°. A self-intersecting regular decagon is known as a decagram.
What type of polygon has 12 sides?
Dodecagon
Regular dodecagon
A regular dodecagon
Type
Regular polygon
Edges and vertices
12
Schläfli symbol
{12} t{6} tt{3}
What is a polygon with 3 sides?
triangle
List of n-gons by Greek numerical prefixes
Sides
Names
3
trigon
triangle
4
tetragon
quadrilateral
5
pentagon
6
hexagon
What is a polygon with 6 sides?
In geometry a hexagon (from Greek ἕξ hex meaning "six" and γωνία gonía meaning "corner angle") is a six-sided polygon or 6-gon. The total of the internal angles of any simple (non-self-intersecting) hexagon is 720°.
What is a shape with 3 sides?
A triangle has 3 sides and 3 corners. A corner is where two lines meet or join. There are different kinds of triangles. These triangles have sides that are all the same length.
What is a shape with 8 sides?
octagon
An octagon is a shape with 8 sides and 8 angles.
What is a 100 sided shape called?
hectogon In geometry a hectogon or hecatontagon or 100-gon is a hundred-sided polygon. The sum of all hectogon's interior angles are 17640 degrees.
What 2d shape has 5 sides?
pentagon
A five-sided shape is called a pentagon. A six-sided shape is a hexagon a seven-sided shape a heptagon while an octagon has eight sides…
What is a 1000000 sided shape called?
Which of the following polygon has 11 sides?
hendecagon In geometry a hendecagon (also undecagon or endecagon) or 11-gon is an eleven-sided polygon. (The name hendecagon from Greek hendeka "eleven" and –gon "corner" is often preferred to the hybrid undecagon whose first part is formed from Latin undecim "eleven".)
What is a 200 sided shape called?
What is the name of a polygon with…?
#
Name of the Polygon + Geometric Drawing
200 sides
dihectogon
300 sides
trihectogon
400 sides
tetrahectogon
500 sides
pentahectogon
What do you call a polygon with 20 sides?
In geometry an icosagon or 20-gon is a twenty-sided polygon. The sum of any icosagon's interior angles is 3240 degrees.
What's a shape with 14 sides?
tetradecagon In geometry a tetradecagon or tetrakaidecagon or 14-gon is a fourteen-sided polygon.
What polygon has 13 sides?
Tridecagon
A 13-sided polygon sometimes also called the triskaidecagon.
What is a shape with 7 sides called?
A heptagon is a seven-sided polygon. It is also sometimes called a septagon though this usage mixes a Latin prefix sept- (derived from septua- meaning "seven") with the Greek suffix -gon (from gonia meaning "angle") and is therefore not recommended.
What is a polygon with 4 sides and 4 angles?
A quadrilateral is a polygon that has exactly four sides. (This also means that a quadrilateral has exactly four vertices and exactly four angles.)Do hexagons have equal sides?
Hexagons are six sided figures and possess the following shape: In a regular hexagon all sides equal the same length and all interior angles have the same measure therefore we can write the following expression.
Do all polygons have equal sides?
Regular Polygon. A regular polygon is a polygon in which all the interior angles are equal and also all the sides are equal. There are different types of regular polygons. … A triangle: An equilateral triangle is a regular polygon with three equal side lengths and three equal angles.
What is pentagon sides?
A pentagon is a geometrical shape which has five sides and five angles. Here "Penta" denotes five and "gon" denotes angle. The pentagon is one of the types of polygons. The sum of all the interior angles for a regular pentagon is 540 degrees.
How Many Sides Does A Polygon Have
Polygons Finding Number of Sides
How many sides does a regular polygon have if given one interior angle | 677.169 | 1 |
Math Labs with Activity – Find the Circumcentre of a Given Triangle
Math Labs with Activity – Find the Circumcentre of a Given Triangle
To find the circumcentre of a given triangle by the method of paper folding.
Materials Required
Three sheets of white paper
A geometry box
Theory The point of intersection of the perpendicular bisectors of the sides of a triangle is called its circumcentre.
Procedure Step 1: We shall first find the circumcentre of an acute- angled triangle. Draw an acute-angled triangle ABC on a sheet of white paper. Step 21). Step 3: Fold the paper along the line that cuts the side AC such that the point A falls on the point C. Make a crease and unfold the paper. Draw a line X2Y2 along the crease. Then X2Y2 is the perpendicular bisector of the side AC (see Figure 19.1). Mark the point O where the lines X1Y1 and X2Y2 intersect. Then, the point O is the circumcentre of ΔABC. What do you observe? Step 4: We shall now find the circumcentre of a right-angled triangle. Draw a right-angled triangle ABC (right angled at C) on another sheet of white paper. Step 52). Step 6: Fold the paper along a2). Mark the point O where the lines X1Y1 and X2Y2 intersect. Then, O is the circumcentre of ΔABC. What do you observe? Step 7: We shall now find the circumcentre of an obtuse¬angled triangle. Draw an obtuse-angled triangle ABC (in which ∠B is obtuse) on the third sheet of white paper. Step 83). Step 9: Fold the paper along the3). Mark the point O where the lines X1Y1 and X2Y2 intersect. Then, O is the circumcentre of ΔABC. What do you observe?
Observations
In an acute-angled triangle, the circumcentre lies inside the triangle.
In a right-angled triangle, the circumcentre lies at the midpoint of the hypotenuse. This can be confirmed by drawing the perpendicular bisector of the hypotenuse. The three perpendicular bisectors meet at the midpoint of the hypotenuse.
In an obtuse-angled triangle, the circumcentre lies outside the triangle.
Result The point O is the circumcentre of the triangle (in each case).
Remarks: The teacher must explain it to the students that since the perpendicular bisectors of all the three sides of a triangle meet at a single point, it is sufficient to find the point of intersection of the perpendicular bisectors of only two sides to obtain the circumcentre of the triangle. | 677.169 | 1 |
Lesson video
Hi, I'm Ms. Kidd-Rossiter and I'm going to be taking you through today's lesson on enlargement by a negative scale factor.
Really great topic, continuing on from all the work that we've done so far on enlargements.
If you can, make sure that you're in a nice, quiet space where you can concentrate and that you're free from distractions.
If you need to, pause the video now so you can get yourself sorted.
If not, let's get going.
So we're starting today's lesson with a try this activity.
On your screen, you can see three shapes.
You've got centre of enlargement, shape A, and shape B, and shape C.
It's your job now to pause the video and have a think about what's the same and what's different.
If you're struggling a little bit with this activity, you might want to think about the size of the shapes, and the orientation.
So pause the video now and have a go at this activity.
Hopefully you had a really good go at that try this activity and you managed to come up with some similarities and differences.
We're now going to build on the work that we did on enlargements from a given point.
On the right-hand side of your screen, you can see that we've got shape A, which is our object and shape B, which is our image.
What was the scale factor of enlargement from, if shape A is the object and shape B is the image, tell the screen now, what's the scale factor? Brilliant.
You should have said that it was three.
What are the two ways that we could have worked that out? Pause the video now and think about that.
Brilliant.
The first way is that we could have compared the length of the side.
So we can see that this one here is two squares wide, and this one here is six squares wide, and we could have done the same for the height of the triangle.
The other way that we could have looked at it is from the given point.
So from our given centre here, P, to this vertex is one arrow, and from the same vertex to this top left corner of B it's three arrows, so that we can see that that is a scale factor three, enlargement.
What about then A and C? So this time A is our object, and C is our image.
Pause the screen now, and have a think about what you think has gone on here.
Okay.
Hopefully you realise that it's still an enlargement.
We've still enlarged our shape.
This has gone from two squares and we've got six squares here.
It can't be an enlargement of scale factor three, because if it was an enlargement of scale factor three, it would have given us this image here.
You can see, that instead of going in the same direction, our rail lines have gone in the opposite direction.
When we enlarge in the opposite direction, this is by a negative scale factor.
So we're enlarging shape A by a negative scale factor.
Our negative scale factor for this enlargement is -3.
So you can see, that from P to this vertex was one arrow, in the opposite direction, we've done it three arrows, one, two, three, and that's given us the corresponding vertex.
Let's look at a second corner.
We can see this one arrow here has taken us to this vertex.
So we're going to stretch by -3 in the opposite direction, one, two, three, to give us this vertex.
And then finally, we've got this bottom vertex here.
When we enlarged by a scale factor of -3, we stretched by three in the opposite direction, one, two, three, and that vertex now becomes the top.
So key things to know about negative enlargements are that they change the orientation of the shape.
If the scale factor is equal to -1, then the size of the shape stays the same, but the orientation changes.
If the scale factor is less than -1, then the size of the shape will increase and the orientation will change.
And if the scale factor is less than zero, but more than -1, then the size of the shape will change and it will become smaller, and the orientation will change.
You're now going to have a go at applying what we've learned to the independent task.
So pause the video, navigate to the independent task and then resume the video when you're ready to go through some answers.
Well done at having a go at that independent task.
We're going to go through some of the answers now.
So the first thing you were asked to do was to copy down the diagram to the left and then enlarge the shape by all the scale factors given on your screen.
We're going to go through the first one and the final one in detail, and I'm going to show you the answers to the others.
So for the first one, enlarge by a scale factor of -1 from 6,4.
Now a scale factor, -1 should be telling me that it's going to stay the same size, but then orientation is going to change.
So the first thing I'm going to do is I'm going to plot my centre, which is 6,4.
So you can see that on the graph there.
Then the next thing I would do is drawing my write lines.
So through the centre and through each vertex of the pentagon.
Then I can see, I can measure from the centre, to each vertex.
And I know I have to go the same distance in the opposite direction for my image.
And you'd repeat that for each vertex.
And then you would see that this is the shape that you end up with.
So really good work if you've got that, if not check your work, maybe listen back to my explanation and give it another go.
If you need to pause the video now and have another go at this task.
There, this will take a moment.
Here's my centre, here's my enlargement.
For part C, see, here's my centre, here's my enlargement.
For part D, here's my centre, and here's my enlargement.
And for part E, we're going to do the same thing, plot the centre, which in this case is 5,3.
So it's actually on the pentagon.
And now we're going to draw in our rate lines.
So here they are.
And we do the same thing, we measure from the centre to each vertex.
And this time it's a scale factor of -2.
So we do it in the opposite direction, multiplied by two.
So from here to this vertex was one square.
So it will go two squares in the opposite direction.
And then we repeat that for each vertex to give us our final answer of this pentagon.
So that is your final answer.
I hope you got that one.
If not go back and check your work and maybe re-listen to my explanation.
For the second task, you were asked to draw the axis, like the ones on the screen, and then enlarge A with the centre P and scale factor -2, which should have given you this shape here, scale factor -1.
5, which should have given you this shape here.
Finally, scale factor -0.
5, which should have given you this enlargement here.
Well done on that task.
It was quite tricky.
So really good for persevering.
We're going to finish off today's lesson with this explore task.
Zaki is saying that he's going to complete an enlargement on the shape on your screen.
He says that the image will have a greater area, and the same orientation as the object.
I would like you to write similar statements, for each of the scale factors that are given here in the boxes.
Pause the video now and have a go at that task.
When you're ready to go through it, resume the video.
For the first one we had with a scale factor of -1.
Hopefully you wrote your answers in full sentences.
I'm just going to go through the key points.
So scale factor of one would have the same area and the same orientation.
So you should have written a sentence that says the image will have the same area and the same orientation as the object.
For a scale factor between zero and one.
So greater than zero and less than one, it will have a smaller area and the same orientation.
So your sentence should have said, the image will have a smaller area and the same orientation as the object.
For a scale factor, which is equal to -1, you should have realised that it had the same area, but a different orientation.
So your sentence should have read, the image will have the same area and a different orientation to the object.
For scale factor greater than one the image will have a greater area, and the same orientation as the object.
So this is actually what Zaki was talking about.
So when you found some examples, you could have had any number that was greater than one With a scale factor, which is greater than -1, but less than zero, the image will have a smaller area and a different orientation to the object.
And finally, for a scale factor, less than -1, the image will have a greater area and a different orientation to the object.
So really well done with those.
I hope that you had a really good go at that activity.
And if not, you've listened to my explanation.
I know that you will have drawn some really great enlargements today, so if you'd like to, please ask your parent or carer to share your work on Twitter, tagging @OakNational and #LearnwithOak. | 677.169 | 1 |
The math::geometry package is a
collection of functions for computations and manipulations on
two-dimensional geometrical objects, such as points, lines and
polygons.
The geometrical objects are implemented as plain lists of
coordinates. For instance a line is defined by a list of four
numbers, the x- and y-coordinate of a first point and the x- and
y-coordinates of a second point on the line.
The various types of object are recognised by the number of
coordinate pairs and the context in which they are used: a list of
four elements can be regarded as an infinite line, a finite line
segment but also as a polyline of one segment and a point set of
two points.
Currently the following types of objects are distinguished:
point - a list of two coordinates representing the x-
and y-coordinates respectively.
line - a list of four coordinates, interpreted as the
x- and y-coordinates of two distinct points on the line.
line segment - a list of four coordinates, interpreted
as the x- and y-coordinates of the first and the last points on the
line segment.
polyline - a list of an even number of coordinates,
interpreted as the x- and y-coordinates of an ordered set of
points.
polygon - like a polyline, but the implicit assumption
is that the polyline is closed (if the first and last points do not
coincide, the missing segment is automatically added).
point set - again a list of an even number of
coordinates, but the points are regarded without any ordering.
circle - a list of three numbers, the first two are the
coordinates of the centre and the third is the radius.
Compute the point which is at relative distance s between the two points and return it as the result of
the command. A relative distance of 0 returns
point1, the distance 1
returns point2. Distances < 0 or > 1
extrapolate along the line between the two point.
Check if two line segments intersect or coincide. Returns 1 if
that is the case, 0 otherwise (in two dimensions only). If an
endpoint of one segment lies on the other segment (or is very close
to the segment), they are considered to intersect
Check whether two polylines intersect, but reduce the
correctness of the result to the given granularity. Use this for
faster, but weaker, intersection checking.
How it works:
Each polyline is split into a number of smaller polylines,
consisting of granularity points each. If a pair of those smaller
lines' bounding boxes intersect, then this procedure returns 1,
otherwise it returns 0.
Return the third point of the rectangular triangle defined by
the two given end points of the hypothenusa. The triangle's side
from point A (or B, if the location is given as "b") to the third
point is the cathetus length. If the cathetus' length is lower than
the length of the hypothenusa, an empty list is returned.
Arguments:
list pa
The starting point on hypotenuse
list pb
The ending point on hypotenuse
float cathetusLength
The length of the cathetus of the triangle
string location
The location of the given cathetus, "a" means given cathetus
shares point pa (default) "b" means given cathetus shares point
pb
Determine if a point is completely inside a polygon. If the
point touches the polygon, then the point is not completely inside
the polygon. Note: this alternative procedure uses the
so-called winding number to determine this. It handles
self-intersecting polygons in a "natural" way.
Determine the points at which the given line intersects the
circle. There can be zero, one or two points. (If the line touches
the circle or is close to it, then one point is returned. An
arbitrary margin of 1.0e-10 times the radius is used to determine
this situation.)
Determine the points at which the given two circles intersect.
There can be zero, one or two points. (If the two circles touch the
circle or are very close, then one point is returned. An arbitrary
margin of 1.0e-10 times the mean of the radii of the two circles is
used to determine this situation.)
Determine the tangent lines from the given point to the circle.
There can be zero, one or two lines. (If the point is on the
cirucmference or very close to the circle, then one line is
returned. An arbitrary margin of 1.0e-10 times the radius of the
circle is used to determine this situation.)
Return the part of the first polyline from the origin up to the
first intersection with the secondReturn the part of the first polyline from the last intersection
point with the second to the endThe coordinate system used by the package is the ordinary
cartesian system, where the positive x-axis is directed to the
right and the positive y-axis is directed upwards. Angles and
directions are defined with respect to the positive x-axis in a
counter-clockwise direction, so that an angle of 90 degrees is the
direction of the positive y-axis. Note that the Tk canvas
coordinates differ from this, as there the origin is located in the
upper left corner of the window. Up to and including version 1.3,
the direction and octant procedures of this package used this
convention inconsistently.
This document, and the package it describes, will undoubtedly
contain bugs and other problems. Please report such in the category
math :: geometry | 677.169 | 1 |
NCERT solutions for class 9 maths chapter 3 Coordinate Geometry is very important in establishing a relationship between Algebra to Geometry. This is why it is essential for every student to understand this topic. This chapter consists of exercises that focus on finding the positions of objects on a two-dimensional plane. In addition, it contains questions relating to the Cartesian system and the plotting of points on the plane using the given coordinates. NCERT solutions class 9 maths chapter 3 Coordinate Geometry provides all the required concepts to study under this topic with appropriate exercise questions.
You will need NCERT solutions for class 9 maths chapter 3. These solutions have been prepared by our best mathematics teachers at eSaral. We have also provided this answer in PDF format. eSaral strives to provide comprehensive chapter-based solutions to students in order to facilitate their comprehension of the concepts. By completing the NCERT solutions for class 9 maths questions, students will have the opportunity to gain a comprehensive understanding of the concepts related to the chapter 3 Coordinate Geometry. With NCERT solutions class 9 free PDF, you will find high-quality answers to all exercise questions of this chapter. Use NCERT solutions to create your techniques of applying the concepts and application to attempt any question successfully.
Important Topics covered in Chapter 3 - Coordinate Geometry
The key topics of this chapter are listed below
3.1 Introduction
3.2 Cartesian System
3.1 Introduction - In this chapter, you'll learn how to find a point on a plane and how to describe a point on a number line. This chapter brings back some of the concepts from previous classes. In the same way, in this section you will learn about the same thing for the coordinate geometry and Cartesian system.
3.2 Cartesian System - You already know that the fixed points of the number line have positive numbers on the right-hand side and negative numbers on the left-hand side. Similarly, in cartesian systems, the number line is represented by the X axis and Y axis on a graph. The point at which the X axis intersects with the Y axis is a fixed point. This fixed point is referred to as the origin.The cartesian system is also divided into four parts called Quadrants.
NCERT Solutions for Class 9 Maths Chapter 3 Exercises
Regular practice of exercises will help the students learn how to answer all kinds of questions and how to solve them step-by-step. Below you'll find an in-depth analysis of exercises solutions of chapter 3 coordinates geometry.
NCERT solutions for class 9 math chapter 3 -coordinate geometry provide many benefits that help you understand the concepts better.
These NCERT solutions to class 9 chapter 3 illustrate a clear comprehension of the fundamental concepts presented in the chapter.
Students will learn how to solve all these NCERT numericals provided by eSaral,quickly and accurately.
NCERT solutions help students to solve Coordinate Geometry problems in a systematic manner. NCERT Solutions guide students through the problem solving process, helping them to understand and follow the steps.
NCERT solutions for class 9 maths chapter 3 provide a variety of practice problems covering different parts of Coordinate geometry. Students can download these solutions to practice on a regular basis.
Frequently Asked Questions
Question 1. How does the NCERT solution for class 9 maths chapter 3 help students ?
Answer 1. The NCERT solutions for class 9 maths chapter 3 can help students to overcome their doubts and get better preparation for exams. The questions provided by the NCERT are not only useful for preparing for exams, but also for various competitive tests.
Answer 2. The eSaral portal is widely regarded as a reliable source of education for class 9 students. The class 9 mathematics chapter 3 solutions prepared by the qualified teachers of eSaral, provide the best way to understand the concept of coordinate geometry. | 677.169 | 1 |
Transcript
Write equations of lines in slope-intercept form given the slope & y-intercept or the graph
Lesson 2
Graph lines using slope-intercept form equations and determine the slope of parallel & perpendicular lines
Lesson 3
Graph and identify perpendicular bisectors given two points
Lesson 4
Identify and use angle bisectors to find missing angle measures and variables
Lesson 5
Use theorems to find missing segments in geometric diagrams
+ knowledge check
+ knowledge check
+ knowledge check
+ knowledge check
+ knowledge check
Equations of Parallel & Perpendicular Lines
Perpendicular Bisectors
Angle Bisectors
Top Picks from Topic 12
Essential Question: How does the equation of a line relate to geometric terms and properties?
OAS Standard(s) Covered: G.2D.1.6 Use coordinate geometry and algebraic reasoning to represent and analyze line segments and polygons, including determining lengths, midpoints, and slopes of line segments.
Informal Knowledge Check Post-Lesson: What do m and b stand for in y=mx+b? What is another way to think of slope?
Informal Knowledge Check Post-Lesson: What is the relationship between slopes of parallel lines? What about slopes of perpendicular lines?
Informal Knowledge Check Post-Lesson: How does the midpoint relate to the perpendicular bisector?
Informal Knowledge Check Post-Lesson: What does the angle bisector tell us about the parts of a whole angle?
Informal Knowledge Check Post-Lesson: If the perpendicular bisector forms two triangles, what do you know about the longest legs of the triangles? | 677.169 | 1 |
[ANSWER] What shape is similar to volute?
[ANSWER] What shape is similar to volute?
What shape is similar to volute shape is similar to volute? question.
The Question ==> What shape is similar to volute?
Helical
Octogonal
Spherical
Pentagonal
Answer: What shape is similar to volute?
The correct answer to the question " What shape is similar to volute? " is: Helical..
Leave a comment if you ACTUALLY got the correct answer before reading the answer. and please don't forget to share this awesome question/answer postwith your family and friends on social media to see if they can answer it. | 677.169 | 1 |
Elements of Geometry, Geometrical Analysis, and Plane Trigonometry: With an ...
Join EB, and draw EO and BO to the centre O. The triangles EOD and BOD, having the side EO equal to
are (I. 3. El.) also equal, and therefore EB is perpendicular to the diameter FH. Wherefore (VI. 9. El.) FA: AH :: FD: DH; but the ratio of FA to AH being given, and consequently that of FD to DH, the point D (VI. 6. El.) is given.
COMPOSITION.
Make (VI. 3. El.) OA : OH :: OH: OD, and then D is the point required. For join OC and OB. Because OH = OC, OA: OC :: OC: OD; wherefore the triangles AOC and COD, having thus the sides about their common angle DOC proportional, are similar; and hence the angle OCA is equal to ODC. In the same manner, it is proved that the angle OBA is equal to ODB. But BOC being an isosceles triangle, the angle OCA is equal to OBA; whence the angle ODC is equal to ODB.
This porism is likewise derived from the local theorem given in Prop. 14. For AC, DC, and AB, DB being inflected in the same ratio, AC: AB :: DC: DB; and consequently (VI. 11. cor. El.) the angle BDC is bisected by DA.
PROP. XXIII. PORISM.
A point being given in the circumference of a circle, another point may be found, so that two straight lines inflected from them to the opposite circumference, shall cut off, on a given chord, extreme segments, whose alternate rectangles shall have a given ratio.
Let the circle ADBE, the point A, and the chord DE, be given in position,-another point C may be found, such that straight lines AB and CB inflected to the opposite circumference, shall form segments containing rectangles DG, FE, and DF, GE, in the ratio of KM to LM.
15. El.) the triangles AFH and GFB are similar, and consequently the angle AHF is equal to FBG; but the angle AHF is given, since the points A, H, and D are given, and, therefore, the chord AC, cutting off from the given circumference, a segment that contains a given angle ABC or FBG is given, and thence the point C.
COMPOSITION.
Produce the chord ED to H in the ratio of KM to LM, join HA, and, at any point B in the circumference, make the angle ABC equal to AHF; C is the point required.
The porism now investigated arises naturally out of this problem :-" From two given points A and C, one of which lies in a given circumference, to inflect straight lines AB and CB, so as to intercept on the chord DE segments that contain rectangles DG, FE and DF, GE, which are in a given ratio." For, the point H being assumed as before, the analysis requires that the angle ABC should be made equal to AHF. Whence, if on AC, a segment of a circle were described containing that angle, its contact or intersection with the given circumference, would determine the point of inflection. Supposing, therefore, the two circles entirely to coincide, the problem will in that case become indeterminate, or admit of innumerable answers.
PROP. XXIV. PORISM.
Two points and two diverging lines being given in position, straight lines, inflected from those points to one of the diverging lines, intercept segments, on the other, from points that may be found, and containing a rectangle which will be likewise assign
able.
Let DF and EF be inflected, from the points D and E, to the diverging line AC; they will cut off segments, on AB, from points I and K which may be found, so that the rectangle IH, GK shall be given,
ANALYSIS.
Join El and EA, DA and DK, and produce ED to meet AC in P. Since A, F, and P are so many points of inflection, it is evident, from the hypothesis, that IA.AK=IH.GK IN.NK; whence IH: IA
LE: NK EM: GN, that is, (VI. 2. El.) ED: DN; hence (VI. 16. El.) the triangles LDE and KDN are similar, and LDK forms one single straight line. Since IA NK, LE: IA :: LE: NK, that is (VI. 2. El.) EO: OL:: ED: DN, and therefore (VI. 1. cor. 1. El.) DO is parallel to AB. But the parallels OD and LM being given in position, the points O and L, and thence I and K, are given, and consequently the rectangle IA, AK is given.
The porism thus investigated follows from this problem : "Two straight lines AB and AC being given in position, with the points I and K, E and D, to find a point F, such that the inflected lines EF and DF shall intercept segments IH and GK, containing a given space:" For, when the points I and K have the position before assigned, the construction becomes indeterminate.
PROP. XXV. PORISM.
Three diverging lines being given in position, a fourth may be found, such that straight lines can be drawn intersecting all these and divided by them into proportional segments. | 677.169 | 1 |
If we can find a polytope with circumradius slightly less than 1, then its pyramid should have very small ditopal angles at the base. Here are the circumradii of regular polytopes:
n-simplex: √[n/(2(n+1))]; approaches √[1/2] as n increases
n-cube: √[n/4]; approaches ∞
n-orthoplex (n≥2): √[1/2]
First I tried the (n-simplex, n-simplex)-duoprism pyramid. The duoprism has circumradius √[n/(n+1)] (which approaches 1, as desired), so the pyramid based on the duoprism has height √[1/(n+1)]. But I calculate the relevant ditopal angle as arccos√[1/(2n+1)], which is actually increasing toward 90°, not decreasing toward 0°. I think this is related to the fact that the inradius of the simplex is decreasing toward 0.
Next I tried the (n-simplex, square)-duoprism pyramid; that is the n-simplex prism prism pyramid. The duoprism has circumradius √[(2n+1)/(2n+2)], so the pyramid has height √[1/(2n+2)]. The ditopal angle in an n-simplex is arccos(1/n). The ditopal angles in an n-simplex prism pyramid, at the base, are arccos√[2/((n+1)(n+2))] and arccos√[(n+1)/(2(n+2))]. The ditopal angles in an n-simplex prism prism pyramid, at the base, are arccos√[1/(n+1)] and arccos√[(n+1)/(n+3)]. The latter is just what we were looking for: it decreases toward 0° as the dimension n increases.
So it is true, the infimum of A∞ is 0°.
Thanks to Klitzing's list, I found another potential example: The rectified n-simplex has circumradius approaching 1, so its pyramid has height approaching 0 . But I haven't calculated the ditopal angles.
mr_e_man wrote:Thanks to Klitzing's list, I found another potential example: The rectified n-simplex has circumradius approaching 1, so its pyramid has height approaching 0 . But I haven't calculated the ditopal angles.
Wouldn't it have at least the height of the n-simplex itself? Since the rectified n-simplex is a Stott expansion of the n-simplex, its height (given a fixed edge length) cannot be less than the n-simplex itself.
Makes me wonder what the structure of the general rectified n-simplex would be. In 3D, the rectified triangle is the dual triangle, so its pyramid is just a tetrahedron, 3+1 triangle faces. In 4D, it's the octahedral pyramid: 8 tetrahedra and 1 octahedron. In 5D, the rectified 5-cell pyramid would have 5 5-cells, 5 octahedral pyramids, and 1 rectified 5-cell. In 6D, the rectified 5-simplex pyramid would have 6 5-simplexes, 6 rectified 5-cell pyramids, and 1 rectified 5-simplex.
Aha, I see it now. In each dimension n, the rectified (n-1)-simplex pyramid would have n (n-1)-simplexes, which have decreasing height with n, and n rectified (n-2)-simplex pyramids, also with decreasing height, and a rectified (n-1)-simplex as facets. Since the lateral facets have decreasing height with n, this series of pyramids in fact has decreasing height that would appear to converge on 0? Not 100% sure but it certainly seems to be so.
FascThe lacing facets here will be simplices and just those very rectified simplex pyramids - of one dimension less.
While the base size of a simplex-pyramid increases slowly and the height decreases faster, it is indeed that the base dihedral between the simplex facets and the base runs down to zero - just as already mentioned. On the other hand, the base size of the other lacing pyramids however increases much faster, that is those lacing facets dig much deeper into the body. Thence there the dihedral between the base and those lacings would have the counterintuitive behaviour to increase with the dimension instead. In fact the base dihedrals have size arccos[(n-2)/n] at the simplex and arccos[1/sqrt(n)] for the facetal rectified simplex pyramids.
So indeed the dihedrals between lacing simplex and base rectified simplex would allow for arbitrary many such components around an ridge, however the neighbouring ridge would prohibite that at the same time!
Klitzing wrote:[...] Thence there the dihedral between the base and those lacings would have the counterintuitive behaviour to increase with the dimension instead. [...] So indeed the dihedrals between lacing simplex and base rectified simplex would allow for arbitrary many such components around an ridge, however the neighbouring ridge would prohibite that at the same time! [...]
Aha, so it can be understood as the analogue to the oblong pyramid situation: take a long rectangle and erect over it a shallow pyramid. The dihedral angle between the narrow triangles and the base will decrease as the long edge of the rectangle lengthens, but the dihedral angle between the wide triangles and the base will increase as the short edge of the rectangle narrows. Thanks to the two dimensions of the rectangle these two things happen simultaneously.
Of course, in our case it's a matter of multiple dimensions in place of the long edge, but the analogy holds.
The most counterintuitive then would be: you always could consider a rectified simplex bipyramid. For if you'd take the limit of n towards infinity of that shape, then some dihedrals at its equator would run towards 0°, while the other ones at the same time would run towards 180°!
For a fixed dimension n, the total number of CRF n-polytopes with small 2-faces (e.g. 50-sided polygons at the largest) is finite. Any finite set of positive numbers cannot approach 0 . So, if we want to find ditopal angles approaching 0, without increasing the dimension, we must consider polytopes with large 2-faces.
The 3D ones are very familiar: prisms and antiprisms. Given edge length 1, an n-gon prism has height 1, and an n-gon antiprism has height
The 4D ones are also known. Let's consider their "poke sections", perpendicular to an n-gon. The section of an (n,m)-duoprism is just an m-gon, with edge length 1. The section of an antiprismatic prism is a rectangle, with lengths 1,h,1,h. The section of a biantiprismatic ring, or antifastegium, is an isosceles triangle, with lengths 1,h,h.
higher-dimensional CRFs 2.png (13.76 KiB) Viewed 26242 times
Here I claim (tentatively) that these are the only possibilities for CRF 4-polytopes with large 2-faces.
Now to 5D. Clearly, any CRF 3-polytope with edge length 1 will be a section of a valid CRF 5-polytope; consider the prism product of an n-gon and that polyhedron. Also the polygons shown above, extruded into prisms with height 1, will be valid sections. Many of the CRF polyhedra have variants where some edges have length h; these may or may not be valid sections.
higher-dimensional CRFs 3.png (114.73 KiB) Viewed 26242 times
How exactly do these correspond to 5-polytopes? Well, each vertex (x,y,z) represents an n-gon with vertices at either
(R cos (2k)180°/n, R sin (2k)180°/n, x, y, z)
or
(R cos (2k+1)180°/n, R sin (2k+1)180°/n, x, y, z)
for various integers k, where R = 1/(2 sin 180°/n) is the circumradius of an n-gon. If two vertices in 3D are connected by an edge with length 1 (depicted blue), then the corresponding n-gons should have the same orientation; both use '2k', or both use '2k+1'. If two vertices in 3D are connected by an edge with length h (depicted orange), then the corresponding n-gons should have opposite orientations; one is the dual of the other.
Is a polyteron made in this way guaranteed to be CRF? It's not clear to me.
...Isn't this construction essentially a "lace hyper-city"?
Anyway, it doesn't look like any of these have angles approaching 0 (though I don't claim to have found all of them, and that pentagonal cupola has a very small angle, from 6.1655° to 10.8123° depending on n).
For 6D, I thought of taking the prism product of an n-gon and a cube, {n}×{4,3} = {n}×{}×{}×{}, and making a segmentotope by aligning vertices (as much as possible) over the centres of the tesseracts: {n}×{4,3} || dual{n}. This continues the dimensional analogy:
{n}||dual{n} = antiprism, with height near √[3/4] {n}×{}||dual{n} = biantiprismatic ring, with height near √[2/4] {n}×{4}||dual{n} = polyteron produced by the square pyramid shown above, with height near √[1/4]
But it turns out that {n}×{4,3}||dual{n} has its squared height approaching 0 from the wrong direction: it's negative! Of course that's because the tesseract has circumradius 1, and can't have a CRF pyramid.
mr_e_man wrote:You were thinking of augmenting existing CRFs, which requires the sum of angles around a ridge to be less than 180°. I was thinking of constructing completely new CRFs, which requires the sum of angles around a peak to be less than 360°. These are different tasks.
Just to state that I recently got the idea for a full series of 7D scaliform CRFs, the "(N,M)-duoantifastegiaprismatic alterprisms":
Consider first the 4D (N,M)-duoprism xNo xMo, which clearly can occur, while keeping the symmetry, within 3 further orientations: xNo oMx, oNx xMo, and oNx oMx. Now place these 4 in a further 3D position space (orthogonally to that 4D structure space) within the vertex positions of a simplex. The respective pairwise distance clearly has to be taken such that it remains CRF, esp. all lacing edges still are unity too. To that end the distance between xNo xMo and xNo oMx clearly ought be the height of an M-antiprism, while the distance between xNo xMo and oNx xMo will be that of an N-antiprism. These four pairwise distances line out a skew tetragon similar to the zig-zag of an antiprism (which is slightly chiral if N and M would be different). The 2 bases of that digonal "antiprism" of position space then would be the lacing prism between xNo xMo and its bidual oNx oMx, i.e. the 5D (N,M)-duoantifastegium.
Because the lacing skew tetragon of position space generally has different side lengths whenever N and M is not the same, it happens that the 2 digonal bases (i.e. those 2 duoantifastegia) would not be exactly orthogonal aligned ahead of each other. This is what made me to put "antiprism" into quotes above, respectively to use "alterprism" within its general name instead.
For sure, as such it is just an ordinary member of the set of general lace simplices, in fact a lace simplex with 3D position space. But on the other hand it somehow is rather special as all 4 "vertices" (of position space) are occupied by essentially the same polytope, but then again all 4 of them are oriented differently! This as well then is a further reason for being named as an "alterprism". (Or here rather "altersimplex" if you'd like.)
Four of its facets are clearly obvious: those would be the four faces of position space, i.e. the trigonics from e.g. xNo xMo, xNo oMx, and oNx oMx. Surely those are no longer scaliform themselves. Additionally those "(N,M)-duoantifastegiaprismatic alterprisms" would have 2M lace simplices consisting of xNo o, oNx o, xNo x, and oNx x, plus further 2N similar ones using the M-symmetry instead.
As it turns out this series exists for any combination of N>2 and M>2 and never becomes somehow degenerate. In fact the side lengths of that lacing simplex of position space not only remain strictly positive always, the lacing skew tetragon in addition will never become flat either. As long as N and M are chosen integral, the whole resulting polyexon will clearly be convex, in fact by means of construction even a CRF. And because the structures situated at the "vertices" of the position space simplex are oriented orthogonal to that position space and further all 4 have the same shape each, just oriented differently (but in a symmetrically indistinguishable way), the whole thingy becomes scaliform. Finally, none of the facet types, mentioned above, themselves would qualify as scaliform. Therefore those such constructed polyexa cannot be uniform for any N and M, hence they are purely scaliform. | 677.169 | 1 |
TRIGONOMETRY in a Sentence
Learn TRIGONOMETRY from example sentences; some of them are from classic books. These examples are selected from a corpus with 300,000 sentences, including classic works and current mainstream media. Some sentences also link to their contexts.
Example sentences for TRIGONOMETRY, such as:
1. But the minds of rats could not understand trigonometry. 2. We may determine the height of a mountain by trigonometry. 3. I worked off a year's trigonometry that summer, and began Virgil alone. | 677.169 | 1 |
Engineering Curves (Problem 1)
Problem -1, Engineering Curves – Draw Ellipse, Parabola and a Hyperbola on the same axis and same directrix. Take distance of focus from the directrix equal to 50 mm and eccentricity ratio for the ellipse, parabola and hyperbola as 2/3, 1 and 3/2 respectively. Plot at least 8 points. Take suitable point on each curve and draw tangent and normal to the curve at that point.
Procedure: Ellipse
Step-1 First draw a vertical line and horizontal line of convenient length. The vertical line is the directrix and the horizontal line is the axis.
Step-2 Mark a point F, Which is the focus point, at the given distance, that is 50 mm from directrix on axis.
Step-3 Divide the distance of 50 mm into 5 equal divisions.
Step-4 Give the numbers 1,2,3 etc. up to 6 on the right hand side of the focus point F on the axis as shown into the figure. And draw vertical lines parallel to the directrix.
Step-5 For the ellipse the ratio of eccentricity is 2/3, so mark a point Ve, which is vertex of ellipse at the distance of 2 units that is 20 mm from F on the left side as per the figure given above.
Step-6 From Ve draw a vertical line and draw an arc with Ve as center and radius equal to VeF, which will intersect with vertical line drawn from the point Ve.
Step-7 Draw a straight line emerging from the point O and passing from the intersection of the vertical line and the arc drawn previously.
Step-89 Give the notations p1, p2, p3 etc. & p1', p2', p3' etc. on both sides as given into the figure.
Step-10 Draw a smooth medium dark free hand curve passing through the points p1, p2, p3 etc. & p1', p2', p3' etc. and the point Ve, which is the curve of an Ellipse.
Procedure: Hyperbola
Step-1 The eccentricity ration of the Hyperbola is 3/2, so mark a point Vh at a distance 3 units, that is 30 mm from the focus point F on the left hand side.
Step-2 From Vh draw a vertical line and draw an arc with Vh as center and radius equal to VhF, which will intersect with vertical line drawn from the point Vh.
Step-3 Draw a straight line emerging from the point O and passing from the intersection of the vertical line and the arc drawn previously.
Step-45 Give the notations r1, r2, r3 etc. & r1', r2', r3' etc. on both sides as given into the figure.
Step-6 Draw a smooth medium dark free hand curve passing through the points r1, r2, r3 etc. & r1', r2', r3' etc. and the point Vh, which is the curve of a Hyperbola.
Procedure: Parabola
Step-1 The eccentricity ration of the Parabola is 1, so mark a point Vp at a distance 2.5 units, that is 25 mm from the focus point F on the left hand side.
Step-2 Take the distance from O as O1, O2, O3 etc. on horizontal axis and with F as center cut the lines 1,2,3 etc. respectively, which are perpendicular to the axis and give the notations q1,q2,q3 & q1', q2', q3'etc. as shown into the figure.
Step-3 Draw a smooth medium dark free hand curve passing through the points q1,q2,q3 & q1', q2', q3'etc. and the point Vp, which is the curve of a Parabola.
Step-4 To draw normal and tangent to of the above curves, mark a point say N on the parabola, from this point N draw a line connecting to the point F, draw a line which is perpendicular to the line FN and intersecting with the directrix, now draw a medium dark line starting from this point and passing through the point N which is Tangent of the curve. And draw a perpendicular line to this tangent and passing through the point N which is normal of the curve. Like in this way the normal and tangent to the Ellipse and Hyperbola are obtained.
Note: Tangent and Normal are perpendicular to each other, so by drawing any one first the other is obtained at perpendicular to the previous one.
Step-5 Give the dimensions by any one method of dimensions and give the name of the components by leader lines wherever necessary. | 677.169 | 1 |
Plane and Solid Geometry
443. The maximum of isoperimetric triangles on the same base is the one whose other two sides are equal.
Hyp. ABC and ABD have equal perimeters, and AC=CB.
Proof. Draw median CE and DF AB, meeting CE in F. Join FA and FB.
Then CE is the perpendicular bisector of AB.
(Why?)
(442)
.. FE <CE.
(311)
or
.. area AFB < area ACB,
(unequal altitudes),
area ADB < area ACB.
Q.E.D.
444. COR. Of all isoperimetric triangles, the equilateral
has the maximum area.
PROPOSITION IV. THEOREM
445. Of all polygons having all sides given but one, the maximum can be inscribed in a semicircle having the undetermined side as diameter.
E
F
Q
B
Hyp. Polygon ABCDEF is the maximum of all polygons having given the sides AF, FE, ED, DC, and CB.
To prove ABCDEF can be inscribed in a semicircle whose diameter is AB.
Proof.
Join any vertex, as D, with A and B.
Then ▲ ADB must be the maximum of all triangles that can be formed with sides AD and DB. For otherwise by making ZADB a right one without changing the sides AD and DB, we could increase A ADB without altering the remaining parts AFED and DCB of the polygon. Or polygon ABCDEF would be increased, which is contrary to the hypothesis, since ABCDEF is a maximum.
.. A ADB is the maximum of all triangles having AD and DB given.
Then
ZADB is a right angle,
(Why ?)
and D is on a semicircumference that can be constructed on AB. For the same reason every vertex of the polygon must lie on the semicircumference.
Q.E.D.
Ex. 1017. To inscribe an angle in a given semicircle so that the sum of its arms is a maximum.
PROPOSITION V. THEOREM
446. Of all polygons constructed with the same given sides, that which can be inscribed in a circle is the maximum.
Hyp. Polygon ABCDE, which is inscribed in a circle, is mutually equilateral with polygon A'B'C'D'E', which cannot be inscribed in a circle.
To prove
area ABCDE> area A'B'C'D'E'.
Proof. From A draw diameter AF, and join F to the two nearest vertices C and D.
On C'D', construct AC'D'F' equal to ▲ CDF, and join A'F'.
Then
A CFDA C'F'D'.
area ABCDE > area A'B'C'D'E'. (Ax. 5.) q.e.d.
PROPOSITION VI. THEOREM
447. Of all isoperimetric polygons of the same number of sides, the equilateral is the maximum.
Hyp. ABCDE is the maximum of all isoperimetric polygons of the same number of sides.
To prove
AB BC= CD = DE = EA.
=
Proof. The polygon must be equilateral, for suppose any two sides AB and BC were not equal.
Then construct on the diagonal AC the isosceles triangle AB'C, isoperimetric with ABC.
Area AB'C would be greater than area ABC.
(443)
.. polygon AB'CDE would be greater than the isoperimetric polygon ABCDE, which would contradict the hypothesis.
Hence
AB =
BC.
Q.E.D.
448. COR. Of all isoperimetric polygons of the same number of sides, the maximum is regular.
Ex. 1018. In a given segment to inscribe an angle so that the sum of its arms is a maximum.
Ex. 1019. In a given semicircle to inscribe a trapezoid whose area is a maximum.
PROPOSITION VII. THEOREM
449. Of two isoperimetric regular polygons, that which has the greater number of sides has the greater
Proof. Let F be any point in CD.
ABCFDE may be considered a hexagon, having one of its angles equal to a straight angle.
.. area H> area ABCDE.
(Why?)
Q.E.D.
450. COR. The area of a circle is greater than the area of any polygon whose perimeter equals the circumference of the circle.
Ex. 1020. In a given circle to inscribe a triangle, having a maximum perimeter.
Ex. 1021. The circumscribed regular polygon has a smaller area than any other polygon circumscribed about the same circle.
Ex. 1022. Of all equivalent parallelograms on the same base, which has the minimum perimeter ? | 677.169 | 1 |
1 Sec 1-6 Concept Polygons Objective Given a figure, we will identify and name it if it is a polygon as measured by a scoring guide 2 Vocabulary Polygon a plane figure that meets these 2 criteria 1. It is formed by three or more line segments called sides 2. each side intersects exactly two sides, one at each endpoint, so that no two sides with a common endpoint are collinear. Polygon SNOW 3 Example 1 Determine which shapes are polygons. If a shape is a polygon, then state if it is concave or convex. A B C D concave concave convex A - yes B - no C - yes D - yes 4 Names of Polygons Handout You have 10 min. to do this handout, then we will go over it 5 Regular Polygon A regular polygon is both equilateral and equiangular 6 Example 2 Each figure is a regular polygon. Expressions are given for two sides lengths. Find the value of x. x25x13 x210x-7 205x 4x 7x34 11x-14 484x 12x 7 Example 3 The expressions (3x63) and (7x-45) represent the measures of two angles of a regular decagon. Find the measure of an angle of the decagon. 3x637x-45 1084x 27x 3(27)63 144 8 Example 4 Draw a figure that fits the description | 677.169 | 1 |
How many edges does a complete graph have.
It's not true that in a regular graph, the degree is $|V| - 1$. The degree can be 1 (a bunch of isolated edges) or 2 (any cycle) etc. In a complete graph, the degree of each vertex is $|V| - 1$. Your argument is correct, assuming you are dealing with connected simple graphs (no multiple edges.)
Expert Answer. 1.1. Find the number of vertices and edges in the complete graph K13. Justify. 1.2. Draw the following graphs or explain why no such graph exists: (a) A simple graph with 5 vertices, 6 edges, and 2 cycles of length 3. (b) A graph with degree-sequence (2, 2, 2, 2, 3) (c) A simple graph with five vertices with degrees 2, 3, 3, 3 ...We would like to show you a description here but the site won't allow us.Draw a planar graph representation of an octahedron. How many vertices, edges and faces does an octahedron (and your graph) have? The traditional design of a soccer ball is in fact a (spherical projection of a) truncated icosahedron. This consists of 12 regular pentagons and 20 regular hexagons. This problem has been solved! You'll get a detailed solution from a subject matter expert that helps you learn core concepts. Question: 4. (a) How many edges does a complete tournament graph with n vertices have? (b) How many edges does a single-elimination tournament graph with n vertices have? Please give a simple example with a diagram of ...
Graphs display information using visuals and tables communicate information using exact numbers. They both organize data in different ways, but using one is not necessarily better than using the other.Contrary to what your teacher thinks, it's not possible for a simple, undirected graph to even have $\frac{n(n-1)}{2}+1$ edges (there can only be at most $\binom{n}{2} = \frac{n(n-1)}{2}$ edges). The meta-lesson is that teachers can also make mistakes, or worse, be lazy and copy things from a website.
5. A clique has an edge for each pair of vertices, so there is one edge for each choice of two vertices from the n n. So the number of edges is: (n 2) = n! 2! × (n − 2)! = 1 2n(n − 1) ( n 2) = n! 2! × ( n − 2)! = 1 2 n ( n − 1) Edit: Inspired by Belgi, I'll give a third way of counting this! Each vertex is connected to n − 1 n − 1 ...Properties of Cycle Graph:-. It is a Connected Graph. A Cycle Graph or Circular Graph is a graph that consists of a single cycle. In a Cycle Graph number of vertices is equal to number of edges. A Cycle Graph is 2-edge colorable or 2-vertex colorable, if and only if it has an even number of vertices. A Cycle Graph is 3-edge colorable or 3-edge ...
vertex-critical graph G which at the same time is very much not edge-critical, in the sense that the deletion of any single edge does not lower its chromatic number. In the following, let us say that such a graph has no critical edges. Dirac's problem for a long time remained poorly understood. It was not before 1992 that Brown [1] Question: Draw complete undirected graphs with 1, 2, 3, 4, and 5 vertices. How many edges does a Kn, a complete undirected graph with n vertices, have? Definition 9.1.11: Graphic Sequence. A finite nonincreasing sequence of integers d1, d2, …, dn is graphic if there exists an undirected graph with n vertices having the sequence as its degree sequence. For example, 4, 2, 1, 1, 1, 1 is graphic because the degrees of the graph in Figure 9.1.11 match these numbers.
aA simpler answer without binomials: A complete graph means that every vertex is connected with every other vertex.
Draw complete graphs with four, five, and six vertices. How many edges do these graphs have? Can you generalize to n vertices? How many TSP tours would these graphs … Tuesday, Oct. 17 NLCS Game 2: Phillies 10, Diamondbacks 0 Wednesday, Oct. 18 ALCS Game 3: Astros 8, Rangers 5. Thursday, Oct. 19 NLCS Game 3: …Create a design with AI. Once you sign in with your account, you will notice a prompt box and graphics samples on the right. The box lets you input a descriptive …ITERATIVEDFS s : ( ) PUSH s ( ) while stack not empty POP if v is unmarked mark v for each edge v, w ( ) PUSH w ( ) Depth-first search is one (perhaps the most common) instance of a general family of graph traversal algorithms. The generic graph traversal algorithm stores a set of candidate edges in some data structure that I'll call a 'bag'.Question: Draw complete undirected graphs with 1, 2, 3, 4, and 5 vertices. How many edges does a Kn, a complete undirected graph with n vertices, have?I Graphs that have multiple edges connecting two vertices are calledmulti-graphs I Most graphs we will look at are simple graphs Instructor: Is l Dillig, ... pair of vertices is …An undirected graph is one in which the edges do not have a direction + 'graph' denotes undirected graph. Gl Undirected graph. V(GI) = {0, 1,2,3} ( VI, v2 ) in E is un-ordered. …SoNov 20, 2013 · Suppose a simple graph G has 8 vertices. What is the maximum number of edges that the graph G can have? The formula for this I believe is . n(n-1) / 2. where n = number of vertices. 8(8-1) / 2 = 28. Therefore a simple graph with 8 vertices can have a maximum of 28 edges. Is this correct? Data visualization is a powerful tool that helps businesses make sense of complex information and present it in a clear and concise manner. Graphs and charts are widely used to represent data visually, allowing for better understanding and ...1 Answer. Sorted by: 2 / 2. Share. Cite.A graph is called simple if it has no multiple edges or loops. (The graphs in Figures 2.3, 2.4, and 2.5 are simple, but the graphs in Example 2.1 and Figure 2.2 are … a
Search Algorithms and Hardness Results for Edge Total Domination Problem in Graphs in graphs. For a graph . Formally, the problem and its decision version is defined as follows:. In 2014, Zhao et al. proved that the Decide-ETDS problem is NP-complete for planar graphs with maximum degree 3.
We would like to show you a description here but the site won't allow us. So Two 1 / 4. Find step-by-step Discrete math solutions and your answer to the following textbook question: An undirected graph is called complete if every vertex shares an edge with every other vertex. Draw a complete graph on five vertices. How many edges does it have?De Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge
A finite graph is planar if and only if it does not contain a subgraph that is a subdivision of the complete graph K 5 or the complete bipartite graph K 3,3 (utility graph). A … Advanced Math. Advanced Math questions and answers. 2a) How many vertices does the network above have? 2b) How many edges will a spanning tree for the above network …Instagram: roderick world harris jrtaper haircut with dreadsdisability supports of the great plainsapformat May Sep sports marketing average salarycareers with finance major ▷ Graphs that have multiple edges connecting two vertices are called multi ... ▷ How many edges does a complete graph with n vertices have? Instructor ... journalism and marketing degree Obviously, Q is a 2 connected graph. Add edges to Q until addition any edge creates a cycle of length at least p + 2. Denote the resulting graph by Q ... If the complete multipartite graph K R is not a complete graph or a star, then we have g R (n 1, c, t) + g R (n 2, c, t) ...1391. The House failed to elect a new speaker on the third ballot Friday morning. One-hundred and ninety-four House Republicans voted in favor of Rep. Jim … | 677.169 | 1 |
triangle ABC above, side AC is extended to point D. What is the value of y-x?
40
75
100
140
Hint:
Hint: To find the value of y-x, we first need to find the value of x and y There are two properties we need to know to find them. The sum of the interior angles of a triangle is 180. And the exterior angle theorem which states that if a triangle's side gets extended, then the resultant exterior angle is always equal to the sum of the two opposite interior angles of the triangle.
The correct answer is: 100
Given, ABC is a triangle, AC is extended to point D. Also, Now, we know that the sum of the interior angles of a triangle is 180°. So, Next, By exterior angle theorem, if a triangle's side is extended, then the resultant exterior angle is always equal to the sum of the two opposite interior angles of the triangle. Here, AC is extended to AD. The exterior angle is The opposite angles are and So Thus, we get the values Hence, the required value of The correct option is C)
Note: It is important to know the basic properties of a triangle to solve such problems. The most important property to remember is; the sum the interior angles of a triangle is 180°. This is often used in solving problems in geometry | 677.169 | 1 |
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Página 3 ... line ; a wall , or a hedge , or a mound of earth , their boundary . The first advance beyond this would be to ... straight line ; the eleventh and twelfth , on the Elements of Solids ; and the thirteenth , on the Regular Solids . To the ...
Página 5 ... straight line . △ triangle . parallel to . square . rectangle . parallelogram . O circle . Oce circumference . A single capital letter , as A , or B , in reference to a Diagram , denotes the point A , or the point B. Two capital ...
Página 7 ... line ) , is the line forming with the base a right angle : lines are perpendicular to each other when at the point of junction they form a right angle . A Figure is applied to a straight line when the line forms one of its boundaries ...
Página 10 ... Line ( linea , a linen thread ) , is length without breadth ; or extension in one direction . ' A mathematical line ... straight line is that which lies evenly between its extreme points . " A straight line is the shortest distance ...
Página 11 ... straight line joining them lies wholly in that surface . -HERON . * " A plane surface is that which lies evenly , or equally , with the straight lines in it . " - EUCLID . " A plane surface is one whose extremities hide all the ...
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Páááágina 12 - When a straight line standing on another straight line makes the adjacent angles equal to one another, each of the angles is called a right angle ; and the straight line which stands on the other is called a perpendicular to it. | 677.169 | 1 |
Learn about geometric shapes such as symmetrical quadrilaterals, deltoids, and symmetrical trapezoids. Understand how to identify these shapes based on their properties and symmetry. Explore the possibility of inscribing a circle around a symmetrical trapezoid. | 677.169 | 1 |
Step Description. This Relationships in Triangles Unit Bundle contains guided notes, homework assignments, three quizzes, a study guide and a unit test that cover the following topics: • Midsegments of Triangles (includes reinforcement of parallel lines) • Inequalities in Triangles: Determine if three sides can form a triangle.
Chapter 5 – Relationships with Triangles Answer Key CK-12 Geometry Concepts 14 5.8 Indirect Proof in Algebra and Geometry Answers Answers will vary. Here are some hints. 1. Assume n is odd, therefore n = 2 a + 1. 2. Use the definition of an equilateral triangle to lead you towards a contradiction. 3.If Test Match Created by Nicholas_Pai Terms in this set (11) Points of Concurrency Triangle Centers: Circumcenter, Incenter, Centroid, Orthocenter Concurrent Lines Lines that intersect at Points of Concurrency: Perpendicular Bisectors (Circumcenter), Angle Bisectors (Incenter), Medians (Centroid), Altitudes (Orthocenter) Circumcenter 9 Transformations Homework 3 Rotations Answers Gina Wilson from lamborghini-islero.com. Texas go math grade 5 answer key unit 1 number and operations: Unit 5 relationships in triangles homework 3 circumcenter and incenter chapter six notes: Indirect proof and inequalities in one triangle i can use the triangle inequality theorem to | 677.169 | 1 |
Class 8 Courses
In the given figure, a circle is inscribed in a quadrilateral ABCD touching its sides Ab, BC, CD and AD at P, Q, R and S respectively given figure, a circle is inscribed in a quadrilateral ABCD touching its sides Ab, BC, CD and AD at P, Q, R and S respectively. If the radius of the circle is 10 cm, BC = 38 cm, PB = 27 cm and AD ⊥ CD then the length of CD is | 677.169 | 1 |
8 1 additional practice right triangles and the pythagorean theorem.
A right triangle has one leg that measures 7 inches, and the second leg measures 10 inches. ... Information recall - access the knowledge you've gained regarding the Pythagorean Theorem Additional8: Pythagorean Theorem and Irrational Numbers. 8.2: The Pythagorean Theorem. 8.2.1: Finding Side Lengths of TrianglesPractice using the Pythagorean theorem to solve for missing side lengths on right triangles. Each question is slightly more challenging than the previous. Pythagorean is important to have a solid understanding of the properties of ...Lesson 8-1: Right Triangles and the Pythagorean Theorem 1. Pythagorean theorem 2. Converse of the Pythagorean theorem 3. Special right triangles Also considerorems 8-1 and 8-2 Pythagorean Theorem and Its Converse Pythagorean Theorem If a triangle is a right triangle, then the sum of the squares of the lengths of the legs isThe Pythagorean Theorem is an important mathematical concept and this quiz/worksheet combo will help you test your knowledge on it. The practice questions on the quiz will test you on your ability ...
Practice using the Pythagorean theorem to solve for missing side lengths on right triangles. Each question is slightly more challenging than the previous. Pythagorean Learn more at mathantics.comVisit for more Free math videos and additional subscription based content!8-1 Additional PracticeRight Triangles and the Pythagorean TheoremFor Exercises 1-9, find the value of x. Write your answers in simplest radical Instagram: pick 3and4 md lottery drawingcambridge learnertraductor de ingles a espanol holapelicula krone mit lilie p291182611nc 12x80 The response to what | 677.169 | 1 |
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After answering the question, we can state that probability in a circle = (angle where B can exist)/(Total Angle) => P = 2/3
Explanation:
What is a circle?
A circle is formed by every point in the plane that is a certain distance away from another point (center). Thus, it is a curve formed by points moving in the plane at a constant distance from a fixed point. It is also rotationallysymmetric about the center at all angles. A circle is a two-dimensional closed object in which every pair of points in the plane is equally spaced from the "center." A specular symmetry line is formed by a line that goes through the circle. It is also rotationally symmetric about the center at all angles.
We now want A and B to be separated by a distance greater than r. As a result, the angle should be greater than 60 degrees because the radius is equal to 60 degrees and decreases as the angle increases. As a result, point B cannot be in the 60-degree zone of A on both sides, that is, to A's left and right. | 677.169 | 1 |
The angle bisectors of a parallelogram form a rectangle
We have to prove that PQRS is a rectangle. We know that the opposite sides of a parallelogram are parallel and congruent. We know that if two parallel lines are cut by a transversal, the sum of interior angles lying on the same side of the transversal is always supplementary. The opposite sides of a parallelogram are equal in length, and the opposite angles are equal in measure. It is proven that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle. About Us.
The angle bisectors of a parallelogram form a rectangle
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CG Vyapam SI. The profit earned when an article is sold for Rs. LMOP is a parallelogram as its one opposite pair of sides is parallel and equal. Tech from Indian Institute of Technology, Kanpur.
A parallelogram is a two-dimensional geometrical shape whose sides are parallel to each other. It is a type of polygon having four sides also called quadrilateral , where the pair of parallel sides are equal in length. The Sum of adjacent angles of a parallelogram is equal to degrees. In geometry, you must have learned about many 2D shapes and sizes such as circles, squares, rectangles, rhombus, etc. All of these shapes have a different set of properties.
The angle bisectors of a parallelogram form a rectangle
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Karnataka SDA. Leaving Soon? Nainital Bank Clerk. The sum of all the angles of a right triangle is -. SBI Apprentice. Odisha Police Driver. Vizag Steel Junior Trainee. RISF Constable. Punjab Pre-Primary Teacher. SSC Scientific Assistant. Therefore, PQRS is a rectangle. Bihar STET. Rajasthan 3rd Grade Teacher. Gujarat Metro JE. If any equilateral triangle is formed by the same length of the wire then what will be the area of the equilateral triangle? | 677.169 | 1 |
January 2017 geometry regents answers.
He then scanned it and shared it online.According to the scoring key and rating guide to the June 2017 geometry Regents, choice No. 2 is the correct answer to question 24.After a diagram, question 24 reads, "Which statement is not sufficient to prove [triangle] ABC is similar to [triangle] EDC?"Students are given four possible answers to choose ...
Answer Key for Geometry Practice Tests for Regents Examinations ,2009-01-15 Answer Key for Past New York State Regents Exams in Geometry. Geometry Common Core Regents Course Workbook Donny Brusca,2017-04-13 Course Workbook for the New York regents exam 0110 answers geometry regents exam 0110 jmap the university of the state of new york regents high jmap regents examination in ... 22 2024 notice to teachers january 2017 regents examination in algebra i chinese edition only questions 19 and 22 only 64 kb notice to teachersBusiness Contact: [email protected] Mathgotserved Unit 1 Foundations1.2 Segment Addition Postulate
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NYS Mathematics Regents Preparation is dedicated to the enhancement of twenty-first century student learning in the study of mathematics, ... Geometry, and Algebra 2.June 2017 Common Core Geometry Regents Part 1 X Why. WebAssign. Www Kixel Co. Decomposing Functions Into Even And Odd Parts « … Apr 19th, 2024 Geometry Regents Answers For June 2014 EXAMPLES PAST PAPERS EXAM QUESTIONS AND STEP BY STEP SOLUTIONS''x Why June 2017 Common Core Geometry Regents Part 1 May …GEJanuary 2017 Regents Examination in Algebra II (211 KB) Scoring Key and Rating Guide (119 KB) Scoring Key (Excel version) (19 KB) Model Response Set (3.2 …
How to answer questions for Geometry Common Core Regents Exam, High School Math, January 2017, examples, step by step solutions and tips
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REGENTS HIGH SCHOOL EXAMINATION GEOMETRY (COMMON CORE) Thursday, January 26, 2017 — 9:15 a.m. to 12:15 p.m., only SCORING KEY AND RATING GUIDE Mechanics of Rating The following procedures are to be followed for scoring student answer papers for the Regents Examination in Geometry (Common Core). More detailed information about
Regents Exam in ELA (Common Core) Rating Guide — Jan. '17 [6] Anchor Paper - Part 2 - Level 6 - A Anchor Level 6-A The essay introduces a precise and insightful claim, as directed by the task (Over the past century DST has undergone several reforms in order to meet the needs of the nation and its people and DST should remain in existence because of the many benefits it brings to thetoget Nys January 2013 Geometry Regents Answer Key Book file PDF. file Nys January 2013 Geometry Regents Answer Key Book Free Download PDF at Our eBook Library. This Book have some digitalformats such us : kindle, epub, ebook, paperbook, and another formats. Here is The Complete PDF Library Geometry Regents January 2013 Answer Key2015January 2019 Geometry, Part III. Each correct answer is worth up to 4 credits. Partial credit is available. Work must be shown. Correct answers without work receive only 1 point. 32. A triangle has vertices A (-2,4), B (6,2), and C (l, -1). Prove that 6.ABC is an isosceles right triangle. [The use of the set of axes below is optional.]Chart for Converting Total Test Raw Scores to Final Exam Scores (Scale Scores) (Use for the January 2017 exam only.) To determine the student's final examination score (scale score), find the student's total test raw score in the column labeled "Raw Score" and then locate the scale score that corresponds to that raw score. questions on the ...Regents Examination in Physical Setting/Physics (100 KB) Answer Booklet (42 KB) Scoring Key and Rating Guide (561 KB) Conversion Chart (46 KB) January 2008 Regents Examination in Physical Setting/Physics (183 KB) Scoring Key and Rating Guide (200 KB) Conversion Chart (304 KB) June 2007 Regents Examination in Physical Setting/Physics (183 KB)
REGENTS HIGH SCHOOL EXAMINATION GEOMETRY (Common Core) Friday, June 16, 2017 ...REGENTS HIGH SCHOOL EXAMINATION GEOMETRY (Common Core) Thursday, January 26, 2017 — 9:15 a.m. to 12:15 p.m., only ... _____ Print your name and the name of your school on the lines above. A separate answer sheet for Part I has been provided to you. Follow the instructions from the proctor for completing the student information on …Instagram: ray and martha's obituaries answer. Note that diagrams are not necessarily drawn to scale. For each statement or question, choose the word or expression that, of those given, best completes the statement or answers the question. Record your answers on your separate answer sheet.[48] Geometry (Common Core) – Jan. '17 [2]January 2013 geometry regents answer key Copy ... examination in geometry common core all editions questions 14 and 22 only 13 kb january 2017 regents examination in geometry 207 kb scoring key and rating guide 79 kb geometry the university of the state new york regents high why does chad ehlers wear tape on fingers Regents Answer Jan 1th, 2024GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Wednesday, January 23, 2019 - 9:15 a.m. to 12:15 p.m., only Student Name: ... the answer sheet, indicating that you had no unlawful knowledge of the questions or answers prior shooting in gallup new mexico REG fruits for sword mains Archive: Regents Examination in Physical Setting/Chemistry (June 2011 and prior) Expand All Collapse All. June 2011. January 2011. August 2010. June 2010. January 2010. August 2009. June 2009.Curious to know how old those big trees are in your yard? We'll tell you how to use geometry to figure out their ages without risking their health. Advertisement You probably learn... frontier homepage powered by yahoo The following are the questions and answers (and commentary) for part of the New York State Algebra Regents exam. ... June 2017, Geometry (Common Core), Part III. 32. ... January 2020 Geometry Regents Part I (Multiple Choice) August 2016 Common Core Geometry Regents, Part 1. June 2018 Common Core Geometry Regents, Part I (multiple choice) chad softley January 2011 geometry regents answers explained ... examination in geometry common core all editions questions 14 and 22 only 13 kb january 2017 regents examination in geometry 207 kb scoring key and rating regents high school examination geometry jmap Feb 10 2024 web jan 27 2011 € geometry the university of the texters see you later 4 letters from Spring 2014 to January 2024 Sorted by State Standard: Topic TABLE OF CONTENTS TOPIC STANDARD SUBTOPIC QUESTION NUMBER TOOLS OF GEOMETRY G.GMD.B.4 G.GMD.B.4 ... Geometry Regents Exam Questions by State Standard: Topic 7 25 A right cylinder is cut perpendicular to its base. The shape of the cross section is aGeometry Regents Exam Questions Donny Brusca 2020-02-29 Contains every Geometry Common Core Regents exam question through the January 2020 exam, organized by topic and aligned to the sections of the Geometry Regents Course Workbook. Answer key available separately at CourseWorkbooks.com. Barron's Regents Exams and Answers Stanley Kaplan 1982-01-01 device is strictly prohibited when ... the answer sheet, indicating that you had no unlawful knowledge of the questions or answers prior oreion side by side for sale Nys january 2013 geometry regents answer key ... notice to teachers june 2017 regents examination in geometry common core all editions question 24 ... questions 14 and 22 only 13 kb january 2017 regents examination in geometry 207 kb scoring key and rating guide 79 kb . sd40ve accessories The following are the worked solutions for the Geometry (Common Core) Regents High School Examination January 2017. Geometry Common Core Regents New York State Exam - January 2017. The following are questions from the past paper Regents High School Geometry, January 2017 graph below represents a dog walker's speed during his 30-minute mild cheese crossword 5. Staying Engaged with geometry regents january 2012 answers Joining Online Reading Communities Participating in Virtual Book Clubs Flilowing Authors and Publishers geometry regents january 2012 answers 6. Navigating geometry regents january 2012 answers eBook Formats ePub, PDF, MOBI, and More geometry regents january 2012 answers ...The University of the State of New York. REGENTS HIGH SCHOOL EXAMINATION. GEOMETRY. Wednesday, January 25, 2023 ... phone for cash kioskOther Results for August 2018 Geometry Regents Answer Key:. … | 677.169 | 1 |
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QUESTIONS LIST: rectangle : opposite sides are congruent and parallel, all four (4) sides are right angles, rhombus : opposite angles are congruent, all four (4) sides are congruent, midpoint : a point on a line segment that divides it into two (2) equal parts, linear pair : a pair of adjacent angles form a linear pair, then they are supplementary, angle bisector : a line that split an angle into two equal angles, parallelogram : a simple quadrilateral with two pairs of parallel sides, quadrilateral : a plane figure that has four (4) sides or edges, four (4) corners and two (2) pairs of parallel sides, trapezoid : a quadrilateral having only one pair of parallel side, perpendicular : are two lines that meet or intersect each other at right angle, theorem : a rule in branches of mathematics expressed by symbols or formulas, can be proven by chain of reasoning, postulate : a statement that is assumed true without proof, segment : a part of the line that is bounded by two (2) district end points, vat : vertical angle theorem, sap : segment addition postulate | 677.169 | 1 |
What is Solid angle: Definition and 68 Discussions
In geometry, a solid angle (symbol: Ω) is a measure of the amount of the field of view from some particular point that a given object covers. That is, it is a measure of how large the object appears to an observer looking from that point.
The point from which the object is viewed is called the apex of the solid angle, and the object is said to subtend its solid angle from that point.
In the International System of Units (SI), a solid angle is expressed in a dimensionless unit called a steradian (symbol: sr). One steradian corresponds to one unit of area on the unit sphere surrounding the apex, so an object that blocks all rays from the apex would cover a number of steradians equal to the total surface area of the unit sphere,
4
π
{\displaystyle 4\pi }
. Solid angles can also be measured in squares of angular measures such as degrees, minutes, and seconds.
A small object nearby may subtend the same solid angle as a larger object farther away. For example, although the Moon is much smaller than the Sun, it is also much closer to Earth. Indeed, as viewed from any point on Earth, both objects have approximately the same solid angle as well as apparent size. This is evident during a solar eclipse.
I have a table of densities of galaxies :
Expected number density of galaxies for photometric survey per unit area and redshift intervals, ##\mathrm{d} N / \mathrm{d} \Omega \mathrm{d} z\left[\mathrm{sr}^{-1}\right]## and the corresponding density of galaxies per ##\operatorname{arcmin}^2## for...
I have a doubt regarding the role of the solid angle when calculating the power(W) with the brightness of the source I'm observing with area Asource. I was given the definition:
with A the Aantenna. If now I take as an example the picture below to calculate W, we conisder as solid angle the one...
I found this paper
We have an interferometer with to arms. The firsr has a couple of HWP's inclened by an angle theta
and the second has the crossed couple. A mixed state is in input.
i look to the figure withe the Bloch sphere. i see 2 paths on it. one...
I'm a bit confused on the derivation above. I understand what the goal of the derivation is, as it derives Gauss's Law using the solid angle, but i was wondering if someone could kind of fill in the steps the author skipped and explain the use of the solid angle.
Hi,
I am trying to simulate necessary flux values on a car due to solar radiance. I'm trying to attain the necessary flux values using a lamp setup. I have the lamp specifications in Watts but I need to convert them into radiance values (W/m2/sr) for my application. I would like to know how to...
Hello,
I am trying to find an analytical expression to determine the solid angle subtended by a disk source onto the face of the cylinder. I will appreciate if someone can provide me directions.
I am aware how to calculate solid angle by a point source to cylinder's face ( omega =...
I googled a lot on proof of Gauss theorem and nearly every other proof (on web and so on books) state that solid angle of closed surface is 4pi but I can't find the proof of this nowhere !
I tried setting up the integral but don't know how to proceed furthur :
Ω=∫(cosθ/r^2)*dA
Also The one...
Hello Everybody!
Concept of Solid Angle was pretty much straight forward until they were on surface patches were taken into account which were visualized as base of cone.
I am having difficult when 3d Objects like Sphere/Cylinder .
We can very easily calculate the respective area and plugin the...
I'm a Physics undergraduate at University and in my labs module I had to recreate the Rutherford Scattering experiment. The basic setup was similar to this:
My problem is that this setup only records data in one plane and not 3D. In reality the particles are scattered in a sort of cone...
Hello everyone! I have a question about angular integration in arbitrary d dimensions. The interest comes from the need to use dimensional regularization. Suppose I start with a 2-dimensional integral and then I have to move to d=2-\epsilon dimension to regularize my integral. Now, suppose...
The problem is to find out the solid angle ω subtended by a disc of radius a at a point P distant z from its centre along its axis. α is the semi-vertical angle of the disc at the point P in question.
The answer is supposed to be ω = 2π (1 - cos α), according to an online text. However, I find...
For a physics problem, I need to calculate the solid angle subtended by an oblique cone (cone where the apex does not lie on the line perpendicular to the cone's base from the center of the base).
Consider the following problem:
I have a 2D disk which emits light in an ever growing hemisphere...
I'm trying to figure out how the element of solid angle transforms under a transformation between two inertial frames moving with velocity v w.r.t. each other under an arbitrary direction. But I should say I disappointed myself! Anyway, some books which contain a brief discussion on this(which...
Show by direct calculation that Eqs. (4-134) and (4-137) in the textbook by Duderstadt and Hamilton hold, i.e.:(a) ∫ dΩΩiΩj= 4π/3 δij; i,j = x,y,z;
4π(b) ∫ dΩΩxΩyΩz = 0, if l, m, or n is odd.
4π
The integrals are over 4π.
This is part of the derivation of the diffusion equation...
When considering a small beam of null-geodesics in spacetime it is possible to define the solid angle spanned by two of the rays at the observer.
At page 111 in "Gravitational Lenses" by P.Schneider et. al. they state with reference to Figure (b) that:
"The dependence of this distance on the...Background: Using Biot-Savart law we proved that for a closed loop with current ##I##, the magnetic field at a point P was equal to ##\vec{B}=-\frac{\mu_{0} I \nabla{\Omega}}{4 \pi}## where ##\Omega(x,y,z)## is a function of the position of P that represents the solid angle at which the loop is...
In deriving the pressure of a gas, my book states that
'if all molecules are equally likely to be traveling in any direction, the fraction whose trajectories lie in an elemental solid angle dΩ is dΩ/4π'.
This initally made sense to me, but then thinking about it, I wrote dΩ=sinθdθdφ and this...
I am somewhat confused about the connection between divergence and solid angle for a beam. I know individually what each term means... but I'm confused as to how (or even if) one can calculate the solid angle of a beam, given the divergence.
I have some notes from a previous lecture series I...
Hi,
It's surprising how little information is available on this topic, considering it seems like such a fundamental problem. The only tutorial I have found is and my university does not have access to the other papers...
I have an interesting question that I'm not sure how to go about solving. This question has a little general relativity and (maybe) a little QM, but I wasn't sure where to post it.
Question:
Imagine that a \pi0 meson traveling along the z-axis (velocity v=0.99c, rest mass M) decays into two...
hello
please see attached snapshot from the book of Ligthmann & al. (problem book in relativity and gravitation).
Can somebody please explain the expression for the derivative of the solid angle ?
here it is given as dvxdvy where the v's are speeds !
How come this is so ?
Thanks,
Hi folks, can someone help explain this in words of one syllable or less? I am looking at a text that compares flux and intensity of a distant source, and it states that
∫∫dΩ = ∏
I know that
dΩ = sinθ dθ d∅
but I don't understand where the given result comes from. What are the...
Hi Everyone,
For an individual inquiry and formal lab report task at school I have chosen to conduct an experiment to find out whether hot shoe mounted flash units follow the inverse square law and how the flash zoom is affected by the inverse square law.
My first question is that In order...
I read that if the cone with apex angle 2α whose central axis is vertical, apex at the origin, then one can use spherical coordinate to calculate the solid angle of the cone
∫02∏∫0αsin\varphid\thetad\varphi
However, what if the central axis is align to y-axis horizontally, instead of...
This is only an example from Kraus Antenna 3rd edition page 404. The question is really a math problem involves calculation of ratio of solid angles. Just ignore the antenna part. this is directly from the book:
Example 12-1.1 Mars temperature
The incremental antenna temperature for the...
Homework Statement
This isn't a homework question, I am writing up my scribbled notes from todays lecture and have got stuck on some calculus, and lost the thread of the argument. Last week, we integrated Plancks law to find
B(T) = ∫ Bv(T) dv
= 2∏4(kT)4 / 15c2h3
Then...
Homework Statement
Hi
I am looking at a unit sphere. Two squares are projected onto the sphere on opposite ends, as shown in figure 1 (the figure only shows one square, the other one is at the opposite end).
There are two more sets of these squares, each set in its own dimension, so there...
[b]1. The solid angle subtended by a 100 cm^2 circular detector at a distance of 1 m is ______steradians.
[b]2. Ω = A/r^2 and A =∏r^2 (area of a circle)
[b]3. I originally tried to find r by solving 100 = ∏r^2 and I got r = 5.6. I then tried to plug into the first equation for Ω only to...
A radiation source at some point through the collimator of a detector.
Modeled as 100 spheres tracing out the image of a circle on a plane some distance above the original position of the source.
What I want to do is collect all the spheres that go through the top of a cylinder and then...
Hi,
I was reading my astrophysics textbook and came across solid angles. I'm not sure I fully understand, for example there was a problem in the book that went as follows.
The attached "math.jpg" shows a light source (yellow) in the centre of an arc. The problem is 2D, but the arc is...
The definition of Beam solid angle: For an antenna with single main lobe, the Beam solid angle defines as:
The solid angle \; \Omega_A\; where all the radiated power would flow with radiation intensity equal to maximum and constant inside the Beam solid angle \; \Omega_A\;.
The book gave...
I understand the surface of the sphere is 4\pi sr. where the area of one sr is r^2.
My question is why d\Omega = sin \theta \;d \theta \;d\phi? Can anyone show me how to derive this.
Is it because surface area dS = (Rd\theta)(R\; sin\;\theta\;d\phi)\;\hbox { so if }\; \Omega = \frac S {R^2}...
So, I was trying to find a rigorous mathematical derivation of gauss's law(please I don't want to hear again any field lines nonsense) and I stumbled upon jackson's proof which uses the solid angle concept and seems a solid enough proof(stupid joke:smile:).The problem is that it's the first time...
hello,
Please attached snapshot of an answer to an Ex. I was stunned at the formula for the derivative of the solid angle which is :
d(solid angle) = dvx * dvy
I would appreciate if somebody can provide some hints on how one can find it ?
Thank you,
Hello,
I was following the derivation of the solid angle of right rectangular pyramid that I found at .
I don't quite understand the step between the 3rd to the 4th equation. In particular how the integral...
[b]1. . A cylindrical detector, similar to the NaI detectors you use in lab, has a diameter of 5 inches and a length of 5 inches. A 60Co source is placed on the cylindrical axis, 20 cm away from the front face of the detector.
a. Determine the solid angle subtended by the detector...
Hello,
it is often written in books that the solid angle \Omega subtended by an oriented surface patch can be computed with a surface integral:
\Omega = \int\int_S \frac{\mathbf{r}\cdot \mathbf{\hat{n}} }{|\mathbf{r}|^3}dS
where r is the position vector for the patch dS and n its normal (see...
Calculate the hieght of the cones by using Solid angle? pleasez help
Homework Statement
Calculate the hieght of the cones by using Solid angle?
Homework Equations
d(omega)
The Attempt at a Solution
what can i do in that question? | 677.169 | 1 |
What Are the Features of a Cube but Not a Sphere? - Comprehensive Guide
Shapes in geometry fall into two broad categories: polygons and non-polygons. Polygons are two-dimensional shapes with predefined angles and a finite number of sides, while non-polygons are three-dimensional shapes with curved surfaces and no set angles or side number. Cubes and spheres are two such common 3D shapes we encounter in our everyday lives. Here, we discuss the features of a cube but not a sphere.
What is a Cube?
A cube is a 3D shape with six square faces all at right angles, a long edge length and the same width and length. All the faces meet to form six equal squares. In other words, a cube is a regular polyhedron with six square faces. A cube is also a cuboid (contains six faces), and a hexahedron (contains six side faces).
What is a Sphere?
A such as a triangle, square, pentagon, hexagon, and so on. The surface of a sphere has no start and end point and no edges or plane-symmetry.
What are the Features of a Cube but Not a Sphere?
Cube has six plane faces: A cube consists of six plane faces merged together to form a perfect cube structure. Every face has three external angles and one internal angle. On the other hand, a sphere has a curved surface without any edges or planes.
Cube has eight vertices: A cube has eight vertices located at the intersections of each of its edges which form a perfect cube structure. On the other hand, a sphere does not have any vertices or edges.
Cube has twelve edges: A cube has twelve edges which join together each of its corners to form the cube shape. On the other hand, a sphere has no edges since its curved surface has no starting and ending points.
Cube has equal length and width: All the edges of a cube are of equal length, forming a perfect cube structure. On the other hand, a sphere has no equal length and width as its curved surface is continuous.
Cube has right angles: All the six faces of the cube are perpendicular to each other, which is due to their equal length and width. On the other hand, a sphere has no right angles as its curved surface has no start and end points.
FAQs
What are the different types of shapes?
There are two broad categories of shapes: polygons and non-polygons. Polygons are two dimensional shapes with predefined angles and a finite number of sides, while non-polygons are three dimensional shapes with curved surfaces and no set angles or side number.
How can you differentiate between a cube and a sphere?
A cube is a 3D shape with six square faces all at right angles, a long edge length and the same width and length. All the faces meet to form six equal squares. A.
What features make a cube unique?
Some features that make a cube unique are its six plane faces, eight vertices, twelve edges, equal length and width, and right angles.
What is a regular polyhedron?
A regular polyhedron is a 3D shape with flat faces, straight edges and equal angles between each face. A cube is an example of a regular polyhedron with six square faces. | 677.169 | 1 |
How do you find the diagonal of an irregular quadrilateral?
Homework Statement. In an irregular quadrilateral ABCD, the length of all sides are AB=a BC=b CD=c DA=d and the length of the diagonal AC is x.
Homework Equations. Cosine formula c2 = a2 + b2 – 2abcosθ
The Attempt at a Solution. I really have no idea how to start.
Are there irregular quadrilaterals?
In other words, an irregular quadrilateral is a quadrilateral with sides that are not all equal in length. Notice that the sides are not all equal in length. Opposite sides have equal length, but they are not all equal. Therefore, this is an irregular quadrilateral.
How many diagonals do quadrilaterals have?
Answer: 2 diagonals are there in a quadrilateral.
What is an irregular quadrilateral shape?
Irregular quadrilaterals are: rectangle, trapezoid, parallelogram, kite, and rhombus. They are symmetrical, but are not required to have congruent sides or angles.
What is diagonal in quadrilateral?
A line segment drawn from one vertex of a quadrilateral to the opposite vertex is called a diagonal of the quadrilateral. For example, AC is a diagonal of quadrilateral ABCD, and so is BD.
How do you find the diagonal of a quadrilateral?
If we suppose the length of a square is L, then the length of the diagonal = L √2. Area of a square = L2. A quadrilateral whose four sides are all congruent in length is a rhombus.
How many diagonals does a convex quadrilateral have?
two diagonals
The two diagonals of a convex quadrilateral are the line segments that connect opposite vertices. The two bimedians of a convex quadrilateral are the line segments that connect the midpoints of opposite sides.
How many regular quadrilaterals are there?
There are 5 types of quadrilaterals – Rectangle, Square, Parallelogram, Trapezium or Trapezoid, and Rhombus.
Does a quadrilateral have two diagonals?
It will be True, A quadrilateral has two diagonals.
How do you calculate irregular area?
How to use irregular area calculator?
Step 1: Measure all sides of the area in one unit (Feet, Meter, Inches or any other).
Step 2: Enter length of horizontal sides into Length 1 and Length 2. And Width of the vertical sides into Width 1 and Width 2.
Step 3: Press calculate button.
Our Formula: Area = b × h.
How many diagonals does a quadrilateral have?
The number of diagonals in a polygon = n (n – 3)/2, where n = number of sides of the polygon. For a quadrilateral, n = 4. The number of diagonals in a quadrilateral = 4 (4 – 3)/2. = 4/2. = 2. How Many Diagonals does a Quadrilateral Have. A quadrilateral is thus found to have two diagonals.
What are the irregular quadrilaterals?
Irregular quadrilaterals are: rectangle, trapezoid, parallelogram, kite, and rhombus. They are symmetrical, but are not required to have congruent sides or angles. Do not despair, though, because a few of them yield to area formulas, just as the square does. In addition to symmetrical, irregular quadrilaterals,…
How many diagonals does a regular triangle have?
So a triangle, the simplest polygon, has no diagonals. You cannot draw a line from one interior angle to any other interior angle that is not also a side of the triangle. A quadrilateral, the next-simplest, has two diagonals. A pentagon, whether regular or irregular, has five diagonals.
How do you find the number of diagonals of a polygon?
You have learned a lot about particularly important parts of polygons, their diagonals. You now know how to identify the diagonals of any polygon, what some real-life examples of diagonals are, and how to use the formula, # of Diagonals = n(n − 3) 2 # o f D i a g o n a l s = n ( n – 3) 2, where n is the number of sides (or vertices) of the | 677.169 | 1 |
The unit circle math ku answers.
Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products. 4The circumference is the distance around a circle (its perimeter!):About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features NFL Sunday Ticket Press Copyright ... …
A unit circle is a circle on the Cartesian Plane that has a radius of 1 unit and is centered at the origin (0, 0). The unit circle is a powerful tool that provides us with easier reference when we work with trigonometric functions and angle measurements. You can apply the Pythagorean Theorem to the unit circle. Noble …
Each student needs this unit circle and set of triangles. It's important that you use these ones because the hypotenuse of the triangles is equal to the radius of the circle. Students will start out the lesson by finding sides lengths for a 30-60-90 triangle and 45-45-90 triangle that both have a hypotenuse of 1. are given in both degrees and radians). Plus, students will be excited to do the math so that they can get to the puzzle!To ... 435. It seems evident from infinitely many primitive pythagorean triples (a, b, c) ( a, b, c) that there are infinitely many rational points (a c, b c) ( a c, b c) on the unit circle. But how would one go about, and show that they are dense, in the sense that for two rational points x x and y y of angles α α and β β on the unit circle, if α ...
new math activity: The Unit Circle, was designed for regular and Honors high school students in trigonometry, precalculus, geometry, and algebra 2. ... Plus, students will be excited to do the math so that they can get to the puzzle!To complete the Math-ku puzzle, students must first answer each question on their activity sheet. As t ...TheDefining Sine and Cosine Functions from the Unit Circle. The sine function relates a real number t t to the y-coordinate of the point where the corresponding angle intercepts the unit circle. More precisely, the sine of an angle t t equals the y-value of the endpoint on the unit circle of an arc of length t. t. In Figure 2, the sine is equal to ... 270 − 225 = 45. Okay, so this is the basic 45-45-90 triangle, whose legs (in the unit circle) have lengths of \frac {1} {\sqrt {2\,}} 21. The hypotenuse is, as always in the unit circle, equal to 1. I'll label the corresponding triangle in the first quadrant: In the third quadrant, the x - and y -values are negative.
The SAT gives you the information that the number of degrees in a circle i s 360 ∘, and the number of radians is 2 π. From this, you can easily convert from radians to degrees, using the fact that 360 ∘ = 2 rad. Here's a problem that asks for a conversion: Answer: 4. To solve this problem, let's start with what's given, 720 ∘.41. $2.00. PDF. Trigonometry Unit Circle Flashcards I have complied a complete set of flashcards for the unit circle. 16 cards testing the conversion of radians to degrees 32 cards testing sin in radians and degrees 32 cards testing cos in radians and degrees 32 cards testing tan in radians and degrees Double s. Since the circumference of the unit circle happens to be (2π) ( 2 π), and since (in Analytical Geometry or Trigonometry) this translates to (360∘) ( 360 ∘), students new to Calculus are taught about radians, which is a very confusing and ambiguous term. Such students are taught that (2π) ( 2 π) radians equals (360∘) ( 360 ∘), and so ...StartingBy The Math Series. In this activity, students will practice finding the domain and range for trigonometric functions as they work through 12 matching questions. More specifically, students will: Determine the domain or range of a sine, cosine, tangent, cosecant, Subjects: Math, PreCalculus, Trigonometry. Grades:Finding the Reference Angle. Converting Radians to Degrees. Period of Sine and Cosine Curves. Free worksheet (pdf) and answer key on Unit Circle. 25 scaffolded questions that start relatively easy and end with some real challenges. Plus model problems explained step by step. For
(Unit Circle) Given a unit circle, what » distinguishes the unit circle from all other circles? Note that the radius is » 1 unit; watch out for a reason why this might be useful. It has a radius of 1 and a • centre (0, 0), and is drawn on a Cartesian plane. Identify the 4 quadrants. »The unit circle math ku answers – Math Concepts An online mean value theorem calculator allows you to find the rate of change of the function and the derivative of a …
StartingExercise 1.2.6. We know that the equation for the unit circle is x2 + y2 = 1. We also know that if t is an real number, then the terminal point of the arc determined by t is the point (cos(t), sin(t)) and that this point lies on the unit circle. Use this information to develop an identity involving cos(t) and sin(t). TheMathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. ... Since the hypotnuse is always 1 in the unit circle sin $\theta$ will equal the height of the triangle and Y coordinate on the circle. I will now read the answers for finding tangent $\theta$ $\endgroup ...Browse unit circule activities resources on Teachers Pay Teachers, a marketplace trusted by millions of teachers for original educational resources.Let us see why 1 Radian is equal to 57.2958... degrees: In a half circle there are π radians, which is also 180°. π radians = 180°. So 1 radian = 180°/π. = 57.2958...°. (approximately) To go from radians to degrees: multiply by 180, divide by π. To go from degrees to radians: multiply by π, divide by 180. Here is a table of equivalent ... Step 1: Identify The Quadrant. Since we're dealing with the unit circle with tan, we will need to use the values we've memorized from sine and cosine, and then solve. First, however, we need to figure out what quadrant we're in so we know whether our answers for sine and cosine will be positive or negative.
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Maximize your travel. Submit it here and you could see the answer in our new Weekly Update, written by Brian Kelly. Oops! Did you mean... Welcome to The Points Guy! There isn't a strict mathematical formula at work here. At some point we'd ...About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features NFL Sunday Ticket Press Copyright ...Level 1 - 2 questions are red, Level 3 - 4 questions are orange, Level 5 - 6 questions are yellow and Level 7 - 8 questions are green. The level of a question can be changed from the suggested level by selecting a new level in the top-right corner of the question preview window. MYP mathematics programmes vary greatly from school to school.(b) Note, for z,w ∈ U, the product zw ∈ U. We say the unit circle U is closed under multiplication. (c) Define the map f :[0,2π)−→ U where f(θ)=eiθ. Then, f is a bijection. (d) In fact, f(x +y) = f(x)f(y) sends sum to the product. Here, addition x+y in [0,2π)is defined "modulo 2π". 6. We discuss the algebra of Roots on Unity Free practice questions for High School Math - The Unit Circle and Radians. Includes full solutions and score reporting.The Unit Circle. The point of the unit circle is that it makes other parts of the mathematics easier and neater. For instance, in the unit circle, for any angle θ, the trig values for sine and cosine are clearly nothing more than sin(θ) = y and cos(θ) = x. Working from this, you can take the fact that the tangent is defined as being tan(θ ... When2. Long horizontal or vertical line =. √ 3. 2. For example, if you're trying to solve cos. π. 3. , you should know right away that this angle (which is equal to 60°) indicates a short horizontal line on the unit circle. Therefore, its corresponding x-coordinate must equal | 677.169 | 1 |
Calculating the Hypotenuse of a Right Angle Triangle
In the realm of geometry, we often encounter the intriguing task of calculating the length of the hypotenuse in a right angle triangle. The hypotenuse, being the longest side opposite the right angle, holds significant importance in understanding the triangle's dimensions and properties. Whether you're a student grappling with geometry concepts, an engineer designing structures, or simply someone fascinated by mathematical exploration, this article aims to provide a comprehensive guide to calculating the hypotenuse using various methods.
Before delving into the calculations, it's worth revisiting the fundamental concept of a right angle triangle. A right angle triangle, as its name suggests, possesses one angle measuring exactly 90 degrees. This definitive feature distinguishes it from other types of triangles and forms the basis for the various methods used to calculate the hypotenuse.
With a solid grasp of the concept, we can now embark on our journey of exploring the methods for calculating the hypotenuse of a right angle triangle. These methods include the Pythagorean Theorem, trigonometric ratios, and geometric relationships. Each method offers a unique perspective on the problem, providing a comprehensive understanding of the hypotenuse's determination.
Calculating a Right Angle Triangle
Explore key points to calculate the hypotenuse:
Pythagorean Theorem: a2 + b2 = c2
Trigonometric Ratios: sin, cos, tan
Geometric Relationships: adjacent, opposite, hypotenuse
Special Right Triangles: 30-60-90, 45-45-90
Similar Triangles: proportions, ratios
Applications: engineering, architecture, surveying
Pythagoras' Legacy: Ancient Greek mathematician
Right Angle Triangle Calculator: Online tools
With these points in mind, you can embark on your journey to conquer the realm of right angle triangle calculations.
Pythagorean Theorem: a2 + b2 = c2
The Pythagorean Theorem stands as a cornerstone of geometry, providing a fundamental relationship between the sides of a right angle triangle. This theorem states that the square of the hypotenuse (the longest side opposite the right angle) is equal to the sum of the squares of the other two sides (the adjacent and opposite sides).
Sides of a Right Triangle:
In a right angle triangle, we have the adjacent side, opposite side, and hypotenuse. The adjacent side is the side adjacent to the right angle and the angle we know the measure of. The opposite side is the side opposite the right angle and the angle we know the measure of. The hypotenuse is the longest side of the triangle and is always opposite the right angle.
Pythagorean Equation:
The Pythagorean equation, a2 + b2 = c2, expresses the relationship between the sides of a right triangle. Here, 'a' represents the length of the adjacent side, 'b' represents the length of the opposite side, and 'c' represents the length of the hypotenuse.
Squaring the Sides:
The theorem involves squaring each side of the triangle. Squaring a number means multiplying it by itself. For example, if the adjacent side is 3 units long, then a2 = 32 = 9.
Pythagorean Triples:
Certain sets of numbers, known as Pythagorean triples, satisfy the Pythagorean equation. A common example is the 3-4-5 triple, where 32 + 42 = 52. These triples can be useful for quickly identifying right triangles.
The Pythagorean Theorem finds extensive applications in various fields, including architecture, engineering, surveying, and navigation. Its simplicity and elegance have captivated mathematicians and scientists for centuries, solidifying its place as a cornerstone of mathematical knowledge.
Trigonometric Ratios: sin, cos, tan
Trigonometric ratios provide another powerful tool for calculating the sides and angles of a right angle triangle. These ratios are defined using the lengths of the sides opposite, adjacent, and hypotenuse to the angle of interest.
Sine (sin):
The sine of an angle is defined as the ratio of the length of the opposite side to the length of the hypotenuse. $$sin\theta = \frac{opposite}{hypotenuse}$$
Cosine (cos):
The cosine of an angle is defined as the ratio of the length of the adjacent side to the length of the hypotenuse. $$cos\theta = \frac{adjacent}{hypotenuse}$$
Tangent (tan):
The tangent of an angle is defined as the ratio of the length of the opposite side to the length of the adjacent side. $$tan\theta = \frac{opposite}{adjacent}$$
Using Trigonometric Ratios:
Given one angle and one side of a right triangle, trigonometric ratios can be used to find the lengths of the other sides or the measure of the other angles. For example, if you know the angle and the length of the opposite side, you can use the sine ratio to find the length of the hypotenuse.
Trigonometric ratios are essential tools in trigonometry and have wide-ranging applications in various fields, including surveying, navigation, engineering, and astronomy. By understanding these ratios, you can unlock a powerful set of techniques for solving problems involving right angle triangles.
Geometric Relationships: adjacent, opposite, hypotenuse
The geometric relationships between the sides of a right angle triangle provide valuable insights into its properties and behavior. Understanding these relationships is crucial for solving various problems involving right triangles.
Adjacent Side:
The adjacent side of a right triangle is the side that is next to the angle we are interested in. It is also the side that is perpendicular to the opposite side.
Opposite Side:
The opposite side of a right triangle is the side that is opposite the angle we are interested in. It is also the side that is perpendicular to the adjacent side.
Hypotenuse:
The hypotenuse of a right triangle is the longest side of the triangle. It is the side that is opposite the right angle.
Pythagorean Theorem:
The Pythagorean theorem, a2 + b2 = c2, expresses the fundamental relationship between the sides of a right triangle. Here, 'a' represents the length of the adjacent side, 'b' represents the length of the opposite side, and 'c' represents the length of the hypotenuse.
These geometric relationships form the foundation for many of the methods used to solve problems involving right triangles. By understanding the properties and relationships between the sides, we can unlock a powerful toolkit for solving a wide range of geometrical problems.
Special Right Triangles: 30-60-90, 45-45-90
In the realm of right angle triangles, certain special triangles possess remarkable properties and relationships that simplify calculations. Two of the most well-known special right triangles are the 30-60-90 triangle and the 45-45-90 triangle.
30-60-90 Triangle:
In a 30-60-90 triangle, the angles measure 30 degrees, 60 degrees, and 90 degrees. The sides of this triangle follow a specific ratio: the hypotenuse is twice the length of the shorter side (opposite the 30-degree angle), and the longer side (opposite the 60-degree angle) is √3 times the length of the shorter side.
45-45-90 Triangle:
In a 45-45-90 triangle, all three angles measure 45 degrees, 45 degrees, and 90 degrees. This triangle exhibits a simple 1:1:√2 ratio for its sides: the hypotenuse is √2 times the length of either of the congruent sides.
Applications:
Special right triangles find extensive applications in various fields, including architecture, engineering, surveying, and trigonometry. Their predefined angle and side ratios make them useful for quickly solving problems and performing calculations.
Simplifying Calculations:
The properties of special right triangles can significantly simplify calculations. For instance, if you know one side of a 30-60-90 triangle, you can easily find the lengths of the other two sides using the specific ratios associated with this triangle.
By understanding the properties and applications of special right triangles, you can enhance your problem-solving skills and tackle a wide range of geometrical challenges with greater ease and efficiency.
Similar Triangles: proportions, ratios
In the realm of geometry, similar triangles exhibit a fascinating property: their corresponding sides are proportional. This means that the ratios of the corresponding sides of similar triangles are equal, regardless of their actual lengths.
Consider two similar right triangles, ΔABC and ΔDEF. If we have the following corresponding sides:
AB corresponds to DE
BC corresponds to EF
AC corresponds to DF
Then, the following proportions hold true:
AB/DE = BC/EF = AC/DF
These proportions provide a powerful tool for solving problems involving similar right triangles.
For example, suppose we know the lengths of two sides of a right triangle and we also know that it is similar to another right triangle with a known hypotenuse. We can use the пропорции to find the length of the hypotenuse of the first triangle.
Here's an illustration:
ΔABC is similar to ΔDEF.
AB = 6 cm, BC = 8 cm, and DF = 10 cm.
To find the length of AC, we can set up the following proportion:
AB/DE = AC/DF
Substituting the known values, we get:
6/DE = AC/10
Solving for AC, we find that:
AC = (6/DE) * 10
Since DE is similar to AB, we know that DE = 8 cm.
Plugging this value into the equation, we get:
AC = (6/8) * 10
Simplifying, we find that:
AC = 7.5 cm
Therefore, the length of the hypotenuse of ΔABC is 7.5 cm.
The concept of similar triangles and their proportional sides is a fundamental tool in geometry and trigonometry. It allows us to solve a wide range of problems involving right triangles and other geometric shapes.
Applications: engineering, architecture, surveying
The ability to calculate the sides and angles of right triangles is a fundamental skill in various fields, including engineering, architecture, and surveying.
Engineering:
Structural engineers rely on right triangle calculations to determine the forces acting on structures and to ensure their stability. They use trigonometry to calculate the angles and lengths of beams, columns, and trusses.
Mechanical engineers use right triangle trigonometry to design and analyze machines, such as engines, pumps, and gears. They calculate the forces, moments, and stresses acting on machine components.
Civil engineers use right triangles to design and construct roads, bridges, and other infrastructure. They calculate slopes, grades, and angles to ensure proper drainage and stability.
Architecture:
Architects use right triangle calculations to determine the dimensions and proportions of buildings and structures. They use trigonometry to calculate roof pitches, window angles, and the placement of structural elements.
Interior designers use right triangles to determine the best layout for furniture and other objects within a space. They calculate angles and distances to create aesthetically pleasing and functional designs.
Surveying:
Surveyors use right triangle trigonometry to measure distances, angles, and elevations. They use theodolites, levels, and other instruments to accurately determine the boundaries of properties, the location of landmarks, and the topography of land.
Surveyors also use right triangles to establish property lines, set grade elevations, and determine the location of underground utilities.
These are just a few examples of the many applications of right triangle calculations in engineering, architecture, and surveying. The ability to accurately and efficiently solve right triangle problems is a valuable skill for professionals in these fields.
In addition to the fields mentioned above, right triangle calculations are also used in navigation, astronomy, and other scientific disciplines. The versatility and practicality of right triangle trigonometry make it an essential tool in a wide range of applications.
Pythagoras' Legacy: Ancient Greek mathematician
Pythagoras of Samos was an ancient Greek mathematician, philosopher, and religious leader who lived in the 6th century BC. He is best known for the Pythagorean theorem, a fundamental relationship between the sides of a right triangle. However, his contributions to mathematics and philosophy extend far beyond this single theorem.
Pythagoras founded a school of thought known as the Pythagoreans. The Pythagoreans believed that mathematics was the key to understanding the universe and that everything in the cosmos could be explained through mathematical relationships. They studied geometry, arithmetic, and music, and they made significant contributions to these fields.
Pythagoras' legacy extends far beyond his own lifetime. His ideas influenced later mathematicians, philosophers, and scientists, including Plato, Aristotle, and Euclid. The Pythagorean theorem remains a cornerstone of geometry and is used in countless applications across various fields.
In addition to his mathematical contributions, Pythagoras is also known for his philosophical teachings. He believed in the transmigration of souls, or the idea that the soul passes through a cycle of birth, death, and rebirth. He also taught that music and mathematics could purify the soul and lead to a higher state of consciousness.
Pythagoras' legacy is complex and far-reaching. He was a brilliant mathematician, philosopher, and religious leader whose ideas have had a profound impact on Western thought and culture. His contributions to mathematics, in particular the Pythagorean theorem, continue to be used and studied to this day.
Right Angle Triangle Calculator: Online tools
In the digital age, a variety of online tools and calculators are available to make calculating the sides and angles of right triangles quick and easy. These tools can be particularly useful for students, engineers, architects, and anyone else who needs to perform these calculations on a regular basis.
One common type of right angle triangle calculator is the Pythagorean theorem calculator. This type of calculator allows you to enter the lengths of two sides of a right triangle and it will calculate the length of the third side using the Pythagorean theorem. Some calculators also allow you to enter the measure of one angle and the length of one side, and they will calculate the lengths of the other two sides and the measure of the remaining angle.
Another type of right angle triangle calculator is the trigonometric calculator. This type of calculator allows you to enter the measure of one angle and the length of one side, and it will calculate the lengths of the other two sides and the measure of the remaining angle using trigonometric ratios. Trigonometric calculators can also be used to find the area and perimeter of a right triangle.
Online right angle triangle calculators are typically easy to use and provide accurate results. They can be a valuable tool for anyone who needs to perform these calculations quickly and easily.
Here are some tips for using online right angle triangle calculators:
Make sure you are using a reputable calculator. There are many different calculators available online, so it is important to choose one that is accurate and reliable.
Enter your values carefully. Double-check your numbers to make sure you have entered them correctly.
Select the appropriate calculation type. Some calculators offer multiple calculation options, so be sure to select the one that is appropriate for your needs.
Interpret the results correctly. Once you have calculated the values, make sure you understand what they mean and how to apply them to your problem.
FAQ
Here are some frequently asked questions about right angle triangle calculators:
Question 1: What is a right angle triangle calculator?
Answer 1: A right angle triangle calculator is an online tool that allows you to calculate the sides and angles of a right triangle. You can use these calculators to find the length of a missing side, the measure of an unknown angle, or the area and perimeter of a right triangle.
Question 2: How do I use a right angle triangle calculator?
Answer 2: Using a right angle triangle calculator is easy. Simply enter the values that you know (such as the lengths of two sides or the measure of one angle) into the calculator, and it will calculate the remaining values for you.
Question 3: What types of right angle triangle calculators are available?
Answer 3: There are two main types of right angle triangle calculators: Pythagorean theorem calculators and trigonometric calculators. Pythagorean theorem calculators use the Pythagorean theorem to calculate the length of the third side of a right triangle, while trigonometric calculators use trigonometric ratios to calculate the lengths of the other two sides and the measure of the remaining angle.
Question 4: Which right angle triangle calculator should I use?
Answer 4: The type of right angle triangle calculator that you should use depends on the information that you know about the triangle. If you know the lengths of two sides, you can use a Pythagorean theorem calculator. If you know the measure of one angle and the length of one side, you can use a trigonometric calculator.
Question 5: Are right angle triangle calculators accurate?
Answer 5: Yes, right angle triangle calculators are accurate, provided that you enter the correct values. Make sure that you double-check your numbers before you enter them into the calculator.
Question 6: Can I use a right angle triangle calculator on my phone?
Answer 6: Yes, there are many right angle triangle calculator apps available for smartphones. These apps allow you to perform the same calculations as online calculators, but you can use them on the go.
Question 7: How to interpret result of right angle triangle calculator?
Answer 7: Once you have calculated the values using a right angle triangle calculator, make sure you understand what they mean and how to apply them to your problem. For example, if you are calculating the length of the hypotenuse, you need to know that the hypotenuse is the longest side of the right triangle.
Closing Paragraph:
Right angle triangle calculators are a valuable tool for anyone who needs to perform these calculations quickly and easily. They are accurate, easy to use, and available online and on mobile devices. If you are working with right triangles, be sure to take advantage of these helpful tools.
In addition to using a calculator, there are a few other things you can do to make calculating right triangle problems easier:
Tips
Here are a few tips for using a right angle triangle calculator effectively:
Tip 1: Choose the right calculator.
There are many different right angle triangle calculators available online and on mobile devices. Choose one that is reputable and easy to use. Make sure that the calculator offers the features that you need, such as the ability to calculate the length of a missing side, the measure of an unknown angle, or the area and perimeter of a right triangle.
Tip 2: Double-check your numbers.
Before you enter your values into the calculator, double-check them to make sure that they are correct. Even a small mistake can lead to an incorrect answer.
Tip 3: Understand what the results mean.
Once you have calculated the values, take a moment to understand what they mean and how to apply them to your problem. For example, if you are calculating the length of the hypotenuse, you need to know that the hypotenuse is the longest side of the right triangle.
Tip 4: Use the calculator as a learning tool.
Right angle triangle calculators can be a helpful learning tool. If you are struggling to understand a concept, try using a calculator to work through some examples. This can help you to visualize the concepts and to see how they work in practice.
Closing Paragraph:
By following these tips, you can use a right angle triangle calculator to quickly and easily solve problems involving right triangles. Whether you are a student, an engineer, an architect, or simply someone who is interested in mathematics, a right angle triangle calculator can be a valuable tool.
In conclusion, right angle triangle calculators are a powerful tool that can be used to solve a variety of problems. By choosing the right calculator, double-checking your numbers, understanding the results, and using the calculator as a learning tool, you can make the most of this valuable resource.
Conclusion
Right angle triangle calculators are a powerful tool that can be used to solve a variety of problems quickly and easily. Whether you are a student, an engineer, an architect, or simply someone who is interested in mathematics, a right angle triangle calculator can be a valuable resource.
In this article, we have explored the different methods for calculating the sides and angles of a right triangle, including the Pythagorean theorem, trigonometric ratios, and geometric relationships. We have also discussed the applications of right triangle calculations in fields such as engineering, architecture, and surveying. Finally, we have provided tips for using a right angle triangle calculator effectively.
By understanding the concepts and methods presented in this article, you will be well-equipped to solve even the most challenging right triangle problems. So, whether you are facing a geometry exam, designing a building, or simply trying to measure the height of a tree, remember that there is a right angle triangle calculator out there to help you.
Closing Message:
Embrace the power of right angle triangle calculators and use them to unlock the secrets of these fascinating geometric shapes. With a little practice, you will be able to solve even the most complex right triangle problems with ease. | 677.169 | 1 |
Unit Adopted from All Things Algebra by Gina Wilson. Unit 7 Test Study Guide (Part 1, Questions 1 - 26)Unit 7 Polygons and QuadrilateralsPart 2:
Home7.1 Angles of Polygons 7.2 Properties of Parallelograms 7.3 Proving That a Quadrilateral Is a Parallelogram 7.4 Properties of Special Parallelograms 7.5 Properties of Trapezoids and Kites 7 Quadrilaterals and Other Polygons Gazebo (p. 365) Amusement Park Ride (p. 377) Window (p. 395) Diamond (p. 406) Arrow (p. 373) Wi d ( 395) Amusement Park ... To Find the Missing Angle: 2 135 1 x Step 1: Find the sum of the interior angles for the polygon (ignore given angles, count number of sides, and use formula) 180 (n – 2) = 180 (7 – 2) = 180 (5) = 900° 3 90 …Geo Unit 7 Polygons and Quadrilaterals Schedule - Google Docs Geometry Classify polygons based on their sides and angles. [7.1a] Find and use the measures of interior …
SHORT ANSWER Directions: Show all your work for each of the following questions. 12. Find, in simplest radical form, the distance between points A(-1, 5) and B(-7, 3). 13 Lisa Davenport. This editable 6th grade math foldable provides an overview for finding the area of 2 -dimensional figures including quadrilaterals, triangles and polygons, organized by the following three tabs:Tab 1: Quadrilaterals (Square, Rectangle, Parallelogram, Trapezoid, Rhombus)Tab 2: Triangles (acute, right, and obtuse)Tab 3: Polygons ...Study with Quizlet and memorize flashcards containing terms like Classify the figure in as many ways as possible., Which statement is true?, Lucinda wants to build a square sandbox, but she has no way of measuring angles. Which of the following explains how she can make sure the sandbox is square by measuring only length. and more …quiz for 10th grade students. Find other quizzes for and more on Quizizz for free! a
Adopted from All Things Algebra by Gina Wilson. Lesson 7.5 Rhombi and Squares (Properties of rhombi and squares)Unit 7 Polygons and Quadrilaterals. test. Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere A | 677.169 | 1 |
About
For elementary questions concerning circles (or disks). A circle is the locus of points in a plane that are at a fixed distance from a fixed point. Use this tag alongside [geometry], [Euclidean geometry], or something similar. Do not use this tag for more advanced topics, such as complex analysis or topology.
A circle is a shape in geometry, defined as the locus of points that have a fixed distance from a certain point, called the centre. The fixed distance from the centre of a circle to any of its points is called the radius. The length of the set of points is called the circumference, and for Euclidean space is related to the length of the radius by: | 677.169 | 1 |
The intersection of an N-plane and an M-plane can be a plane of any dimension up to whichever of N and M is lower. So, for instance, a 4-D (or 5-D or 6-D or…) plane intersecting with our own space (a 3-plane) could be a point, a line, a (2-D) plane, or fill the entire space. You can't get any interesting shapes, unless the planes you're intersecting already have interesting shapes.
A hyperplane intersecting with an N-d shape can only produce either the entirety of the shape, or the portion of the shape obtained by intersecting it with a plane of dimension less than N. So, the possibilities are: the whole shape, an (N-1)-dimensional slice of the shape, an (N-2)-dimensional slice of the shape, …, a 1-dimensional slice of the shape, a point, and nothing at all.
So a 4-d hyperplane intersecting with a 2-d circle can only produce the same shapes as a 2-d hyperplane intersecting with a 2-d circle can: the whole circle, or a pair of points, or a single point, or nothing. [Or, if you mean the whole disk inbetween the circular boundary: the whole disk, or a line segment, or a single point, or nothing].), or a 1-d slice of it (which will be a pair of points [and the line segment inbetween them, if you include the whole ball and not just its surface sphere]), or a single point, or nothing at all.
It'll either annihilate the universe, or create a unicorn. You can never tell what you'll get.
Couldn't you perform a boolean intersecting operation of a 4D plane and a 3D sphere or 2D disc? You'd end up with some interesting topology and shapes, but it'd be arbitrary as it depends on the arrangement / orientation and intersection of the hyperplane relative to the sphere/disc.
Of course a plane has no volume, but a hypercube and a sphere intersection could create a shell that would morph in weird ways as you rotated the hypercube in 4 dimensions.
I'm not sure if I'm getting quite the result I was hoping for. So the hyperplane can intersect the shape, which is great. But would it also be covering area outside the shape as well?
Basically I was hoping to prove that a hyperplane going down to lower dimensions can equal a shape in that lower dimension.
In general, it can't. The intersection of two affine spaces has to be an affine space, which means that you're going to get either a point, a line, a plane, or some higher dimensional equivalent. You're not going to get circles, spheres or anything else interesting.
In other words: The intersection of two (or more) hyperplanes is always a hyperplane, of some dimension or another.
As a lay answer, this is probably the clearest, but it's no good mathematically. A hyperplane in a vector space V is defined to be an affine space of codimension one, which means that a hyperplane in V is not going to be a hyperplane in any larger vector space containing V.
Ah, like Chronos, I was taking "hyperplane" simply to mean "affine space", with no need for consideration of any ambient space in which it may be embedded. But re-reading the OP, this isn't the way the terminology is being used (e.g., the OP uses "3-d hyperplane" to refer to a standard plane with 2 intrinsic dimensions), and you are correct as to how it is being used instead.
Still, I wouldn't say it's "no good mathematically". It's just a matter of terminology to make it fine. The intersection of two (or more) affine spaces is always an affine space, of some dimension or another.
Ah, like Chronos, I was taking "hyperplane" simply to mean "affine space", with no need for consideration of any ambient space in which it may be embedded.
Or rather, I was taking "hyperplane" simply to mean "affine subspace", with no need for consideration of the specific dimension of the ambient space in which it may be embedded. Of course, for determining intersections, the actual embeddings into a common superspace matter a great deal.)…
While this is true, wouldn't it have been better to be a bit more specific here and just say "circle" instead of "ellipse"? | 677.169 | 1 |
Parallel Lines: Definition and Properties;color:#000000;} .ft02{font-size:18px;font-family:ArialMT;color:#000000;} .ft03{font-size:18px;line-height:22px;font-family:Arial;color:#000000;} Parallel lines ● Parallel and perpendicular lines are defined as follows:● Parallel lines- two or more lines that do not intersect with each other, no matter what distance you move them. ● Perpendicular lines- two or more lines that intersect at a 90 degree angle. What if we don't have a line that's exactly equal to 0? Rephrasing, instead of x + 1/2 y = 0, what if we had x + 1/2 y = 1? Consider having the line x plus 1/2 y being equal to 0. And then another line, x plus 1/2 y being equal to 1. | 677.169 | 1 |
How to Measure the Length of a Path Between Two Points - Comprehensive Guide
Developers may need to calculate the distance between two points multiple times while developing apps or solutions. Whether you're creating an app that uses coordinates or plotting a route on a map, the distance between two points is an essential piece of information. In this guide, we'll provide a comprehensive overview of what distance calculations are, how they work, and a step-by-step guide on how to measure the length of a path.
What is a Distance Calculation?
Distance is a measure of the difference between two points. Distance calculations are mathematical formulas that can be used to determine the exact amount of space between two points or objects. By using a distance calculation, you can find the exact length of a path, map routes, and measure the space between objects in real-time.
How Do Distance Calculations Work?
Distance calculations are often done using numerical formulas and formulas. These formulas can determine the distance between two points, as well as calculate the time and distance of a path. Some of the most common methods for finding the distance between two points include the Pythagorean Theorem and the Haversine Formula.
Step-by-Step Guide for Calculating the Distance Between Two Points
There are a few different ways to measure the length of a path, depending on the type of coordinates you're trying to measure. In this section, we'll provide a step-by-step guide on two of the simplest and most effective methods:
Using the Pythagorean Theorem
Identify the coordinates of the two points you want to measure.
Determine the difference between the x-coordinates.
Determine the difference between the y-coordinates.
Square each of these differences.
Add the two squared values together.
Take the square root of the sum from the previous step.
This is the distance between the two points.
Using the Haversine Formula
Determine the coordinates for both points.
Convert the coordinates to radians.
Calculate the difference between the longitudes of the two points.
Calculate the difference between the latitudes of the two points.
Calculate the haversine of the difference between the latitudes and longitudes.
Calculate the angular distance using the haversine of the differences from Step 4 and 5.
Convert the angular distance to a linear distance.
FAQ
What Is an Easy Way to Calculate Distance?
The easiest way to calculate distance is to use the Pythagorean theorem. This formula is simple to compute and can give you an accurate measurement of the space between two points.
Can Distance Calculations Be Used in Real-Time?
Yes, distance calculations can be used in real-time. You can use numerical formulas or calculations to determine the exact distance between two objects in a very short span of time.
Are Distance Calculations Accurate?
Yes, distance calculations are generally accurate. By using the Pythagorean theorem or the Haversine Formula, you can get accurate measurements of the distance between two points.
How Do You Find the Difference Between Two Sets of Coordinates?
The easiest way to find the difference between two sets of coordinates is to subtract one set from the other. For example, if you're comparing an x-coordinate of 9 and a y-coordinate of 7, you would subtract the x-coordinate from the y-coordinate to get the difference of -2.
What Are the Most Common Methods for Calculating Distance?
The most common methods for calculating distance are the Pythagorean theorem and the Haversine Formula. Both are relatively simple to use and can give you accurate measurements of the distance between two points. | 677.169 | 1 |
What are the key principles behind W.D. Gann angle measurement?
What are the key principles behind W.D. Gann angle measurement? W.D. Gann's angle theorems have been used since he first published them in 1908 for drawing a line from a fixed point directly through the endpoint of an acute angle. W.D. Gann's angle theorems also determine the orientation of a line through a 3D point and the endpoint of an acute angle. They also help to determine the orientation of a line through a 3D point and the perpendicular drawn from that point out to the endpoint of an obtuse angle. This will be demonstrated in this tutorial. In this Tutorial visite site will help you figure out Gann's Theorem by providing Visualisations and solutions as a series of animated explainer videos. Suppose you need to find a perspective, or some other device, that will help you draw a line from any fixed object, not necessarily the vertex of an acute angle, directly through a fixed object, or some fixed point on a wall for a building. As the object moves away from the fixed objects, the line will begin to diminish in size when it intersects the fixed object.
Ephemeris
As illustrated in the left graph of this video, you will see that you get the equivalent angle on both sides of the fixed points, so it does not matter whether the fixed object is on the right-hand side, or the left according to the position of the fixed point. The angle is only determined by the position of the fixed object. The fixed point does not need to be the vertex of a vertex angle to work out your angle. This might be the area of a building on a flat surface with no windows. Note in this process, that although I used a triangle to measure the angle, in my video, you may instead use any regular acute angle, or any fixed point object for measuring your angle. For a deeper understanding and the reasons behind Gann's theorem, you canWhat are the key principles behind W.D. Gann angle measurement?If you measure your shot with some degree of precision, regardless of the marker, on a level playing surface at a 70 foot distance using W.D. Gann's Principles for shooting Accuracy using the Shooter's Manual technique and the line of sight technique you will ALWAYS GET THE SAME RESULTS… As far as that is concerned, this particular author doesn't give definitive answers of their own (as far as I know), but the method was outlined in shooting for accuracy at the NFT (W.
Vibration Numbers
D. Gann) 1. Take a shot – or, in practice form the shot 2. Apply the shooter's manual (however you feel you are doing it), and fire the shot as if you were executing a real W.D.Garnet shot 3. If you repeat that process over and over, you will eventually end up with the same exact hit 1. Take a shot – or, in practice form the shot 2. Apply the shooter's manual, and use your mental time line to move from that entry point (the point you enter from the target area) 3. Keep moving "in" through the target for a continuous path 4. Keep moving back, ideally always in the direction of the side plane Here's another simple illustration, to make it more clear. Click to view larger As you make a turn to the right you are moving to the left, keep moving in, as would be expressed as moving left right left By doing so, you are on a straight, non-linear path and when you return back, you are still on a linear path BUT, if you start to develop curves, the path you took would appear to be curved, but when you are on a surface, the surface itself is a straight path,and that is what the shooter's manual calls the true path (that true path is the one that is never changing, regardless of how you move your line of sight, it's always straight) However, if you start to move in arcs, you are also moving on a linear path (remember, your true path is always straight and never changes And, if you start moving in circles, it will be called a non-linear path Now, as a sidenote, Gann says there is no one rule that says you must move completely in the line of sight, but it helps when done correctly, but you can be unpredictable and continue to move in and out of the line of sight, and that is the secret sauce I believe in using for some of shooting precision, and that's exactly what the shooter's manual is telling you to do, practice makes perfect I'm not sure why anyone would care to argue about which method is more desirable, but the basic point should be that if you get the time of your shot to the exact minute you make yourWhat are the key principles behind W.D.
Ephemeris Points
Gann angle measurement? Answer:I took a very accurate ruler and a protractor and found the w.d – Gann triangle using the following rule: x2 – x3= o (so to be correct and more precise my other rule is x2 + x3=o) Then using trial by error (that is using the other rulers degrees of measure (to o) starting from 0 and then with each 30 seconds measure to o)and then 1/2.. 1/4.. 1/8.. until 1/16 up. Each line of the other triangle that is resulting from the other rule can be measured in a regular way (using a ruler ). Use the W.D – Gann angle to determine the depth of a structure in relation to a known reference point or to the Sun's Rising and Falling. As you can see, it's a pretty cool way to measure the distance relative to the sun. So start at 12 o Clock and count in 30 second intervals.
Financial Timing
You will find that this way taking the decimal part of each angle and doing the math for each one is easy… -A general Rule of thumb is to start in the morning when angle is negative and rise very, very slowly- The ratio of the measure of the side and the angle the next morning is the size of the gap between the structure and the shadow (or Gann) cast. Positive, negative and zero are signifying the amount of the structure above or is in or just below the projection area. But is inaccurate as it gives a bit to much if the Gann shadow is really that big in that one hour. So do not forget that it has to rise a little further if the Gann's at noon, or later as it maybe, with the sun much higher. To get the right Angle in place of the one hour in W.D – Gann angle There is the general rule which you may use yourself, which has a slight variation in each solar hour and are easy to remember as the direction of the sun moves from East to West. The variation of the angle to the West depends on the time of the day, the angle of the sun, the phase of the moon and the movement of the houses. The main goal is also that the shadow of the Gann should be cast outside your house or between houses. It can not be cast in front of buildings or if it is in a street around the house, as this cast shadow is where the Gann will be. If the Gann is cast behind you in the house or something is cast – i.
Eclipse Points
e.- behind you and over your head. So count from that point to 100 and look at the clock. Using this method the most important piece of question regarding which gann's shadow cast the projection area is solved already. Determines the shadow cast | 677.169 | 1 |
The Triangle Inequality Theorem Quiz
15 Questions
Which of the following statements is true regarding the triangle inequality?
The triangle inequality states that the sum of the lengths of any two sides must be greater than the length of the third side.
Which of the following is a correct representation of the triangle inequality using vectors and vector lengths?
$|\mathbf{x} + \mathbf{y}| \leq |\mathbf{x}| + |\mathbf{y}|$
Which of the following is a correct statement about the triangle inequality?
The triangle inequality does not apply to degenerate triangles.
Which of the following is a correct statement about the triangle inequality in Euclidean geometry?
The triangle inequality states that the sum of the lengths of any two sides must be greater than the length of the third side.
If x, y, and z are the lengths of the sides of a triangle, which of the following inequalities represents the triangle inequality?
$z \leq x + y$
Define the triangle inequality in mathematics.
The triangle inequality states that for any triangle, the sum of the lengths of any two sides must be greater than or equal to the length of the remaining side. In equation form, this can be expressed as $z \leq x + y$, where $x$, $y$, and $z$ are the lengths of the sides of the triangle.
What does the triangle inequality state about the possibility of equality?
The triangle inequality permits the inclusion of degenerate triangles, but some authors, especially those writing about elementary geometry, will exclude this possibility, thus leaving out the possibility of equality.
How is the triangle inequality written using vectors and vector lengths?
In Euclidean geometry and some other geometries, the triangle inequality is a theorem about distances, and it is written using vectors and vector lengths (norms): $|| \mathbf{x} + \mathbf{y} || \leq || \mathbf{x} || + || \mathbf{y} ||$, where the length $z$ of the third side has been replaced by the vector sum $\mathbf{x} + \mathbf{y}$.
What is the significance of equality in the triangle inequality?
Equality in the triangle inequality only occurs in the degenerate case of a triangle with zero area.
What are the lengths of the sides of a triangle in relation to the triangle inequality?
If $x$, $y$, and $z$ are the lengths of the sides of the triangle, with no side being greater than $z$, then the triangle inequality states that $z \leq x + y$.
What is the triangle inequality?
The triangle inequality states that for any triangle, the sum of the lengths of any two sides must be greater than or equal to the length of the remaining side.
What is the condition for equality in the triangle inequality?
The condition for equality in the triangle inequality is only in the degenerate case of a triangle with zero area.
How is the triangle inequality written using vectors and vector lengths?
The triangle inequality using vectors and vector lengths is written as $||\mathbf{x} + \mathbf{y}|| \leq ||\mathbf{x}|| + ||\mathbf{y}||$.
What does the triangle inequality theorem state in Euclidean geometry?
In Euclidean geometry, the triangle inequality is a theorem about distances and it states that for any triangle, the sum of the lengths of any two sides must be greater than or equal to the length of the remaining side.
What does the triangle inequality permit in terms of triangles?
The triangle inequality permits the inclusion of degenerate triangles.
Test your knowledge of the triangle inequality theorem and its applications with this quiz! Explore the relationship between the lengths of the sides in a triangle and discover when equality is possible. | 677.169 | 1 |
...meet together, but are not in the same straight line. And when a straight line standing on another makes the adjacent angles equal to one another, each of the angles is called a right angle. A right angle is divided into 90 equal parts called degrees, a degree is divided into 60 equal partsmeeting or cutting each other in a point, as c (10) " When a straight line standing on another straight line makes the adjacent angles equal to one another,...stands on the other is called a perpendicular to it." The angles AB c, ABD, are right angles. Obs. 1. Because DB is in the same straight line with B c, the...
...st. line makes the adj. ¿_s equal to each other, each of these /.s is called a rt. ¿_ ; and the st. line which stands on the other is called a perpendicular to it. Ax. 1. Things which are equal to the same thing are equal to one another. EI 2 CI 2 Dat. Quass. Pst....
...ECF; (I. 8.) And they are adjacent angles. 5. But when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of the angles is called a right angle; (def. 10.) 6. Therefore each of the angles DCF, ECF, is a right angle. Conclusion. — Therefore from...
...ADG must be equal to the angle GDB ; (L, 8) but ' when a straight line standing upon another straight line makes the adjacent angles equal to one another, each of the angles is a right angle;' (i. def. 10) therefore the angle GDB must be a right angle ; but FOB is a right angle;...
...line makes the adjacent angles equal to each other, each of these angles is called a right angle; D and the straight line which stands on the other is called a perpendicular to it. , The Anglo CDA being equal to angle CDB, each of them is a right angle, and CD is perpendicular toAB....
...line standing on another straight line makes the adjacent angles equal to one another, each of these angles is called a right angle ; and the straight line which stands on tbo other is called a perpendicular to it —!— XI. An obtuse angle is that which is greater than...
...line standing on another straight line, makes the adjacent angles equal to one another, each of these angles is called a right angle ; and the straight...stands on the other is called a perpendicular to it. XI. An obtuse angle is that which is greater than a right angle. xn. An acute angle is that which is | 677.169 | 1 |
What is a shape with 9 sides?
nonagon
A nine sided shape is a polygon called a nonagon. It has nine straight sides that meet at nine corners. The word nonagon comes from the Latin word "nona", meaning nine, and "gon", meaning sides.
What is the shape with 8 sides?
octagon
An octagon is a shape with 8 sides and 8 angles.
What is a shape with 100 sides called?
In geometry, a hectogon or hecatontagon or 100-gon is a hundred-sided polygon. The sum of all hectogon's interior angles are 17640 degrees.
What kind of shape has 10 sides?
decagon
In geometry, a decagon (from the Greek δέκα déka and γωνία gonía, "ten angles") is a ten-sided polygon or 10-gon….Decagon.
Regular decagon
A regular decagon
Type
Regular polygon
Edges and vertices
10
Schläfli symbol
{10}, t{5}
What's a 10 sided shape?
Answer: A 10 sided polygon is called a decagon. The Greek word 'Polygon' consists of poly meaning 'many' and gon meaning 'angle. ' Explanation: A decagon is a ten-sided polygon with ten vertices and ten angles.
What is a 1 billion sided shape called?
Regular megagon
Megagon
Regular megagon
Coxeter diagram
Symmetry group
Dihedral (D1000000), order 2×1000000
Internal angle (degrees)
179.99964°
Dual polygon
Self
What do a heptagon look like?
The heptagon shape is a plane or two-dimensional shape comprised of seven straight sides, seven interior angles, and seven vertices. A heptagon shape can be regular, irregular, concave, or convex. All heptagons can be divided into five triangles. All heptagons have 14 diagonals (line segments connecting vertices)
What does a heptagon look like?
Which shape has 7 sides and 7 angles?
Definition of heptagon : a polygon of seven angles and seven sides mathematics : a flat shape that has seven sides and seven angles : a closed geometric figure having seven angles and seven sides
What shape has got 7 sides?
The heptagon has got 7 sides. A regularheptagon has got 7 sides of equal length. Can you name two coins that are heptagons? Answer Both the 20p and the 50pcoins are heptagons.
Which shape has 7 vertices?
So 7 sided polygon has 7 sides and 7 vertices 7 sided polygon is known as heptagon. · Equilateral polygon is the geometric shapes which consists of all the same sides.
What do you call shape with eight sides?
"Octa-" is a prefix from Latin language meaning "eight.". That is why a shape, with eight sides, is called an octagon. It is a geometric figure from the group of regular polygons. It has eight angles and sides. Types of octagon: There are regular and irregular octagons | 677.169 | 1 |
Polygon Geometry: Pentagons, Hexagons, and Dodecagons
Polygon geometry, also referred to as plane geometry, is a branch of mathematics that deals with the study of closed two-dimensional figures with straight lines called polygons. Polygons are essential in many fields including architecture, design, and engineering. They come in different shapes and sizes, with the most common being pentagons, hexagons, and dodecagons.
Pentagons
A pentagon is a polygon with five sides and five angles. It is one of the most familiar and easily recognizable polygons in geometry. Each angle of a regular pentagon measures 108 degrees, and the sum of all the angles is 540 degrees. In architecture, this shape is often used in building design, from the pentagonal tower of Shanghai World Financial Center to the pentagonal vents on the roof of the U.S. Capitol.
Hexagons
A hexagon is a polygon with six sides and six angles. It is a common shape found in nature, including honeycombs, snowflakes, and the basalt columns of Giant's Causeway in Ireland. Each angle of a regular hexagon measures 120 degrees, and the sum of all the angles is 720 degrees. These shapes are commonly used in the design of tiles, mosaics, and flooring, as seen in the pattern of a honeycomb.
Dodecagons
A dodecagon is a polygon with twelve sides and twelve angles. It is less commonly seen in architecture and design but still features as a prominent shape in some buildings like the landmark bronze statue of Sun Yat-sen in Taipei, Taiwan. Each angle of a regular dodecagon measures 150 degrees, and the sum of all angles is 1800 degrees. Dodecagons feature in many puzzles and games, including dice and board games.
Conclusion
Polygons are not just a static shape but also have applications in various fields. The regularity of these polygons makes them ideal for design, pattern-making, and even generating electrical power through wind turbines. The properties of polygons are vital in geometry and other areas of mathematics. The concepts continue to help us understand and solve real-world problems today, and it is fascinating to see how these polygons persist in our daily lives. Overall, understanding pentagons, hexagons, and dodecagons provides a window into the field of plane geometry, and their practical applications in science, art, and architecture | 677.169 | 1 |
Since , we can write its coordinates as . The equation of line is then .
Similarly, since , and , we can see that the equations of and respectively are and
Multiplying the three together yields the solution to the equation:
Dividing by yields:
, which is equivalent to Ceva's theorem
QED
Trigonometric Form
The trigonometric form of Ceva's theorem states that cevians concur if and only if
Proof
First, suppose concur at a point . We note that
,
and similarly,
.
It follows that
.
Here, the sign is irrelevant, as we may interpret the sines of directed angles mod to be either positive or negative.
The converse follows by an argument almost identical to that used for the first form of Ceva's theorem. ∎
Problems
Introductory
Suppose , and have lengths , and , respectively. If and , find and . (Source)
Intermediate
In are concurrent lines. are points on such that are concurrent. Prove that (using plane geometry) are concurrent. (Ceva I.2) | 677.169 | 1 |
math4finance
The distance from the centroid of a triangle to its vertices are 16cm, 17cm, and 18cm. What is the l...
4 months ago
Q:
The distance from the centroid of a triangle to its vertices are 16cm, 17cm, and 18cm. What is the length of the shortest median
Accepted Solution
A:
Answer:[tex]24[/tex] [tex]\text{cm}[/tex]Step-by-step explanation:Given: The distance from the centroid of a triangle to its vertices are [tex]16\text{cm}[/tex], [tex]17\text{cm}[/tex], and [tex]18\text{cm}[/tex].To Find: Length of shortest median.Solution:Consider the figure attachedA centroid is an intersection point of medians of a triangle.Also,A centroid divides a median in a ratio of 2:1.Let G be the centroid, and vertices are A,B and C.length of [tex]\text{AG}[/tex] [tex]=16\text{cm}[/tex]length of [tex]\text{BG}[/tex] [tex]=17\text{cm}[/tex]length of [tex]\text{CG}[/tex] [tex]=18\text{cm}[/tex]as centrod divides median in ratio of [tex]2:1[/tex]length of [tex]\text{AD}[/tex] [tex]=\frac{3}{2}\text{AG}[/tex] [tex]=\frac{3}{2}\times16[/tex] [tex]=24\text{cm}[/tex]length of [tex]\text{BE}[/tex] [tex]=\frac{3}{2}\text{BG}[/tex] [tex]=\frac{3}{2}\times17[/tex] [tex]=\frac{51}{2}\text{cm}[/tex]length of [tex]\text{CF}[/tex] [tex]=\frac{3}{2}\text{CG}[/tex] [tex]=\frac{3}{2}\times18[/tex] [tex]=27\text{cm}[/tex]Hence the shortest median is [tex]\text{AD}[/tex] of length [tex]24\text{cm}[/tex] | 677.169 | 1 |
NCERT Solutions Class 9 Maths Chapter 10 Exercise 10.1 Circles
NCERT Solutions for Class 9 Maths Chapter 10 Exercise 10.1 Circles is based on terms and the properties related to circles. This exercise involves questions on some very basic properties of circles such as the center of a circle, radius, diameter, chord, semicircle, sector, arc, segment, etc. NCERT solution class 9 maths chapter 10 exercise 10.1 contains problems like fill in the blanks and true or false type of questions, based on these terms related to the circles. Practicing these questions will help promote students' basic understanding to get well-versed in these key concepts.
The knowledge of these terms will enable students to differentiate between the specific properties of circles and their related concepts. Class 9 maths NCERT solutions chapter 10 is a comprehensive guide that delivers the knowledge of core concepts to minimize mistakes. Students can study and practice with Exercise 10.1 as given below.
NCERT Solutions Class 9 Maths Chapter 10 Exercise 10.1 Tips
Regular practice of questions covered in the NCERT Solutions Class 9 Maths Chapter 10 Circles Exercise 10.1 will make it simple for students to grasp all the basics related to this two-dimensional geometry shape. By practicing the questions based on the properties covered in exercise 10.1 of the NCERT solutions class 9 maths, students will attain the skills required for advanced geometry studies.
Reading through the content like examples and formulas available in these solutions will enable students to develop the logic and reasoning to step by step solve the questions. It is better to thoroughly revise all the terms and definitions before answering the questions provided in the NCERT solutions class 9 maths chapter 10 exercise 10.1. This will also benefit the students to reinforce all their knowledge. | 677.169 | 1 |
What is the sine of 60 degrees.
Apr 23, 2019 · The sine of 60° is √3/2. What is sine of an angle? The ratio between the hypotenuse and the leg opposite the angle, when viewed as a component of a right triangle, is the trigonometric function for an acute angle. Given. height = √3. hypotenuse = 2. sin θ = height/ hypotenuse. sin θ = √3/2. To know more about sine of an angle refer to : For sin 69 degrees, the angle 69° lies between 0° and 90° (First Quadrant ). Since sine function is positive in the first quadrant, thus sin 69° value = 0.9335804. . . Since the sine function is a periodic function, we can represent sin 69° as, sin 69 degrees = sin (69° + n × 360°), n ∈ Z. ⇒ sin 69° = sin 429° = sin 789°, and so on.sin (60°) = 0.866025. sin (75°) = 0.965926. sin (90°) = 1. Powered by mymathtables.com. More Trigonometric Pages. Table of Cotangent 0° to 90°. Table of Cotangent 91° to 180°. Table of Cotangent 181° to 270°.Online degree programs enable you to further your knowledge from home. They offer flexibility and are a great choice for parents. If you didn't have the chance to go to college, th...
Online degrees allow busy people to continue their education. Find out what employers think of online degrees and how to evaluate online degree programs. Advertisement Earning a de...How to use the trig ratios of special angles to find exact values of expressions involving sine, cosine and tangent values of 0, 30, 45, 60 and 90 degrees? Example: Determine the exact values of each of the following: a) sin30°tan45° + tan30°sin60°. b) cos30°sin45° + sin30°tan30°. Show Video Lesson.
Step 1. a unit circle is a circle of unit radius—that is, a radius of 1. 1) What is the radius of the unit circle? 2) Identify the sine, cosine and tan for either 30,45 , or 60 degrees in the 1st Quadrant using exact values NOT decimal approximations. 3) What angle in each quadrant has the same reference angle as chosen in step 2?There's a school of thought in the United States that says the only practical path through higher education is to earn a degree that immediately makes money. Or, at least, one that...
Are you in need of your degree certificate download? Whether you are a recent graduate or someone who misplaced their physical copy, obtaining your degree certificate online has ne...How do you find the value of #sin 60#? Trigonometry Right Triangles Trigonometric Functions of Any Angle. 1 Answer Gió Apr 25, 2018 I tried this: Explanation: Have a look: Answer link. Related questions. How do …The triangle shown is an equilateral triangle. An equilateral triangle has sides lengths a. What is the area of the equilateral triangle with the length of each side equal to a? One-half a sine (60 degrees) 3 a sine (60 degrees) One-half a squared sine (60 degrees) a squared sine (60 degrees)So this was the sine of 60 degrees. This whole thing is going to evaluate to cosine of angle ABC is 15 over 17 times cosine of 60 degrees is one half. So times one half. And then, we're going to subtract sine of ABC, which is 8 over 17. And then, times sine of 60, which is square root of 3 over 2.
Sin 90 degrees is equal to one. This degree value can also be expressed in radians as sin(?/2) = 1. This value of the sine function corresponds to one-fourth of the complete arc di...
sin-60°. = cos (90° + 60°) = cos 150°. = sin (180° + 60°) = sin 240°. -sin-60°. = cos (90° – 60°) = cos 30°. = sin (180° – 60°) = sin 120°. Note that sin-60° is periodic: sin (-60° + n …# What is inverse sine? Inverse sine is the inverse of basic sine function. In the sine function, value of angle θ is taken to give the ratio opposite/hypotenuse. However, inverse sine function takes the ratio opposite/hypotenuse and gives angle θ. sin-1 (opposite/hypotenuse) = θ Inverse sine symbol. Inverse sine is represented as sin-1 or ...The Trignometric Table of sin, cos, tan, cosec, sec, cot is useful to learn the common angles of trigonometrical ratios from 0° to 360°. Select degrees or radians in the drop down box θ) I'm skeptical. Please show me an example.The sine graph or sinusoidal graph is an up-down graph and repeats every 360 degrees i.e. at 2π. In the below-given diagram, it can be seen that from 0, the sine graph rises till +1 and then falls back till -1 from where it rises again. The function y = sin x is an odd function, because; sin (-x) = -sin x.
Use our sin(x) calculator to find the sine of 10 degrees - sin(10 °) - or the sine of any angle in degrees and in radians. ... Type a value like: 60, -30, pi/3, 3pi/2, etc. Angle: Calculator use. To use this calculator, just type a value for the angle, then press 'Calculate'.Oct 27, 2012 ... Comments7 · 06 - Review of Essential Trigonometry (Sin, Cos, Tangent - Trig Identities & Functions) · 05 - Sine and Cosine - Definition & Mea...Jun 7, 2021 · Find an angle θ with 0∘ < θ < 360∘ that has the same: sine as 30°:∅= degrees cosine as 30°:∅= degrees The sine of a 30 degree angle is equal to the cosine of a _____ degree angle. 30 45 15 60 The table shows Sin[] and Cos[] in surds, for angles that are integer multiples of 3° or of 5⅝° = 90°/16. The surds are shown in several formats.Related Concepts. Tr ...Triangle (B) c 2 = a 2 + b 2-2 a b. cos (C) Step 2: Click the blue arrow to submit.
... sine, cosine and tangent of an angle. A ... 60 and 30 degrees. You can then use SOHCAHTOA to find the desired values. To find the sin, cos and tan of a 45 degree ...
Evaluate sin(-60 degrees ) Step 1. Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the fourth quadrant. Step 2. The exact value of is . Step 3. The result can be shown in multiple forms. Exact Form:The law of sines says that a / sin(30°) = b / sin(60°) = c / sin(: c = 2 × a and b = √3 × a. Law of sines calculator finds the side lengths and ... Tr Sine Calculator. In mathematics, the sine is a trigonometric function of an angle. The sine of an acute angle is defined in the context of a right triangle: for the specified angle, it is the …Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below. Solution for Find an angle 0 with 0° < 0 < 360° that has the same: Sine function value as 220° 8 - degrees Cosine … To find the value of sin 56 degrees using the unit circle: Rotate 'r' anticlockwise to form a 56° angle with the positive x-axis. The sin of 56 degrees equals the y-coordinate(0.829) of the point of intersection (0.5592, 0.829) of unit circle and r. Hence the value of sin 56° = y = 0.829 (approx) ☛ Also Check: sin 60 degrees; sin 45 degreesThe 30-60-90 and 45-45-90 triangles are used to help remember trig functions of certain commonly used angles. For a 30-60-90 triangle, draw a right triangle whose other two angles are approximately 30 degrees and 60 degrees. The sides are 1, 2 and the square root of 3. The smallest side (1) is opposite the smallest angle (30 degrees).
Click here 👆 to get an answer to your question ️ The sine of 30 degrees is 1/2. What is the cosine of 60 degrees? See what teachers have to say about ... Which of the following explains why Cosine 60 degrees = sine 30 degrees using the unit circle? verified. Verified answer. The sine of a 30 degree angle is equal to the cosine ...
π /3 radians is 60° degrees through the cycle, which is about 2/3 of the way to the top ( π /2) of the +5V cycle, making it closest to +4.3 volts. What we are looking for is instantaneous voltage of a sine wave at a specified angle. In this case it is answer = 5 ∗ sin (60°) since the peak voltage is 5 and the angle is 60° degrees.
-sin(60°) = sin(-60°) -sin(60°) = sin(300°) Referencing the unit circle, we can see that sin(60°)= , so -sin(60°)= , and sin(-60°) is equivalent to sin(-60° + 360°) = sin(300°), which is equal to .The sine of 60 degrees, denoted as sin 60°, is equal to 0.866025404. To calculate the sine of an angle, you can use a scientific calculator or refer to a trigonometric table. However, …Cos 30°= Sin 60° = √3/2. Cos 45° = Sin 45° = 1/√2. Cos 60° = Sin 30° =½. Cos 90° = Sin 0° = 0. Tangent: Tan 0° = Sin 0°/Cos 0° = 0. Similarly, Tan 30° =1/√3. Tan 45° = 1. Tan 60° = √3. Tan 90° = ∞. For more information on sin 60° and other values of sin, cos, and tan, visit Vedantu's website and get practice questions ...Cosine function, along with sine and tangent, is one of the three most common trigonometric functions. In any right triangle, the cosine of an angle is the length of the adjacent side (A) divided by the length of the hypotenuse (H) What is vaue of Cosine 0°? = 1. What is vaue of Cosine 60°? = 0.5. What is vaue of Cosine 90°? = 0 …Sep 14, 2016 ... 30 and 60 Degrees | GEOMETRICAL PROOF | Trigonometric Ratios | sin cos tan |. 74K views · 7 years ago ...more ...Sine calculator to easily calculate the sine function of any angle given in degrees or radians. Calculate sin(x) with this trigonometry calculator. Sin angle calculator with degrees and radians. CalculatorsFor sin 42 degrees, the angle 42° lies between 0° and 90° (First Quadrant ). Since sine function is positive in the first quadrant, thus sin 42° value = 0.6691306. . . ⇒ sin 42° = sin 402° = sin 762°, and so on. Note: Since, sine is an odd function, the value of sin (-42°) = …To convert degrees to radians, multiply by π 180° π 180 °, since a full circle is 360° 360 ° or 2π 2 π radians. The exact value of sin(60) sin ( 60) is √3 2 3 2. Multiply √3 2 ⋅ π 180 3 2 ⋅ π 180. Tap for more steps... Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework ...There's a school of thought in the United States that says the only practical path through higher education is to earn a degree that immediately makes money. Or, at least, one that...
sinsamuelonum1. Answer: Sine 60°= √3/2. =1.732/2. 0.8660. Step-by-step explanation: In mathematics, the sine is a trigonometric function of an angle. The sine of an acute angle is defined in the context of a right triangle: for the specified angle, it is the ratio of the length of the side that is opposite that angle to the length of the ...Answer: sin (60°) = 0.8660254038. sin (60°) is exactly: √3/2. Note: angle unit is set to degrees. Use our sin (x) calculator to find the exact value of sine of 60 degrees - sin (60 °) - or the sine of any angle in degrees and in radians.Instagram: 2017 gmc acadia freon typeweather in albrightsville pennsylvaniamaricopa county arizona assessor property searchdarcy mcqueeny Therefore, Sin 30 degree equals to the fractional value of 1/ 2. Sin 30° = 1 / 2. Therefore, sin 30 value is 1/2. In the same way, we can derive other values of sin degrees like 0°, 30°, 45°, 60°, 90°,180°, 270° and 360°. Below is the trigonometry table, which defines all the values of sine along with other trigonometric ratios. abco federal credit union routing numberboone and scenic valley railroad Trigonometry Examples. Popular Problems. Trigonometry. Find the Exact Value cos(60 degrees ) Step 1. The exact value of is . Step 2. The result can be shown in ...Aug 27, 2015 · sin(60^@) = sqrt(3)/2color(white)("XX")csc(60^@) = 2/sqrt(3) cos(60^@) = 1/2color(white)("XXXX")sec(60^@) = 2 tan(60^@) = sqrt(3)color(white)("XXX")cot(60^@)=1/sqrt(3) Use the basic trigonometric definitions and the diagram below. Note: only the left half triangle is directly relevant; both sides combine to form an equilateral triangle from which (with the help of the Pythagorean Theorem) the ... cemu games not showing up Jan 18, 2024 · As the arcsine is the inverse of the sine function, finding arcsin(1/2) is equivalent to finding an angle whose sine equals 1/2. On the unit circle, the values of sine are the y-coordinates of the points on the circle. Inspecting the unit circle, we see that the y-coordinate equals 1/2 for the angle π/6, i.e., 30°. Jul 23, 2023 ... Sin 60° = √3/2 but why? || लेकिन कैसे ...270° to 360° — fourth quadrant. In this case, 250° lies in the third quadrant. Choose the proper formula for calculating the reference angle: 0° to 90°: reference angle = angle, 90° to 180°: reference angle = 180° − angle, 180° to 270°: reference angle = angle − 180°, 270° to 360°: reference angle = 360° − angle. | 677.169 | 1 |
Different lines in math
May 4, 2022 | Hillsboro
Lines are the building blocks of geometry. Lines are 2D objects, with no width, that can stretch on indefinitely, or join together with other lines to create shapes. There are many different lines to memorize and understand perpendicular, horizontal, etc. So below you can find a smorgasbord of all the different lines that make the shapes we know!
Types of Lines
There are five types of lines.
Vertical Line: A vertical line can be thought of as a standing line and has an undefined slope. We can draw a graph for a vertical line by plotting x = n, where n is equivalent to any sort of real number. All points in the vertical line have the same x coordinate and are parallel to the y-axis of the coordinate plane.
You'll see vertical lines in architectural pillars, the trunks of trees, the lines of a skyscraper, and so many other places!
Horizontal Line:
A horizontal line, also called a sleeping line, is a line in which all points have the same y - coordinate. It is parallel to the x-axis of the plane, and the slope of the horizontal line is zero. It is referred to as the sleeping line because it rules left to right without ever crossing the X-axis. Examples of horizontal lines include the ocean and the sky (the horizon), lines on a sheet of notebook paper, and the edge of a table.
The horizontal line equation is y = b. Here y is any point in the line of x coordinates, and b is the y-intercept. It is independent of x.
Parallel Line:
Two lines are parallel when the distance between the two straight lines is the same at all points. These two lines do not intercept or meet at any point. The symbol for parallel lines is ||, and you can find them in striped clothing, fence posts, or any other pair of lines.
Perpendicular Line:
Perpendicular lines are formed when a horizontal and a vertical line meet at a point. They are two lines that form a congruent adjacent angle with each other, clocking in at ninety degrees. The symbol for perpendicular lines is ⊥, and you're most likely to see them on a grid or where lines come together to make a T shape. Clocks are the best examples of perpendicular lines at 3'O clock and 9'O clock.
Other Geometric Lines
So far, we have learned about four different types of straight lines. However, there are even more types of lines that you'll find in geometry!
intersecting lines - lines that meet at a point at any angle (unlike perpendicular lines, which only meet at 90°)
secant lines - lines that intersect a curve at two or more points
skew lines - lines that have no intersections with each other but are not equidistant (unlike parallel lines)
tangent lines - lines that touch a curve at one point
Line Segments and Rays
Some people get confused about the differences between lines, line segments and rays. Basically, the differences in their meanings are:
line - continues for infinity
line segment - has two endpoints
ray - has one endpoint; the other side continues for infinity
You can measure the length of a line segment, which is a part of a line, and all the points inside. You can't measure a line or a ray, as they each have at least one side that continues without an endpoint.
So, the line can be defined as a straight one-dimensional figure with no thickness and extends endlessly in both directions.
Lines are one of the important aspects and introduction to Geometry. With the types of lines introduced in this blog, kids will apply their ideas to real life and explore their thoughts.
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Class 11 Maths MCQ – Trigonometric Functions – Angles
1. If the initial side is overlapping on the terminal side, then angle is ________
a) 0°
b) 180°
c) 90°
d) 270° View Answer
Answer: a
Explanation: The angle is formed if we start to rotate from initial side till terminal side comes. If they both overlap then angle is said to be 0°.
2. If we start to rotate and after completing one revolution again initial side overlap with terminal side, then the angle formed is _________
a) 0°
b) 180°
c) 90°
d) 360° View Answer
Answer: d
Explanation: The angle is formed if we start to rotate from initial side till terminal side comes. If we start to rotate and after completing one revolution again initial side overlap with terminal side, then the angle formed is 360° | 677.169 | 1 |
Problem 4, 1975 USA Math Olympiad and Isosceles Triangles
The following problem has been offered at the 1975 USA Mathematics Olympiad:
Two given circles intersect in two points P and Q. Show how to construct a segment AB passing through P and terminating on the two circles such that AP×PB is a maximum.
The problem admits a simple trigonometric solution. However, Hubert Shutrick discovered several engaging properties of this configuration which led to a synthetic solution of the problem. The applet below illustrates some of these properties. Hubert's solution is presented on a separate page.
We shall denote the given circles C(E) and C(F). Let C(O) be the circumcircle of EFQ, with center O.
Besides point Q, C(O) meets C(E) in L and C(F) and M. Let S be the intersection of AL and BM. We first show that S lies on C(O).
∠ABS = ∠PBM = ∠PQM,
∠BAS = ∠PAL = ∠PQL.
However, in ΔABS, ∠ABS + ∠BAS + ∠ASB = 180°. It follows that ∠LQM + ∠LSM = 180° so that the quadrilateral LSMQ is cyclic.
Conversely, if S is on C(O) and SL and SM meet C(E) in A and C(F) in B then AB passes through P. Indeed, ∠QLS is supplementary,complementary,supplementary,obtuse,270° of ∠QMS and ∠ALQ = ∠APQ and ∠BMQ = ∠BPQ.
Now observe, that, since the arcs LQ and MQ are fixed for a given pair of circles C(E) and C(F), inscribed angles QAS and ASQ do not depend on the position of A. Similarly, angles QBS and BSQ do not depend on the position of point B. For convenience then choose A so that to have AQ a diameter of C(E). (In passing, this makes AB the longest segment through P with the endpoints on C(E) and C(F).) In this case, LQ ⊥ AS and, therefore, QS is a diameter of C(O) which, in turn, implies that MQ ⊥ BS so, similarly, BQ is a diameter of C(F). E, O, F being the centers of C(E), C(O), C(F), divide the corresponding diameters,angles,circles,diameters,triangles in half making, say, triangles ASQ and EOQ similar,equal,identical,congruent,similar. Since the latter is isosceles, so is the former, implying AS = QS. By the same token, BS = QS, so that, finally AS = BS.
ΔASB is also isosceles so that C(O) is the locus of the centers of the circumcircles ABQ.
Also, we see that ∠ABS = ∠BAS, from which ∠PQM = ∠PQL, meaning that QP is the bisector of angle LQM. | 677.169 | 1 |
10 Examples of Pyramids
Pyramids are cool 3D shapes. They start with a flat shape at the bottom, like a square or triangle. Then, they have sides that slant and come together to a point at the top. This pointy top is called the apex.
In this article, we will discuss ten examples of pyramids in mathematics. | 677.169 | 1 |
What is a quadratic form?
What is a quadratic form?
What is a quadratic form? A: Let us take an example of $\mathbb{R}^3$, the Riemannian 3-sphere. The circle is the base of a circle of radius $r=3$. The point $z$ is the origin in the unit circle. The radius of the circle is $r=2\pi$. For a quadratically symmetric 2-sphere $S$ the number of points in the sphere equals $r^3$. The sphere is the base for the circle, and the circle is the center of the sphere. The radius of the sphere is $r^2$ (equivalent to $2\pi$ in the example), and the radius of the root is $r$. The radius of a quadrically symmetric 2 space is $r^{-2}$, and the radius is $r$, which is the radius of a half-plane. The sphere is the center for the circle. A third example: Given an $n$-dimensional simplex, the sphere of radius $n$ is the base. Here's another example of a quadric in which the 2-component circle is the root. The circle, the center of a hyperbolic 2-sphenoid, is the base (though the sphere is the root). A quadric and a pentagon are both hyperbolic. The sphere is hyperbolic, and the sphere is hyperboloid. A quadratic is hyperbola. The sphere of radius n is the base, and the base is the center. What is a quadratic form? A: The truth of $\mathbb{R}^3$ is that the real form $\mathbb R^3$ visit this page logarithmic derivatives, but the real form has real roots, so the real form does not have a right answer. This is how we define the real and the imaginary part of a complex number. For simplicity, we assume that $\mathbb C$ is a real number and that $\mathrm{Im}(z)$ is nonzero. Then $\mathbb R^3$ becomes a real number if and only if its imaginary part is zero.
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A real number is called a quadrature at a point. A real number is a real quadrature of $x$ and $y$. The real part of the positive real part of an imaginary quadrature $x$ is the imaginary part minus the imaginary part plus the zero-point. The real part of a real number $x$ can be written as a sum of real and imaginary parts. We can do this directly. For every complex number $x$, let $A(x,y)$ denote the real part of $x$. A nonzero real quadrative integer $n$ is called a $n$-term quadrature. A $n$th term is the quadrature: $$A^n(x, y) = \int_0^\infty e^{-s} x^n(s, y) dy.$$ The following is well-known. $A^n$ is a $n \times n$-matrix for $n \ge 2$ with elements $A(0, y)$ and $A(1, y)$. The first term of the formula above can be seen as the first $n$ terms of the form $A(y,y)$. The second term is the logarithm of the first $2$ terms. Let $x^n_1, x^n_2, \ldots, x^2_n$ be two nonzero real numbers. Then the formula above is also a quadrative formula, i.e. $$\sum_{i,j=1}^n (A^i A^j)^2 = \sum_{i=1}^{n-1} A^i A^{i+1} = \sum_1^n A^n A^{n+1} + \sum_2^n A(x^n, x^1),$$ where the sum is over all $x_1,\ldots,x_n$ and the sums over $i$ and $j$ are the same as in the first formula. There is a general argument that you can use to prove that the sum of the second and the first in the formula above should be zero. You can do this using a basis of the complex plane. To see why this is correct you need to consider the complex plane: $\mathbb C^2 = (0,0)$ $\{ 0, 0\} = (0,-1)$ Then the real part is $0$ because we have $\mathbb A^2 = 0$ but the imaginary part is $1$: $$\mathbb R = \{ 0\} \cup \{ 0 \} \cup (0,1) \cup (1,0) \cup \ldots \cup (n,0)$$ We can write the real part as: $$A(0=x^0, x^0+x^1=0) = A(0+x, x^-x+x^2=0)$$ and the imaginary parts as: $$\mathbb A(x) = \mathbb A + \mathbb R$$ A positive quadrature means that the real part has determinant $\pm 1$. In order to find the real part, we have to find a pair of complex numbers $x^{\pm}$ such that $A(a+x, b+x) = A^{\pm(x)}$ for some real constant $a, b$.
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This can be seen easily from the formula above. We have to find $\mathrm R$ such that the real parts of $x^\pm$, the real parts and the imaginary components are equal. You can see that the real and imaginary part of the first two terms are zero. The real part is zero because the real part can be written like: $$A(0-x,x^-x-x^2)=A(0+\cos(What is a quadratic form? A quadratic forms is a pair of two forms which have the same name. Sometimes the names are different, depending on the type of quadratic formula. A square form is a pair that has the same name, but it does not have the same value. 3. A quadratic equation is a pair where all quadratic equations have the same solution. 4. A quadric equation is a set of quadratically equal equations that have the same expression. 5. The following equation is a quadric equation: 6. A quadrant equation is a triplet that has the following values: 7. A quadregated equation is a solution of the quadrant equation: E 8. A quartic equation is a complex quadratic sequence that has the value of the square: 9. A quartet equation is a function of 3 variables. 10. A quintic equation is both a solution of a quadratically equivalent equation and a solution of all linear equations. 11. A quintuple equation is a real quadratic element of a set of real go to this web-site additive equations.
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There are many quadratic formulas for this type of equation. 12. A quadrupole equation is a family of complex quadratically defined polynomials. 13. A quadruple equation is a linear combination of a family of quadrately defined polynomial. 14. A quartix equation is a sequence of a family that has the values of the squares: 15. A quartial equation is a combination of a quadrupole and a family of square polynomially defined polynopoles. 16. A quintix equation is both an additive family of quadrupole, and a quartix family of cheat my medical assignment roots. 17. A quarteter equation is a triples of | 677.169 | 1 |
ISRO Junior Personal Assistant 2010 Paper-II
Two ships are sailing in the sea on the two sides of a light house. The angles of elevation of the top of the light house as observed from the two ships are 30 deg and 45 deg respectively. If the light house is 100 meter high, the distance between the two ships is: | 677.169 | 1 |
Midpoint And Distance Formula Worksheet
Midpoint And Distance Formula Worksheet. Level up discovering the midpoint of a line section whose endpoints are situated on different quadrants of a coordinate grid. Do you want a fun way for faculty kids to follow the midpoint and distance method with minimal prep? The user-friendly drag&drop user interface makes it simple to add or move areas. These worksheets are vital for high school students.
Forms 10/10, Features Set 10/10, Ease of Use 10/10, Customer Service 10/10. Experience a quicker approach to fill out and sign forms on the internet. Access probably the most extensive library of templates out there. Enter all necessary info within the required fillable areas. The user-friendly drag&drop consumer interface makes it easy to add or move areas.
NotaryCoordinate airplane The midpoint and distance formulation. Find the distance between two points on a coordinate aircraft. Let's check out the gap formulation.
In the project students will use the formulation to discover out distances between states utilizing a map converted right into a coordinate plane. This assignment is nice for center school and highschool geometry classes. The introduction of fractional coordinates in our distance formulation worksheets gives a brand new lease of life to your practice in making use of the formulation, and computing how far two factors are. These pdf midpoint formula worksheets have been meticulously curated for highschool college students. Geometry distance and midpoint worksheet answers.
The Distance Formula Worksheet Solutions
She has long hair, however she always pulls it to the aspect. Try the free Mathway calculator and problem solver beneath to follow varied math topics. Try the given examples, or type in your own downside and verify your answer with the step-by-step explanations. Challenge college students to determine out the ordered pairs of a degree that is at a fractional distance from one other indicated level. Also, ask them to search out the lacking coordinates of points on geometric figures.
Derive the gap method and assist your learners by displaying them the logic behind the space formula. This sensible video ought to help conceptual understanding and is suitable for in-class… Pupils work in pairs to find the distances between landmarks with a map. The groups determine how far to journey on the streets and the size between the landmarks with the distance formulation and compare the 2. Educator Edition Save time lesson planning by exploring our library of educator reviews to over 550,000 open academic sources . Decide whether the triangle formed utilizing the given vertices is a scalene, equilateral, isosceles, or right triangle.
Distance And Midpoint
Students complete assigned issues and… Find out how far it is from here to there. Scholars learn more vocabulary coping with line segments and discover out the relationship between equal distances and congruent segments. Pupils then use the space and midpoint formulation to… Support young mathematicians with learning the ideas of midpoint and distance with this foldable resource. Offering each graphic examples and written equations, this reference clearly demonstrates for faculty students how these key phrases…
The midpoint of a line segment could be found utilizing the ordered pair of each endpoint and dividing by 2. Watch this video to see the graph of a line phase and see how the trainer uses a formula to search out the… Your learners connect the new ideas of transformations within the coordinate airplane to their previous data using the stable vocabulary improvement on this unit. Like a international language, arithmetic has its own set of vocabulary terms… The midpoint formula – which Sal maintains is "silly to memorize" since you simply common – is the primary focus of this video. Viewers will respect the seamless transition between algebraic and geometric concepts in this lecture.
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The worksheet accommodates a secret message that the students will uncover based mostly on the proper answers to each problem.
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Knowing the properties of quadrilaterals is a prerequisite. Use the formulation to search out the side lengths and prove if the coordinates are vertices of a sq., rectangle, parallelogram, or rhombus. Displaying all worksheets related to – Midpoint And Distance Formula. The endpoints and the midpoint of a line phase are given with an unknown worth.
Distance Formula And Midpoint Method
Identify the two given points, add them together, after which divide by two and that is the answer. So you need to find the midpoint between two cities on a map. If you understand the coordinates of the two cities then you should use a method to determine the midpoint. Check out this video to see the formula.
I taught this lesson on a block day, so I had plenty of time for actions and evaluation after the lesson. The distance formula always seems to be somewhat extra troubling for faculty students, so I had them full aDistance Formula Partner Worksheetto follow. When they had been finished, they completed theMidpoint and Distance Formula Stations Maze. This was the first time that I've used a stations maze this 12 months, so I defined the instructions very slowly and clearly. I use them on a regular basis, so I will not have to give instructions again.
These worksheets are a should have for highschool students. Find the coordinates of the midpoint of the road segment CD with endpoints C(-2, -1) and D . Find the coordinates of the midpoint of the road segment AB with endpoints A(-2, 3) and B(2, -2). There are two given values on a number line, what is the midpoint?
1) x y −4 −2 2 4 −4 − ) x y −4 −2 2 four −4 − ) x y −4 −2 2 4 −4 −2 2 4. Some of the worksheets for this idea are 3 the midpoint formulation utilizing midpoint and. Who study slower should this doc with the distance worksheet for each geometrical shapes missing coordinates are you more over the method to spam. The Midpoint Formula is the formula to calculate the midpoint between two factors on the coordinate plane.
Equipping students with our free printable midpoint formulation worksheets equates to providing them access to probably the most valuable apply tools. While applying the midpoint formula, it is essential to reinstate that the midpoint of a segment is nothing but the average of the x-coordinates and y-coordinates of the two endpoints. Instant validation is no more a distant dream with our reply keys. Count on our printable midpoint and distance worksheets if testing utility abilities is on your thoughts. The free pdf worksheets comprise geometrical shapes like triangles, quadrilaterals, polygons, circles with workouts to search out the midpoint of the edges, length of the diagonals, chords, or sides.
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Multiplying Mixed Numbers Worksheet. Student versions, if current, include only the query page. Worksheet for multiplying combined numbers, with an example and solutions too. Teachers/scorers not acquainted enough with the new CCSS standards have to be uncovered to conceptual understandings beyond what was discovered in previous years. Just click any of the bundles below for plot of a narrative or play. Interactive resources...... | 677.169 | 1 |
Describe van Hiele Level 1 (analysis), Level 2 (informal deduction) and Level 3 (deduction). Demonstrate the difference by using a scalene, isosceles and equilateral triangle as an example to explain learner thinking at the three different levels. | 677.169 | 1 |
Practice Set 2.1
Question 1.
In the adjoining figure, each angle is shown by a letter. Fill in the boxes with the help of the figure.
Corresponding angles.
(1) ∠p and [ ] (2) ∠q and [ ]
(3) ∠r and [ ] (4) ∠ s and [ ]
Interior alternate angles.
(5) ∠s and [ ] (6) ∠w and [ ]
Answer:
• Given: Line q is transversal is to line m and line l.
• To find corresponding angles of
1) ∠ p
2) ∠ q
3) ∠ r
4) ∠ s
• Explanation: For ∠p, ∠w is the angle which is in the same side and same direction of transversal so ∠w is the corresponding angle to ∠p.
2) For ∠q, ∠x is the angle which is in the same side and same direction of transversal so ∠x is the corresponding angle to ∠q.
3) For ∠r, ∠y is the angle which is in the same side and same direction of transversal so ∠r is the corresponding angle to ∠y.
4) For ∠s, ∠z is the angle which is in the same side and same direction of transversal so ∠s is the corresponding angle to ∠z.
Now5) For ∠s the angel which is in the inner side as well as on the opposite side of transversal and it's arm show opposite direction is ∠x. So ∠s and ∠x form pair of Interior Alternate angel.
6) For ∠w the angel which is in the inner side as well as on the opposite side of transversal and it's arm show opposite direction is ∠r. So ∠w and ∠r form pair of Interior Alternate angel.
Question 2.
Observe the angles shown in the figure and write the following pair of angles.
(1) Interior alternate angles
(2) Corresponding angles
(3) Interior angles
Answer:
• Given: Line q is transversal is to line m and line l.
• To find: (1) Interior alternate angles
(2) Corresponding angles
(3) Interior angles
(1) Now1) For ∠b the angle which is in the inner side as well as on the opposite side of transversal and it's arm show opposite direction is ∠h. So ∠b and ∠h form pair of Interior Alternate angel.
2) For ∠c the angel which is in the inner side as well as on the opposite side of transversal and it's arm show opposite direction is ∠e. So ∠c and ∠e form pair of Interior Alternate angel.
(2) Corresponding anglesFor ∠a, ∠e is the angle which is in the same side and same direction of transversal so ∠a is the corresponding angle to ∠e.
2)For ∠b, ∠f is the angle which is in the same side and same direction of transversal so ∠b is the corresponding angle to ∠f.
3)For ∠d, ∠h is the angle which is in the same side and same direction of transversal so ∠d is the corresponding angle to ∠h.
4)For ∠c, ∠g is the angle which is in the same side and same direction of transversal so ∠c is the corresponding angle to ∠g.
(3) Interior angles
A pair of angles which are on the same side of the transversal and inside the given lines is called a pair of interior angles.
So, we get only two such pairs of angels.
1) ∠b has ∠e on the same side of transversal and inside the given line. So ∠b and ∠e form pair of interior angels.
2) ∠c has ∠h on the same side of transversal and inside the given line. So ∠c and ∠h form pair of interior angels.
Practice Set 2.2
Question 1.
Choose the correct alternative.
In the adjoining figure, if line m || line n and line p is a transversal then find x.
A. 135°
B. 90°
C. 45°
D. 40°
Answer:
• Given: Line m || line n and line p is a transversal
• To find: The value of x.
Now in the given figure we have 3x and x. 3x and x form a pair of interior angle.
Now by the property of interior angels we know that, each pair of interior angles formed by two parallel lines and their transversal is of supplementary angles i.e. 180°.
∴ x+3x=180(∵ Property of interior angles.)
⇒ 4x=180
⇒
∴ The value of x is 45°.
Question 2.
Choose the correct alternative.
In the adjoining figure, if line a || line b and line l is a transversal then find x.
A. 90°
B. 60°
C. 45°
D. 30°
Answer:
• Given: Line a || line b and line l is a transversal
• To find: Value of x.
Now from the figure we can see
∠GHB=∠CHF (∵ opposite angles are same)
⇒ ∠GHB=2x
∠AGB+∠BHG=180° (∵ (∵ Property of interior angles.)
∴ 4x+2x=180°
⇒ 6x=180°
⇒
∴ The value of x is 30°
Option(D)
Question 3.
In the adjoining figure line p || line q. Line t and line s are transversals. Find the measure of ∠x and ∠y using the measures of angles given in the figure.
Answer:
• Given: Line p || line q, line t and line s are transversals.
• To find: The measure of ∠x and ∠y.
Here we can see
∠KLD=∠HLD (∵ Opposite angles are equal)
⇒ ∠ KLD=70°
∠KLI+∠JIL=180° (∵ (∵ Property of interior angles.)
∴ 70+Y=180°
⇒ Y=110°
∴ The value of y is 110°
Also,
∠ BKL+∠ JKL =180° (Linear pair)
⇒ x+∠JKL = 180°
∴ ∠ JKL= 180-x…(1)
∠ KJI+∠ AJI =180° (Linear pair)
⇒ 40+∠KJI = 180°
∴ ∠ KJI= 140°…(2)
Now,
∠KJI+∠JKL=180° (∵ Property of interior angles.)
140+180-x=180 (From 1 and 2)
⇒-x=180-180-140
⇒ -x=-140°
∴ x=140°
∴ The value of x is 140°.
Question 4.
In the adjoining figure. line p || line q. line l || line m. Find measures of ∠a, ∠b, and ∠c, using the measures of given angles. Justify your answers.
Answer:
• Given: Line p || line q, line l || line m.
• To find: The measure of ∠a, ∠b and ∠c.
Now in this figure
∠CIJ+∠AJI=180° (∵ Exterior angles are supplementary)
⇒ 80° +a=180°
⇒ a = 100°
Also,
∠AJI+∠IJL=180° (Linear pair)
⇒100° +∠IJL=180°
∴ ∠IJL=80°
∠BLK=∠LJI=b(corresponding angles are equal)
∴ b=80°
∠EIK=∠CIJ=80° (Opposite angles are equal)
∠GKD=∠EIK=c(corresponding angles are equal)
∴ c=80°
∴ Values of a,b and c are 100° , 80° ,80° respectively.
Question 5.
In the adjoining figure, line a || line b. line l is a transversal. Find the measures of ∠x, ∠y, ∠z using the given information.
Answer:
• Given: Line a || line b, line l is transversal.
• To find: The measure of ∠x, ∠y and ∠z.
In, the figure above
∠AGE=z
Also,
∠AGE+∠EGB=180° (Linear pair)
z+105=180°
z=75°
∠GHD=x
∠EGB=∠GHD(corresponding angles are equal)
∴ x=105°
∠DHF=y
Also
∠DHF=∠GHD (Opposite angles are equal)
∠DHF=105°
∴ Values of x, y and z are 105° ,105° ,75° respectively.
Question 6.
In the adjoining figure, line p || line l || line q. Find ∠x with the help of the measures given in the figure.
Answer:
• Given: line p || line l || line q.
• To find: Value x
Now, In the above figure ∠GHD=x
Also,
∠GHD=∠GHD+∠IHD
∠AGH=∠GHD (∵ Alternate angles are equal)…(1)
∠EIH=∠DHI (∵ Alternate angles are equal)…(2)
From (1) and (2) we get,
∠GHD=40°
∠DHI=30°
We know
∠GHD=∠GHD+∠IHD
∠GHD= 40°+30°
∠GHD= 70°
But,
∠ GHD = x
∴ x=70°
∴ The value of x is 70°.
Practice Set 2.3
Question 1.
Draw a line l. Take a point A outside the line. Through point A draw a line parallel to line l.
Answer:
Here we need draw a line l then take a point A point A.
3) Now take another perpendicular of same length as of AP, and in same direction.
4) Draw a line through those points.
5) This line is parallel to given line l.
Question 2.
Draw a line l. Take a point T outside the line. Through point T draw a line parallel to line l.
Answer:
Here we need draw a line l then take a point T point T.
3) Now take another perpendicular of same length as of TP, and in same direction.
4) Draw a line through those points.
5) This line is parallel to given line l.
Question 3.
Draw a line m. Draw a line n which is parallel to line m at a distance of 4 cm from it.
Answer:
Method : Draw a line parallel to line l at a distance 4 cm.
Steps of construction :
(1) Draw line l.
(2) Take two points A and B on the line l.
(3) Draw perpendiculars to the line l from points A and B.
(4) On the perpendicular lines take points P and Q at a distance of 4cm from | 677.169 | 1 |
math4finance
What is the perimeter of a polygon with vertices at (-2,1) (-2,7) (1,11) (4,7) and (4,1)
4 months ago
Q:
What is the perimeter of a polygon with vertices at (-2,1) (-2,7) (1,11) (4,7) and (4,1)
Accepted Solution
A:
Answer:28 units.Step-by-step explanation:Consider vertices of the polygon are A(-2,1), B(-2,7), C(1,11), D(4,7) and E(4,1).Distance formula:[tex]D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]Using distance formula we get[tex]AB=\sqrt{\left(-2-\left(-2\right)\right)^2+\left(7-1\right)^2}=6[/tex][tex]BC=\sqrt{\left(1-\left(-2\right)\right)^2+\left(11-7\right)^2}=5[/tex][tex]CD=\sqrt{\left(4-1\right)^2+\left(7-11\right)^2}=5[/tex][tex]DE=\sqrt{\left(4-4\right)^2+\left(1-7\right)^2}=6[/tex][tex]AE=\sqrt{\left(4-\left(-2\right)\right)^2+\left(1-1\right)^2}=6[/tex]The perimeter of a polygon is[tex]Perimeter=AB+BC+CD+DE+AE[/tex][tex]Perimeter=6+5+5+6+6[/tex][tex]Perimeter=28[/tex]Therefore, the perimeter of the polygon is 28 units. | 677.169 | 1 |
Two triangles are congruent if two angles and the side included between them in one of the triangles are equal to the two angles and the side included between them of the other triangle. This is known as the
Q. Line segments ab and cd bisect each other at o.ac and bd are joined forming triangles aoc and bod.state three equality relations between the parts of the two triangles, that are given or otherwise known. Are the two triangles are congruent. State in symbolic form . Which congruence do you use
Q. A carom board (4cm×4cm) has the queen at the centre. When the queen is hit by the striker, it moves to the front edge, rebounds and goes into the hole behind the striking line. The displacement of the queen is . | 677.169 | 1 |
Sets the sweep angle of the arc, measured counter-clockwise up from the horizontal. Note: 'sweepAngle' is a t angle. The arc is first converted to an arc by center point before the 'sweepAngle' is set. Refer to the @Arc (function) in the FME Functions and Factories manual for a detailed definition of 'sweepAngle'. | 677.169 | 1 |
Law of sines word problems worksheet pdf
Students will practice deciding when to apply the law of cosines vs the law of sines to calculate the side length of a triangle and to calculate the measure of an angle. Angle of elevation and depression word problems trigonometry, finding sides, angles, right triangles duration. Law of sine and cosine word problems practice quiz1. You could not by yourself going taking into account ebook buildup or library or borrowing from your associates to right of entry them.
Since we do not know an opposite side and angle, we cannot employ the formula. Law of sines and cosines word problems worksheet kuta free. At the birthday party, there was only one balloon bundle set up and it was in the middle of everything. Of sines worksheet answers, law of sines and cosines worksheet with key pdf. Sine cosine tangent word problems displaying top 8 worksheets found for this concept some of the worksheets for this concept are sine cosine and tangent practice, work trigonometric ratios sine cosine and tangent, sine cosine and tangent practice, trigonometric ratios date period, sohcahtoa work, maths module 8, law of sinescosines word problems, extra. Solve for all missing sides and angles in each triangle. Chapter 6 6 part 2 the cosine law word problems from law of sines worksheet, source. Law of sines applicationword problem, ex 1 youtube. Our mission is to provide a free, worldclass education to anyone, anywhere. D sine law and cosine law find each measurement indicated. Displaying all worksheets related to law of sines word problems.
Law of sines word problems word problems worksheets. Model problems in the following example you will find the possible measures of an angle given the sine of the angle. The law of sines is also known as the sine rule, sine law, or sine formula. Two beautiful law of sines problems basic mathematics. This worksheet includes word problems as well as challenging bonus problems. Students will practice applying the law of sines to calculate side lengths and angle measurements. V worksheet by kuta software llc extra practice s v2j0d1z1w ik nu itta w vseoyfpt awha jr rer 7l clgc k. Points a an b are on opposite sides of the grand canyon. In this example, the reader will notice that the american spelling of the word hi is ha. Students always need more practice with word problems, especially when it comes to. Ue tshe law of sines to solve oblique triangles aas or asa. Law of sines word problems word problems lesson worksheets.
Word problems involving tangent, sine and cosine please support my channel by becoming a. Now that i know all the angles, i can plug it into a law of sines formula. Give your answers with lengths rounded to 4 significant digits, angles in degreeminutesecond form rounded to whole numbers. Juan and romella are standing at the seashore 10 miles apart. To approximate the length of a lake, a surveyor starts at one end of the lake and walks 245 yards. Find the if y shown in quadrant i is angle a with a sine of. Model problems in the following example you will find the length of a side of a triangle using law of cosines.
Chapter 2 epilogue trig word problems math with mr. General triangle word problems practice khan academy. A researcher wants to determine the width of a pond from east to west, which cannot be done by actual measurement. Then they will have to use sine, cosine, tangent, law of sines and law of cosines.
Law of sines and cosines word problems worksheet with. Trigonometry law of sines sine rule maths tutorials, geometry and trigonometry. The law of sines can be used to compute the remaining sides of a triangle when two angles and a side are known aas or asa or when we are given two sides and a nonenclosed angle ssa. Then i am going to cut apart the problems and have each card laminated. Two ranger stations located 10 km apart on the southwest and southeast corners of a national park. Both a pdf and word document are included as well as worked out solutions. Some of the worksheets for this concept are find each measurement round your answers to the, extra practice, law of sines practice work, find each measurement round your answers to the, law of sines law of cosines, sine cosine and tangent practice, law of sinescosines word problems, law of sines activity. E t2q0n1 f2z 9kmuvt5a j ts4o8fjtqw waxrmep nlylcc b. Ambiguous case of law of sines worksheet pdf with answer key sines work. The picture below illustrates a case not suited for the law of sines. During a figure skating routine, jackie and peter skate apart with an angle of 15 between them. Law of sines application word problem, ex 3 duration. Ue tshe law of sines to solve oblique triangles ssa. Apr 03, 2019 law of sines calculator from law of sines and cosines word problems worksheet with answers, source the third example from the law of sines and cosines worksheet is the reverse speech problem. Review the law of sines and the law of cosines, and use them to solve problems with any triangle. Angle of elevation and depression word problems trigonometry, finding. Law of sines pdf free printable which includes the formulas, detailed steps to solve oblique triangles, and 2 practice problems. Ambiguous case of the law of sines explained in a video tutorial, with pictures, practice problems as well as a free pdf worksheet with answer key. A post is supported by two wires one on each side going in opposite directions creating an angle of 80 between the wires. Works.
You could not by yourself going taking into account ebook buildup or library or borrowing from. For find the length of to the nearest whole degree, given, and. Electronic equipment allows sw ranger to determine that the camper is at a location that makes an angle of 61 with the southern boundary. Use the law of sines and law of cosines to find missing dimensions. This worksheet will give students practice with solving word problems by first drawing a picture to see what the problem is saying. Law of sines and law of cosines word problems author.
See more ideas about law of sines, word problems and law of cosines. File type pdf law of sines worksheet answers law of sines worksheet answers getting the books law of sines worksheet answers now is not type of challenging means. For instance, in exercise 44 on page 438, you can use the law of sines to determine the length of the shadow of the leaning tower of pisa. Oct 12, 2015 law of sines application word problem, ex 3 duration. We can also write the law of sines or sine rule as. The law of sines, unlike the law of cosines, uses proportions to solve for missing lengths.
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Work through the word problems that follow to consolidate your understanding of the sine law and its application to real world contexts. To numerically describe which dinosaur in figure 6. Problems such as this, in which we know the measures of three sides of a triangle and we need to find the measurement of a missing angle, cannot be solved by the law of sines. Extra practice sine law and cosine law squarespace. When copying these word problems, law of sines and cosines word problems, i am going to make a different colored set for each team in the class. Law of sines and cosines word problems betterlesson. Sinc over the opposite side, c is equal to sin a over its opposite side, a. Law of sines and cosines, and areas of triangles she loves math. Substitute the values into the appropriate formula do not solve. Worksheets are extra practice, find each measurement round your answers to the, law of sines work, find each measurement round your answers to the, law of sines, law of sineslaw of cosines word problems, law of sines practice work, chapter 14 packet trigonometric applications. Sine law displaying top 8 worksheets found for sine law.
State whether the law of sines or law of cosines is the best choice to solve for x for the given figure. Displaying all worksheets related to law of sines word problems word problems. Law of sines law of cosines worksheet set up and label a diagram. Law of sines calculator from law of sines and cosines word problems worksheet with answers, source the third example from the law of sines and cosines worksheet is the reverse speech problem. To use the law of sines, you need to know one opposite angleside pair measurements.
Law of sines or sine rule solutions, examples, videos. Law of sines worksheet pdf with answer key and model problems. These two law of sines problems below will show you how to use the law of sines to solve some real life problems. Law of cosines worksheets answer to the nearest tenth. Ue tshe law of sines to model and solve reallife problems.
Law of sines and cosines word problems worksheet kuta. The second part of the sheet focuses on problems that require using the formulas more than once law of cosines to get side, then law of sides to get angle etc. A triangle has two sides with lengths of 20 and 15. In the two previous cases, asa and saa, the law of sines can be used to solve the triangles without any problems. Apr 09, 2020 some of the worksheets below are law of sines and cosines worksheet in pdf, law of sines and law of cosines. The law of sines one method for solving for a missing length or angle of a triangle is by using the law of sines. A farmer wants to purchase a triangular shaped land with sides 120 feet and 60 feet and the angle included between these two sides is 60. Law of sines formula, how and when to use, examples and. Why you should learn it you can use the law of sines to solve reallife problems involving oblique triangles. Math 3 the uses in degrees, the same as law of sines homework help this helps build terminology, creed, smart technologies ulc, inverse notes, the practice problems take on itunes. Law of sines and cosines word problems worksheet with answers. Law of sines and cosines worksheet pdf dsoftschools.
In this example, the reader will notice that the american spelling of the word. Ambiguous case of law of sines worksheet pdf with answer. Then show the equations you can use to solve the problem. From the ground, she measures the angle of elevation to the top of. Use basic trigonometric functions to solve word problems involving right triangles or use law of sines and cosines to solve word problems use nonright triangles.
Law of sineslaw of cosines worksheet cac mathematics. The angle of elevation from bailey is 600 and from riley is 750. Two people looked up in the sky and saw superman flying. The ends of the wires are 12m apart on the ground with one wire forming an angle of 40 with the ground. Word problems posted by jessica heitfield on 4202018. Word problems with sine and cosine law iannizzis math. | 677.169 | 1 |
Elements of Geometry and Trigonometry
From inside the book
Results 1-5 of 58
Page 9 ... line is length without breadth , or thickness . The extremities of a line are called points : a point , there- fore , has neither length , breadth , nor thickness , but position only . 3. A straight line is the shortest distance from ...
Page 10 Adrien Marie Legendre Charles Davies. 9. When two straight lines , AB , AC , meet each other , their inclination or ... line AB meets another straight line CD , so as to make the adjacent angles BAC , BAD , equal to each other , each of ...
Page 13 ... line AB , which expresses the distance between the points A and B. The expression Ax ( B + C - D ) represents the ... straight line can be drawn . 12. Through the same point , only one straight line can be drawn which shall be parallel ...
Page 14 Adrien Marie Legendre Charles Davies. PROPOSITION I. THEOREM . If one straight line meet another straight line , the sum of the two adjacent angles will be equal to two right angles . Let the straight line DC meet the straight line AB at ...
Page 15 ... ... straight A lines ,. BOOK I. | 677.169 | 1 |
The Definition and Properties of Parallelograms
When approaching a zebra crossing, one might notice that its design consists of closed shapes with four sides, two of which are equal and parallel. These shapes are known as parallelograms, a special type of quadrilateral. In this article, we will delve into the definition and properties of a parallelogram. | 677.169 | 1 |
Tag: calculus
Amare the ant is traveling within Triangle ABC, as shown below. Angle A measures 15 degrees, and sides AB and AC both have length 1.
Amare must:
Start at point B.
Second, touch a point — any point — on side AC.
Third, touch a point — any point — back on side AB.
Finally, proceed to a point — any point — on side AC (not necessarily the same point he touched earlier).
What is the shortest distance Amare can travel to complete the desired path?
I solved the problem in two different ways. The elegant solution: [Show Solution]
Let's solve a slightly more general version of the problem. Suppose the triangle has angle $\theta$ at point $A$, and suppose Amare's next three waypoints are at $D$, $E$, and $F$, as shown below.
Reflect the triangle about the line $AC$, so $B \mapsto B'$ and $E \mapsto E'$, as shown:
Now reflect the new triangle about the line $AB'$, so $C\mapsto C'$ and $F \mapsto F'$:
The key insight is that since the reflections preserve lengths, the path $BD+DE+EF$ followed by Amare has the same length as the path $BD+DE'+E'F'$ shown in red below:
If we move the points $E$ and $F$, then the points $E'$ and $F'$ move accordingly, and we obtain another possible path:
Rather than picking $D,E,F$, we can instead pick $D,E',F'$. Since the goal is to minimize the total distance, it's clear that we should place $F'$ such that $AF' \perp BF'$, and $D$ and $E'$ should be placed so that all three points lie on a line. This produces the figure:
So the shortest distance Amare can travel can be found by examining the right triangle $ABF'$. Since $AB$ has length $1$, we conclude that that
$\displaystyle
\text{Minimum distance } = \sin(3\theta)
$
In the case where $\theta = 15^\circ$, we get a length of $\sin(45^\circ) = \frac{1}{\sqrt{2}} \approx 0.7071$.
Note that this solution only works if $3\theta \lt 90^\circ$, i.e. $0 \lt \theta \lt 30^\circ$. In the case that $\theta \geq 30^\circ$, we obtain the degenerate solution $D = E = F = A$, so Amare should head directly to point $A$ and the total distance traveled is $1$.
We can also use calculus to solve the problem. Let's start with the same picture as before:
Suppose $|AD|=x$, $|AE|=y$, and $|AF|=z$. From the law of cosines:
\begin{align}
|BD| &= \sqrt{1+x^2-2x\cos(\theta)} \\
|DE| &= \sqrt{x^2+y^2-2xy\cos(\theta)} \\
|EF| &= \sqrt{y^2+z^2-2yz\cos(\theta)}
\end{align}Let $f(x,y,z)$ be the sum of these three distances, which is the total distance traveled by Amare. We want to find $x,y,z$ such that $f(x,y,z)$ is minimized. A necessary condition for minimality is that the partial derivatives of $f$ with respect to $x,y,z$ should be zero. Let's start with $z$:
\[
\frac{\partial}{\partial z}f(x,y,z) = \frac{z-y \cos (\theta )}{\sqrt{y^2+z^2-2 y z \cos (\theta )}}
\]Setting this equal to zero, we conclude that $z=y\cos(\theta)$. We can now substitute this value of $z$ into the definition for $f$ and we obtain a simpler expression in only two variables:
\begin{align}
g(x,y) &= f(x,y,y\cos(\theta)) \\
&= \sqrt{1+x^2-2x\cos(\theta)} + \sqrt{x^2+y^2-2xy\cos(\theta)} + y\sin(\theta)
\end{align}To make sure there is no funny business going on, let's plot this function to see what it looks like for $\theta=15^\circ$. Here is a contour plot:
If you read the first solution, then it should come as no surprise that the total distance is also equal to $\sqrt{1-z^2}$, since (based on the last figure of the first solution), we have $z=|AF|=|AF'|$ and $|BF'|^2 + |AF'|^2 = 1$ by the Pythagorean theorem.
This week's Riddler Classic is a pursuit problem with a twist. Here is the problem, paraphrased.
You are walking in a straight line (moving forward at all times) near a lamppost. Your evil twin begins opposite you, hidden from view by the lamppost, as illustrated in the figure below.
Assume your evil twin moves precisely twice as fast as you at all times, and they remain obscured from your view by the lamppost at all times. What is the farthest your evil twin can be from the lamppost after you've walked the 200 feet as shown?
Let's place the lamppost at the origin of a cartesian plane. We start at $(-1,-1)$ and move toward $(1,-1)$. Here, we are using units of 100 feet. The problem does not say anything about our speed, just that we move forward at all times. So let's assume our position is given by $(q(t),-1)$, where $-1\leq q(t)\leq 1$ and $q(t)$ is an increasing function of $t$.
Let's assume our evil twin's position is $(r(t),\theta(t))$ in polar coordinates. We are told the evil twin always remains occluded by the lamppost. This means the angle we make with the lamppost must match that of the evil twin. In other words,
$$
\cot(\theta) = -q.
$$Next, our evil twin is $k$ times faster than us. Since an increment of position in polar coordinates is given by $(\mathrm{d}r, r\,\mathrm{d}\theta)$, we conclude that:
$$
\dot{r}^2 + r^2 \dot\theta^2 = k^2 \dot q^2,
$$where the dot indicates derivative with respect to $t$. In the two equations above, $q,r,\theta$ are functions of time $t$, but I omitted the $(t)$'s to make the equations look neater. Substituting $q = -\cot(\theta)$ into the second equation, we obtain:
$$
\dot r^2 + r^2 \dot \theta^2 = k^2 \csc^4(\theta) \dot\theta^2.
$$We also have that $\theta$ starts at $\frac{\pi}{4}$ with $r(\frac{\pi}{4}) = \sqrt{2}$, which is the point $(1,1)$, where the evil twin starts, and we continue until $\theta=\frac{3\pi}{4}$. Simplifying the ODE by solving for $\frac{\mathrm{d}r}{\mathrm{d}\theta}$, we obtain:
This ODE has everything we need to compute the trajectory of the evil twin. Note that the ODE does not actually depend on how fast we are moving (we have eliminated $p(t)$ and $t$ completely). Nevertheless, solving this is not so simple. There are two cases to consider:
If $k^2 \csc^4(\theta) < r^2$, the right-hand side of the equation is negative, so there are no solutions! This happens when the evil twin is too far away from the lamppost and it becomes impossible for them to stay hidden because they cannot move fast enough.
If $k^2 \csc^4(\theta) > r^2$, there are two possible values for $\frac{\mathrm{d}r}{\mathrm{d}\theta}$, which correspond to choosing either the positive or negative square root. In general, we don't have to pick just one of them; we can make a different choice for each $\theta$, which yields infinitely many possible trajectories.
Since our goal is to finish with the largest possible $r$ (largest distance from the lamppost), one might suspect that we should pick the positive square root every time, so that $\frac{\mathrm{d}r}{\mathrm{d}\theta} > 0$. This greedy approach fails, however, because the evil twin gets too far from the lamppost too quickly, and ultimately gets stuck, unable to remain hidden behind the lamppost.
The ODE has no closed-form solution as far as I can tell, so I opted for a numerical/graphical approach. We can plot the slope field for the ODE, and we obtain the solution below:
We start at $(1,1)$, and at every location in space, we can follow the blue arrow or the orange arrow (either positive or negative square root), and we stop once we reach the finish line $y=-x$. The greedy solution (picking the positive square root the entire time) leads to a dead end, as the evil twin hits the infeasibility boundary (where blue and orange arrows coincide) and will no longer remain occluded by the lamppost. Along the line $y=2$ for $x\lt 0$, the blue lines are horizontal and the orange ones point downwards. So it is impossible to cross this boundary. Therefore, $(-2,2)$ is the farthest possible point from the lamppost achievable on the finish line, and we can achieve it by being greedy until we hit $y=0$, and then moving horizontally until we hit the finish line.
We conclude that the farthest distance is $2\sqrt{2}$, or about 282.84 feet, in the problem's original units.
Generalizations
What happens if we vary $k$ (how much faster the evil twin is)?
When $k < 1$, there is no solution; the evil twin can hide for a little while, but eventually, they will be revealed because they are simply not fast enough to remain hidden.
When $k=1$, the optimal solution is for the evil twin to mirror perfectly, i.e. move horizontally and end at the point $(-1,1)$. Although it is possible to go farther from the lamppost initially, this leads to a dead end.
When $1 \lt k \lessapprox 4.4$, the solution is similar to the one for $k=2$ derived above; the evil twin should be greedy until they reach the line $y=k$, then they should move horizontally until they reach the point $(-k,k)$, so the final distance from the lamppost is $k\sqrt{2}$.
When $k\gtrapprox 4.4$, the evil twin is fast enough that being greedy the entire time pays off. However, the flow field in this case flattens out anyway, and we still end up at the same $(-k,k)$ optimal point, with a final distance of $k\sqrt{2}$.
Hames Jarrison has just intercepted a pass at one end zone of a football field, and begins running — at a constant speed of 15 miles per hour — to the other end zone, 100 yards away.
At the moment he catches the ball, you are on the very same goal line, but on the other end of the field, 50 yards away from Jarrison. Caught up in the moment, you decide you will always run directly toward Jarrison's current position, rather than plan ahead to meet him downfield along a more strategic course.
Assuming you run at a constant speed (i.e., don't worry about any transient acceleration), how fast must you be in order to catch Jarrison before he scores a touchdown?
If the football field is $L$ long and $w$ wide, and Jarrison has a speed of $v_0$, then in order to catch him just in time, our speed $v$ should satisfy
\[
\frac{v}{v_0} \gt \frac{1+\sqrt{4\frac{L^2}{w^2}+1}}{2\frac{L}{w}}.
\]For the data given in the problem, we have $L=100$ and $w=50$, so $\frac{v}{v_0} \gt \frac{1+\sqrt{17}}{4}\approx 1.28$. Therefore, we would need to run at a speed of at least $19.21\,\text{mph}$ to catch Jarrison. Here is a plot of $\frac{v}{v_0}$ as a function of $\frac{L}{w}$.
In the limit $\frac{L}{w} \to \infty$, the field becomes infinitely long. Therefore, we spend most of the time running almost parallel to Jarrison. So we only need to be a tiny bit faster in order to eventually catch him, and we have $\frac{v}{v_0}\to 1$.
In the limit $\frac{L}{w} \to 0$, the field becomes infinitely wide. We must cover an increasing amount of lateral ground to catch Jarrison, so $\frac{v}{v_0} \to \infty$.
Fun fact: if the field is square ($L=w$), we must be $\varphi$ times faster than Jarrison to catch him, where $\varphi = \frac{1+\sqrt{5}}{2} \approx 1.618$ is the golden ratio!
We can also ask: what is the shape of the path that we will follow? (This is called a pursuit curve.) Using the coordinate system depicted below, our trajectory will satisfy:
\[
\frac{x}{w} = \frac{1}{2}\left[ \frac{\left(1-\frac{y}{w}\right)^{1+\frac{v_0}{v}}-1}{1+\frac{v_0}{v}}-\frac{\left(1-\frac{y}{w}\right)^{1-\frac{v_0}{v}}-1}{1-\frac{v_0}{v}}\right]
\]
Note: This is the solution I came up with… Admittedly, it's a bit tedious and not particularly intuitive. If you have a more direct or more elegant solution approach, I would love to hear about it!
Based on the diagram above, Jarrison starts at $(0,w)$ and runs toward $(L,w)$ at a speed $v_0$. Therefore, Jarrison's position as a function of time is $(v_0t, w)$. We start at $(0,0)$ and we run at a speed $v$. Let's also suppose our position at time $t$ is $(x(t),y(t))$.
We run in such a way that our velocity always points toward Jarrison. Therefore, we have:
\[
\frac{\dot y}{\dot x} = \frac{w-y}{v_0t-x},
\]where the dots denote time derivatives, i.e. $\dot x = \frac{\mathrm{d}}{\mathrm{d}t}x(t)$. To keep notation simple, we'll omit the $(t)$ when writing $x$ or $y$. We also know that our speed is constant at $v$, therefore, we have:
\[
\dot x^2 + \dot y^2 = v^2
\]These two coupled differential equations, together with the initial conditions $x(0)=y(0)=0$, completely describe our motion. Our task is to solve these equations, and then find the value of $v$ such that the solution passes through the point $(L,w)$, i.e., we catch Jarrison right as he scores a touchdown.
To solve this problem, we'll use the change of variables $u = \frac{\dot x}{\dot y}$. Substitute this into both equations. For the first equation, also isolate $t$ and differentiate so that $t$ no longer appears explicitly. Ultimately, we obtain:
\[
\dot u = \frac{v_0}{w-y}
\quad\text{and}\quad
\dot y = \frac{v}{\sqrt{u^2+1}}
\quad\text{with: }\begin{cases}y(0)=0 \\ u(0)=0\end{cases}
\]Combining these equations, we obtain the single ODE (in differential form)
\[
\frac{\mathrm{d}y}{w-y} = \frac{v}{v_0} \cdot \frac{\mathrm{d}u}{\sqrt{u^2+1}}
\]Integrating from $t=0$ ($y=u=0$) to an arbitrary later point, we obtain:
\[
\log\left( \frac{w}{w-y} \right) = \frac{v}{v_0} \log \left| \sqrt{u^2+1}+u \right|
\]Therefore,
\[
\frac{w}{w-y} = \left| \sqrt{u^2+1}+u \right|^{v/v_0}
\]Now note that if $\sqrt{u^2+1}+u = f$, we have $u = \frac{1}{2}\left( f-f^{-1} \right)$. Therefore, we can solve the above equation and obtain:
\[
u
\]Now recall the definition $u = \frac{\dot x}{\dot y} = \frac{\mathrm{d}x}{\mathrm{d}y}$. Therefore,
\[
\mathrm{d}x\mathrm{d}y
\]Integrating from $t=0$ ($y=u=0$) to an arbitrary later point, we obtain:
\[
\frac{x}{w} = \frac{1}{2}\left[ \frac{\left(1-\frac{y}{w}\right)^{1+v_0/v}-1}{1+\frac{v_0}{v}}-\frac{\left(1-\frac{y}{w}\right)^{1-v_0/v}-1}{1-\frac{v_0}{v}}\right]
\]This tells us $x$ as a function of $y$. The only thing we're missing is $v$. To find $v$, we use the fact that the curve must pass through $(L,w)$, which leads to
\[
\frac{L}{w} = \frac{v_0 v}{v^2-v_0^2}.
\]Solving for $v$ yields:
\[
\frac{v}{v_0} = \frac{1+\sqrt{4\frac{L^2}{w^2}+1}}{2\frac{L}{w}}.
\]Note that the right-hand side is always greater than $1$, regardless of the value of $L$ or $w$. This makes sense; we need to be going faster than Jarrison if we ever hope to catch him.
For each different method, let's call $x_n$ the amount of cake remaining after the first $n$ friends have taken their share. Since we start with a full cake, $x_0=1$. The amount of cake remaining after all friends have taken their share is therefore $\lim_{n\to\infty} x_n$.
Method 1
In this method, Friend number $n$ takes $\frac{1}{n+1}$ of what remains, which is $\frac{1}{n+1}x_{n-1}$. After this happens, the amount of cake remaining is:
\[
x_n = x_{n-1}-\frac{1}{n+1}x_{n-1} = \left(1-\frac{1}{n+1}\right)x_{n-1}.
\]The net result is therefore:
\begin{align}
x_n &= \left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\cdots\left(1-\frac{1}{n+1}\right) \\
&= \left(\frac{1}{2}\right)\left(\frac{2}{3}\right)\left(\frac{3}{4}\right)\cdots\left(\frac{n}{n+1}\right) \\
&= \left(\frac{1}{\bcancel{2}}\right)\left(\frac{\bcancel{2}}{\bcancel{3}}\right)\left(\frac{\bcancel{3}}{\bcancel{4}}\right)\cdots\left(\frac{\bcancel{n}}{n+1}\right) \\
&= \frac{1}{n+1}
\end{align}This is an example of a telescoping series; all the terms in the middle cancel out and we are left with $\frac{1}{n+1}$ of the cake after all friends have gone. Unfortunately, this tends to zero as $n\to\infty$, so with this method, there will be no cake left after all friends have gone!
Method 2
In this method, Friend number $n$ takes $\frac{1}{(n+1)^2}$ of what remains. Using an approach similar to the previous method, we are left with:
\begin{align}
x_n &= \left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)\cdots\left(1-\frac{1}{(n+1)^2}\right) \\
&= \left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)\left(\frac{4^2-1}{4^2}\right)\cdots \left(\frac{(n+1)^2-1}{(n+1)^2}\right) \\
&= \left(\frac{1\cdot 3}{2 \cdot 2}\right)\left(\frac{2\cdot 4}{3 \cdot 3}\right)\left(\frac{3\cdot 5}{4\cdot 4}\right)\cdots \left(\frac{n\cdot(n+2)}{(n+1)(n+1)}\right) \\
&= \left(\frac{1\cdot \bcancel{3}}{2 \cdot \bcancel{2}}\right)\left(\frac{\bcancel{2}\cdot \bcancel{4}}{\bcancel{3} \cdot \bcancel{3}}\right)\left(\frac{\bcancel{3}\cdot \bcancel{5}}{\bcancel{4}\cdot \bcancel{4}}\right)\cdots \Biggl(\frac{\bcancel{n}\cdot(n+2)}{\bcancel{(n+1)}(n+1)}\Biggr) \\
&= \frac{n+2}{2(n+1)}.
\end{align}In the third step, we used the difference of squares formula $n^2-1 = (n-1)(n+1)$ to again obtain a telescoping series. This time, however, the sum is $\frac{n+2}{2(n+1)}$, for which the limit $n\to\infty$ is $\frac{1}{2}$. So half of the cake is left after all friends have gone.
Method 3
In this method, Friend number $n$ takes $\frac{1}{(2n)^2}$ of what remains. Again using an approach similar to the previous methods, we obtain.
\begin{align}
x_n &= \left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{4^2}\right)\left(1-\frac{1}{6^2}\right)\cdots\left(1-\frac{1}{(2n)^2}\right).
\end{align}While it looks deceptively similar to the two previous examples solved above, this infinite product is substantially more complicated since there is no telescoping. There is a way to approach this using calculus, but I will instead show a proof using complex analysis that I think is a bit more intuitive.
Every polynomial $p(z)$ of degree $n$ can be factored as $p(z)=a(z-c_1)\cdots(z-c_n)$ where the $c_k$ are zeros of the polynomial (they may be complex numbers). This result can be extended to the case where $p(z)$ is an infinite polynomial with infinitely many zeros. This is known as the Weierstrass factorization theorem. Roughly speaking, if $f(z)$ is an entire function (it must have a power series expansion that is valid everywhere), then we can write it as an infinite product $f(z)=a(z-c_1)(z-c_2)\cdots$ where the $c_k$ are zeros of the infinite polynomial.
Consider the function $f(z) = \frac{\sin(\pi z)}{\pi z}$. This is an entire function because $\sin(\pi z)$ has a power series that is valid everywhere and it has no constant term (recall $\sin(x) = x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\cdots$) so we can divide through by $\pi z$ and we get another valid power series. The places where $f(z)=0$ correspond to $z \in \{\pm1,\pm2,\pm3,\dots\}$. Note also that $f(0) = \lim_{z\to 0} \frac{\sin(\pi z)}{\pi z} = 1$. By the Weierstrass factorization theorem (I'm skipping some technical details here), we can write:
\begin{align}
f(z) &= a(1-z)(1+z)\left(1-\frac{z}{2}\right)\left(1+\frac{z}{2}\right)\left(1-\frac{z}{3}\right)\left(1+\frac{z}{3}\right)\cdots \\
&= a\left(1-z^2\right)\left(1-\frac{z^2}{2^2}\right)\left(1-\frac{z^2}{3^2}\right)\cdots
\end{align}Since $f(0) = 1$, it must be the case that $a=1$. So we have the series:
\[
\frac{\sin(\pi z)}{\pi z} = \left(1-z^2\right)\left(1-\frac{z^2}{2^2}\right)\left(1-\frac{z^2}{3^2}\right)\cdots.
\]This is known as the Euler infinite product representation for the sinc function. Substituting $z=\frac{1}{2}$, the infinite product becomes exactly the one we wanted to evaluate for how much cake is left:
\[
\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{4^2}\right)\left(1-\frac{1}{6^2}\right)\cdots = \frac{2}{\pi} \approx 0.6366.
\]This infinite product (or its inverse, rather) also has a name; it's called the Wallis product.
Bonus: Odd-numbered terms
As a side note, if we only keep the odd-numbered terms of the product, we obtain = \frac{\pi}{4} \approx 0.7854.
\]This can be proved using a similar approach, this time with the function:
\[
\cos\left(\frac{\pi z}{2}\right) = (1-z^2)\left(1-\frac{z^2}{3^2}\right)\left(1-\frac{z^2}{5^2}\right)\left(1-\frac{z^2}{7^2}\right)\cdots
\]It then follows that
= \lim_{z\to 1} \frac{\cos\left(\frac{\pi z}{2}\right)}{1-z^2} = \frac{\pi}{4}
\]As a sanity check, we found that using the even terms gives $\frac{2}{\pi}$ and using the odd terms gives $\frac{\pi}{4}$. So using all the terms gives $\frac{2}{\pi}\cdot \frac{\pi}{4} = \frac{1}{2}$, which is what we found using "Method 2"!
Since we have a fixed amount of crust that we will shape into a cylinder, the problem is one of maximizing the volume for a fixed area. This is known as the isovolume problem (which is a misnomer; it should really be called the isoarea problem).
Let's suppose the crust has an area of $A$. Given this area, we would like to maximize the volume of the cylinder. Let's suppose the radius of the base is $r$ and the height is $h$. Then the formulas are:
\[
A = 2\pi r^2 + 2\pi r h \qquad\text{and}\qquad V = \pi r^2 h.
\]Since the area $A$ is fixed, we are not free to choose $r$ and $h$ independently. Once $r$ is chosen, $h$ is uniquely determined by the equation for $A$. Solving for $h$ in terms of $A$ from the area equation, we obtain:
\[
h = \frac{A-2\pi r^2}{2\pi r}
\]Substituting this value of $h$ into the equation for $V$, we obtain:
\[
V= \frac{1}{2} r (A-2\pi r^2)
\]Here is a plot of what this function $V(r)$ looks like for a value of $A=1$:
We can maximize $V$ by looking for an $r$ such that $\frac{\mathrm{d}V}{\mathrm{d}r}=0$ and $\frac{\mathrm{d}^2V}{\mathrm{d}r^2}\lt 0$. In other words, at the optimal choice of $r$, the tangent to the curve $V(r)$ is flat and the curvature is negative (curves downward). This leads to the equations:
\[
\frac{1}{2}(A-6\pi r ^2) = 0 \qquad\text{and}\qquad -6\pi r \lt 0
\]Since $r\geq 0$ (radius can't be negative), the second equation is always satisfied, and the first equation implies that $r=\sqrt{\frac{A}{6\pi}}$. If $A=1$ as in the plot above, this leads to $r=0.23033$, which looks about right! With the optimal choice of $r$, the corresponding choice of $h$ becomes $h=\sqrt{\frac{2A}{3\pi}}$ and the corresponding optimal volume is $V=\frac{A^{3/2}}{3\sqrt{6\pi}}$. Interestingly, we can observe that with these choices,
\[
2r = 2\sqrt{\frac{A}{6\pi}} = \sqrt{\frac{4A}{6\pi}} = \sqrt{\frac{2A}{3\pi}} = h
\]So in order to maximize the area, the diameter should be equal to the height! In other words, when viewed from the side, our pie should have a square shape; we'll be making more of a cake rather than a pie.
The fraction of crust used for the base of the optimal pie is:
\[
\frac{\pi r^2}{A} = \frac{\pi \left(\frac{A}{6\pi}\right)}{A} = \frac{1}{6}.
\]
$\displaystyle
\begin{aligned}
&\text{So we should use $\tfrac{1}{6}$ of the crust for the base, $\tfrac{1}{6}$ for the top,}\\
&\text{and the remaining $\tfrac{2}{3}$ for the sides.}
\end{aligned}$
One way to explain the shape of this tall pie is to think back to the isovolume problem discussed earlier. The solution to the isovolume problem is a sphere. So in order to be as efficient as possible, the pie's shape should be as close to a sphere as possible, hence the tall shape. For a sphere, $A=4\pi r^2$ and $V=\frac{4}{3}\pi r^3$, so by eliminating $r$, we obtain $V=\frac{A^{3/2}}{6\sqrt{\pi}}$.
If we take the ratio of the volume of the sphere and the volume of the optimized cylinder, we obtain:
\[
\frac{V_\text{cylinder}}{V_\text{sphere}} = \frac{ \frac{A^{3/2}}{3\sqrt{6\pi}} }{ \frac{A^{3/2}}{6\sqrt{\pi}} } = \sqrt{\frac{2}{3}} \approx 0.8165
\]So the optimized cylindrical pie will have a volume which is about 82% of the largest possible volume.
Bonus: Different bottom and top area
If the top and bottom areas of the pie have different radii, then the shape is no longer a cylinder, but rather a frustom of a cone. The formulas are now a bit more complicated, because they depend on two radii ($r$ and $R$) and the height $h$:
\begin{align}
A &= \pi (r+R) \sqrt{h^2+(R-r)^2}+\pi \left(r^2+R^2\right)\\
V &= \frac{1}{3} \pi h \left(r^2+r R+R^2\right)
\end{align}Using a similar procedure as before (solving for $h$ in the area equation and substituting this into the volume equation), we obtain the following volume formula that depends only on the radii of the top and bottom
\[
V = \frac{\left(r^2+r R+R^2\right) \sqrt{\left(A-2 \pi r^2\right) \left(A-2 \pi R^2\right)}}{3 (r+R)}
\]A necessary condition for optimality is that both partial derivatives are zero, namely $\frac{\partial V}{\partial r} = \frac{\partial V}{\partial R} = 0$. This leads to the equations:
\begin{align}
\frac{\partial V}{\partial r} &= -\frac{r \left(A-2 \pi R^2\right) \left(2 \pi \left(4 r^2 R+2 r^3+2 r R^2+R^3\right)-A (r+2 R)\right)}{3 (r+R)^2 \sqrt{\left(A-2 \pi r^2\right) \left(A-2 \pi R^2\right)}} \\
\frac{\partial V}{\partial R} &= -\frac{R \left(A-2 \pi r^2\right) \left(2 \pi \left(2 r^2 R+r^3+4 r R^2+2 R^3\right)-A (2 r+R)\right)}{3 (r+R)^2 \sqrt{\left(A-2 \pi r^2\right) \left(A-2 \pi R^2\right)}}
\end{align}Much messier than the ordinary cylinder case… But we can simplify this. It must be true that $A\gt 2\pi R^2$ and $A\gt 2\pi r ^2$, since the area of the smaller circle plus the area of the sides must be at least the area of the larger circle (they are only equal when the pie is flat). So if $\frac{\partial V}{\partial r} = \frac{\partial V}{\partial R} = 0$, we can cancel a bunch of terms and we are left with:
\begin{align}
2 \pi \left(4 r^2 R+2 r^3+2 r R^2+R^3\right)-A (r+2 R) &= 0& &(1) \\
2 \pi \left(2 r^2 R+r^3+4 r R^2+2 R^3\right)-A (2 r+R) &= 0& &(2)
\end{align}This still looks rather messy to solve, but we can make a clever observation. If we subtract one equation from the other $(2)-(1)$ and factor, we obtain:
\[
(R-r) \left(A+2 \pi r^2+6 \pi r R+2 \pi R^2\right)=0
\]The right factor consists of a sum of positive terms, and therefore must be positive. So we conclude that $R=r$. Put another way, the only way we'll have an optimized volume is if the top and bottom circles are of equal size. This reduces our more complicated case to the simpler case we already solved above.One morning, it starts snowing. The snow falls at a constant rate, and it continues the rest of the day.
At noon, a snowplow begins to clear the road. The more snow there is on the ground, the slower the plow moves. In fact, the plow's speed is inversely proportional to the depth of the snow — if you were to double the amount of snow on the ground, the plow would move half as fast.
In its first hour on the road, the plow travels 2 miles. In the second hour, the plow travels only 1 mile.
We'll assume the snow starts at $t=0$ hours. So after $t$ hours, the depth of snow on the ground is proportional to $t$. Since the plow's speed $v$ is inversely proportional to the amount of snow on the ground, we have $v = \tfrac{c}{t}$, where $c>0$ is a constant of proportionality. We'll assume the constant is in units of miles so that $v$ is in miles per hour.
Suppose the plow starts at $t=x$ hours. During the first hour, the plow travels 2 miles. Since distance is the integral of velocity with respect to time, we have:
\[
2\text{ miles} = \int_{x}^{x+1} v\,\mathrm{d}t = \int_x^{x+1} \frac{c}{t}\,\mathrm{d}t = c\,\log\left( \frac{x+1}{x} \right)
\]During the second hour, the plow travels 1 mile. So we also have:
\[
1\text{ mile} = \int_{x+1}^{x+2} v\,\mathrm{d}t = \int_{x+1}^{x+1} \frac{c}{t}\,\mathrm{d}t = c\,\log\left(\frac{x+2}{x+1} \right)
\]Dividing the first equation by the second, the units and constants of proportionality cancel. Simplifying, we obtain:
\[
2\,\log\left(\frac{x+2}{x+1}\right) = \log\left( \frac{x+1}{x} \right)
\]Exponentiating both sides to cancel out the logs,
\[
\left(\frac{x+2}{x+1}\right)^2 = \frac{x+1}{x}
\]Finally, clearing fractions and simplifying, we obtain:
\[
x^2+x-1 = 0
\]Since we must have $x\gt 0$, we can discard the negative solution to this equation, and keep only the positive one. This leads us to:
\[
x = \frac{\sqrt{5}-1}{2} \approx 0.618034
\]Incidentally, this is the inverse of the Golden Ratio, though that's more of a coincidence than anything meaningful… If the plow started at noon, then the snow started $x$ hours before noon, which is 11:22.55 AM (11 hours, 22 min, 55 sec).
In this Riddler problem, the goal is to spread out settlements in a circle so that they are as far apart as possible:
Antisocial settlers are building houses on a prairie that's a perfect circle with a radius of 1 mile. Each settler wants to live as far apart from his or her nearest neighbor as possible. To accomplish that, the settlers will overcome their antisocial behavior and work together so that the average distance between each settler and his or her nearest neighbor is as large as possible.
At first, there were slated to be seven settlers. Arranging that was easy enough: One will build his house in the center of the circle, while the other six will form a regular hexagon along its circumference. Every settler will be exactly 1 mile from his nearest neighbor, so the average distance is 1 mile.
However, at the last minute, one settler cancels his move to the prairie altogether (he's really antisocial). That leaves six settlers. Does that mean the settlers can live further away from each other than they would have if there were seven settlers? Where will the six settlers ultimately build their houses, and what's the maximum average distance between nearest neighbors?
I approached this problem from a modeling and optimization perspective. If there are $N$ settlers and we imagine the coordinates of the settlers are $(x_i,y_i)$ for $i=1,\dots,N$ and $d_i$ is the distance between the $i^\text{th}$ settler and its nearest neighbor, then we can model the problem as follows:
\[
\begin{aligned}\underset{x,y,d\,\in\,\mathbb{R}^N}{\text{maximize}} \quad & \frac{1}{N}\sum_{i=1}^N d_i \\
\text{such that} \quad & x_i^2 + y_i^2 \le 1 &&\text{for }i=1,\dots,N\\
& (x_i-x_j)^2 + (y_i-y_j)^2 \ge d_i^2 &&\text{for }i,j=1,\dots,N\text{ and }j\ne i
\end{aligned}
\]The objective is to minimize the average distance between each settler and its nearest neighbor (the average of the $d_i$). The first constraint says that each settler must be within the circle (a distance of at most $1$ from the origin). The second constraint says that the $i^\text{th}$ settler is a distance at least $d_i$ from each of the other settlers.
The game is then to find $(x_i,y_i,d_i)$ that satisfy these constraints and maximize the objective. Broadly speaking, this is an example of a mathematical optimization problem. Unfortunately, the problem as stated is nonconvex problem because of the form of the second constraint. This means that the problem may have local maximizers that are not globally optimal. There are two ways forward:
First, we could use local search: start with randomly placed settlers, wiggle their positions in ways such that the objective continues to increase, and then stop once we can no longer improve the objective. Examples of such approaches include hill-climbing and interior-point methods. In general, such approaches only find local maxima, so we should try many random starting positions to ensure we find the best possible answer.
Second, we could use global search: these are methods that attempt to find a globally optimal solution by considering many configurations simultaneously and adding random perturbations that can "kick" a solution out of a local maximum. Examples include simulated annealing and particle swarms. These approaches tend to be much slower than local search, but they have a much better chance of finding a global optimum.
Ultimately, I used local search with several random initializations. Here is how the maximum average distance scales with the number of settlements.
The bar plot also shows how well we would do if we distributed the settlements evenly on the circumference ("all on the border") and if we put one in the middle and the rest evenly distributed on the circumference ("one in the center"). We can also examine what the settlement distributions actually look like. Here they are:
The case $N=6$ is particularly interesting because it has a settlement that is almost in the center, but not quite. Indeed, the average minimum distance to neighbors for this case is $1.002292$, so we benefit ever so slightly by placing one settlement off-center.
We can solve for the $N=6$ case analytically as well, but the solution isn't pretty (fair warning!) It turns out the settlements are distributed throughout the circle in the following way:
Here, $x$ is the distance between the origin and the point $A$. The other distances satisfy $z \lt y \lt x+1$, and the average distance to the nearest neighbor is therefore $\tfrac{1}{6}(1+x+2y+3z)$. Note that once we set $x$, this uniquely determines the positions of all the other points. After some messy trigonometry, we obtain the following equations relating $x$, $y$, $z$ and two auxiliary variables $p$ and $q$:
\begin{align}
y^2&=1+x-x^2-x^3\\
z^2&=1-x^2+2x\left(pq-\sqrt{1-p^2}\sqrt{1-q^2}\right) \\
p&= \tfrac{1}{2}(1-2x-x^2)\\
q&= \tfrac{1}{2}(1-x+x^2+x^3)
\end{align}We can eliminate $\{y,z,p,q\}$ and solve for the average distance $\bar d = \frac{1+x+2y+3z}{6}$ as a function of $x$ alone:
\[
\bar d = \frac{x+1}{12} \left(2+4 \sqrt{1-x}+3 \sqrt{2}\sqrt{1-x} \sqrt{\left(x^3+2 x^2-x+2\right)-x \sqrt{x+3} \sqrt{x^3+x^2-x+3}} \right)
\]We can then plot this function of $x$, and we get the following curve:
Interestingly, this function reaches its maximum just a bit after $x=0$. Zooming in, we can see more clearly:
The maximum occurs at about $x=0.0555108$ and the maximum value is $1.00229$, just as we found before. If we try to solve for this analytically by taking the derivative of this function, setting it equal to zero, and eliminating rationals, we obtain the following result… That the optimal $x$ a root of the following $30^\text{th}$ order polynomial:
\begin{align}
x^{30}&+\tfrac{152}{9} x^{29}+\tfrac{9259}{81} x^{28}+\tfrac{6851}{18} x^{27}+\tfrac{383363 }{648}x^{26}+\tfrac{1999495} {5832}x^{25}\\
&+\tfrac{43478263}{52488} x^{24}+\tfrac{75726373}{23328} x^{23}+\tfrac{3282369305}{1679616} x^{22}-\tfrac{31642768891}{7558272} x^{21}\\
&+\tfrac{367775655235}{136048896} x^{20}+\tfrac{392027905801}{34012224} x^{19}-\tfrac{2016793647847}{136048896} x^{18}-\tfrac{1258157703385}{68024448} x^{17}\\
&+\tfrac{4545130566235}{136048896} x^{16}-\tfrac{22059424877}{17006112} x^{15}-\tfrac{8448216227365}{136048896} x^{14}+\tfrac{2878062707167}{68024448} x^{13}\\
&+\tfrac{7398046956073}{136048896} x^{12}-\tfrac{312655708903}{3779136} x^{11}+\tfrac{17266856539}{1679616} x^{10}+\tfrac{63670102441}{839808} x^9\\
&-\tfrac{29827845749}{559872} x^8-\tfrac{541401565}{23328} x^7+\tfrac{619158287}{23328} x^6+\tfrac{6168305}{2592} x^5\\
&-\tfrac{156285}{32} x^4-\tfrac{5153}{36} x^3+\tfrac{3923}{12} x^2+\tfrac{45}{2} x-\tfrac{35}{16}
\end{align}Yuck! Here is what you get if you plot this polynomial; as expected, there is a root at $x=0.0555108$.
As you may know, having one child, let alone many, is a lot of work. But Alice and Bob realized children require less of their parents' time as they grow older. They figured out that the work involved in having a child equals one divided by the age of the child in years. (Yes, that means the work is infinite for a child right after they are born. That may be true.)
Anyhow, since having a new child is a lot of work, Alice and Bob don't want to have another child until the total work required by all their other children is 1 or less. Suppose they have their first child at time T=0. When T=1, their only child is turns 1, so the work involved is 1, and so they have their second child. After roughly another 1.61 years, their children are roughly 1.61 and 2.61, the work required has dropped back down to 1, and so they have their third child. And so on.
(Feel free to ignore twins, deaths, the real-world inability to decide exactly when you have a child, and so on.)
Five questions: Does it make sense for Alice and Bob to have an infinite number of children? Does the time between kids increase as they have more and more kids? What can we say about when they have their Nth child — can we predict it with a formula? Does the size of their brood over time show asymptotic behavior? If so, what are its bounds?
Let's call $t_k$ the time at which Alice and Bob have their $k^\text{th}$ child. The first child is born at time zero, so $t_1 = 0$. Since the work due to a child is one divided by their age, the work due to child $k$ at time $t$ is given by:
\[
\text{work at time }t = \begin{cases}
\frac{1}{t-t_k} & \text{if }t > t_k \\
0 & \text{otherwise}
\end{cases}
\]We also know that each new child is born once the total work due to all previous children equals $1$. Therefore, we have the following equations:
\begin{align}
t_2\text{ satsifies:} && \frac{1}{t_2-t_1} &= 1 \\
t_3\text{ satisfies:} && \frac{1}{t_3-t_2} + \frac{1}{t_3-t_1} &= 1 \\
t_4\text{ satisfies:} && \frac{1}{t_4-t_3} + \frac{1}{t_4-t_2} + \frac{1}{t_4-t_1} &= 1 \\
\cdots\qquad \\
t_k\text{ satisfies:} && \sum_{i=1}^{k-1} \frac{1}{t_k-t_i} &= 1
\end{align}and so on. We also know that $t_1 \lt t_2 \lt t_3 \lt \ldots$ since the children are born sequentially. The nice thing about these equation is that they can be solved sequentially. We know that $t_1=0$. Substituting this into the first equation yields $t_2=1$. Substituting these values into the second equation yields:
\[
\frac{1}{t_3-1}+\frac{1}{t_3} = 1
\]Rearranging yields $t_3^2-3t_3+1=0$. This equation has two solutions, but we take the one that satisfies $t_3 \gt t_2$, and we obtain $t_3=\tfrac{3+\sqrt{5}}{2} \approx 2.618$. We can continue in this fashion, solving each subsequent equation by substituting the previous values… but things get complicated in a hurry. The next equation becomes:
\[
t_4^3-\left(\tfrac{11+\sqrt{5}}{2}\right)t_4^2+\left(\tfrac{13+3\sqrt{5}}{2}\right)t_4-\left(\tfrac{3+\sqrt{5}}{2}\right)=0
\]Once again, we find that there is a unique solution that satisfies $t_4 \gt t_3$. This solution is approximately $t_4\approx 4.5993$. There doesn't appear to be any automatic way to continue this process. Every time we try to compute the next $t_k$, we have to find the root of an even more complicated polynomial!
We'll take a numerical approach, but let me first address one issue: how can we be sure that these ever-more-complicated polynomials always have a root that satisfies our constraints? For example, what if the polynomial for $t_4$ had no roots larger than $t_3$? What if it had many such roots? Thankfully, there is a simple way we can guarantee that there is always exactly one admissible root to each polynomial by leveraging the intermediate value theorem. Specifically, the function
\[
f_k(t) = \frac{1}{t-t_1} + \frac{1}{t-t_2} + \cdots + \frac{1}{t-t_{k-1}}
\]with $t \gt t_{k-1} \gt \ldots \gt t_1$ is a continuous and strictly decreasing function. Moreover, $\lim_{t\to {t_{k-1}}^+}f_k(t) = +\infty$ and $\lim_{t\to\infty} f_k(t) = 0$. Therefore, by the intermediate value theorem, there must be some point $t_k \in (t_{k-1},\infty)$ such that $f_k(t_k)=1$. The strict monoticity of $f_k$ ($f_k$ is strictly decreasing) implies that this $t_k$ must be unique.
The root may be unique, but how can we find it efficiently? Here, we can leverage a further property of $f_k$; namely convexity. The fact that $f_k$ is convex (and smooth) means that we can leverage some efficient methods from convex analysis that make use of derivative information. There are many ways to get the answer from here… What I ended up doing was to use the L-BFGS algorithm on the objective $(f_k(t)-1)^2$. My implementation in Julia using the Optim.jl package took about 40 seconds to evaluate birth dates of the first 1,000 children.
Note: there are many ways to formulate this as an optimization problem. One way is to note that solving $f_k(t)=1$ is equivalent to extremizing the antiderivative of $f_k(t)-1$. In other words, we could solve:
\[
t_k = \underset{t \gt t_{k-1}}{\text{arg min}} \left( t-\sum_{i=1}^{k-1}\log(t-t_i) \right)
= \underset{t \gt t_{k-1}}{\text{arg max}}\,\, e^{-t}\prod_{i=1}^{k-1}(t-t_i)
\]Incidentally, this last expression for $t_k$ makes it clear that there should be a unique $t_k$. When $t\gt t_{k-1}$, the product is a polynomial that is strictly increasing, and it's multiplied by $e^{-t}$, which will drive it to zero eventually. There must therefore be a unique maximizer. I tried implementing both of these approaches and both turned out to be slower than directly minimizing $(f_k(t)-1)^2$.
If you're just interested in the answers to the questions, here they are: [Show Solution]
Does it make sense for Alice and Bob to have an infinite number of children? Biological limitations aside, yes! We proved that for each $t_k$, there exists a unique $t_{k+1} \gt t_k$ that satisfies the constraints of the problem. So by induction, Alice and Bob can have as many children as they like.
Does the time between kids increase as they have more and more kids? The answer is yes, and I will provide a "proof by picture". After finding $t_k$ for $k=1,2,\dots$, we know the birth dates of each child. Then it remains to compute $(t_{k+1}-t_k)$ for each $k$ (the time between births). Here is the plot
The first point on the left says that we wait $1$ year after child $1$ is born, we wait about $1.6$ years after child $2$ is born, and so on. Plotted on a log scale as above, we see the curve is almost a perfect straight line. Here is a formula we can use to approximate the number of years to wait:
So yes, the time between children increases (without bound), but it does so slowly. By the time Alice and Bob have their 1,000th child, they will be waiting approximately 7.5 years between kids.
What can we say about when they have their $N^\text{th}$ child? Is there a formula we can write down? I couldn't find an analytic expression, but I did find a pretty good approximation. First, let's make a plot to see what we're working with. Here is what the birth years is like for the first 19 children
We can see here that the growth rate is slightly faster than linear. Since the time between children fit so well to a curve of the form $a+b\log x$, it makes sense to fit the birth dates themselves to the integral of this form. In an effort to fit the simplest possible curve, I did a bit of experimenting and found that the form $a x + x \log x$ does a really good job. I found the equation of best fit using linear regression and put the equation in the plot above. If we extend the time horizon, we get a similar result:
This is a pretty good fit; the plot actually depicts the true data with the fit overlaid on top. The fit is so good that you can't even see the data underneath! So for large $N$, a good approximation is:
The final question is whether the size of the brood over time shows asymptotic behavior. We already plotted the time of birth vs child number, and found the relationship to be approximately: $y \approx -0.315 N + N\log N$. So here, we need to look at the inverse relationship and ask what $N$ does as a function of $y$. Since $y$ grows a little faster than linearly, this means $N$ will grow a little more slowly than linearly.
A farmer owns a circular field with radius R. If he ties up his goat to the fence that runs along the edge of the field, how long does the goat's tether need to be so that the goat can graze on exactly half of the field, by area?
There are many ways to solve this problem, so I figured I would pick an unorthodox way: using calculus! Consider the diagram below.
The circular field is centered at $O$. The goat is tethered at $C$ and the tether has length $r$. Let $B$ be the point diametrically opposed to $C$ and let $A$ be the farthest point along the perimeter that the goat can reach. Let $\theta$ be the angle $BOA$, as shown. If $r$ changes a little bit, by some amount $dr$, then $\theta$ will change by some amount $d\theta$. Likewise, the fraction of the total area reachable by the goat will change by $dA$. Let's calculate how these quantities are related.
First, note that $dA$ is simply the area between the two circular arcs centered at $C$ and passing through $A$ and $A'$, respectively. We must also divide by the total area of the field in order to get a ratio. As $dr\to 0$, we have:
\[
dA = \frac{\tfrac{1}{2} \theta\, d(r^2)}{\pi R^2}
\]Our next task is to find out how $d\theta$ is related to $d(r^2)$. Look at the triangle $OAC$. By the law of cosines, we have:
$r^2 = 2R^2(1 + \cos(\theta))$. Taking differentials, we obtain:
\[
d(r^2) = -2R^2 \sin(\theta)\,d\theta
\]Substituting back into our expression for $dA$, we obtain:
\[
dA = -\tfrac{1}{\pi} \theta\,\sin(\theta)\,d\theta
\]We can now integrate! Note that $A=1$ when $\theta=0$. Therefore, we have:
\begin{align}
A(\theta) &= 1 + \int_{0}^\theta \tfrac{1}{\pi} t\, \sin(t)\,dt \\
&= \frac{\pi + \theta\,\cos(\theta)-\sin(\theta)}{\pi}
\end{align}This tells us the ratio of areas as a function of $\theta$. If we want this as a function of $r$ instead, we can use the law of cosines again to eliminate $\theta$ and write $A$ as a function of $r$. This yields:
\[
A(\rho) = 1+\frac{\left(\frac{\rho ^2}{2}-1\right) \cos ^{-1}\left(\frac{\rho ^2}{2}-1\right)}{\pi }-\frac{\rho \sqrt{4-\rho ^2}}{2\pi}
\]where I made the substitution $\rho := r/R$. We can plot this quantity as a function of $\rho$, and we get:
As we might expect, when $r=0$, $A=0$. And the area ratio grows monotonically with $r$ until we reach $r=2R$, at which point the goat can reach the farthest point (which is point $B$) and therefore can reach the entire field. There is no closed-form expression for the exact tether length that leads to a particular area ratio, since that would require solving the above $A(\rho)$ for $\rho$. However, we can easily solve the equation numerically. Doing so, we obtain: | 677.169 | 1 |
Area of Triangle Problem | AMC-10A, 2009 | Problem 10
Area of the Triangle- AMC-10A, 2009- Problem 10
Triangle $ABC$ has a right angle at $B$. Point $D$ is the foot of the altitude from $B$, $AD=3$, and $DC=4$. What is the area of $\triangle ABC$?
\(8\)
\(7\sqrt 3\)
\(8\sqrt 3\)
Key Concepts
Geometry
Triangle
similarity
Check the Answer
Answer: \(7\sqrt 3\)
AMC-10A (2009) Problem 10
Pre College Mathematics
Try with Hints
We have to find out the area of \(\triangle ABC\).now the given that \(BD\) perpendicular on \(AC\).now area of \(\triangle ABC\) =\(\frac{1}{2} \times base \times height\). but we don't know the value of \(AB\) & \(BC\).
Given \(AC=AD+DC=3+4=7\) and \(BD\) is perpendicular on \(AC\).So if you find out the value of \(BD\) then you can find out the area .can you find out the length of \(BD\)?
Can you now finish the problem ..........
If we proof that \(\triangle ABD \sim \triangle BDC\), then we can find out the value of \(BD\) | 677.169 | 1 |
Lattice constant
Physical dimensions of unit cells in a crystal / From Wikipedia, the free encyclopedia
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A lattice constant or lattice parameter is one of the physical dimensions and angles that determine the geometry of the unit cells in a crystal lattice, and is proportional to the distance between atoms in the crystal. A simple cubic crystal has only one lattice constant, the distance between atoms, but in general lattices in three dimensions have six lattice constants: the lengths a, b, and c of the three cell edges meeting at a vertex, and the angles α, β, and γ between those edges.
Unit cell definition using parallelepiped with lengths a, b, c and angles between the sides given by α, β, γ[1]
The crystal lattice parameters a, b, and c have the dimension of length. The three numbers represent the size of the unit cell, that is, the distance from a given atom to an identical atom in the same position and orientation in a neighboring cell (except for very simple crystal structures, this will not necessarily be distance to the nearest neighbor). Their SI unit is the meter, and they are traditionally specified in angstroms (Å); an angstrom being 0.1 nanometer (nm), or 100 picometres (pm). Typical values start at a few angstroms. The angles α, β, and γ are usually specified in degrees. | 677.169 | 1 |
Unit Circle: Sine and Cosine Functions
This sketch shows how we can create a function for sine and cosine.
As a point goes around the unit circle, we can see how the different ratios change based on the angle, given here as and measured in radians.
First, select Sine and then Animate. At each of the points on the traced out curve, the x-coordinate represents the angle and the y-coordinate represents the sine of the angle.
Notice that the y-coordinates of and the point we are tracing are always the same.
In this way, we can see how to graph the function:
Cosine can be similarly graphed, but here we are plotting the -coordinate of for each angle . | 677.169 | 1 |
Understanding the Concepts of Regular Polygons
The Beauty of Pentagons
Pentagons are fascinating polygons with five sides and five vertices. The shape and classification of pentagons can vary, with some being convex and others concave. Convex pentagons have interior angles that...
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The Beauty of Pentagons
Pentagons are fascinating polygons with five sides and five vertices. The shape and classification of pentagons can vary, with some being convex and others concave. Convex pentagons have interior angles that are all less than 180°, while concave pentagons have at least one angle greater than 180°. When all sides and interior angles of a pentagon are equal, it is considered regular. However, if a pentagon has equal sides but is concave, it is called equilateral. Any pentagon that does not fall under these classifications is irregular.
Types of pentagons
The sum of the internal angles of a pentagon, regardless of its classification, is always constant at 540°. This can be proved by dividing the pentagon into three non-overlapping triangles. No matter the arrangement of these triangles, their internal angles will always add up to 540°.
Exploring the Properties of Regular Pentagons
Symmetry: An Axis of Beauty
Regular pentagons possess a unique symmetry. They have five axes of symmetry that pass through a vertex and the midpoint of the opposite edge. These axes intersect at the center of the pentagon, which is also its center of gravity or centroid.
Axes of symmetry of regular pentagon
Interior and Central Angles: A Perfect Pair
In regular pentagons, all interior angles are equal. The sum of these interior angles is always 540°. Therefore, each interior angle, denoted as φ, measures 108°. Furthermore, by drawing lines from the center of the pentagon to each vertex, we create five identical triangles. The central angle, denoted as θ, of each triangle is 72°.
In each triangle, the remaining two angles are identical, measuring 54°. The sum of all angles in the triangle is 180° (72° + 54° + 54°), which is also half of the interior angle φ (108°/2 = 54°). Importantly, the sum of the interior angle φ and the central angle θ is always 180°. In other words, φ and θ are supplementary angles.
Interior and central angle of a regular pentagon
Circumcircle and Incircle: Encircling the Beauty
A regular pentagon possesses two significant circles. The circumcircle, or the circumscribed circle, passes through all five vertices of the pentagon. Its center coincides with the center of the pentagon, where all axes of symmetry intersect. On the other hand, the inscribed circle, or the incircle, touches tangentially to all five edges of the pentagon at their midpoints. The centers of both circles are the same, lying at the center of the pentagon. The radius of the circumcircle is called the circumradius, while the radius of the incircle is the inradius.
Circumcircle and incircle of a regular pentagon
The relationships between the side length (a) of a regular pentagon and its circumradius (Rc) and inradius (Ri) can be determined. Using basic trigonometry, the following expressions can be derived for any regular polygon:
Rc = a / (2 sin(θ/2))
Ri = a / (2 tan(θ/2))
Ri = Rc cos(θ/2)
For the regular pentagon specifically, with θ = 72°, the approximations for these expressions are:
Rc ≈ 0.851a
Ri ≈ 0.688a
Ri ≈ 0.809Rc
Area and Perimeter: Measuring the Beauty
To find the area of a regular pentagon, we divide it into five identical isosceles triangles. Each triangle has one side (a) and two sides equivalent to the circumradius (Rc). The height of each triangle, cast from the vertex lying at the pentagon center, is equal to the inradius (Ri). Therefore, the area of each triangle is (1/2) a Ri. The total area of the five triangles is:
A = (5/2) a Ri
A ≈ 1.720 * a^2
The perimeter of any N-sided regular polygon is simply the sum of the lengths of all sides: P = N * a. For the regular pentagon, the perimeter is 5a.
Area and perimeter of a regular pentagon
The Bounding Box: The Perfect Fit
The bounding box of a regular pentagon is the smallest rectangle that completely encloses the shape. While the dimensions of the bounding box can be intuitively drawn, finding the exact measurements requires some calculations.
Height: Reaching New Heights
The height (h) of a regular pentagon is the distance from one vertex to the opposite edge. It is perpendicular to the opposite edge and passes through the center of the pentagon. By definition, the distance from the center to a vertex is the circumradius (Rc), and the distance from the center to an edge is the inradius (Ri). Therefore, the expression for the height is:
h = Rc + Ri
The height can be expressed in terms of the circumradius (Rc), inradius (Ri), or side length (a) using the respective analytical expressions. The following formulas can be derived:
h = Rc * (1 + cos(θ/2))
h = Ri * (1 + (1/cos(θ/2)))
h = (a/2) * (1 + cos(θ/2)) / sin(θ/2)
Approximating the value of θ to 72°, we obtain the following approximations:
h ≈ 1.809 * Rc
h ≈ 2.236 * Ri
h ≈ 1.539 * a
Width: Expanding Horizons
The width (w) of a regular pentagon is the distance between two opposite vertices (the length of its diagonal). To find this distance, we can use a right triangle formed by extending one side of the pentagon. The hypotenuse of this right triangle is the side length (a) of the pentagon. By considering the adjacent interior angle φ, which is supplementary to the central angle θ, we can calculate the length (w1) of the triangle side:
w1 = a * cos(θ)
Finally, the total width (w) can be obtained by adding twice the length (w1) to the side length (a), as the triangle to the right of the pentagon is identical to the one examined:
w = a + 2 a cos(θ)
Approximating θ to 72°, we get the following approximation:
w ≈ 1.618 * a
Drawing a Regular Pentagon: Unleash Your Creativity
Drawing a regular pentagon with a given side length (a) is achievable using simple drawing tools. Follow these steps:
Draw a line segment with a length equal to the desired side length (a).
Extend the line segment to the left.
Construct a circular arc with its center at the right end of the line segment and a radius equal to the segment's length.
Repeat the previous step, but this time with the center point at the left end of the line segment.
Draw a line perpendicular to the line segment, passing through the intersection of the two arcs. The line should cross the line segment at its midpoint.
Draw another line perpendicular to the line segment, passing through the left end of the line segment. Mark the intersection point with the circular arc drawn in step 4.
Draw a new circular arc by placing one needle of the compass at the midpoint of the line segment (found in step 5) and the drawing tip at the intersection marked in step 6. Rotate the compass until it intersects with the extended line segment drawn in step 2. Mark this new intersection.
Draw another circular arc by placing one needle of the compass at the right end of the line segment and the drawing tip at the intersection marked in step 7. Rotate the compass clockwise, marking two intersections: one with the arc drawn in step 4, and another with the line drawn in step 5. These two intersections represent two vertices of the pentagon.
Using the second intersection as the compass needle and the first one as the drawing tip (both intersections marked in the previous step), draw a circular arc that intersects with the arc drawn in step 3. Mark this new intersection as another vertex of the pentagon.
The two ends of the line segment, along with the three intersections marked in steps 8 and 9, represent the five vertices of the regular pentagon. Connect them with straight lines to complete the shape.
Drawing a regular pentagon given its side length (a)
Please note that the described procedure is not strictly a "ruler and compass" construction. Steps 5 and 6 involve the use of a triangle to draw perpendicular lines. However, if a strict geometric drawing by "ruler and compass" is required, the use of a triangle can be replaced with the method for constructing perpendicular lines using a ruler and compass alone.
The Beauty of Regular Polygons: A Quick Reference
Here is a concise list of the main formulas and helpful approximations related to regular pentagons:
Circumradius (Rc) ≈ 0.851a
Inradius (Ri) ≈ 0.688a
Height (h) ≈ 1.539a
Width (w) ≈ 1.618a
Area (A) ≈ 1.720a^2
Refer to this quick reference to explore the many possibilities and properties of regular pentagons | 677.169 | 1 |
17 Angles
Angle Measure
Angle measurement is important in construction, surveying, physical therapy, and many other fields. We can visualize an angle as the figure formed when two line segments share a common endpoint.
We can also think about an angle as a measure of rotation. One full rotation or a full circle is , so a half rotation or U-turn is , and a quarter turn is . We often classify angles by their size relative to these and benchmarks.
When lines cross, they form angles. No surprises there. If we know the measure of one angle, we may be able to determine the measures of the remaining angles using a little logic.
Exercises
Find the measure of each unknown angle.
Angles in Triangles
If you need to find the measures of the angles in a triangle, there are a few rules that can help.
The sum of the angles of every triangle is .
If any sides of a triangle have equal lengths, then the angles opposite those sides will have equal measures.
Exercises
Find the measures of the unknown angles in each triangle.
Angles and Parallel Lines
Two lines that point in the exact same direction and will never cross are called parallel lines. If two parallel lines are crossed by a third line, sets of equally-sized angles will be formed, as shown in the following diagram. All four acute angles will be equal in measure, all four obtuse angles will be equal in measure, and any acute angle and obtuse angle will have a combined measure of .
Exercises
Find the measures of angles , , and .
Degrees, Minutes, Seconds
It is possible to have angle measures that are not a whole number of degrees. It is common to use decimals in these situations, but the older method—which is called the degrees-minutes-seconds or DMS system—divides a degree using fractions out of : a minute is of a degree, and a second is of a minute, which means a second is of a degree. (Fortunately, these units work exactly like time; think of degree as hour.) For example, . | 677.169 | 1 |
Basic Proportionality Theorem (BPT) – If a line is parallel to a side of a triangle which intersects other two sides in distinct points, then the line divides other two sides in proportion.
Pythagorean Theorem – If a line is parallel to a side of a triangle which intersects other two sides in distinct points, then the line divides other two sides in proportion.
PERIMETER OF TRIANGLE – The perimeter of a triangle is the total distance covered by a triangle which can be calculated by adding all the sides of the triangle. P = a + b +c
AREA OF TRIANGLE – The area of a triangle can be defined as the total space or region which is enclosed inside any types of triangle.
Triangles (Fashion Themed) Worksheets
This is a fantastic bundle which includes everything you need to know about Triangles across 21 in-depth pages. These are ready-to-use Common core aligned 5th and 6 | 677.169 | 1 |
A Regular Polygon with Interior Angle of 165 Degrees
A regular polygon is a polygon where all sides are equal in length and all angles are equal. The sum of the interior angles of any polygon can be calculated using the formula (n-2) x 180 degrees, where n is the number of sides in the polygon.
In the case of a regular polygon with an interior angle of 165 degrees, we can determine the number of sides by dividing 360 degrees by the interior angle. Therefore, 360 degrees ÷ 165 degrees ≈ 2.182. Since a polygon cannot have a fraction of a side, we round this number down to the nearest whole number, which means our polygon has 2 sides or, more commonly expressed, is a line segment.
A polygon with an interior angle of 165 degrees is not a usual case for regular polygons, as such a large interior angle would result in less than 3 sides for the polygon, making it theoretically impossible to form a closed shape. Regular polygons typically have interior angles that are more common, such as 60, 90, 120, or 144 degrees.
For a more practical scenario, let's consider a regular polygon with an interior angle of 165 degrees. In this case, the closest common interior angle for a regular polygon would be 150 degrees for a pentagon (5 sides). A regular pentagon has interior angles of 108 degrees, so a polygon with interior angles of 165 degrees would be irregular, given the sum of the interior angles must still equal (n-2) x 180 degrees.
Properties of a Regular Polygon with an Interior Angle of 165 Degrees
Let's delve into the properties of a regular polygon with an interior angle of 165 degrees in more detail.
1. Definition
A regular polygon with an interior angle of 165 degrees is a closed geometric shape where all sides are equal in length and all interior angles are equal to 165 degrees.
2. Unique Characteristics
The sum of all interior angles in a regular polygon with 165 degrees is still a multiple of 180 degrees.
The exterior angle of the polygon would be 15 degrees (180 degrees – 165 degrees).
The number of sides can be calculated by dividing 360 by 165, yielding a non-whole number.
3. Construction Challenges
Constructing a regular polygon with an interior angle of 165 degrees poses a challenge due to the unusual angle measurement, making it difficult to create a symmetrical shape with equal sides and angles.
4. Real-world Applications
Regular polygons with uncommon interior angles like 165 degrees are not commonly encountered in real-world applications due to their irregular nature. However, irregular polygons can be found in architecture, art, and design.
FAQs:
Can a regular polygon have an interior angle of 165 degrees?
While theoretically possible, a regular polygon with an interior angle of 165 degrees would not fit the standard definition of a regular polygon due to the unusual angle measurement.
What is the formula to calculate the sum of interior angles in a polygon?
The formula is (n-2) x 180 degrees, where n represents the number of sides in the polygon.
How many sides would a regular polygon with an interior angle of 165 degrees have?
By dividing 360 degrees by 165 degrees, the result is approximately 2.182 sides, which is not feasible for a polygon.
What is the closest common interior angle for a regular polygon with 165 degrees?
The closest common interior angle would be 150 degrees, often seen in a regular pentagon with 5 sides.
What practical applications involve irregular polygons with unusual interior angles?
Irregular polygons with uncommon interior angles can be found in artistic designs, architectural layouts, and specialized geometric constructions where symmetry is not a primary concern | 677.169 | 1 |
rhombus formula ssc
Only then you can score marks in all the topics. Mensuration formulas PDF Download is available here. A: Area of a rhombus = (d1.d2)/2, where d1 and d2 are the lengths of diagonals of the rhombusPerimeter of rhombus = 4 x Side of rhombus. Solution: Mensuration formulas based questions regularly feature in CAT, SSC CGL and other exams. Ex: square, rectangle, triangle. The diagonals of a rhombus are unequal and bisect each other at right angles. In this article, we have provided all the important properties of rhombus along with formulas related to rhombus. Formula of Perimeter of Rhombus. Also for TNPSC exams are also it will be useful. The reason is they may don't know the formula. You can read the properties here or download them as a PDF provided below for offline access. A: The opposite angles of a rhombus are equal to each other. Area = ah. For 2d geometric shapes, usually there are formula for area and perimeter. In trapezium, one set of opposite sides are parallel and unequal. A: If you understand the concept of geometric shapes, then it is easy for you to memorize all mensuration formulas. Latest Syllabus 2019 - 2020. Altitude or height =. This is one of the special properties of rhombus which is very helpful in many mathematical calculations. Diagonals bisect vertex angles. The properties of rhombus for Class 9 is one of the most important topics for CBSE Class 9 students as they are asked frequently in the final examination. As in the previous blog of SSC CGL Tier-1 30 Days Study Plan, we have discussed the time allocation strategy during the exam.. The length of AC and BD is d1 and d2 respectively. Latest Syllabus 2019 - 2020. every year the cut-offs are reaching sky-high. There are mainly 6 types of quadrilateral: A rhombus is a special type of parallelogram whose all four sides are equal. Cylinder, Circles, Polygons, Rectangles and Squares, Trapezium, Parallelogram and Rhombus, Area and Perimeter, Cube and Cuboid Formulas is very important to crack SSC Competitive … All the sides of a rhombus are equal in length. Also the opposite sides are parallel to each other. It is a special kind of parallelogram whose diagonals intersect each other at 90°. Here we have added some FAQs regarding the mensuration formulas pdf. A blog about 9th, 10th, 11th and 12th Maharashtra, Tamilnadu, CBSE Board. It is a 3d shape of a rectangle. Here, d1 and d2 are the lengths of the diagonals. Here we have provided some practice questions related to rhombus for you to practice. So we have provided mensuration formulas pdf. Mensuration Formulas … The sum of the interior angles of a quadrilateral is equal to 360°. Here, l = length of rectangle and b = breadth of rectangle. You will get solid with two cones attached to their bases when the rhombus is revolving about the longer diagonal as the axis of rotation. It involves Mensuration Formula And Questions For 2D and 3D Shapes: Check out the basic maths mensuration formulas along with the important questions. Solution: Area of rhombus = 121 cm2 (Given) d1 = 22 cm.Area of the rhombus, A = (d1 x d2)/2, we get121 = (22 x d2)/2121 = 11 x d2or 11 = d2So, the length of the other diagonal is 11 cm. So study all in mensuration formula PDF. Mensuration Formula And Questions Mensuration is the branch of mathematics that deals with measurements of different figures and shapes of geometry. Solution: Given: Area of rhombus = 121 cm 2 and Lets say d 1 = 22 cm. Find the distance between each of the following pairs of points. So, mensuration formulas pdf is essential. So memorize these mensuration formulas PDF for CAT, TNPSC, banking, SSC & other exams. The main factor is that all 4 sides are equal in the rhombus. The perimeter of a rhombus = 4 × a The perimeter of a rhombus = 4 × 5 The perimeter of a rhombus = 20 cm There are 3 ways to find the area of Rhombus. Take the rhombus ABCD: Sides: AB, BC, CD, and ADLength of Each Side: a Diagonals: AC, BDLength of Diagonals: d1, d2. Aspirants preparing for the various competitive exams will search for mensuration all formulas. a is the length of the side. Where d1 and d2 are two-diagonals. Before we discuss rhombus, let us understand what is a quadrilateral. This is one of the most important properties of parallelogram that is helpful in solving many mathematical problems related to 2-D geometry. The quadrilateral will have totally 4 sides. Rhombus has: All … A rhombus is either an equilateral triangle or a slanting square whose sides are equal and the area can be calculated by multiplying both diagonals together and divide the value by two. Many candidates will skip the mensuration questions in the exams. Sometimes you see a rhombus with two sides horizontal, as if a square has been run into by a bus and tilted over (that is a handy mnemonic to remember its name: run, bus; rhombus). Product of its diagonals. There are various types of triangles. The plural form of a rhombus is rhombi or rhombuses. Find the area of the rhombus having each side equal to 17 cm and one of its diagonals equal to 16 cm. if you know the formula, mostly all the mensuration problems are easy to solve. Nowadays, competitive exams are very difficult to crack. h = height of parallelogram. In group 4 exams, the mensuration questions are in basic and easy level. Averages An average is the sum of a list of entities divided by the number of entities in the list. In that mensuration is a topic that involves a lot of formula. since d 2 2 since d 2 2 Perimeter (P) of a rhombus = 4a, i.e. Students can make use of NCERT Solutions for Maths provided by Embibe for your exam preparation. | 677.169 | 1 |
2 Answers
2
Two arguments, but not exactly like atan2()
The first thing you should understand is that although ARCTAN is called in 3 places in the Apollo code, in every case it is used for converting rectilinear coordinates (x, y, z) into spherical coordinates (r, latitude, longitude).
I will describe how ARCTAN is used in the LAT-LONG subroutine, in file LATITUDE_ LONGITUDE_ SUBROUTINES.agc, although the use is similar in the other callers. The first step is to compute gamma = $\sqrt{x^2+y^2}$ (lines 77-82). We then divide x and y each by gamma; x/gamma is stored in global variable COSTH and y/gamma is stored in global variable SINTH (lines 83-87). We are then ready to call the ARCTAN subroutine (line 88). After it returns, the result is stored in global variable LAT (line 89). A similar set of calculations determines the longitude.
Like atan2(), ARCTAN takes two arguments. However, unlike atan2(), the arguments must already be pre-divided. In other words, they must be in the range of -1 to +1, and their squares must add to 1. The arguments are taken from the global variables COSTH and SINTH. As the names suggest and because of the pre-division, these arguments are the cosine and sine of the angle we are looking for. The bulk of the subroutine checks the +/- sign of the arguments, adjusting the result for the appropriate quadrant. The main calculation is done by calling ASIN with argument SINTH (line 219).
ASIN is a macro for the ARCSIN subroutine in file INTERPRETER.agc. In turn, it is computed from the arccosine, using the relation ARCSIN(X) = PI/2 - ARCCOS(X). (Lines 2704-2705. Yeah, even though we already have the cosine stored in COSTH.) ARCCOS is calculated using the "Hastings approximation" and a 7th degree polynomial.
The result of each of the inverse trig subroutines are angles in the range of -1/2 to +1/2. To convert to radians, you must multiply by $\pi$ .
So, there are two arguments, but they must be pre-divided, and the result must be multiplied by $\pi$ .
$\begingroup$@MagicOctopusUrn: RVQ is Return via QPRET. Returns to the location specified in the QPRET register. This is the normal way to return from a subroutine, although if the QPRET register had previously been saved to a memory location, then GOTO can be used instead. BZE seems to use the previously computed value gamma = sqrt(x^2+y^2) to find out if both x and y are zero.$\endgroup$
$\begingroup$@Hobbes There are different assembly languages, at least one per type of processor. I'm not sure what language the code is using, but it's not one I'm familiar with, and I've been around for a while. I can't tell what algorithm the code is using.$\endgroup$
$\begingroup$RobinC welcome to Stack Exchange! It's a big different than other sites you may have used. You can get a feeling for what kinds of answers are well-received by looking around at other answers and have a look at How to Answer as well. The four-quadrant correct atan2 is really important in engineering applications and so it is certainly implemented somehow If ARCTAN doesn't do that (you have not yet said if it does or doesn't) then this will be handled by whatever calls ARCTAN and so your answer should expand on that.$\endgroup$
$\begingroup$Right now it's called a "link-only" answer and this is strongly discouraged (and down-voted) because links break and then the answer becomes useless and the question becomes unanswered again. The idea is to write answers that are helpful to future readers as well as the OP.$\endgroup$ | 677.169 | 1 |
The Many Properties of a Kite
A kite is a geometric shape that has many properties that make it unique. In this blog post, we will explore some of those properties and how they can be used in geometry. Specifically, we will look at the properties of angle bisectors, perpendicular bisectors, and medians.
Angle Bisectors
An angle bisector is a line that passes through the vertex of an angle and bisects (divides) the angle into two equal parts. The angle bisector theorem states that the ratio of the lengths of the two parts of the line segment is equal to the ratio of the lengths of the corresponding sides of the triangle.
Perpendicular Bisectors
A perpendicular bisector is a line that passes through the midpoint of a line segment and is perpendicular to that line segment. The perpendicular bisector theorem states that if a point is on the perpendicular bisector of a line segment, then it is equidistant from the two endpoints of the line segment.
Medians
A median is a line segment that connects a vertex of a triangle to the midpoint of the opposite side. The median theorem states that if a point is on the median of a triangle, then it is equidistant from the two sides of the triangle.
Conclusion
Kites have many properties that make them useful in geometry. In this blog post, we explored three of those properties: angle bisectors, perpendicular bisectors, and medians. We also looked at how those properties can be used in geometry. By understanding these properties, students will be better equipped to tackle problems involving kites.
FAQ
What are the properties of a kite geometry?
There are many properties of kite geometry, but some of the most notable ones include the angle bisector theorem, the perpendicular bisector theorem, and the median theorem. | 677.169 | 1 |
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Geometry Reflections Worksheet
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Drawing Reflections Worksheet
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Transformation Geometry Worksheets 2nd Grade
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Reflections and Rotations INB Pages Mrs. E Teaches Math
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Reflections Worksheet 1 Answers
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Translation And Reflection Worksheet
Draw reflections of the following shapes. Each printable worksheet has eight practice problems. Web geometry reflection worksheets help students to identify the reflection of the given point from the given options. Reflections in geometry is an important concept from a mathematical point of view. Draw the reflected image using a protractor and ruler.
Web geometry reflections worksheet 3,700+ results sort: Web this transformations worksheet will produce problems for practicing reflections of objects. In a coordinate plane, an image will reflect through a line to. These worksheets focus on the line of symmetry, reflection, scale and nature.activities. Ad software for math teachers that creates custom worksheets in a matter of minutes. Create the worksheets you need with infinite geometry. Part two (25 minutes) hand out the reflections worksheet. Web geometry reflection worksheets help students to identify the reflection of the given point from the given options. Practicing these worksheets is also helpful for students to. Web free printable math worksheets for geometry created with infinite geometry stop searching. Web these symmetry in nature worksheets are perfect for your transformation math lessons. Reflection over the line y = x 14. Web translate, reflect or rotate the shapes and draw the transformed image on the grid. Reflections reflecting points reflect points determining reflections determine reflections determining reflections. Web reflections (flips) reflections (flips) grade 4 geometry worksheet draw reflections of the following shapes. Web results for reflections geometry 3,400 + results sort by: High school geometry > unit 1 lesson 5: Transformation of triangles draw the. Instruct students to work in groups of. List hs geometry transformations workbook ~ translations, rotations, & reflections created. | 677.169 | 1 |
What is a trapezium short answer?
A trapezium is a closed shape or a polygon, that has four sides, four corners/vertices and four angles. Anyone pair of opposite sides of a trapezium are parallel to each other.
What is a trapezium in math?
A trapezoid, also known as a trapezium, is a flat closed shape having 4 straight sides, with one pair of parallel sides. The parallel sides of a trapezium are known as the bases, and its non-parallel sides are called legs. A trapezium can also have parallel legs.
What is trapezium with example?
A two-dimensional quadrilateral that has a pair of non-adjacent parallel sides and a pair of non-parallel sides is referred to as a trapezium shape. It looks like a triangle that is cut from the top.
How many sides make a trapezium?
four sided
In the US (for some) a trapezium is a four sided polygon with no parallel sides; in the UK a trapezium is a four sided polygon with exactly one pair of parallel sides; whereas in Canada a trapezoid has an inclusive definition in that it's a four sided-polygon with at least one pair of parallel sides – hence …
What is trapezium Brainly?
Brainly User. Answer: the quadrilateral in which one pair of opposite sides is parallel is called a trapezium.
Why is a shape a trapezium?
A trapezoid (also known as a trapezium) is a flat 2D shape with four straight sides. It has one pair of parallel sides which are usually the top and bottom sides. The parallel sides are called the bases, while the non-parallel sides are called the legs.
Is square a trapezium?
The statement 'All squares are Trapezium' is true. A square has all the sides equal and a trapezium is a quadrilateral with at least one set of parallel sides. Hence all the squares are trapeziums.
What is trapezium draw its figure too?
a quadrilateral with one pair of sides parallel is known as trapezium…
Are all parallelograms rhombus?
A rhombus is a quadrilateral (plane figure, closed shape, four sides) with four equal-length sides and opposite sides parallel to each other. All rhombuses are parallelograms, but not all parallelograms are rhombuses.
What is trapezium and its properties?
Convex polygonTrapezoid / Properties
What are properties of a trapezium?
Is rhombus a trapezium?
The opposite parallel sides are referred to as the base and the non-parallel sides are referred to as legs of the trapezium. It is a closed plane shape having four sides and four corners. A rhombus has all properties of a trapezium. A rhombus can be a trapezium but not vice-versa.
What is a trapezium ks2?
A trapezium is a quadrilateral. It has one pair of parallel sides – the other two sides are not parallel. Sometimes, a trapezium is an isosceles trapezium, with two equal angles along one side.
What is a trapezium look like?
A trapezium is a shape that is 2D and is known by its property of having two opposite parallel sides, rather than all four like a square | 677.169 | 1 |
Usually the Cartesian coordinate system is applied to manipulate equations for planes, straight lines, and circles, often in two and sometimes three dimensions. Geometrically, one studies the Euclidean plane (two dimensions) and Euclidean space. As taught in school books, analytic geometry can be explained more simply: it is concerned with defining and representing geometric shapes in a numerical way and extracting numerical information from shapes' numerical definitions and representations. That the algebra of the real numbers can be employed to yield results about the linear continuum of geometry relies on the Cantor–Dedekind axiom. | 677.169 | 1 |
Unit vector
Unit vector
In mathematics, a unit vector in a normed vector space is a vector (often a spatial vector) whose length is 1 (the unit length). A unit vector is often denoted by a lowercase letter with a "hat", like this: (pronounced "i-hat").
In Euclidean space, the dot product of two unit vectors is simply the cosine of the angle between them. This follows from the formula for the dot product, since the lengths are both 1.
The normalized vector or versor of a non-zero vector is the unit vector codirectional with , i.e.,
where is the norm (or length) of . The term normalized vector is sometimes used as a synonym for unit vector.
The elements of a basis are usually chosen to be unit vectors. Every vector in the space may be written as a linear combination of unit vectors. The most commonly encountered bases are Cartesian, polar, and spherical coordinates. Each uses different unit vectors according to the symmetry of the coordinate system. Since these systems are encountered in so many different contexts, it is not uncommon to encounter different naming conventions than those used here.
Contents
Cartesian coordinates
In the three dimensional Cartesian coordinate system, the unit vectors codirectional with the x, y, and z axes are sometimes referred to as versors of the coordinate system.
These are often written using normal vector notation (e.g. i, or ) rather than the caret notation, and in most contexts it can be assumed that i, j, and k, (or and ) are versors of a Cartesian coordinate system (hence a tern of mutually orthogonal unit vectors). The notations , , , or , with or without hat/caret, are also used, particularly in contexts where i, j, k might lead to confusion with another quantity (for instance with index symbols such as i, j, k, used to identify an element of a set or array or sequence of variables). These vectors represent an example of a standard basis.
When a unit vector in space is expressed, with Cartesian notation, as a linear combination of i, j, k, its three scalar components can be referred to as direction cosines. The value of each component is equal to the cosine of the angle formed by the unit vector with the respective basis vector. This is one of the methods used to describe the orientation (angular position) of a straight line, segment of straight line, oriented axis, or segment of oriented axis (vector).
Cylindrical coordinates
The unit vectors appropriate to cylindrical symmetry are: (also designated or ), the distance from the axis of symmetry; , the angle measured counterclockwise from the positive x-axis; and . They are related to the Cartesian basis , , by:
=
=
It is important to note that and are functions of φ, and are not constant in direction. When differentiating or integrating in cylindrical coordinates, these unit vectors themselves must also be operated on. For a more complete description, see Jacobian matrix. The derivatives with respect to φ are:
Spherical coordinates
The unit vectors appropriate to spherical symmetry are: , the direction in which the radial distance from the origin increases; , the direction in which the angle in the x-y plane counterclockwise from the positive x-axis is increasing; and , the direction in which the angle from the positive z axis is increasing. To minimize degeneracy, the polar angle is usually taken . It is especially important to note the context of any ordered triplet written in spherical coordinates, as the roles of and are often reversed. Here, the American "physics" convention[1] is used. This leaves the azimuthal angle φ defined the same as in cylindrical coordinates. The Cartesian relations are:
The spherical unit vectors depend on both φ and θ, and hence there are 5 possible non-zero derivatives. For a more complete description, see Jacobian. The non-zero derivatives are:
Curvilinear coordinates
In general, a coordinate system may be uniquely specified using a number of linearly independent unit vectors equal to the degrees of freedom of the space. For ordinary 3-space, these vectors may be denoted . It is nearly always convenient to define the system to be orthonormal and right-handed:
where δij is the Kronecker delta (which is one for i = j and zero else) and is the Levi-Civita symbol (which is one for permutations ordered as ijk and minus one for permutations ordered as kji).
vector space — Math. an additive group in which addition is commutative and with which is associated a field of scalars, as the field of real numbers, such that the product of a scalar and an element of the group or a vector is defined, the product of two… … Universalium | 677.169 | 1 |
Quiz 10 1 intro to circles. 1 pt. In a circle, if a radius or diameter is perpendicular to a chord, then it____ the chord and its arc. equals. bisects. arcs. circles. 2. Multiple Choice. 30 seconds.
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