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A is a point at a distance 13 cm from the center ΔABC.
Given :Two tangents are drawn from an external point A to the circle with center O, Tangent BC is drawn at a point R.
Radius of circle equals 5cm and OA = 13 cm
OA = 13 cm
To Find :Perimeter ofΔABC.
∠OPA = 90°
[tangent at any point of a circle is perpendicular to the radius through the point of contact]
(OA)2 = (OP)2 + (PA)2
(13)2 = (5)2 + (PA)2
169 - 25 = (PA)2
(PA)2= 144
PA = 12 cm
As we know that, Tangents drawn from an external point to a circle are equal.
So we have
PB = BR [1] [Tangents from point B]
CR = QC [2] [Tangents from point C]
Now Perimeter of Triangle PCD
= AB + BC + CA
= AB + BR + CR + CA
= AB + PB + QC + CA [From 1 and 2]
= PA + QA
Now,
PA = QA = 12 cm as tangents drawn from an external point to a circle are equal | 677.169 | 1 |
6 ... AC is equal to AB ; ( def . 15 ) and because the point B is the centre of the circle ACE , therefore BC is equal to BA . But it has been proved that CA is equal to AB ; therefore CA , CB are each of them equal to AB ; but things which are ...
УелЯдб 7 ... equal to BC . K H D B E CONSTRUCTION From the point A to B draw the straight ... AL shall be equal to BC . DEMONSTRATION Because the point B is the centre of ... AL is equal to the remainder BG . ( ax . 3. ) But it has been shown that BC ...
УелЯдб 8 ... equal to C , the less . D B A E F CONSTRUCTION From the point A draw the straight line AD equal to C ; ( 1. 2 ) and ... AC , equal to the two sides , DE , DF , each to each , viz . AB to DE , and AC to DF ; and the angle BAC equal to the ...
УелЯдб 9 ... equal to DE , therefore the point B shall coincide with the point E ; and AB coinciding with DE , and the angle BAC being equal to the angle EDF , ( hyp . ) therefore AC shall coincide with DF ; and because AC is equal to DF , wherefore ...
УелЯдб 10 ... equal to one another ; and if the equal sides be produced , the angles upon the other side of the base shall be equal . ( References - Prop . I. 3 , 4 ; ax . 3. ) Let ABC be an isosceles triangle , of which the side AB is equal to AC | 677.169 | 1 |
Question 2.
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Solution:
Let AB be a chord which is equal to radius. Join OA and OB, where O is the centre of the circle.
∵ AO = OB = AB (given)
∴ AOB is an equilateral triangle.
⇒ ∠AOB = 60°
Reflex ∠AOB = 360° – 60° = 300°
∠AOB = 2 × ∠ACB [By theorem 10.8]
⇒ 60° = 2 × ∠ACB
⇒ ∠ACB = \(\frac{60^{\circ}}{2}\) = 30°
and Reflex (∠AOB) = 2 × ∠APB [By theorem 10.8]
⇒ 300° = 2 × ∠APB
⇒ ∠APB = \(\frac{300^{\circ}}{2}\) = 150°
Hence, angle subtended by the chord at a point on the minor arc is 150° and major arc is 30°.
Question 7.
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Solution:
Given: Diagonals AC and BD of cyclic quadrilateral are thediameters of the circle through the vertices of the quadrilateral ABCD.
To prove: Quadrilateral ABCD is a rectangle.
Proof: Since BD is a diameter.
∠C = ∠A = 90°
[Angle in a semicircle is 90°]
Similarly, AC is a diameter.
∠B = ∠D = 90°
[Angle in a semicircle is 90°]
Thus, in a quadrilateral ABCD
∠A = ∠B = ∠C = ∠D = 90°
Hence, quadrilateral ABCD is a rectangle
Proved.
Question 9.
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig. 10.40). Prove that ∠ACP = ∠QCD.
Solution:
∠ABP = ∠ACP ……(i)
[Angles in a same segment are equal]
∠QBD = ∠QCD …..(ii)
[Angles in a same segment are equal]
∠ABP = ∠QBD …..(iii)
(Vertically opposite angles)
From (i), (ii) and (iii), we get
∠ACP = ∠QCD. Hence proved
Question 10.
If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Solution:
Given: Two circles are drawn with sides AB and AC of ΔABC as diameters. The circles intersect each other at D.
To prove: Point D lies on BC i.e., BDC is a straight line.
Construction: Join AD.
Proof: Since AB and AC are diameters of two circles.
∴ ∠ADB = 90° …..(i)
[Angle in semicircle is 90°]
and ∠ADC = 90° …..(ii)
[Angle in a semicircle is 90°]
Adding (i) and (ii), we get
∠ADB + ∠ADC = 90° + 90°
⇒ ∠ADB + ∠ADC = 180°
⇒ BDC is a straight line.
Hence, D lies on BC. Proved. | 677.169 | 1 |
One reply on "Antennas above and below"
Let A, B, C denote (equal length) vectors from the center of the circle to those points. Then the vector sum A+B+C = H. (To show this we need to show that (H-A) is perpendicular to B-C, that H-C is perp to A-B, and that H-B is perp to A-C. But H-A = (A+B+C)-A = B+C, etc. So, we need to show that (A-B) is perp to A+B. But (A-B).(A+B) = A^2 – B^2 = |A|^2 – |B|^2 =0 because the vectors A, B, C all have the same length. Similarly, we show that (B-C).(B+C)=0 and (C-A).(C+A)=0.)
The antenna on the hill is represented by the vector H-A = B+C.
The vector of the "antenna in the ground" is B-E = B – (-C) = B+C. | 677.169 | 1 |
often encountered in more specialized applications forms the bottom of the cone the smallest face of the examples, showcasing their practical applications and relevance. Here are a few examples:
Ice Cream Cone
An ice cream cone is a classic example of a cone. The ice cream serves as the base, the cone itself forms the lateral surface, and the pointed top represents the apex. This delicious treat perfectly exemplifies the three faces of a cone.
Traffic Cone
A traffic cone, also known as a safety cone, is another common example of a cone. The circular base provides stability, while the bright-colored, conical shape ensures visibility. The apex of the traffic cone allows for easy stacking and transportation.
Volcano
A volcano is a natural example of a cone. The base represents the surface of the Earth, while the lateral surface forms the sides of the volcano. The apex is the opening at the top of the volcano, through which lava and gases are released during eruptions.
Conclusion
In conclusion, a cone has a total of three faces: the base, the lateral surface, and the apex. Whether it is a right cone with a perpendicular axis or an oblique cone with a slanted axis, the number of faces remains the same. Understanding the anatomy and properties of cones is essential in various fields, including mathematics, engineering, and everyday life.
Q&A
1. Can a cone have more than three faces?
No, a cone cannot have more than three faces. The base, lateral surface, and apex are the three fundamental faces of a cone.
2. Are all cones symmetrical?
No, not all cones are symmetrical. While right cones are typically symmetrical due to their perpendicular axis, oblique cones can be asymmetrical depending on their orientation.
3. Can a cone have a square base?
No, a cone cannot have a square base. The base of a cone is always circular in shape.
4. Are all pyramids considered cones?
No, not all pyramids are considered cones. While both pyramids and cones have a single point and a base, the shape of the lateral surface differs. A cone has a curved lateral surface, while a pyramid has a flat or polygonal lateral surface.
5. Can a cone have more than one apex?
No, a cone can only have one apex. The apex is the single pointy end of the cone, opposite to the base. | 677.169 | 1 |
360 Degree Chart
Teachers need easy way to explain angles and directions. A 360 degree chart helps visualize these concepts, but hard to find good one for class. They're looking for a tool, something printable, that fits this teaching moment.
We design printable 360 degree charts for easy tracking and visualization of angles. They come in handy for both educators teaching geometry and DIY enthusiasts working on projects needing precise angle measurements. Each chart is made to be clear and straightforward to use, helping you accurately understand and apply angle concepts without hassle. Handy for classroom demonstrations or personal projects, these charts aim to simplify complex measurements.
Origin of 360 Degrees in a Circle
The 360 degrees in a circle originated from the Mesopotamians about 6000 years ago who found the number 60 easily divisible. This metric, also used in early math and trade, passed to the Ancient Egyptians. They applied it to divide a circle into 360 degrees, leading to notable inventions such as calendars and time measurement systems.
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The printable 360-degree chart allows for easy visualization and analysis of data in a comprehensive and interactive way, enabling users to make informed decisions and effectively communicate information.
The Printable 360 Degree Chart is a useful and practical resource for visualizing data. It's convenient and easy to use, making it a great tool for organizing information | 677.169 | 1 |
The Elements of Euclid: With Many Additional Propositions, & Explanatory Notes, Etc, Part 1
From inside the book
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Page 9 ... equal ( c ) , and the lines AB and CB being both radii of the same circle , ACE are equal ; then , because the lines ... DB until it meets the circumference in G ( d ) . From the center D , at the distance DG , describe the circle GKL ( c ) ; ...
Page 12 ... DB . Now it is evident that , in the triangles ABC and ABD , we have the two sides AB and BC equal to the two AB and BD , and the opposite angle A the same in both ; and yet the two triangles are not equal . The fifth and sixth cases ...
Page 13 ... equal to the side AB ( d ) , the side AF to the side AG ( e ) , and the angle A is common to both ; therefore the ... DB equal to AC the less ( a ) , and draw the line DC . Then in the triangles ABC and DBC , because the side AC is equal ...
Page 17 ... equal parts ; that is , into any number of equal parts which can be ... DB ( d ) , and the straight line AB is bisected in the point D. ( a ) I. 1 ... equal to AD ( a ) ; and upon DE construct an equilateral triangle DEF ( b ) ; then a ...
Page 43 ... equal to a given rectilineal angle ( D ) . SOLUTION . Produce AB to E , and upon BE construct a parallelogram BEFG ... DB dividing the given figure into triangles . Then construct a parallelogram FH equal in area to the triangle ABD | 677.169 | 1 |
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Cosine Rule
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Cosine Rule Revision
Cosine Rule
When we first learn the cosine function, we learn how to use it to find missing side-lengths & angles in right-angled triangles. The cosine rule is an equation that can help us find missing side-lengths and angles in any triangle.
Cosine Rule Example Questions
Firstly, we need appropriately label the sides of this triangle. Firstly, we set a=x, and therefore we get that A=19, since it is the angle opposite. It doesn't matter how we label the other two sides, so here we'll let b=86 cm and c=65 cm. Now, subbing these values into the cosine rule equation, we get
As always, we must label our triangle. Firstly, assign the thing we're looking for to be A=x, and therefore make the side opposite to it a=6 mm. Then, it doesn't matter how we choose the other two sides, so we will let b=5 mm and c=7 mm.
Here, we will use the rearranged version of the formula that looks like
Firstly, we need appropriately label the sides of this triangle. Firstly, we set a=x, and therefore we get that A=40\degree, since it is the angle opposite. It doesn't matter how we label the other two sides, so here we'll let b=6 cm and c=4.5 cm. Now, substiuting these values into the cosine rule equation, we get | 677.169 | 1 |
emma is on a 50 m high bridge and sees two boats anchored below from her position boat a has a bearing of 230 and boat b h
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Emma243
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Answer to a math question EmmaIn this diagram, point A represents Emma's position on the bridge, point B represents boat A, and point C represents boat B. We know that the height of the bridge is 50 meters, and the angles of depression to boats A and B are 38° and 35°, respectively. We also know that the bearings of boats A and B are 230° and 120°, respectively.
We can use the tangent function to find the distances from Emma to boats A and B. For boat A, we have:
tan(38°) = 50 / d_A
Solving for d_A, we get:
d_A = 50 / tan(38°) = 64.1 meters
Similarly, for boat B, we have:
tan(35°) = 50 / d_B
Solving for d_B, we get:
d_B = 50 / tan(35°) = 71.4 meters
Now that we know the distances from Emma to boats A and B, we can use the Law of Cosines to find the distance between the two boats. The Law of Cosines states that:
c^2 = a^2 + b^2 - 2ab * cos(C)
where c is the distance between the two points, a and b are the distances to the two points from a third point, and C is the angle between the lines connecting the third point to the two other points. In this case, we have:
c = d_A + d_B
a = 50 meters
b = 50 meters
C = 110°
Substituting these values into the Law of Cosines, we get:
(d_A + d_B)^2 = 50^2 + 50^2 - 2 * 50 * 50 * cos(110°)
Solving for c, we get:
c = sqrt((d_A + d_B)^2 - 50^2 - 50^2 + 2 * 50 * 50 * cos(110°))
Plugging in the values for d_A and d_B, we get:
c = sqrt((64.1 + 71.4)^2 - 50^2 - 50^2 + 2 * 50 * 50 * cos(110°))
Evaluating this expression, we get:
c = 111.4 meters
Therefore, the distance between the two boats is approximately 111 meters to the nearest meter. | 677.169 | 1 |
Transforms and Euler Angles
According to Introduction to Robotics: Mechanics, and Control, Second Edition by John J Craig, orientation can be described by a set of three angles with a total of 24 different angle set conventions.
Vortex® uses a static angle set and a rotating angle set convention.
Note The default convention is the rotating angle set. In the static angle set, the convention rotates around the global static X axis, then around the global static Y axis, and, finally, around the global static Z axis.
Vortex Angle Set Convention: XYZ CounterClockwise Static
The default angle set convention used by Vortex rotates around X, then around the new Y', and, finally, around the new Z''.
Vortex Angle Set Convention: XYZ CounterClockwise Rotating
You need to be careful with the convention with respect to quaternions. Quaternions are composed of four numbers: a directional vector, consisting of 3 components; and W, the cos half angle. The direction vector has its component in the X,Y,Z order. However, the W can be either at the beginning or the end of a quaternion. For example, | 677.169 | 1 |
Angles whose vertices are on the circumference of a circle or formed by tangent lines and chords.
When we say an angle is on a circle, we mean the vertex is on the edge of the circle. One type of angle on a circle is the inscribed angle (see Inscribed Angles in Circles). Another type of angle on a circle is one formed by a tangent and a chord.
Chord/Tangent Angle Theorem: The measure of an angle formed by a chord and a tangent that intersect on a circle is half the measure of the intercepted arc.
Figure \(\PageIndex{1}\)
\(m\angle DBA=\dfrac{1}{2}m\widehat{AB}\)
If two angles, with their vertices on the circle, intercept the same arc then the angles are congruent.
An angle is inside a circle when the vertex lies anywhere inside the circle.
Intersecting Chords Angle Theorem: The measure of the angle formed by two chords that intersect inside a circle is the average of the measures of the intercepted arcs.
What if you were given a circle with either a chord and a tangent or two chords that meet at a common point? How could you use the measure of the arc(s) formed by those circle parts to find the measure of the angles they make on or inside the circle | 677.169 | 1 |
I am working on the last details to the math map I am making, and the last touch it that I need to add things around the map to help them find the answers to this:
I need to know how to hide them somewhere or like, where I can hide them.
Maybe make it a puzzle involving the angles? If they find one angle, they can walk in a straight line through the orgin to a secret room and find another answer for the angle they just entered, like so: | 677.169 | 1 |
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M217 Section 1.Point Line and Plane Postulates • A postulate is a statement that is accepted without proof. • Through any 2 pts there is exactly 1 line • Through any 3 noncollinear pts there is exactly one plane containing them • When two lines intersect, the intersection is a point • When two planes intersect, the intersection is a line
Drawing Planes • Plane G containing two lines that intersect at point J • Two planes that intersect at line SD | 677.169 | 1 |
Pre-Algebra Unit 5 Geometry
Oct 23, 2014
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Pre-Algebra Unit 5 Geometry. Quiz 5B Review. Name that Quadrilateral. - I have 2 pairs of congruent, parallel sides. Parallelogram. Name that Quadrilateral. - I have 4 right angles and 2 pairs of congruent, parallel sides. Rectangle. Name that Quadrilateral. - I have 4 congruent sides.
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Pre-Algebra Unit 5 Geometry | 677.169 | 1 |
Show that the distance between two nonparallel lines is given by $$\frac{|(p_2-p_1)\cdot (a_1\times a_2)|}{|| a_2\times a_1||}$$ where $p_1, p_2$ are any two points on the lines $l_1$ and $l_2$, and $a_1$ and $a_2$ are the directions of $l_1$ and $l_2$.
Answer
Let $P_1, P_2$ be the two parallel planes passing through $l_1, l_2$, respectively. Then the vector \[n=a_1 \times a_2\] is a normal vector for $P_1$ and $P_2$. Notice that The distance between the two lines is the same as the distance between the two parallel planes $P_1, P_2$. Since $p_1 \in P_1$ and $p_2 \in P_2$, the distance between the planes is the length of the orthogonal projection of the vector $(p_2-p_1)$ onto the normal vector $n$, i.e. \[\text{Distance}=|| \frac{(p_2-p_1)\cdot (a_1\times a_2)}{||a_1\times a_2||^2} (a_1\times a_2)||\] \[=| \frac{(p_2-p_1)\cdot (a_1\times a_2)}{||a_1\times a_2||^2}| || (a_1\times a_2)||\] \[=\frac{|(p_2-p_1)\cdot (a_1\times a_2)|}{||a_1\times a_2||}.\] | 677.169 | 1 |
by
Charles, Randall I.
Answer
$x = 1$
Work Step by Step
Theorem 6-20 states that in an isosceles trapezoid, diagonals are congruent. Let's set the two diagonals equal to one another:
$QS = RP$
Let's plug in the expressions for each diagonal:
$x + 5 = 3x + 3$
Subtract $5$ from each side of the equation to move constants to the right side of the equation:
$x = 3x - 2$
Subtract $3x$ from each side of the equation to move the variable to the left side of the equation:
$-2x = -2$
Divide each side by $-2$ to solve for $x$:
$x = 1$ | 677.169 | 1 |
Youll likely encounter a problem that asks about rigid motions on the Geometry Regents exam. Fortunately, this term defines itself! Rigid tells us that the figure will maintain its size and shape, and motion tells us that the figure will move to a different position or direction. There are three types of rigid motions youll need to remember:
Reflections refer to a figure being flipped over a given line of reflection.
Rotations refer to a figure being rotated a given number of degrees around a center of rotation.
Translations refer to a figure shifting in a given direction.
Rigid motions create figures that are congruent to one another. Note that a dilation does not count as a rigid motion because it creates figures that are different sizes and thus similar, not congruent.
Example question:
Break Up Your Studying
Instead, you should space out your studying over several weeks leading up to exam day. In addition to working on past exam questions, you should review your Geometry notes, practice problems, quizzes and tests as well.
One of the benefits of spacing out your studying is that it will give you opportunities to ask your geometry teacher for help before or after school. If you wait until the last minute to study, you will not have this option.
Take Advantage Of Free Resources
There is no shortage of helpful, free resources to help you prepare for the Geometry Regents.
Pro Tip: When you come across a practice question that you are struggling to solve, write down whatever questions you may have and flag the question until you can share it with your teacher or tutor the next time they are available.
Geometry Regents Graphing Calculator Tips And Tricks
According to the official Regents administration directions, students will have access to a graphing calculator for the duration for the Geometry Regents exam. That means its time for you to get comfortable using the different functions your graphing calculator has to offer.
Weve crafted a list of graphing calculator tips and tricks to show you how to make the most of your calculator during the Geometry Regents exam. Take some time getting familiar with these functions before the exam so that you show up on test day ready to go!
Most Common Topic: Prove Theorems About Triangles
Pro Tip: This topic is frequently assessed as a multiple-choice question but can also appear as a constructed-response question on Regents Geometry exams.
Math Standard: HS.G.CO.10 // Prove theorems about triangles. Theorems include: measures of interior angles of a triangle sum to 180° base angles of isosceles triangles are congruent the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length the medians of a triangle meet at a point.
Geometry Final Exam With Study Guide
This editable Geometry Final Exam and Study Guide is meant to be taken at the end of a high school geometry course.
geometry standards. The topics that are not
*law of sines and cosines
*parabola, hyperbola, and ellipse conic sections
Please see preview to view all study guide and final exam questions! The final exam is 50 multiple choice questions. The study guide has 4-6 questions per unit.
This test and study guide are EDITABLE, however the pictures are NOT editable. The pictures can be removed. Also, the font on the editable version is different than the preview so that it will work on everyones version of PowerPoint. The PDF versions are as shown above.
Terms of Use:
This product should only be used by the teacher who purchased it. This product is not to be shared with other teachers. Please buy the correct number of licenses if this is to be used by more than one teacher.
Most Common Topic: Dilations
Description: Understand dilations of a line or line segment.
Frequency: This topic has been assessed in 100% of recent exams.
Pro Tip: This topic is frequently assessed as a multiple-choice question in Part I of the Regents Geometry exam.
Math Standard: HS.G.SRT.1 // Verify experimentally the properties of dilations given by a center and a scale factor. a) A dilation takes a line not passing through the center of the dilation to a parallel line, and leaves a line passing through the center unchanged. b) The dilation of a line segment is longer or shorter in the ratio given by the scale factor.
Dont Get Stuck On A Single Problem
Most students have plenty of time to finish all the questions on the Geometry Regents exam. In fact, most students finish with many minutes leftover on the 3-hour time limit!
However, this doesnt mean you should stay stuck on a single multiple-choice question as you work through the exam. Its important to stay confident and focus on the questions you DO know really well. You can always go back to tricky questions and spend extra time thinking through different options. In fact, we often think of new ideas on how to solve tricky problems just by working on other problems elsewhere on the exam!
We generally suggest about 3 minutes for each multiple-choice question. For more details on how to pace yourself on the exam, check out our full Geometry Regents Review Guide.
Key Geometry Regents Question Vocabulary Cheat Sheet
Perhaps more so than any other mathematics exam, the Geometry Regents requires students to recognize and understand a variety of vocabulary terms and mathematical symbols. Memorizing all of these words and symbols can get confusing, so weve studied released Regents exams to create a cheat sheet for the most important Geometry terms.
Take time to review this section and commit each vocabulary term to memory. It might help you to create a deck of flashcards to practice these terms each day.
Review Past Geometry Regents Exams
Every Geometry Regents exam from the past several years are available for free online. You can practice taking these exams at home to assess your readiness and determine areas of weakness that you can focus on while studying.
Practicing these old exams is great way to familiarize yourself with the format of the exam, what kind of questions will be asked, and what your responses need to look like.
How To Ace Your Math Final Exam
Are you getting ready for math finals maybe even your first math final ever? Are you uncertain how to study and prepare?
Or, if youre a parent, not a student, does the thought of getting through your kids math final fill you with dread?
This week, since Im right in the middle of preparing several of my students for their final math exams, I want to share all my best math studying tips with you whether youre a student or a parent so you can happily survive finals week with a minimum of stress.
This is the exact same process I walk my students through, and also the exact same powerful tips I share with my private clients!
And Im not just going to share my own tips Im also going to share some of my students tips, too!
First, just some basic overall tips about the big picture of taking your math final, especially if its your first one:
1. Breathe. If thinking about your math final, or your kids math final, sends you into a panic, keep breathing. Just keep breathing deeply. If you forget, you can start again right now. Take a deep breath. Right now. You can do this. Take three deep breaths. Yes. Thats right!
2. Eat. Make sure that youre getting really good meals all throughout finals week. You want to keep nourishing your brain with high quality protein! Also, you will be less stressed, more receptive when studying, and find it easier to retain what youre learning when your blood sugar isnt careening all over the place.
During the final
Sending you love,
Preparing For The Test
1Look over your class notes. After school, look over the notes you took in class that day for 15 to 20 minutes.XExpert SourceDaron CamAcademic TutorExpert Interview. 29 May 2020. As a test approaches, review your notes for the entire unit or chapter more thoroughly. Pay special attention to the example problems the teacher provided in class, since these will help break down how a given procedure or formula works. If you don't have any class notes, ask your classmate for notes.XResearch source
2Do problems similar to those that were assigned for homework. Suppose you were assigned odd numbers for homework because the even numbers answers are in the back of the book. Work on those even-numbered problems, then check your answers to see where your strengths and weaknesses are.XResearch source
Ask your teacher if your math book has an online website. Sometimes online textbooks can help by providing quizzes and additional instructional material.
Try solving for different parts of the equation. It's very important to learn to apply what you're learning to different cases. For instance, if you're studying the Pythagorean theorem, don't just solve it for one sideafter you finish, solve it again for the other side, then for the hypotenuse.
Use The Process Of Elimination
Whenever youre presented with a wide range of answer options, it can be helpful to eliminate any answer choices that you know cannot be correct. This is called the process of elimination.
For the Geometry Regents exam, youll be given 4 possible answer choices on all multiple-choice questions. Its very likely you can eliminate at least one and maybe even two of the answer choices pretty easily. As you go through each possible answer choice, literally draw a line through any answer choice that must be wrong.
If you cant decide on a final answer, take an educated guess or try a new strategy. Whatever you do: do NOT leave any question blank on your Regents exam. You do not lose credits for a wrong answer, so its much better to just guess instead of leaving something blank.
What makes the process of elimination such a great trick?
You originally start with a 25% chance of randomly guessing the right answer . If you can eliminate two of the answer choices, that means youve doubled your chances of getting the question correct to 50% awesome!
How To Prepare For A Geometry Test
ASHLEY LORELLE
Studying for and passing geometry in high school and middle school may be overwhelming for some kids, and simple for others. As with any subject in school, the more you study for a test, the higher your chances are of passing. Mathematics does not involve simply memorizing facts like history class does, so studying for a geometry test takes more focus and preparation. Most of all, passing a geometry test takes practice. Once you feel confident in your ability to solve most geometry problems on your homework and in review, then you will do fine on the test.
Memorize the formulas and theorems you will need to use on the geometry test. A lot of geometry involves plugging numbers into a formula to find the area, distance, volume or diameter of a square, rectangle, or parallelogram. Also, fractions may have to be used to find the square units of the area. Your teacher will provide all the formulas you will need to use for the test. If they have not, then ask them to.
Practice problems from your homework. If you come across a problem that you find particularly challenging or that you do not understand, than see your teacher after class or during study time to get some extra help before the exam. You should be able to solve your homework problems with confidence. Practice the same with old quizzes. Have your teacher help you work through questions you previously got wrong.
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The Top 5 Most Common Topics On The Geometry Regents Exam
Geometry Final Exam Review
While each Regents Geometry exam has different questions, there are trends in what topics are most often assessed. Are you curious about the specific topics you should review and practice the most before test day?
Weve got you covered!
We tracked hundreds of official questions from the most recent Geometry Regents exams and found the patterns in what specific topics are often assessed. Below are the five most commonly-assessed topics on the Geometry Regents exam:
Geometry: Final: Exam: Study: Guide
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If Your Prompt Says Parallelogram Think About Some Special Properties
The definition of a parallelogram is simple enough: a quadrilateral with two pairs of parallel sides. Examples of parallelograms include squares, rectangles, and rhombuses. For the Geometry Regents exam, youll also need to remember certain properties of parallelograms:
If Your Prompt Says Similar Think Same Angles Different Size
We all know what similar means in everyday English, but this term takes on a whole new meaning in the context of the Geometry Regents exam. When two shapes are similar, it means they have congruent angles but different side lengths. Symbol Spotlight: Remember that the \cong symbol means congruent and \sim means similar. Congruent shapes, angles, and measurements are exactly the same, while similar shapes are proportional to one another.
It also helps to remember that when a figure is dilated, it creates a figure similar to the original. That means that it will have congruent angles, but a different size.
Example question:
Double Check Radical Expressions With The Click Of A Button
We can guarantee that youll be dealing with a radical expression at some point during the Geometry Regents exam. One way that radical expressions frequently come up is in the distance formula for finding the distance between two points:
d=\sqrt
When using this formula, dont let a math mistake bring you to the wrong answer. Instead, substitute all values into your calculator to get a decimal answer. Then, check the decimal values of the answer choices to see which is equivalent.
What Are Helpful Practice Resources For Geometry Regents
Albert has a number of Geometry Regents practice tests for your test prep!
Unique from other Regents prep sites, Albert not only provides access to some of the previously released Regents tests, but also includes original New York Geometry Regents practice questions. Create your free account today.
If Your Prompt Says Altitude Think About A Perpendicular Line
Altitude is another word that means something different in day-to-day conversations and on the Geometry Regents exam. The altitude of a triangle is a line drawn from a vertex to the opposite side such that it is perpendicular to a line containing the opposite side.
Symbol Spotlight: The symbol \perp means perpendicular. The altitude of a triangle is perpendicular to the line containing the opposite base of that triangle, which means it creates 90^\circ angles.
A couple things to remember about altitudes:
An altitude to the base of an isosceles triangle bisects both the vertex angle and the length of the base.
An altitude to the hypotenuse of a right triangle creates two triangles that are both similar to one another and to the original right triangle. | 677.169 | 1 |
4. Radius and Diameter
It's worth noting that these two terms have multiple meanings. Diameters can also denote the longest paths in a graph, just as a radius can be any path whose length is equal to the graph's minimum eccentricity.
In our example, the diameter is , and the radius is :
4.1. The Relationship Between Radius and Diameter
We can prove that the diameter is bounded from above by twice the radius:
This follows from the triangle inequality. Let and be two distinct nodes whose distance is equal to the diameter:
Let be a central node of . The triangle inequality tells us that:
Since is a central node, it holds that . Therefore:
5. Center and Periphery
A node is peripheral if its eccentricity is equal to the graph's diameter. All the nodes with that property constitute the graph's periphery:
On the other hand, the center of a graph is comprised of the nodes whose eccentricity is equal to the graph's radius:
In our example, the periphery contains , , and , while the only central node is :
So, the peripheral nodes are at the opposite ends of the graph, while the central nodes are in between.
An example of when these concepts may be useful in determining the optimal placement of facilities on a graph. If we model a city map as a graph, it makes sense to build a facility at a central node because that will minimize the longest distance a citizen will have to travel to reach the facility.
Conversely, if the goal is to maximize the distance between two objects, we'd place them on the locations of two peripheral nodes on opposite sides of the town.
6. Conclusion
In this article, we explained several concepts from graph theory: eccentricity, radius, diameter, center, and periphery.
The diameter is always smaller than twice the radius, and the center minimizes the worst-case distances to other nodes in a graph.
Comments are open for 30 days after publishing a post. For any issues past this date, use the Contact form on the site. | 677.169 | 1 |
1.1 Points Lines And Planes Worksheet Answer Key
Displaying top 8 worksheets found for plane line point. Plane s sample answers are given.
Sin And Cosine Worksheets Law Of Cosines Worksheets Trigonometry Worksheets
Points nr and s lie in plane a but point w does not.
1.1 points lines and planes worksheet answer key. Sheet 1 1 points a b and c 2 points x y and z 3 points k l m and n collinear not collinear 4 points e f and g not collinear. Key vocabulary line perpendicular to a plane a line is a line perpendicular to a plane if and only if the line intersects the plane in a point and is perpendicular to every line in the plane that intersects it at that point. Some of the worksheets for this concept are unit 1 tools of geometry reasoning and proof geometry unit 1 workbook geometryunit 1 tools of geometry chapter 7 quiz 1 geometry identify points lines and planes 1 introductionto basicgeometry geometry chapter 2 reasoning and proof answers to geometry unit 1 practice.
Answer key a write whether the given points are collinear or not collinear. Consider two lines in a plane. Points lines and planes gina wilson answer key displaying top 8 worksheets found for this concept.
Some of the worksheets for this concept are points lines and planes exercise 1 points lines and planes 1 chapter 1 lesson 1 points and lines in the plane identify points lines and planes 3 points in the coordinate name answer key points lines and planes 1 chapter 4 lesson1 0 points line segments lines and rays. Use postulates involving points lines and planes. If points on are the same plane they are coplanar.
Use the figure to name each of the following. If points are on the same line they are collinear. This quiz is incomplete.
2 4 use postulates and diagrams obj. To play this quiz please finish editing it. A line containing point a.
16 questions show answers. If the lines are if the lines parallel three intersect four regions are regions are determined. Postulate in geometry rules that are accepted without proof are.
Some of the worksheets for this concept are identify points lines and planes work section 3 1 parallel lines and transversals use the figure to name each of the unit 1 tools of geometry reasoning and proof the segment addition postulate date period geometry unit 1 workbook finding. Section 1 1 worksheet 4 understanding points lines and planes lines in a plane divide the plane into regions. B write true or false.
Points lines and planes description figure symbol. The number of regions depends on the relationship between the lines. Plane and line point lines segments line and point plane | 677.169 | 1 |
Distance Calculator
Distance Calculator Tool
Table of Contents
Many folks on the internet ask "What is the distance between these two points?" – We've got the answer for you. And we can do one better than just a simple two points in a two-dimensional (2D) space, we've got a three-dimensional calculator as well. With our simple and straight forward processes, you can have the distance measured and get your result in near real-time. Some might say that these calculators are "Zippy!". We hope you enjoy our humorous approach and can use the calculators to better your life! If you like this one and want to check out more from our STEM Collection, Click Here .
Calculating the Distance Between Two Points: A Complete Guide
Introduction
The calculation of distance between two points is a crucial concept in geometry, physics, and various other scientific and practical fields. Whether you're plotting coordinates on a map or figuring out the quickest route between two locations, this fundamental task plays an indispensable role in our daily life.
Quick summary:
This article will explore the history of measuring distances.
Discuss the links between calculating distances and various fields.
Highlight the importance of understanding distance calculations in everyday life.
Provide common reasons for calculating distance between two points.
History of Calculating Distance Between Two Points
The endeavor to calculate distances between two points dates back to ancient times. Historically, measuring distances has been carried out using physical objects like ropes or measuring sticks. However, with the advent of math and geometry, more sophisticated methods were developed. For example, the Greek mathematician, Euclid formulated the foundation for calculating distances in his seminal work, "Elements."
The Link Between Calculating Distance and Fundamental Geometry
In geometry, the concept of calculating the distance between two points is core to various theories and principles. It's this calculation that underpins concepts as simple as line lengths and as complex as calculating the radius of a circle or the hypotenuse of a right-angled triangle. Without the understanding of distance calculation, our understanding of three-dimensional space would be drastically hindered.
Knowing how to calculate distances is vital in our everyday life – from navigating directions to planning construction projects, distance calculation comes in extremely handy.
Common Reasons to Calculate Distance Between Two Points
There are numerous reasons why one might want to calculate the distance between two points. Here are a few:
Making travel plans: You can calculate the shortest distance between two places.
Road trip planning: Knowing the distance can help manage time and resources efficiently.
In architecture or construction: Distance calculations are crucial for accurate planning and execution.
In sports: Distances are used to measure the performance of athletes.
For scientific calculations: Astronomy, physics, and engineering all rely heavily on distance calculations.
Do You Know?
Here are some intriguing facts about measuring and calculating distances:
The longest straight-line distance you can travel on earth, without hitting a major body of water, is 8,222 km from Liberia on the west coast of Africa, to China.
The distance between Earth and the farthest known galaxy is approximately 32 billion light-years.
The term ?mile? came from the Roman ?mille passuum,? which means 1,000 paces.
Distance, time, and speed form the foundation for much of physics.
The diameter of a proton is about 0.000000000000001 meters.
The concept of 'parallax' allows astronomers to calculate distances to nearby stars.
Distance plays a vital role in cartography, the practice of making maps.
The Great Wall of China stretches over 21,196 kilometers, making it the longest man-made structure.
We measure astronomical distances in 'light-years,' the distance that light can travel in a year, approximately 5.88 trillion miles.
Remember to remain accurate and concise in discussions about distances, and always bear in mind the profound impact of distance calculations on our understanding of the world around | 677.169 | 1 |
Taking the arctan of the result of the formula above returns the equivalent of the Common Rafter Pitch Angle.
This angle will vary with the Framing Point selected on the ellipse.
The Plan Angle is constant for a given pair of intercepting barrels.
The purlins in the images are set with their wide faces
tangent to the curve of the ellipse.
To cut the compound angle from this face:
The Saw Miter Angle formula is the same as the Jack Rafter Side Cut Angle formula.
The same miter angle is required to cut a purlin intercepting a Valley rafter created by the given combination of Common Rafter Slope and Plan Angle.
Saw Miter Angle = arctan (cos Common Rafter Pitch Angle ÷ tan Plan Angle)
The formula for the Angle on the Purlin or Angle on the Stick is the same as the formula for the
Sheathing Angle or Plywood Angle.
The angle is the same as on a purlin intercepting a Valley rafter created by the given combination of Common Rafter Slope and Plan Angle.
Angle on the Purlin = arctan (tan Plan Angle ÷ cos Common Rafter Pitch Angle)
or Angle on the Purlin = 90° – Saw Miter Angle
The Saw Blade Bevel Angle is the equivalent of the Backing Angle.
The same blade bevel is required to cut a purlin intercepting a Valley rafter created by the given combination of Common Rafter Slope and Plan Angle.
Saw Blade Bevel Angle = arcsin (sin Common Rafter Pitch Angle × cos Plan Angle) | 677.169 | 1 |
Cofunction Identities
The reciprocal identities were already introduced in previous lessons. Refer back to lessons 4-02 and 4-03. The other identities are explained below.
Quotient Identities
On the unit circle, sin θ = y and cos θ = x. Divide these equations.
sinθcosθ=yx=tanθ
Thus,
tanθ=sinθcosθ
And, the reciprocal is also true.
cotθ=cosθsinθ
Pythagorean Identities
Think of drawing a right triangle on the unit circle so that one leg is on the x-axis, the other leg is vertical, and the hypotenuse is a radius of the circle.
Figure 2: Right triangle on the unit circle.
The horizontal leg is the x-coordinate of the point, but on the unit circle cos θ = x. The vertical leg is the y-coordinate of the point, but on the unit circle sin θ = y. The hypotenuse is a radius of the unit circle, so its length is 1. Apply the Pythagorean Theorem.
x2 + y2 = 12
cos2θ + sin2θ = 1
Divide this by either sin2θ or cos2θ to get the other two Pythagorean identities.
Cofunction Identities
Imagine a right triangle where one acute angle measure is u. The two acute angles in a right triangle are complementary. If the other acute angle is v, see figure 3, then
u + v = 90°
Solve for v.
v = 90° − u
Figure 3: Right triangle
If the sides of the triangle are a, b, and c as in figure 3, then sinu=ac and cosv=ac. Thus sin u = cos v. Since v = 90° − u,
Example 2: Use Trigonometric Identities
Solution
We know cosecant and want to find sine, so use an identity with both of those.
sinα=1cscαsinα=12
We know cosecant and want to find cotangent, so use an identity with both of them.
cot2α+1=csc2αcot2α+1=22cotα=3
Try It 1
Let β be an acute angle and tan β = ½. Find a) cot β and b) sec β.
Answers
2; 52
Applications with Right Triangles
Some real-world problems can be solved by drawing right triangles and finding unknown lengths. Other problems use angles of elevation and depression.
Angles of Elevation and Depression
The angle of elevation is the angle between the horizontal up to an object.
The angle of depression is the angle between the horizontal down to an object.
Figure 4: Angles of Elevation and Depression
Example 3: Solve a Problem with a Right Triangle
A 8-foot step ladder actually is not 8-feet high. The size of a step ladder is actually the length of the rails that the steps are attached to. When the ladder is in use the rails are slanted so the height is less. If the rails of an 8-foot step ladder make an angle of 50° with the ground, how high is the top of the ladder from the ground?
Solution
Draw a right triangle using the rails of the ladder as the hypotenuse. The hypotenuse is 8 feet long and the angle with the ground is 50°.
Figure 5
The hypotenuse and one angle are known and the opposite leg is the unknown. The formula of sine from lesson 4-03 has those three parts. Use sine to solve the problem.
sinL=opphypsin50°=h88sin50°=hh≈6.12
The height of the ladder is actually about 6.12 feet.
Try It 2
Find the length of side a in figure 6.
Figure 6
Answer
8.57
Example 4: Angle of Elevation
Figure 7
To find the height of a tree, a person walks to a point 30 feet from the base of the tree. She measures an angle of elevation of 57° between a line of sight to the top of the tree and the ground. Find the height of the tree.
Solution
The adjacent leg and angle are known. The height is the unknown. Use a trigonometric formula that has those three quantities.
tanθ=oppadjtan57°=h3030tan57°=hh≈46.20
The height of the tree is about 46 feet.
Example 5: Angle of Elevation
A hiker is standing on mesa 80 feet above the desert floor. The angle of depression to a creek is 50°. Farther away, the angle of depression to a horse is 30°. How far apart are the horse and the creek?
Solution
Draw a picture to visualize the situation.
Figure 8: α = 30° and β = 50°
This will be a two part problem. First, since AG‾ and DF‾ are parallel, then ∠α and ∠ADF are congruent. So are ∠β and ∠AEF.
Figure 9
Start by finding the length of DF using ΔADF.
tanu=oppadjtan30°=80DFDF=80tan30°DF≈138.56
Now find the length of EF using ΔAEF.
tanu=oppadjtan50°=80EFEF=80tan50°EF≈67.13
Now subtract to find x.
x = DF − EF x = 138.56 − 67.13 = 71.43
The horse is approximately 71 feet from the creek.
Try It 3
A student is observing a radio tower. Two sets of guy wires are attached at the same spot on the ground 50 feet away from the tower. The other ends of the guy wires are attached to the tower at different points. If the angles of elevation of the guy wires are 50° and 70°, how far apart are they attached on the tower?
Practice Exercises
Let θ be an acute angle. Use the given function value with trigonometric identities to evaluate the given function.
If sin θ = 0.9, find a) cos θ and b) csc θ.
If sin θ = 0.25, find a) sin(90° − θ) and b) tan θ.
If sec θ = 1.45, find a) cos θ and b) tan θ.
If cos θ = 0.6, find a) sin θ and b) cot θ.
If csc θ = 10, find a) sin θ and b) csc (90° − θ).
Problem Solving
A 20-ft ladder leans against a building so that the angle between the ladder and the ground is 75°. How high up the building does the ladder reach?
A 30-ft ladder leans against a building so that the angle between the ladder and the ground is 70°. How far from the building is the base of the ladder?
The angle of elevation to the top of the Willis Tower is 33.2° when you are a half-mile from the base of the tower. How high is the tower?
*If the Empire State Building is 1250 ft high and the angle of the elevation to the top is 52°, how far from the building are you?
A group of civil engineers wants to build a bridge over a canyon, but they do not know how wide the canyon is. They raise different tall objects up beside the canyon until one of them casts a shadow to the other side of the canyon. The height of the object is 80 ft and they estimate the angle of elevation of the sun is 35°. Roughly, how wide is the canyon? (Ben P)
A tall pine tree grows vertically. If Sam is 50 feet from the tree and measures the angle of elevation as 80°, how tall is the tree?
A large advertising banner hangs on the side of a building. Duane works in a neighboring building 75 feet away and measures to angle of elevation to the top of the banner as 50° and the angle of depression to the bottom as 20°. How long is the banner?
Marie is standing on a platform waiting to ride a roller coaster. She measures the angle of depression to the bottom of the long hill as 13° and the angle of elevation to the top of the hill as 52°. If she is 110 feet away, how high is the hill?
A steeple is on top of a church. Marco stands 52 ft from the church and measures the angle of elevation to the base of the steeple as 44°. He measures the angle of elevation to the top of the steeple as 56°. How tall is the steeple?
Philip is standing on Inspiration Point in Arcadia Scenic Turnout 800 feet above Lake Michigan. He can see two ship, one behind the other. If the angle of depression to the closer ship is 18° and the farther ship is 15°, how far apart are the ships?
Mixed Review
(4-03) Use a special right triangle to evaluate a) tan 30° and b) secπ4.
(4-03) Evaluate the six trigonometric functions for the given angle.
(4-02) Evaluate the six trigonometric functions for 4π3 using the unit circle.
(4-01) A car with a 30-inch diameter wheels is traveling at 50 mi/h. Find the angular speed of the wheels in rad/min. How many revolutions per minute do the wheels make?
(3-02) What is the intensity of a loud stereo blaring music at 95 dB?
Answers
For example, the sine of an angle is equal to the cosine of its complement; the cosine of an angle is equal to the sine of its complement. | 677.169 | 1 |
Angle of Elevation and find Height & Distance
To find height and distance we use Tan θ = Opposite Side / Adjacent Side
To find length or hypotenuse we use Sin θ = Opposite Side / Hypotenuse
Example 1 : A person is standing 5m away from tree, the angle of elevation of the top of tree is 60° find the height of 5m
∠BCA = 60°
Right angle is at point B
AB (Height) = ?AB = 5 x 1.73 = 8.65
AB = 8.65m is the height of tree
Example 2 :Height of tree is 8.65m, the angle of elevation of the top of tree is 60° find the distance at which the person is standing away from ?
∠BCA = 60°
Right angle is at point B
AB (Height) = 8.65m8.65 = CB x 1.73
CB = 8.65 / 1.73
CB = 5
so the person is standing 5m away from the tree
Example 3 : A tree is 5m high. If angle of elevation is 45° find the hypotenuse
Let AB is the tree. B is the foot and A is the top of tree. C is the point on the ground where a person is standing and which the angle of elevation of 45°
If we draw a picture it will look like as below -
so we get
AC = ?
∠ BCA = 45°
AB = 5m
A right angle triangle △ ABC is formed in which
AB is the Opposite Side from angle of elevation
AC is the Hypotenuse Side
We know that Sin θ = Opposite Side / Hypotenuse
sin 45° =
AB AC
1 √ 2
=
5 AC
AC = 5 x √ 2
Value of √ 2 is 1.41 so,
5 x 1.41 = 7.05
AC = 7.05m is the hypotenuse
Study More Solved Questions / Examples
At a point 20 60° find the height of building
A tower is 10m high. A steel wire is tied at the top of pole and is affixed at a point on the ground. If the steel wire makes an angle of 45° find the length of steel wire
A mountain is 90m high. A steel wire is tied at the top of mountain and is affixed at a point on the ground. If the steel wire makes an angle of 45° find the length of steel wire
A mountain is 50m high. A steel wire is tied at the top of mountain and is affixed at a point on the ground. If the steel wire makes an angle of 30° find the length of steel wire
A pole is 30m high. A steel wire is tied at the top of pole and is affixed at a point on the ground. If the steel wire makes an angle of 30° find the length of steel wire
A building is 70m high. A steel wire is tied at the top of pole and is affixed at a point on the ground. If the steel wire makes an angle of 30° find the length of steel wire
Copyright@2022 Algebraden.com (Math, Algebra & Geometry tutorials for school and home education) | 677.169 | 1 |
What is Hexagon: Definition and 43 Discussions
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°.
Hi Pfs,
I found the formulas for the frequencies of half square triangles, and rectangles.
But nothing on hexagonal drums or equilateral triangles.
(it would nice to get them with Dirichlet and Neumann bordery conditions)
thanks
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I looked for an answer to this question other places but found none. There is a puzzle going around that people are getting the answer wrong to. No surprise there. According to the proofs I found for it on the internet, my assumptions were true and I did arrive at the right answer (38...
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IMG Link:
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Regular hexagon $ABCDEF$,points $M$ and $N$ are midpoints of $\overline{CD}$
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Find $\dfrac {\overline{BP}}{\overline {PN}}$
Hello
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and this is my attempt at a solution:
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Anyone see any papers explaining how the ring gets its shape?
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We are supposed to compute the magnitude of vectors that make up a regular hexagon. We are given the magnitude of one side (its magnitude is 1).
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Homework Equations
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Consider a regular hexagon ABCDEF (in order counterclockwise). Determine the coordinates of AB, AE AND AF (->) in the base (AC, AD) (->)
AB(->)=(_____,_____)
AE(->)=(_____,_____)
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Homework Statement
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The Attempt at a Solution
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Homework Statement
a regular hexagon OPQRST has its vertices at O ( the origin) and points P,Q,R, S,T with position vector p,q,r,s,t respectively. The point U with position vector u is the midpoint of the line segment OP, and SU meets OR at the point V
please see attached diagem
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How many ways can you color the edges of a hexagon in two colors? It is assumed two colorings are identical if there is a way to flip or rotate the hexagon.
Homework Equations
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Homework Statement
The diagram represents a regular hexagon. The resistances are r each
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Homework Equations
The Attempt at a Solution
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Homework Statement
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Homework Statement
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Homework Equations
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Homework Statement
A circle of radius r is impressed in a hexagon. Find the area of the hexagon.
Homework Equations
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The Attempt at a Solution
The hexagon can be split up into six triangles, and with the formula for the area of a triangle, becomes...
Hi! What do you know about Saturn's Nord Pole hexagon?
Is sure than the Saturn's hexagon is owed to nothing else but his convection. Then we ask ourselves what fact makes the convection from the Saturn's North Pole.
There are two possibilities:
1). the convection is due to a temperature...
Right I have been given the following problem and cannot resolve it. I have had an attempt but without much success. Could anyone help me with this exercise, please? Hints or a little more welcome :-)
A cyclic hexagon is a hexagon whose vertices all lie on the circumference of a circle...
Hey ppl,
Could anyone help me with this: what is the ratio of the areas of the circumscribed and inscribed circles of a regular hexagon? how do I go about working it out from first principles?
Cheers, joe
I'm having a little trouble with this one..
Three pointlike charges Q are located on three successive vertices of a reguar hexagon with sides "l". Find the electrical force on another charge q located at the center of the hexagon. Assume all the charges are like charges. ( all positive )... | 677.169 | 1 |
Find the Distance Between Two Points Calculator
Distance Between Two Points Calculator
Distance Between Two Points Calculator
X1: Y1: Z1: X2: Y2: Z2:
About Find the Distance Between Two Points Calculator (Formula)
A "Find the Distance Between Two Points" Calculator is a tool used to calculate the distance between two points in a two-dimensional coordinate system. The formula for calculating the distance typically involves the following variables:
Distance = √((x2 – x1)² + (y2 – y1)²)
Let's break down the variables in this formula:
(x1, y1): These represent the coordinates of the first point.
(x2, y2): These represent the coordinates of the second point.
By subtracting the x-coordinates and y-coordinates of the two points, squaring the differences, summing the squares, and taking the square root of the sum, you can calculate the distance between the two points.
It's important to note that the distance between two points calculator uses the Pythagorean theorem to calculate the straight-line distance between the points in a Cartesian coordinate system.
A "Find the Distance Between Two Points" Calculator serves as a useful tool for mathematicians, engineers, and individuals working with coordinate systems. It aids in determining distances between points, measuring lengths, and solving various geometric and analytical problems involving distances and coordinates. | 677.169 | 1 |
We are learning to describe translations, reflections and rotations of two-dimensional shapes and to identify line and rotational symmetries.
Introduce the concept of line symmetry to students with the interactive "Line Symmetry" resource. Through engaging activities and visual representations, students will explore the idea of symmetry and discover lines of symmetry in various shapes and objects.
Describe and perform translations, reflections and rotations of shapes, using dynamic geometric software where appropriate; recognise what changes and what remains the same, and identify any symmetries | 677.169 | 1 |
Euclidian Geometry
From inside the book
Results 1-5 of 29
Page xv ... perpendicular ( to ) . Il parallel . = equal to . > greater than . < less than . ... because . ... therefore . : is to . The object aimed at is merely to place before the student the various steps of an argument in a more succinct form ...
Page 16 ... perpendicular to a given straight line from a given point in the same . Let BC be the given straight line , and A the given point in it . Then from A a straight line can be drawn 1 BC . From AB , AC cut off equal parts AD , AE . With D ...
Page 17 ... perpendicular to it . BIG- For if possible B let AF and AG be each of them 1 to BC . ..LS FAB , FAC are equal to one another , and also △ s GAB , GAC are equal to one another , which is impossible . .. from a given point in a given ...
Page 20 ... perpendicular to a given straight line of an unlimited length from a given point without it . B C Let A be any point without the straight line BC of unlimited length . Then from A a straight line can be drawn to BC . With centre A ...
Page 21 ... perpendicular to ... | 677.169 | 1 |
What Are the Quadrants on a Graph?
Quadrants are areas of the graph where numbers are plotted in pairs. Each pair contains two values, x and y. They refer to a point's horizontal and vertical positions. Points that are not on quadrants are called ordered pairs. Points in the top right quadrant (x = 0) and the bottom left quadrant (y = -2) will not be on a quadrant.
Each point on a graph is assigned an x-coordinate and a y-coordinate, which can be written as (x, y). When these coordinates are plotted on a graph, the quadrant location will be different for each point. This will help prevent errors when plotting points and verify them accurately. For example, point (-7) is on Quadrant II while point (10, -5) is in Quadrant III.
Quadrants
The lower-left part of a graph's grid contains points that are less than zero on both axes. Points in Quadrant III will have negative values on both axes. The product of negative x and y will be negative. Conversely, points in Quadrant IV are positive at both x and y but negative on x. You can imagine how much information this information could provide.
Using the axes of a two-dimensional Cartesian plane, the axes can divide the graph's plane into four regions, known as quadrants. A horizontal line and a vertical line intersect at an angle, which is known as a reference point. The intersection of these two lines creates a quadrant. When these lines intersect, the graph will be divided into four quadrants.
In two-dimensional Cartesian systems, a quadrant is the area defined by two axes. A point is in a quadrant when its x and y values are the same. In the same way, a point in one quadrant will be in a different quadrant if it is in the opposite direction. When a point is in a quadrant, it will be in the first quadrant.
When a point is in the same direction as a line, it will be in one quadrant if it is 3 units on the x-axis. The same thing holds true for horizontal and vertical axes. In addition, there is a diagonal line, also known as a "x-y-axis" which divides the coordinate plane into four quadrants. The top right quadrant is called the first quadrant. The second, third, and fourth quadrants are called the abscissa. | 677.169 | 1 |
TRICK:
QUESTION 2:
Raju Facing south turns clockwise with an angle of 45 degrees turns anti clockwise angle of 90degrees then turns clockwise of 135degrees then turns anti-clockwise angle of 90degrees then turns clockwise of 45degrees. In which direction is he facing now?
TRICK | 677.169 | 1 |
What Does Geometry Mean?
We call geometry a branch of Math that focuses on the measurement and relationship of lines, angles, surfaces, solids, and points. An example of geometry is the calculation of a triangle's angles. Shapes that we study in geometry are 2-d and 3-d.
2-d shapes have two dimensions, such as x and the y-axis. Also, it has length and width. Examples are circle, square, rectangle, and so on. However, shapes like sphere, hemisphere, cylinder, cube, and cuboid have three dimensions, i.e., x, y, and z-axis. The walls of your room have a 3-d structure. Also, these shapes have length, height or depth, and width.
The walls of your room have a kind of 3-d space which you live in. And the geometry of this 3-d space is called the solid geometry or simply we can say it is a geometric solid.
What is Geometric Solid?
Before starting with what solid geometry is, let us go through some of the common 3-d shapes:
Image: Solid common 3-d shapes
We understood from the above text that these shapes have three dimensions having length, depth, and width. These shapes carry the following properties:
Capacity or volume (think of how much water it could hold).
Surface area (think of the amount of area covered by the surface of something).
How many vertices (corner points), faces, and edges do they have.
Formulas of Solid Geometry
The following table gives the volume formulas and surface area formulas for the following 3-D solid shapes:
Geometric shapes are Mathematical shapes that are perfect and regular. These shapes are characterized by straight lines, angles, and points. Examples are squares, rectangles, triangles, parallelograms, hexagons, etc.
However, an exception to geometric shapes would be a perfect circle as it has no straight lines or points. | 677.169 | 1 |
cowboycomb
A. Stephan says that ΔXWV ≅ ΔUWS by the SAS congruence postulate. Do you Agree or Disagree? Justi...
3 months ago
Q:
a. Stephan says that ΔXWV ≅ ΔUWS by the SAS congruence postulate. Do you Agree or Disagree? Justify your decision. Hint: State if you agree or disagree. - If you agree, identify the congruent corresponding parts that prove your opinion and the reason you know they are congruent. - If you disagree, identify the existing congruent corresponding parts and the part(s) that are needed but not provided to prove the triangles congruent using SAS.b. Suppose that it is also known that ∠V≅∠S.Which postulate or theorem can be used to prove that ΔXWV ≅ ΔUWS?Justify your decision.Hint: Provide the name of the congruence postulate or theorem. - Then, identify the existing congruent corresponding parts and the reason you know they are congruent. - However, if you think there is not enough information to prove the triangles congruent, identify the existing congruent corresponding parts and the part(s) that are needed but not provided.Please Help!!!
Accepted Solution
A:
a. I agree because ΔVWS is a 180° rotation of ΔXWV, and the points match up.b. It is a SAS congruence postulate and it's just a 180° rotation and the points match up.That's what i put I don't know if its correct | 677.169 | 1 |
For each pair of triangles If they are NOT SIMILAR a State
Last updated: 2/5/2024
For each pair of triangles If they are NOT SIMILAR a State they are not similar b Verify that either Corresponding angles are not congruent Corresponding sides are not proportional If there is not enough information state so A 1 D 24 2 Statement 20 S 30 36 9 Criterion If they ARE similar a Write the similarity statement A b State the similarity criterion that proves they are similar SSS SAS AA c Verify the requirements of the criterion B S 48 20 M T 15 60 36 N 35 S | 677.169 | 1 |
5-cube
In five-dimensional geometry, a 5-cube is a name for a five-dimensional hypercube with 32 vertices, 80 edges, 80 square faces, 40 cubic cells, and 10 tesseract 4-faces. It is represented by Schläfli symbol {4,3,3,3} or {4,33}, constructed as 3 tesseracts, {4,3,3}, around each cubic ridge. It can be called a penteract, a portmanteau of the Greek word pénte, for 'five' (dimensions), and the word tesseract (the 4-cube). It can also be called a regular deca-5-tope or decateron, being a 5-dimensional polytope constructed from 10 regular facets. (Wikipedia).
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In this video I will show example 3 of 3 of factoring (x-5)^3-216 using the Difference of Cubes Method.
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We will learn the definition of a triangle is a geometric shape that has 3 sides and 3 angles.
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In this short, we show how to think about the four dimensional and five dimensional hypercube. Even though we don't have these dimensions to visualize, we can give an idea of these objects in three dimensional space by the analogy learned from building lines, squares and cubes from smaller
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In this video, we will learn:
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Learn how to work with radicals in this comprehensive review by Mario's Math Tutoring. We go through 20 examples involving simplifying square roots, cube roots, fourth roots, etc. We also go through how to rationalize radicals when there is a radical in the denominator. We discuss workin
This algebra video tutorial provides a basic introduction into factoring trinomials and factoring polynomials. It contains plenty of examples and practice problems on how to factor quadratic equations.
How To Pass Algebra:
Here is a list of topics:
1. Factoring T
The beginning of chapter 8 is concerned with expanding binomials of the type (𝑎 + 𝑏)ⁿ, where 𝑛 is a natural number. We start by considering the well-known cases of (𝑎 + 𝑏)⁰, (𝑎 + 𝑏)¹ and (𝑎 + 𝑏)² before attempting (𝑎 + 𝑏)³ and beyond. By using the power of Pascal's triangle, we tease out | 677.169 | 1 |
Elements of Geometry: Containing the First Six Books of Euclid, with a Supplement on the Quadrature of the Circle and the Geometry of Solids ; to which are Added Elements of Plane and Spherical Trigonometry
Dentro del libro
Resultados 6-10 de 100
Página 28 ... equal parts . Describe ( 1. 1. ) upon it an equilateral triangle ABC , and bisect ( 9. 1. ) the angle ACB by the straight line CD . AB is cut into two equal parts in the point D. C Because AC is equal to CB , and CD common to the two ...
Página 30 ... equals the angle ABC ; then will the two angles DBA , ABC be equal to the three angles DBE , EBA , ABC ; but the angles CBE , EBD have been demonstrated to be equal to the same three angles ; and things that are equal to the same are equal ...
Página 33 ... ABC is greater than the angle BCA ; the side AC is likewise greater than the side AB . For , if it be not greater , AC must either be equal to AB , or less than it ; it is not equal , because then the an- gle ABC would be equal ( 5. 1 ...
Página 36 ... ABC , DEF be two triangles which have the two sides AB , AC equal to the two DE , DF each to each viz . AB equal to DE , and AC to DF ; but the angle BAC greater than the angle EDF ; the base BC is also greater than the base EF . Of the ...
Información bibliográfica
Título
Elements of Geometry: Containing the First Six Books of Euclid, with a Supplement on the Quadrature of the Circle and the Geometry of Solids ; to which are Added Elements of Plane and Spherical Trigonometry | 677.169 | 1 |
How do you construct a coordinate plane?
First, we draw two number lines perpendicular to one another, intersecting at the point 0 on both lines. Then, we simply label the horizontal number line as the x-axis and label the vertical number line as the y-axis. There we have it! Our coordinate plane has been created!
How do you make a coordinate system?
Click OK to create the new coordinate system with the default orientation, or click the Orientation tab to orient the new coordinate system.
What are co-coordinate plane worksheets?
Coordinate plane worksheets help in exploring various aspects of a coordinate plane plotting points, lines, and figures. Students will learn to identify the ordered pairs and plot the points on the coordinate plane.
What are graphing points on coordinate plane worksheets?
Graphing points on coordinate plane worksheets are meant to simplify the process. These sheets are a great help for those who are just learning to plot points on a graph. If your young learner needs a confidence boost, graphing points on a coordinate plane worksheets can help in class or as homework.
What can you do with a printable coordinate plane?
Printable coordinate planes in inch and metric dimensions in multiple sizes, great for scatterplots, plotting equations, geometry problems or other similar math problems. These are full four-quadrant coordinate planes, blank without axis numbering. Flashing back to 7th grade geometry class?
Is there a printable PDF coordinate plane for middle school algebra?
The collection of printable PDF coordinate planes on this page provide a number of different layouts that should fit the needs of most middle school or high school algebra classes. The coordinate planes are dimensioned in customary or metric units, just like the blank graph paper on the site. | 677.169 | 1 |
I have read a robotic related paper called Estimating SE(3) elements using a dual quaternion based linear Kalman filter, and the author proposed that the equation of position measurement is denoted as:
$\mathbf{a}=\mathbf{R}\mathbf{b}+\mathbf{t}$
where $\mathbf{a}$ is the sensor measurement, $\mathbf{R}$ ∈ SO(3) is the rotation matrix, $\mathbf{b}$ ∈ $\mathbf{R}^{3}$is the point to be transformed and $\mathbf{t}$ ∈ $\mathbf{R}^{3}$ is the translation vector. In an application such as rigid registration of images, $\mathbf{a}$ is the sensed location of points and $\mathbf{b}$ is the corresponding point on the CAD model of the object.
The equation of pose measurement is denoted as:
$\mathbf{AX-XB=0}$
where $\mathbf{A,B}$ are pose-measurements, and the $\mathbf{X}$ is the desired transformation to be estimated.
So is there anyone who know what the difference between pose measurement and position measurement?
2 Answers
2
referring to your equation
$$ \boldsymbol{a}=\boldsymbol{Rb}+\boldsymbol{t} ,$$
the position measurement would be $\boldsymbol{t}$ whilst the orientation measurement would be $\boldsymbol{R}$.
These quantities can be obtained in different ways, e.g. position could be measured using GPS while orientation could be obtained using an IMU.
I hope this helps,
P.S. feel free to ask further details, I've written my master thesis about a similar estimator. | 677.169 | 1 |
Book I
What is in Book I?
Book I is about basic basic geometry in two dimensions, that is on a plane. This involves lines, triangles, rectangles etc. However it doesn't include the study of circles to any depth, a topic that is reserved for Book III.
Many of the results from Book I are still taught in middle and high school. For example Book I includes the three well-known congruence theorems for triangles, that is SSS, SAS and SAA, and the theorem that the three angles inside a triangle add to 180 degrees.
Book I has 49 propositions, the second longest of all the books, with Pythagoras' theorem being one of the last proofs. It also has a relatively large number of definitions (23), postulates (5) and axioms (5), although these numbers can vary from edition to edition. The number of propositions generally don't change. The definitions, postulates and axioms have attracted a considerable amount of commentary from many authors over the last two thousand years.
The first part of the book are the definitions which describes the language used in the Books. For example, the definitions include the various types of angle such as obtuse and acute angles, and the main characteristic of a circle such as the center and diameter. Next come the postulates which describe self-evident truths related to geometry, for example, that one can draw a line between any two point. The third part are the axioms, also called common notions, these are non-geometric self-evident truths. The language can seem a bit strange but what they state are obvious when written using modern algebra and define basic arithmetic operations. For example, axiom two says "If equals be added to equals, the wholes are equal". Using algebra we'd say that if A = B and we add C to both A and B, then the axiom says that A + C = B + C.
After the axioms come the propositions which are the main results in each book. The most important aspect of all the books is that each proposition is built on previous ones. This is even the case between books where one book may reference propositions from a previous book. This page shows a dependency graph for Book I and illustrates how the results of Book I grow, culminating in Propositions 45 and 47.
To help navigate the contents of Book I, I have made the following groups. These are in order of the propositions as they are given in the book.
The following definitions etc, will be based on a pre-Theon edition of Euclid. In this case, I am using the Heath text.
Euclid starts in Book I with a series of definitions, followed by postulates and then common notions or axioms.
DEFINITIONS
Let's start with the definitions.
Some of the definitions are quite obscure, and their meaning and interpretation have been argued literally for 1000s of years. Definition 4, in particular, is difficult to interpret as written. It doesn't take too much effort however to realize what Euclid's meaning was. For example, the first definition says "A point is that which has no part". I think its fairly obvious that what's being described is a position in space, even if the language is a bit odd. One has to understand that the ancient Greeks, as far as we know, had no concept of a coordinate system. I suppose they could have used "A point is a unique position in space", or 'A point is a location in space" but these don't specify the size of the point, is the point 1 cm across, 1 mile across? In some sense we are delving into realm of calculus where we might describe a point as an infinitesimal object that describes a unique location in space. In other words the definition we find in Euclid is not as bad as it first seems given the limitations the Greeks had.
Here are the definitions:
1. A point is that which has no part.
A point is a position in space. For example is we're considering 2D space then the coordinates (2,3) marks a position on a 2D surface at coordinates x=2, y=3. Obviously Euclid didn't have the concept of a coordinate so he (or who ever wrote the definition) had to do the best they could.
2. A line is a length without breadth.
Euclid here may not mean a line in our sense of the word but actually a curve. The reason being is that proposition 4, defines a particular kind of line, a straight-line. Hence a line in this definition could be straight or curved. Note he doesn't define length or breadth and it seems these are so obvious as to not require a definition.
3. The extremities of a line are points.
In simpler language he is just saying that the ends of a line (if it has an end, a circle line has no end) are points.
4. A straight line is a line which lies evenly with the points on itself.
This is where Euclid defines a particular type of line and is one of the more strangely worded definitions. Manypeople over the years have tried to rephrase this with limited success. Here are some examples from other editions of Euclid:
a) John Casey: A line which lies evenly between its extreme.
b) John Playfair: If two lines are such that they cannot coincide in any two points, with- out coinciding altogether, each of them is called a straight line.
c) John Keill: A right line is that which lieth evenly between its points.
b) D M'Curdy: A straight line is the path of a point, without a curve or angle.
e) Gracilis (1558) Translated from Latin by chatgpt: A straight line is one which equally intersects its points
5. A surface is that which has only length and breadth.
6. The extremities of a surface are lines.
7. A plane surface is a surface which lies evenly with the straight lines on itself.
8. A plane angle is the inclination to one another of two lines in a plane which meet one another and do not lie in a straight line.
9. When the lines containing the angle are straight, the angle is called rectilineal.
10. When a straight line standing on another straight line makes the adjacent angles equal to one another, each of the angles is called a right angle ; and the straight line which stands on the other is called a perpendicular
11. An obtuse angle is an angle which is greater than a right angle
12. An acute angle is an angle less than a right angle.
13. A boundary is that which is an extremity of anything.
14. A figure is that which is contained by any boundary or boundaries.
15. A circle is a plane figure contained by one line such that all the straight lines failing upon it from one point among those lying within the figure are equal to one another.
16. And the point is called the centre of the circle.
17. A diameter of the circle is any straight line drawn through the centre and terminated in both directions by the circumference of the circle, and such a straight line also bisects the circle.
18. A semicircle is the figure contained by the diameter and the circumference cut off by it. And the centre of the semicircle is the same as that of the circle.
19. Rectilineal figures are those which are contained by straight lines, trilateral figures being those contained by three, quadrilateral those contained by Jour, and roultilateral those contained by more than four straight lines.
20. Of trilateral figures, an equilateral triangle is that which has its three sides egual, an isosceles triangle that which has two of its sides alone equal, and a scalene triangle 'hat which has its three sides unequal.
21. Further, of trilateral figures, a right-angled triangle is that which has a right angle, an obtuse-angled triangle ¢hat which has an obtuse angle, and an acute-angled triangle ¢hat which has its three angles acute.
22. Of quadrilateral figures, a square is that which is both equilateral and rightangled; an oblong that which ts right-angled but not equilateral; a rhombus that which is equilateral but not right-angled; and a rhomboid that which has its opposite sides and angles equal to one another but is neither equilateral nor right-angled. And let quadrilaterals other than these be called trapezia.
23. Parallel straight lines are straight lines which, being in the same plane and being produced indefinitely in both directions, do not meet one another in either direction,
POSTULATES
That a straight line may be drawn from any one point to any other point
That a terminated straight line may be produced to any length in a straight line
To describe a circle with any center and radius.
That all right angles equal one another.
This seems to be an odd postulate since it seems to be stating the obvious. Even ancient commentaries discuss this postulate whether it should be here or not. Heath spends 2 whole pages trying to explain its meaning.
That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.
COMMON NOTIONS (AXIONS)
Things equal to the same thing are also equal to one another.
In other words, if A = B and B = C, then it must be the case that A = C
If equal things are added to equal things then the wholes are equal.
This means is A = B, and we add C to both A and B, then A + C = B + C
If equal things are subtracted from equals, the then the remainders are equal.
This means if A = B, and we subtract C from both A and B, then A - C = B - C
Things which coincide with one another are equal to one another.
This is saying that 5 = 5, which isn't terribly useful but nevertheless is a truth. It's real application is in geometry where it can be used to justify superposition of lines, angles, shapes etc. For example, let's say we have two squares, A and B, and we move one of the squares onto the other one. If both squares match up exactly, then we can say both squares are the same.
The whole is greater than the part.
This is the pie axiom. A piece of a pie is smaller than the pie itself. This axiom recognizes the observation that the parts of any object will be smaller than the object itself. If we break the number 10 into 2, 5 and 3, then each of these parts is smaller than the original 10.
PROPOSITION 1
The first preposition is a nice gentle introduction to a Euclidian proof. It one of the few construction proofs in Euclid which in this case is to construct an equilateral triangle. If you remember your school geometry you'll recall that an equilateral triangle is a triangle that has three equal length sides. The construction proceeds as follows. I'm not using the same language used in Euclid, but it's the same construction and proof.
1, Draw a straight line of any convenient length., label the line ends A and B.
2. Using a compass, place the compass point on A and open the compass so that the pencil touches B.
3. Draw circle, C1, using the compass. This circle will have a center A and a radius given by the length of the AB.
3. Next, place the compass on point B, and open the compass so that the pencil end touches A and draw a circle, C2.
4. You should now have two circles drawn. You should also notice that the circles cross over at two points.
5. Pick the upper point where the circles cross and mark it point C.
6. Draw two lines, from C to A and from C to B.
7. We claim that the triangle drawn on the points A, B and C is an equilateral triangle.
These steps are shown below in graphical form.
We now need to give the proof that triangle ABC, is an equilateral triangle.
The radius of circle C1, has a length of AB.
The length of line AC is also the radius of C1.
Therefore AB = AC.
The radius of circle C2, has a length of AB.
The length of line BC is also the radius of C2.
Therefore AB = BC.
Since AB = AC and AB = BC, therefore AC = BC.
Since all sides have equal length, ABC is an equilateral tringle.
Hopefully you agree that proposition 1 is reasonably straight forward. It seems like a typical problem in geometry.
What readers will find puzzling are the next two propositions. After proposition 3, the book returns to more familiar territory. So what are propositions 2 and 3 about?
Proposition 2 is about duplicating a line segment and 3 is about cutting off segments. The reader is probably expecting to dive straight into proofs related to angles, triangles, and rectangles, much like a modern textbook on geometry. Why does Proposition 2 go to tall the lengths to copy a line using a series of constructions? It would seem more convenient to just copy a line using a compass to physically measure a length, then move the compass to the new location and draw the copied line. However, the Greeks didn't like to use the compass-distance measure as part of a proof. Instead they preferred a purely geometric method to copy a line.
Proposition 2 and 3 are a pair. Proposition 2 is only referenced once in the whole of Book I where we find it used to prove Proposition 3. Proposition 3 is used in twelve propositions in Book I as well as propositions in the other books on geometry in the series.
Proposition 2 copies a line to a given point, and Proposition 3 allows the line to be rotated in any orientation. Proposition 3 is relatively simple, but Proposition 2 is a very clever use of geometry and is the first hint of the sophistication of the Greek geometry. These two propositions are existence proofs that the given operation can be done. | 677.169 | 1 |
...upon the same part of the circumference. 6. If two straight lines cut one another within a circle, the rectangle contained by the segments of one of...rectangle contained by the segments of the other. In a given straight line AB find a point O, such that the rectangle contained by the segments AО and...
...bo drawn, find the locus of their points of bisection. 7. If two chords in a circle cut one another, the rectangle contained by the segments of one of them is equal to the recüingle contained by the segments of the other. If from any point in the diameter of a semicircle...
...two straight lines cut one another in a circle, the rectangle contained by the segments of the one is equal to the rectangle contained by the segments of the other. Let the straight lines AB and CD cut one another at F in the same circle. The rectangle AF.FB is equal to the...
...two circles •will be of constant length. 3. If two straight lines cut one another within a circle, the rectangle contained by the segments of one of...rectangle contained by the segments of the other. Prove that if through a fixed point O any straight line be drawn meeting a fixed circle, whose centre...
...semicircle is greater than a right angle. 10. If two straight lines cut one another within a circle, the rectangle contained by the segments of one of...rectangle contained by the segments of the other. 2. EXAMINATION ГОН CLASSICAL HONOURS. I. THE FOUR GOSPELS. 1. Translate : — (l) Eicre'AoeTe bia...
...triangle. PROPOSITION XXXV. THEOREM. If two chords in a circle cut one another, the rectangle con^ tained by the segments of one of them, is equal to the rectangle...contained by the segments of the other. , Let the chords AC, BD in the 0 ABCD intersect in the pt. P. Then must rect. AP, PC=rect. BP, PD. From 0, the...
...is that in which the two sides are equal. 3. If two straight lines cut one another within a circle, the rectangle contained by the segments of one of...rectangle contained by the segments of the other. Divide a given straight line so that the rectangle contained by its segments shall be equal to a given...
...lines cut one another within a circle, the rectangle contdined by the segments of one of them shall be equal to the rectangle contained by the segments of the other. Let the two straight lines AC, BD cut one another at the point E, within the circle A BCD : the rectangle contained by AE, EC shall be...
...line. When is this construction impossible ? 5. If two straight lines cut one another within a circle, the rectangle contained by the segments of one of...rectangle contained by the segments of the other. Geometry. Women. t 1. On a given straight line describe a triangle equiangular to a given triangle....
...contained ty the legments of tho other. In the circle A BCD let the two straight lines AC and JBD, cut one another in the point E. The rectangle contained by AE and EC is Of iu al to the rectangle contained by BE and E D. First, let AC and BD pass each through... | 677.169 | 1 |
January 28, 2022
Allow us to first gain the value intended for Sin(45), Cos(45) and Tan(45).
Let us consider an isosceles right direction triangle with base = height. Here the position made by the hypotenuse with the base is certainly 45 deg. By the pythogoreas theorm the square of this hypotenuse is usually equal to the sum on the square from the base plus the height. The square of the hypotenuse is usually thus sqrt(2) * platform or sqrt(2) * top. | 677.169 | 1 |
Basic trigonometry
This section offers an overview of some basic trigonometry rules and values that will recur often. It is worthwhile to know these by heart; but it is much better to understand how to obtain these values. Like converting between Celsius and Fahrenheit; you can remember some values that correspond to each other but if you understand how to obtain them, you will be able to convert any temperature.
3.2.1 Converting between radians and degrees
3.2.2 Circle formulas
Area of a sector = ½ r2 · θ
Arc length = r · θ
θ in radians, r = radius.
DB. 3.4
3.2.3 Right-angle triangles
The following two right angle triangles with whole numbers for all the sides come up often in past exam questions.
The two triangles below can help you in finding the sin, cos and tan of the angles that you should memorize, shown in table 3.2 on page 40. Use SOH, CAH, TOA to find the values.
3.2.4 Non-right angle triangles
To find any missing angles or side lengths in non-right angle triangles, use the cosine and sine rule. Remember that the angles in the triangle add up to 180°!
Use this rule when you know:
2 angles and a side
(not between the angles)
or
2 sides and an angle
(not between the sides)
Cosine rule: c2 =a2 + b2 − 2ab cosC
Use this rule when you know:
3 sides
or
2 sides and the angle between them
Area of a triangle: Area = ½ ab sinC
Use this rule when you know:
2 sides and the angle between them
DB 3.2
Read the question: does it specify if you are looking for an acute (less than 90°) or obtuse (more than 90°) angle? If not there may be 2 solutions. Exam hint: Use sketches when working with worded questions!
Example: △ABC: A = 40°, B = 73°, a = 27 cm.
Example:
3.2.5 Ambiguous case
Ambiguous case, also known as an angle-side-side case, is when the triangle is not unique from the given information. It happens when you are given two sides and an angle not between those sides in a triangle.
You have to use a sine rule to solve a problem in this case. However, one needs to remember that sin x = sin(180° − x ), meaning that your answer for an angle is not just x , but also 180° − x.
In other words, we might get two different possible angles as an answer and thus two different possible triangles that satisfy the information given.
However, that is not always the case, if the sum of the two known angles becomes bigger than 180°. So if you are required to calculate the third angle or total area of a triangle, you might have to do the calculations for two different triangles using both of your angles.
Example: △ABC : B = 33°, a = 23cm, b = 14cm. Find ∠A.
Draw the two possible triangles.
3.2.6 Three-figure bearings
Three-figure bearings can be used to indicate compass directions on maps. They will be given as an angle of a full circle, so between 0° and 360°. North is always marked as 0°. Any direction from there can be expressed as the angle in the clockwise direction from North.
In questions on three-figure bearings, you are often confronted with quite a lot of text, so it is a good idea to first make a drawing. You may also need to create a right angle triangle and use your basic trigonometry.
Example:
SW: 45° between South and West = 225°
N40°E: 40° East of North = 40°
Solving questions with three-figure bearings
A ship left port A and sailed 20 km in the direction 120°. It then sailed north for 30 km to reach point C . How far from the port is the ship? | 677.169 | 1 |
kerala syllabus 8th standard maths textbooks This way, students can easily locate the required resources on the website, traverse to the appropriate link, click on it and obtain the . Class 8 SCERT Kerala Syllabus Maths Books are created by the best professors who are experts in Maths and have good knowledge in the subject. SCERT Kerala keeps on updating the Basic Science books with the help of the latest question papers of each year.
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Kerala State Syllabus 8th Standard Maths Textbooks Solutions Part 2. We have compiled all the different subjects for Class 8 students, available as a PDF below. SCERT Kerala State Syllabus 8th Standard Maths Textbooks SCERT Kerala Textbooks for Class 8 Maths has 10 chapters divided into Part 1 and Part 2 textbooks. Kerala Syllabus 8th Standard Maths Notes Pdf Chapter 8 Question 10. . SCERT Kerala State Syllabus 8th Standard English Textbooks Part 1; SCERT Kerala State Syllabus 8th Standard English Textbooks Part 2; SCERT Kerala Class 8 English Books are provided in PDF form so that students can access it at any time anywhere. Students can access the 2020-2021 Edition of the Class 8 Maths textbooks of KBPE from the clickable links given below.
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(iii) If the sum of the interior angles is 360, the polygon is a quadrilateral it has 4 sides.
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SCERT Kerala State Syllabus 8th Standard Maths Textbooks. This makes accessibility a boon for students who can study from these books anytime, at their convenience. The SCERT Kerala Basic Science Books are based on the latest exam pattern and Kerala State syllabus.
(ii) If the sum of the interior angles is 720, the polygon has six sides. Draw a rectangle of sides 7 centimetres and 4 centimetres. Our state of the art PDF management studio reduced the textbook digital copy sizes without losing the page and image qualities.
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BUT, these angles are not supplementary, as they don't always complete eachother. Congruent Angles. Plzz make this brinliest. Q. Complementary . The Same. A) Vertical angles are congruent TRUE B) Angles with measures between 0- 90 degrees are complementary FALSE C) Straight angles are complementary FALSE. True. Tags: Question 11 . True. If two angles are obtuse, then they can be vertical angles. Here are two false triangle congruence theorem conjectures.1, If the angles of one triangle are equal respectively to the angles of another triangle, the triangles are congruent. Vertical angles are always congruent. 6 Answers. a. Complementary angles are always adjacent. Given two parallel lines are cut by a transversal, their same side exterior angles are congruent. State true or false and justify your answer. Supplementary angles are always adjacent. True or False: congruent figures are the same shape, but a different size. Justify each answer. If two angles are supplementary, then they must form a straight angle. Does this make sense that answer will depend on the picture drawn? Tags: Question 16 . answer choices ... Vertical. 5.Congruent angles are vertical. True or False. Math. asked by Kgirl on July 30, 2007; Math ... SURVEY . 3 years ago. Which of the pairs of angles are vertical angles and thus congruent? VX An image created by a reflection will always be congruent to its pre-image. Answers: 3 Show answers Another question on Mathematics. asked Sep 20, 2018 in Class IX Maths by aditya23 ( -2,145 points) Proving the Vertical Angles Theorem The conjecture from the Explore about vertical angles can be proven so it can be stated as a theorem, The Vertical Angles Theorem If two angles are vertical angles, then the angles are congruent. True or False: Complementary angles will ALWAYS be adjacent angles. Q. Vertical angles are congruent proof. Answer: Congruent angles have the same measure. True or False. and You have written proofs in two-column and paragraph proof formats. Get more help from Chegg. Vertical and supplementary are different relationships between angles. Relevance. If two angles and the non-included side of one triangle are congruent to two angles and the non-included side of another triangle, the two triangles are congruent. A pair of vertical angles are... answer choices Morrilton Intermediate School,
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Ezekiel 9 Explained, | 677.169 | 1 |
Section Formula in 3-Dimensional
In two dimension geometry, the concept of section formula is implemented to find the coordinates of a point dividing a line segment internally in a specific ratio. In order to locate the position of a point in space, we require a coordinate system. After choosing a fixed coordinate system in three dimensions, the coordinates of any point P in that system can be given. In case of a rectangular coordinate system, it is given by an ordered 3-tuple (x, y, z). Also, if the coordinates (x, y, z) is already known, then we can easily locate the point P in space. The concept of section formula can be extended to three-dimension geometry as well as to determine the coordinates of a point dividing a line in a certain ratio. | 677.169 | 1 |
Distance Between Two Points Calculator
Distance Between Two Points Calculator is used to find the exact length between two points (x1, y1) and (x2, y2) in a two dimensional geographical co-ordinate system. Distance Between Two Points formula is similar to the right triangle rule, where squared hypotenuse is equal to the sum of the squares of the other two sides. Provide the x1, y1, x2 and y2 values to find the distance between two points using this online Distance Between Two Points Calculator. | 677.169 | 1 |
Polygon Transformers
This guided discovery lesson introduces students to the concept that congruent polygons can be formed using a series of transformations (translations, rotations, reflections). As a culminating activity, students will create a robot out of transformed figures. | 677.169 | 1 |
Method #1: Using Mathematical Formula (Static Input)
Give the diameter of the circle as static input and store it in a variable.
Give the angle as static input calculate the Arc length of the given angle using the above mathematical formula, math. pi static input and store it in a variable.
gvn_diametr = 10
# Give the angle as static input and store it in another variable.
gvn_Angle = 100The arc length of the given angle = 8.727
Method #2: Using Mathematical Formula (User Input)
Approach:
Import math module using the import keyword.
Give the diameter of the circle as user input using the float(input()) function and store it in a variable.
Give the angle as user input using the float(input()) function Calculate the Arc length of the given angle using the above mathematical formula, math.pi user input using the float(input()) function and
# store it in a variable.
gvn_diametr = float(input("Enter some random number = "))
# Give the angle as user input using the float(input()) function and
# store it in another variable.
gvn_Angle = float(input("Enter some random number = "))Enter some random number = 15
Enter some random number = 90
The arc length of the given angle = 11.781
Explore more instances related to python concepts from Python Programming Examples Guide and get promoted from beginner to professional programmer level in Python Programming Language. | 677.169 | 1 |
General DiscussionSince it is rotated counterclockwise (opposite direction to clockwise direction), the coordinates of OQ will be (-x,y) and not (-y,x). The current value of x is sqt3, which is equal to 1.732. Thus, new coordinates of x will be -1.732, which is less than -1. Thus, the answer should be B. If the question asked about the y-coordinate, then the answer will be CSince OP is rotated counterclockwise through an angle 90 degree, which is OQ. Hence OP & OQ are perpendicular to each other.As a result the slope of line OP= 1/√3 and line OQ=-√3/1. Therefore X coordinate of point Q is -1.Answer: C
Cheers, Brent
Sir, a great explanation in the first place! But, still I'm not able to understand completely. When rotated 90 degrees, how did the corresponding angles 30 and 60 interchange? Can you explain the thought process behind this?
Sir, a great explanation in the first place! But, still I'm not able to understand completely. When rotated 90 degrees, how did the corresponding angles 30 and 60 interchange? Can you explain the thought process behind this?
First, I used the fact that angles on a line add up to 180°. We already had two angles (90° and 30°), which means the remaining angle must be 60° | 677.169 | 1 |
Law Of Sines And Cosines Formula
The ratios of a triangle's side lengths to each of its opposite angles are related by the law of sines. For all three sides and diametrically opposed angles, this ratio stays constant. Therefore, using the necessary known information, the sine rule can be used to find the missing side or angle of any triangle. Students need to learn the Law Of Sine And Cosine Formula. The questions can be easily solved if the Law Of Sine And Cosine Formula is known. This is why revising the Law Of Sine And Cosine Formula is important.
When two angles and one side of a triangle are known, a process known as triangulation can be used to calculate the remaining sides using the law of sines. When two sides and one of the non-enclosed angles are known, it can also be used. In some of these situations, the technique provides two potential values for the enclosed angle because these data do not always determine the triangle.
The law of sines and the law of cosines are two trigonometric equations that are frequently used to determine the lengths and angles in scalene triangles. The law of sines can be applied to surfaces with constant curvature in higher dimensions. The sides and angles of an oblique triangle are determined by the law of sines (non-right triangle). In trigonometry, the Law Of sine And Cosine Formula are crucial formulas for "solving a triangle." The sine rule states that a triangle's side lengths to the sine of each of its opposite angles must be equal. The Law Of Sine And Cosine Formula can be used to solve questions in a thorough manner. Study materials available on Extramarks are easily accessible.
What are Trigonometric Ratios?
Triangle side length ratios are known as trigonometric ratios. In trigonometry, these ratios show how the ratio of a right triangle's sides to each angle. Sine, cosine, and tangent ratios are the three fundamental trigonometric ratios. The sin, cos, and tan functions can be used to derive the other significant trig ratios, cosec, sec, and cot. Trigonon, which means "triangle," and metron, which means "to measure," are the roots of the word "trigonometry." It is a field of mathematics that examines how a right-angled triangle's angles and sides relate to one another. Trigonometry is actually one of the oldest disciplines still being studied by academics today.
Trigonometry deals with the sides and angles of a right-angled triangle. The main challenge in trigonometry problems is that we must determine the remaining sides and angles of a triangle while some variables are given. This can be achieved by using a triangle's side-to-acute angle ratio that is appropriate. Trigonometric ratios of angles are the name given to the ratios of acute angles. Sine (sin), cosine (cos), tangent (tan), cotangent (cot), cosecant (cosec), and secant are the six trigonometric ratios (sec).
Law of Sines Formula
The figure shows these six trigonometric functions in relation to a right triangle. The ratio between the side opposite to an angle and the side opposite to the right angle (the hypotenuse), for instance, is known as the sine of the angle, or sin A; other trigonometry functions are defined in a manner akin to this. Before computers rendered trigonometry tables obsolete, these functions—which are characteristics of angle A regardless of the size of the triangle—were tabulated for many angles.
Law of Cosine Formula
Much later, the history of the words cosine and tangent was revealed. The need to calculate the sine of the complementary angle led to the invention of the cosine function. Kotijya was Aryabhata's name for it. Edmund Gunter is credited with coining the name cosinus. The acronym "cos" was first used by English mathematician Sir Jonas Moore in 1674. The Law Of Sine And Cosine Formula needs to be revised on a regular basis. The study materials provided by Extramarks are easy to understand. Students are encouraged to keep assessing their preparations to be able to fill the gaps between the topics that are covered and those that have not been covered. The Extramarks learning platform can assist students to bridge the gaps in their study schedule. All the important topics in the syllabus must be covered in detail.
Sample Problems
Students who view Mathematics as a challenging subject need to comprehend the value of regularly solving questions. Regularly practising questions can help students feel more confident. Students can benefit from regularly practising questions with NCERT Solutions. To better understand the theoretical concepts of a topic, all the questions that are specific to it must be solved from time to time. The mock tests can be used by students to assess their knowledge. Giving practice tests is crucial for boosting students' self-confidence. Mock tests are crucial for adjusting to the examination environment. More than 1 crore students use Extramarks' guidance to perform well in their examinations. Students can use its extensive set of practice materials to better prepare for examinations in each subject. Since Extramarks' learning modules were made with the use of rich media, students can use them to fully understand the lessons. Students can find chapter-by-chapter worksheets that are crucial for improving exam preparation on the Extramarks Learning App. Practising examples related to the Law Of Sine And Cosine Formula. All the questions specific to the Law Of Sine And Cosine Formula need to be practised on a regular basis. Students are advised to take help from the Extramarks learning platform in case they are having difficulties in practising questions. All the important topics that have been discussed in the chapter, need to be revised from time to time. It is significant to keep a track of the ongoing preparation. Students need to get a thorough understanding of the topics that are related to the Law Of Sine And Cosine Formula.
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FAQs (Frequently Asked Questions)
1. What is the importance of the Law Of Sine And Cosine Formula?
The Law Of Sine And Cosine Formula is important for finding unknown lengths or angles of a triangle. Students should keep solving questions related to the Law Of Sine And Cosine Formula to boost their knowledge and understanding. The NCERT solutions provided by Extramarks are necessary for solving questions based on the Law Of Sine And Cosine Formula on a regular basis.
2. Where can students find exact solutions to questions based on the Law Of Sine And Cosine Formula?
The questions asked on the basis of the Law Of Sine And Cosine Formula can be easily practised by taking assistance from the Extramarks website and mobile application. The NCERT solutions accessible on Extramarks can be used to get exact solutions to questions involving the Law Of Sine And Cosine Formula. | 677.169 | 1 |
The First Six, and the Eleventh and Twelfth Books of Euclid's Elements: With ...
PROP. XXII. PROB.-To describe a triangle of which the sides shall be equal to three given straight lines; but any two of these must be greater (I. 20) than the third.
Let A, B, C be three given straight lines, of which any two are greater than the third: it is required to make a triangle of which the sides shall be equal to A, B, C, each to each.
From
E
D G
H
K
Take an unlimited straight line DE, and let F be a point in it, and make (I. 3) FG equal to A, FH to B, and HK to C. the centre F, at the distance FG, describe (I. post. 3) the circle GLM, and from the centre H, at the distance HK, describe the circle KLM. Now, because (hyp.) FK is greater than FG, the circumference of the circle GLM, will cut FE between F and K, and therefore the circle KLM cannot lie wholly within the circle GLM. In like manner, because (hyp.) GH is greater than
C
M
HK, the circle GLM cannot lie wholly within the circle KLM. Neither can the circles be wholly without each other, since (hyp.) GF and HK are together greater than FH. The circles must therefore intersect each other: let them intersect in the point L, and join LF, LH; the triangle LFH has its sides equal respectively to the three lines A, B, C.
Because F is the centre of the circle GLM, FL is equal (I. def. 30) to FG; but (const.) FG is equal to A: therefore (I. ax. 1) FL is equal to A. In like manner it may be shown that HL is equal to C; and (const.) FH is equal to B: therefore the three straight lines LF, FH, HL are respectively equal to the three A, B, C: and therefore the triangle LFH has been constructed, having its three sides equal to the three given lines, A, B, C: which was to be done.*
PROP. XXIII. PROB.-At a given point in a given straight line, to make a rectilineal angle equal to a given
one.
Let AB be the given straight line, A the given point in it, and C the given angle: it is required to make an angle at A, in the straight line AB, that shall be equal to C.
*It it evident that if MF, MH were joined, another triangle would be formed, having its sides equal to A, B, C. It is also obvious that in the practical construction of this problem, it is only necessary to take with the compasses FH equal to B, and then, the compasses being opened successively to the lengths of A and C, to describe circles or arcs from F and H as centres, intersecting in L; and lastly to join LF, LH.
The construction in the text is made somewhat different from that given in Simson's Euclid, with a view to obviate objections arising from the application of this proposition in the one that follows it.
In the lines containing the angle C, take any points D, E, and join them, and make (I. 22) the triangle AFG, the sides of which AF, AG, FG, shall be equal to the three straight lines CD, CE, DE, each to each. Then, because FA, AG are equal to DC, CE, each to each, and the base FG to the base DE, the angle A is equal (I. 8) to the angle C. Therefore, at the given point A in the given straight line AB, the angle A is made, equal to the given angle C: which was to be done.*
АА
D
PROP. XXIV. THEOR.-If two triangles have two sides of the one equal to two sides of the other, each to each, but the angles contained by those sides unequal: the base of that which has the greater angle is greater than the base of the other.
Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two DE, DF, each to each; but the angle BAC greater than EDF: the base BC is also greater than the base EF.
Of the two sides DE, DF, let DE be that which is not greater than the other, and at the point D, in the straight line DE, make (I. 23) on the same side with EDF_the_angle EDG equal to BAC; make also DG equal (I. 3) to AC or DF, and join EG, GF.t Because AB is equal to DE, and AC to DG, the two sides BA, AC are equal to the two ED, DG, each to each; and the angle BAC is equal to the angle EDG: therefore the base BC is equal (I. 4) to the base EG. Again, because DG is equal to DF, the angles (I. 5) DFG, DGF are equal; but the angle DGF is greater
D
H
G
E
*The construction of this proposition is most easily effected in practice, by making the triangles isosceles. In doing this, circular arcs are described with equal radii from C and A as centres, and their chords are made equal.
It is evident that another angle might be made at A, on the other side of AB, equal to C.
+ Hence the angle DGE is not greater (I. 5, and 18) than DEG; but DHG is greater (I. 16) than DEG: therefore DHG is greater than DGH, and (I. 19) DG or DF is greater than DH; and consequently the point H lies between D and F, and the line EG above EF. If, on the contrary, a line equal to DE the less side, were drawn through D, making with DF, on the same side of it with DE, an angle equal to A, the extremity of that line might have fallen on FE produced, or above it or below it; and thus the proof would require three cases, while the method here given, which is that of Simson, is universally applicable.
We should also have an easy proof by bisecting the angle FDG by a straight line cutting EG in a point, which call K, and joining DK and KF. For KG would be equal (I. 4) to KF; and by adding EK, we should have EG equal to EK, KF, and, therefore (I. 20) EG greater than EF.
Let the student compare this proposition and the following with the fourth and eighth of this book.
(I. ax. 9) than EGF: therefore the angle DFG is greater than EGF; and much more is the angle EFG greater than the angle EGF. Then, because the angle EFG of the triangle EFG is greater than its angle EGF; the side EG is greater (I. 19) than the side EF: but EG is equal to BC; and therefore also BC is greater than EF. Therefore, if two triangles, &c.
PROP. XXV. THEOR.-If two triangles have two sides of the one equal to two sides of the other, each to each, but their bases unequal; the angle contained by the sides of that which has the greater base, is greater than the angle contained by the sides equal to them, of the other.
Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two DE, DF, each to each; but the base BC greater than the base EF: the angle A is likewise greater than the angle D.
For, if it be not greater, it must either be equal to it or less; but the angle A is not equal to D, because then (I. 4) the base BC would be equal to EF; but it is not.
Neither is it less; because then (I. 24) the base BC would be less than EF; but it is not: therefore the angle A is not less than the angle D; and it has been shown, that it
D
Δ A
B
C
E
is not equal to it: therefore, the angle A is greater than D. Wherefore, if two triangles, &c.
PROP. XXVI. THEOR.-If two angles of one triangle be equal to two angles of another, each to each, and if a side of the one be equal to a side of the other similarly situated with respect to those angles; (1) the remaining sides are equal, each to each; (2) the remaining angles are equal; and (3) the triangles are equal to one another.
Let ABC, DEF be two triangles which have the angles ABC, BCA equal to the angles DEF, EFD, each to each, viz. ABC to DEF, and BCA to EFD; also one side equal to one side: and, First, let those sides be equal to which the angles are adjacent, that are equal in the two triangles; viz. BC to EF: the other sides are equal, each to each, viz. AB to DE, and AC to DF; and the remaining angles BAC, EDF are equal; also the areas of the triangles are equal.
G
B
D
E
F
For, if AB be not equal to DE, one of them must be the greater. Let AB be the greater, and make (1.3) BG equal to DE, and join GC; then because GB is equal to DE, and (hyp.) BC to EF, the two sides GB, BC are equal to the two DE, EF, each to each; and (hyp.) the angle GBC is equal
to the angle DEF: therefore the angle GCB is equal (I. 4, part. 3) to DFE. But DFE is (hyp.) equal to ACB: wherefore, also the angle GCB is equal to ACB, a part to the whole, which (I. ax. 9) is impossible: therefore AB is not unequal to DE, that it, it is equal to it. Then, since BC is equal to EF, the two AB, BC are equal to the two DE, EF, each to each; and the angle ABC to DEF: therefore (I. 4) the base AC is equal to the base DF, the triangle ABC to the triangle DEF, and the third angle BAC to the third angle EDF.
A A
B
H C
F
Next, let the sides which are opposite to equal angles in each triangle be equal to one another, viz. AB to DE; likewise in this case the other sides are equal, viz. BC to EF, and AC to DF; also the triangle ABC to DEF, and the third angle BAC to the third angle EDF. For, if BC be not equal to EF, let BC be the greater, and make (I. 3) BH equal to EF, and join AH. Then, because BH is equal to EF, and (hyp.) AB to DE; the two AB, BH are equal to the two DE, EF, each to each; and they contain equal angles: therefore (I. 4, part 3) the angle BHA is equal to EFD; but EFD is equal (hyp.) to BCA: therefore also the angle BHA is equal to BCA, that is, the exterior angle BHA of the triangle AHC is equal to its interior and remote angle BCA, which (I. 16) is impossible: wherefore BC is not unequal to EF, that is, it is equal to it. Then, since AB is equal (hyp.) to DE, and the angle B to the angle E: therefore (I. 4) the base AC is equal to the base DF, the third angle BAC to the third angle EDF, and the triangle ABC to the triangle DEF.* Therefore, if two triangles, &c.
*By the fifth corollary to the 32d proposition of this book, if two angles of one triangle be equal to two angles of another, their remaining angles are also equal. This proposition, therefore, if it be postponed till the student has learned the 32d, may be demonstrated in a single case, the same as either the first or second in Euclid's method, given above; and it may be so postponed without impropriety, as the 32d does not depend on it, either directly or indirectly. If the proposition be taken thus, the enunciation may be as follows:
If two triangles be equiangular to one another, and if a side of the one and a side of the other, which are opposite to equal angles, be equal; then (1) the remaining sides are equal, each to each, viz. those which are opposite to equal angles; and (2) the triangles are equal.
This is the third of the propositions in which triangles are demonstrated to be in every respect equal. They are proved to be such in the fourth proposition, when two sides and the contained angle of the one are respectively equal to two sides and the contained angle of the other: in the eighth, when the three sides of the one are equal to the three sides of the other, each to each: and in the twenty-sixth, when two angles and a side of the one are respectively equal to two angles and a side, similarly situated, of the other. There is only one other case in which two triangles are in all respects equal; that is, when a side and the opposite angle of one of them are respectively equal to a side and the opposite angle of the other, and a second side in the one equal to a second side in the other, the angles opposite to these latter sides being of the same kind; that is, both right angles, both acute, or both obtuse. This proposition, which is a case of the 7th of the sixth book, is of little use, except that it completes the theory; but it may be a proper exercise for the student to prove it.
It will be seen from these remarks, that two triangles are in every respect equal, when of the sides and angles any three, except the three angles in one of them, are
B
PROP. XXVII. THEOR.-If a straight line falling upon two other straight lines, in the same plane, make the alternate angles equal to one another, these two straight lines are parallel.
Let the straight line EF, which falls upon the two straight lines AB, CD, in the same plane, make the alternate angles, AEF, EFD, equal to one another; AB is parallel to CD.
A
C
H
Б
B
G
F
D
For, if it be not parallel, AB and CD, being produced, will meet either towards B, D, or towards A, C: let them be produced and meet towards B, D, in the point G. Therefore GEF is a triangle, and its exterior angle AEF is greater (I. 16) than the interior and remote angle EFG; but (hyp.) it is also equal to it, which is impossible: therefore, AB and CD, being produced, do not meet towards B, D. In like manner, it may be demonstrated, that they do not meet towards A, C. But those straight lines which are in the same plane, and which meet neither way, though produced ever so far, are parallel (I. def. 11) to one another: AB therefore is parallel to CD.* Wherefore, if a straight line, &c.
K
PROP. XXVIII. THEOR.-Two straight lines are parallel to one another, (1) if a straight line falling upon them make the exterior angle equal to the interior and remote, upon the same side of that line; and (2) if it make the interior angles upon the same side together equal to two right angles.
Let the straight line EF, which falls upon the two straight lines, AB, CD, make the exterior angle EGB equal to the interior and remote angle GHD upon the same side of EF, or make the interior angles on the same side, BGH, GHD, together equal to two right angles; AB is in each case parallel to CD.
respectively equal to the corresponding sides or angles, in the other, with only the one limitation in the proposition just enunciated, that in it the angles opposite to two equal sides must be of the same kind: and from this it appears, that of the sides and angles of a triangle, three must be given to determine the triangle, and these three cannot be the three angles. Were only the three angles given, the sides, as will appear from the 32d proposition of this book, might be of any magnitudes whatever. The student may exercise himself in proving this proposition by superposition; and also by producing BA and BC, instead of cutting off parts of them.
* In this proof ABG and CDG must be regarded as straight lines. It is not necessary, indeed, to make the actual construction here given; for the student will see that AB and CD cannot meet on either side, since, if they did, a triangle would thus be formed, and the exterior angle would be (I. 16) greater than the interior and remote angle, which is contrary to the hypothesis. The alternate angles as they are understood by Euclid, or the interior alternate angles, as they might be called with perhaps more propriety, are the interior remote angles on opposite sides of the line which falls on the other two. AEH and KFD, and also BEH and CFK, may be called exterior alternate angles; and it is easy to prove that if these be equal, the lines are parallel. | 677.169 | 1 |
Knowing the (x,y) coordinates of A, B, and C; how would one find the point D? D being the point where line CD cuts the triangle into two smaller right triangles?
I feel like this is elementary, but searching for this is just giving me formulas to find the length of the lines, not the point I am trying to find.
EDIT:
Someone asked before deleting their post if this is a right triangle. Well; it COULD be, but not necessarily.
I should clarify why I need this. I have a quadratic bezier curve, and I want to find this point D on the triangle created by the 3 points (start, end, and control) of the bezier curve. I ultimately intend to check the angles formed by line CD at point C to restrict the "sharpness" of the curve (for road creation)
3 Answers
3
First you need to find the slope of the line from A to B, then find the perpendicular slope. Lastly, find where the lines intersect.
The trick is that this is not actually a real triangle problem. Instead, you are trying to find the point on a line that is closest to another point (point C). The point on the line that is closest to the point will /always/ be perpendicular to the line, so yeah. Here's a tutorial on that if you're still confused:
Next time, I believe this would go in the "Mathematics" stack exchange forum. | 677.169 | 1 |
Satellites
The radius of earth is 6440 Kilometer. There are many Satellites and Asteroids moving around the earth. If two Satellites create an angle with the center of earth, can you find out the distance between them? By distance we mean both the arc and chord distances. Both satellites are on the same orbit (However, please consider that they are revolving on a circular path rather than an elliptical path).
Input
The input file will contain one or more test cases.
Each test case consists of one line containing two-integer s and a, and a string 'min' or 'deg'. Here s is the distance of the satellite from the surface of the earth and a is the angle that the
satellites make with the center of earth. It may be in minutes (′) or in degrees (◦). Remember that the same line will never contain minute and degree at a time.
Output
For each test case, print one line containing the required distances i.e. both arc distance and chord distance respectively between two satellites in Kilometer. The distance will be a floating-point value with six digits after decimal point.
Sample Input
500 30 deg
700 60 min
200 45 deg
Sample Output
3633.775503 3592.408346
124.616509 124.614927
5215.043805 5082.035982 | 677.169 | 1 |
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Definition, Formula, and Examples of Arc Length
Mathematically, any object's curved edge is called an arc. An arc is a continuous curve segment or a circle's circumference. The length of an arc is the distance around a circle, curve, or another geometric shape.
A circle can be divided into two semicircles or one major arc and one minor arc by choosing two spots. An arc's length is greater than any line connecting its endpoints. In this post, we will examine the concept of an arc, and several forms of the length of an arc formula.
What is Arc Length?
A relay race is being held on the precisely round indoor track of a newly opened gym to mark the occasion. Due to the round nature of the track, each runner will travel in a curved line from the point. Five people make up each team, and each person only runs a piece of the entire track.
They receive the baton to the point at which they pass it to the following runner. As a result, each runner must go over an arc, a curved line connecting two sites. The arc length is the distance along an arc from one point to the next.
Arc Length Formula
The Arc Length Formula is frequently used to calculate distances along the curved path that makes up an arc. An arc length is a distance traversed by a circle's curved line. If a straight line and two endpoints of a circle are compared, the straight line will always have a shorter length.
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Depending on the arc's central angle unit, many formulae can be used to calculate an arc's length. The center angle is frequently measured in radians or degrees. The formula for arc length for a circle is known to be times the circle radius.
General Approach to the Formula for Arc Length
By joining a limited number of its points, one may approximate a curve in the plane. This is accomplished by creating a polygonal route using line segments. Calculating the length of each linear segment is rather simple. The Pythagorean theorem, for instance, may be applied to Euclidean space.
Additionally, the chordal distance, a measure of the approximate length, may be calculated by adding each linear segment's length. Sometimes a polygonal route may not already exist for the curve. Here, employing an increasing amount of shorter segment lengths is necessary. Approximations are undoubtedly more accurate and efficient as a consequence.
The subsequent approximations' lengths won't become shorter. Additionally, they could keep growing eternally. However, they will undoubtedly gravitate to a finite limit for smooth curves. This happens as the segment lengths are arbitrarily reduced. There are several curves where the value of L is the smallest. Any polygonal approximation's length is capped at this amount.
Angle Subtended by Arc
The phrases arc and arc length most frequently are about circles. The length between any two locations on a circle's circumference may consider the arc length because a circle is constantly curved. The entire distance traveled around a circle is its circumference.
It may be calculated using the equation C=pi d, where C stands for circumference, d for diameter, and {eq}\pi {/eq} is the constant in mathematics. Practical circle-related issues often pertain to a circle's radius r rather than its diameter. Since the diameter is always two times the radius. The equation for circumference may be written as $$C=2pi r $$, which is the more usual form.
Determine Arc Length
It is important to remember that every arc is a circle. It has a length smaller than the circumference to comprehend how circumference and arc length are related. This implies that an additional scaling factor must be used to determine arc length. This omitted component is the subtended angle, sometimes referred to as the angle subtended by the arc.
Two rays can be added to the circle to determine the subtended angle. Each of the two rays starts at the circle's origin and travels through a separate endpoint on the measured arc. The angle between the two beams is thus the subtended angle.
The technically accurate term for the method above determining the angle is "subtended." The length of the related arc is directly proportional to the subtended angle. The magnitude of the subtended angle increases with increasing arc length and vice versa.
Degrees and Radians
The arc length may calculate using either a formula that employs degrees or radians, technically speaking. The inner angle of a circle is often expressed in degrees, although it may also be expressed in radians.
The arc length measurement in a unit circle is mathematically identical to the radian measurement of the angle that the arc length subtends. The number of radii of the circle that might fit in the arc that the angle creates may regard as the measure of an angle in radians. An angle of 1 rad, therefore, subtends a radius-length arc.
Arc Measure vs. Arc Length
An arc is a common and helpful figure in geometry. Any smooth curve is typically referred to as an arc. The arc length is the distance along the curve from the beginning to the endpoint. The term "arc" refers to a section of a circle along its circumference.
The length of the arc or the magnitude of the angle it subtends at its center is typically used to determine the arc's size. The angle subtended at the center is sometimes referred to as the arc measure or the angle measure of an arc. It is expressed in radians or degrees.
The length of an arc or the angle of an arc is used to determine its size. Arc length is the distance along the curve, whereas arc angle measure is the angle that an arc subtends at its center. Whereas the angle of measure is measured in angles, the arc length is measured in length units.
Conclusion
The length of an arc is the distance along the circle's circumference. An arc length is any small or long distance along the curve line. The center angle is frequently measured in radians or degrees. It is consistently regarded as a circumference. In addition, arc length denotes the separation of two points along a curve's section. | 677.169 | 1 |
An array of vertices, wound in counterclockwise order, of the size specified by the vertexCount parameter.
normal
The normal direction of the polygon.
triangle
An array to which the triangles are written. This must be large enough to hold n − 2 triangles, where n is the number of vertices specified by the vertexCount parameter.
base
A base vertex index that is added to all of the indexes written to the triangle array.
Description
The TriangulatePolygon function takes an arbitrary planar polygon specified by n vertices, triangulates it, and returns the number of triangles that were generated. Except in cases in which there are degenerate vertices, the number of triangles generated will be n − 2.
The input polygon is specified by the vertexCount and vertex parameters and may be any planar shape that is not self-intersecting. Convex polygons, concave polygons, and polygons containing sets of collinear vertices are all valid. The set of vertices must be specified in order and must be wound counterclockwise about the normal direction specified by the normal parameter. The minimum number of vertices is 3, in which case a single triangle is generated.
The results are undefined if the input polygon is self-intersecting or if the vertices do not lie in a plane that is perpendicular to the normal direction specified by the normal parameter. If the array of vertices contains degenerate entries (a pair of vertices that are very close to each other), then the result may be a subset of the fully triangulated polygon, indicated by the return value being less than n − 2. | 677.169 | 1 |
Introduction
Hello, mathematicians-in-the-making! We will talk about a fascinating part of geometry – Angle Measurements. Have you ever wondered how we determine that a circle has 360 degrees? Or why is the right angle always 90 degrees? These concepts all come from studying angle measurements. So, fasten your seatbelts and prepare to explore the mysterious world of angles!
Grade Appropriateness
This topic is typically introduced to students ages 8 to 12. It is suitable for children in 3rd to 7th grade, depending on their math curriculum and their pace of learning.
Math Domain
Angle Measurements fall under the domain of Geometry in Mathematics. Geometry uses shapes and space to help us understand and explain the world around us.
Definition of the Topic
An 'angle' in geometry is the space or region between two intersecting lines or surfaces at the point where they meet. Angle measurements tell us the size of the angle and are usually measured in degrees (°).
Key Concepts
There are several types of angles based on their measurements:
Acute Angle: This is an angle less than 90°.
Right Angle: This is an angle of exactly 90°, often marked by a small square.
Obtuse Angle: This is an angle between 90° and 180°.
Straight Angle: This is an angle of 180°.
Reflex Angle: This is an angle greater than 180° but less than 360°.
Full Angle: An angle of 360° makes a complete circle.
Discussion with Illustrative Examples
In geometry, an angle is the figure created when two rays intersect at a point known as the vertex. Angle measures the amount of turn of its rays in degrees.
Arms: The two rays joining to form an angle are called arms of an angle. Here, WH and WM are the arms of the ∠HWM. Vertex: The vertex is the common endpoint of the two rays where they meet. Here, the point W is the vertex of ∠HWM.
Types of Angles Based on Measurements
Acute Angles: These angles are less than 90° Right Angles: These angles measure exactly 90° Obtuse Angles: These angles measure more than 90° but less than 180°. Straight Angles: These angles measure exactly 180° Reflex Angles: These angles measure more than 180° but less than 360° Full Rotation Angles: These angles measure exactly 360°
Let's make this more fun with a few examples:
Acute Angle: Imagine you open a book slightly. The angle between the open pages forms an acute angle.
Practice Test
2. Tommy drew an angle of 150 degrees. What type of angle did Tommy draw?
3. Sara opens her book to an angle of 80 degrees. What type of angle does this create?
4. If an angle measures 360 degrees, what is its classification?
Answers:
1. 90 degrees – Right Angle,
200 degrees – Reflex Angle
45 degrees – Acute Angle
180 degrees – Straight Angle
10 degrees – Acute Angle
2. Obtuse Angle
3. Straight Angle
4. Full Rotation
Frequently Asked Questions (FAQs)
Why do we measure angles in degrees?
We measure angles in degrees because it is a way to break down a circle into equal parts. The concept comes from ancient Babylonians, who used a counting system based on the number 60.
Can an angle be negative?
Yes, an angle can be negative. Negative angles are measured in the opposite direction of positive angles. Hence, if a positive angle measures the counterclockwise direction, the negative angle measures the clockwise direction.
What is a protractor, and how do I use it?
A protractor is a semi-circle-shaped tool used to measure angles. To use a protractor, you place the center point at the vertex of the angle (where the two lines meet) and align one
line along the zero of the protractor. The number that the other line crosses is the measure of the angle in degrees.
Is a right angle only 90 degrees?
Yes, a right angle is always 90 degrees. It forms a perfect 'L' shape.
What is the smallest possible angle?
The smallest possible angle in a plane is 0 degrees, which occurs when two lines overlap precisely.
Well, that's it for our exploration of angle measurements! Remember that understanding and identifying various angles will become easier as you practice. Keep up the excellent effort, and remember to look around you for more real-world instances of | 677.169 | 1 |
Geometry, Color Code, EUCLID ,
linear algebra and geometry,…
the basic objects of geometry are lines and planes represented by linear equations.
branch of computer science that studies the algorithms which can be stated in terms of geometry
normally used for statistics/optimization to recognize patterns over time
used on video games to design how the player will control their view, through isometric graphs, how the terrain is design through the use of polygons and determining how each character moves with the use of arcs and angles
Calculus
SOH-CAH-TOA
Tangent Lines
Geometry in construction
Building houses, stairs, etc.
Bridges and roadways
Cutting down trees
Activites
Pool
Angles for you to hit the cue ball to hit the other ball in the pocket
Axiom 2: To produce (extend) a finite straight line continuously in a straight line.
Axiom 3: To describe a circle with any centre and distance (radius).
Axiom 4: That all right angles are equal to one another.
Axiom 5: Parallel Postulate
Power of a Point
real number h that reflects the relative distance of a given point from a given circle. Specifically, the power of a point P with respect to a circle O of radius r is defined by. where s is the distance between P and the center O of the circle | 677.169 | 1 |
Step 2: Radii OK and NL are perpendicular to OM because of the radius-tangent theorem. By definition of perpendicular, angles KOM and LNM are right angles. This means that triangles KOM and LNM are right triangles. Angle LMN is common to both right triangles, so by the _________________________, triangles KOM and LNM are similar.
Answer by Guest
Answer:
Angle Angle Angle (AAA) property
Step-by-step explanation:
A right angled triangle is a triangle that has one of its angles to equal . It could be in the form of an isosceles triangle or an acute angled triangle.
The given question compares the congruence properties of two triangles, KOM and LNM.
Thus,
<KOM = <LNM (right angle theorem)
<LMN = <KMO (common angles to both triangles)
⇒ <OKM = <NLM (property of a triangle i.e sum of angle in a triangle is )
ΔKOM = ΔLNM (congruence property)
Therefore, by angle angle angle (AAA) congruence property, the two triangles are similar.
Rate answer
Wrong answer?
If your question is not fully disclosed, then try using the search on the site and find other answers on the subject Mathematics. | 677.169 | 1 |
...the Angle AFE j. g ti fhall bef equal to the Angle BFE. But when a Right Line {landing upon a Right Line makes the adjacent Angles equal to one another, each of the equal Angles is ia Right Angle. Wherefore AFE, or BFE, is a Right Angle. And therefore the Right Line CD drawn thro*...
...Therefore the Angle * 8- i. AD G is * equal to the Angle GD B. But when a Right Line ftanding upon 'a Right Line, makes the adjacent Angles equal to one another, each of the 10. i. equal Angles will f be a Right Angle. Wherefore the Angle GDB is a Right Angle. But FDB is alfo...
...therefore the angle ADG is equal c to the angle GDB. but when a ftraight line ftanding upon another ftraight line, makes the adjacent angles equal to one another ,each of the angles is a right angle *. therefore the angle GDB is a right angle. but FDB is like- d.io.Def.i. wife...
...will be equal to the angle BFE. But fmce a right line flanding upon a right line makes the adjoining angles equal to one another, each of the equal angles is a right angle : therefore each of the angles AFE, BFE is a right angle, and fo the right line c D drawn thro' A\...
...the Angle AFE fliall be f equal to the Angle BFE. But when af 8. t. Right Line Standing upon a Right Line makes the adjacent Angles equal to one another, each of the equal Angles is J a Right Angle. Wherefore AFE, orJ0«/. 10. i« BFE, is a Right Angle. And therefore the Right •Line...
...is equal' to the angle CHG; and they are adjacent angles ; but when a ftraight line ftanding cet on a ftraight line makes the adjacent angles equal to one another, each of them is a right angle, and the ftraight line which flands upon the other is called a perpendicular...
...equal e to the angle CHG; and they are adjacentc. 8. i. angles. but when a ftraight line ftanding on a ftraight line makes the adjacent angles equal to one another, each of them is aright angle, and the ftraight linewhich ftands upon the other is called a perpendiculartoit....
...by a 'letter placed at that point; as the Angle at E.' .X. When a ftraight line flanding on another ftraight line makes the adjacent angles equal to one another, each of the angles is called a right angle ; and the ftraight line which ftands on the other is called a perpendicular...
...Therefore the Angle ADG is * equal to the Angle GD B. But when 3*8.,. Right Line ftandmg upon a Right Line makes the adjacent Angles equal to one another, each of the equal Angles will f be a Right Angle. Wherefore fZ?»/.io.» F the the Angle GDB is a Right Angle. But FDB is alfo...
...divided. DKFINITION X. When a Right Line Handing on a Right Line, makes the fucceffive Angles on each fide equal to one another, each of the equal Angles, is a Right Angle ; and the infifting Right Line, is called a PERPENDICULAR to that upon which it ftands. DEFINITION Xf. An OBTUSE... | 677.169 | 1 |
Join the diagonal BD and prove as above in triangles
BCD and ABD.
Thus, this property also holds true for a non-convex quadrilateral.
Ex 3.1 Class 8 Maths Question 4.
Examine
the table. (Each figure is divided into triangles and the sum of the angles deduced
from that).
What can you say about the angle sum of a convex
polygon with number of sides? (a) 7(b) 8(c) 10(d) n
Solution: From
the given table, we can conclude that the sum of the angles of a convex polygon
with side 'n' = (n – 2) × 180° (a) If the number of sides = 7 Then the angle sum = (7 – 2) × 180° = 5 × 180° =
900° (b) If the number of sides = 8 Then the angle sum = (8 – 2) × 180° = 6 × 180° =
1080° (c) If the number of sides = 10
Ex 3.1 Class 8 Maths Question 5.
What is a regular polygon? State the name of a
regular polygon of
(i) 3 sides(ii) 4 sides(iii) 6 sides
Solution: A polygon, whose all sides and all angles are equal,
is called a regular polygon.
(i) A polygon with 3 sides is called an equilateral triangle.
(ii) A polygon with 4 sides is called a square.
(iii) A polygon with 6 sides is called a regular hexagon. | 677.169 | 1 |
Vector Addition Calculator Magnitude and Angle Online
Vector addition is a fundamental concept in fields such as physics, engineering, and computer science, enabling the combination of multiple vectors into a single vector known as the resultant vector. The Vector Addition Calculator Magnitude and Angle is a specialized tool designed to effortlessly calculate the magnitude (length) and direction (angle) of this resultant vector, streamlining tasks ranging from simple homework problems to complex engineering projects.
Formula
Understanding the calculation process is crucial for anyone working with vectors. The formulas for determining the magnitude and angle of the resultant vector are straightforward yet powerful:
In these formulas, Vx represents the sum of the x-components of the vectors, while Vy is the sum of the y-components. The atan2 function calculates the angle by finding the arctangent of the quotient of Vy and Vx, providing the direction of the resultant vector in radians or degrees, depending on your calculation settings.
Vector Addition Scenarios and Guidelines
Scenario
Description
General Approach
Wind Vectors
Combining two wind vectors to find the resultant wind direction and speed.
Use vector addition to combine the magnitudes (speeds) and directions (angles) of the two wind vectors.
Force Vectors
Determining the net force on an object by adding multiple force vectors.
Sum up the components of all forces acting in the x and y directions separately, then use these sums for magnitude and angle calculations.
Displacement Vectors
Calculating the total displacement of an object moving along two or more paths.
Add the x and y components of each displacement vector, then find the resultant vector's magnitude and direction.
Velocity Vectors
Finding the resultant velocity of an object when it has multiple velocity vectors due to different movements.
Combine the velocity vectors by adding their x and y components, then calculate the magnitude and direction of the resultant velocity.
Example
Let's apply the formulas to a real-world example. Imagine you have two vectors: A (3, 4) and B (1, 2). Using the magnitude and angle formulas, we can calculate the resultant vector's magnitude and direction, illustrating the simplicity and efficiency of the Vector Addition Calculator. | 677.169 | 1 |
Reading Angles Worksheet Ks2
Reading Angles Worksheet Ks2. Web use these measuring angles for ks2 worksheets for your students to practise drawing and measuring angles. Maths for early years age range:
Angles interactive worksheet for Grade 5 from
It's also known as a 'quarter turn' because it is a quarter of a full turn, which measures 360˚. Worksheet/activity file previews docx, 13.47 kb docx, 13.25 kb docx, 12.85 kb docx, 13.11 kb docx, 25.12 kb a collection of worksheets involving angles for year 4. Web examine the relationship and space between two intersecting lines with our primary resources on angles.
Source:
These worksheets will help your classes to understand the difference between acute, obtuse, reflex and mixed angles. Measuring angles up to 360 degrees;
Web this worksheet deals with acute and obtuse angles for ks2! This ks2 maths angle worksheet demonstrates the three main types of angle:
Questions Of Increasing Difficulty Which Should Enable Students To Understand How To Read Angles Off And Draw An Angle On Printed Protractors As.
Web reading angles on a protractor part 3. Parallel lines cut by a transversal. Web examine the relationship and space between two intersecting lines with our primary resources on angles.
4.5 (28 Reviews) Last Downloaded On.
Follow this link for same flipcharts in powerpoint format (your wish is my command. Web recognising and identifying angles subject: First read the definitions, then see if your children can label each angle correctly.
Web Use These Handy Worksheets To Teach Your Students How To Use A Protractor To Measure Different Angle Types Within 180°. | 677.169 | 1 |
Naming angles worksheets naming angles worksheets provide adequate practice beginning with using three points to name an angle followed by familiarizing students of grade 4 and grade 5 with the 4 ways to. Whether Students classify acute obtuse and right angles in these geometry worksheets.
Fourth grade and fifth grade kids practice identifying the arms and vertex of angles with this batch of parts of an angle pdf worksheets. Worksheets math grade 4 geometry classifying angles. The angles worksheets are randomly created and will never repeat so you have an endless supply of quality angles worksheets to use in the classroom or | 677.169 | 1 |
\(\blacksquare\)\(\utilde{i}\) and \(\utilde{j}\) are vectors of magnitude \(1\) unit that parallel to axis and axis respectively.
\(\blacksquare\) If \(A \begin{pmatrix} x_1, y_1 \end{pmatrix}\) is a point on a Cartesian plane, the vector formed from the origin \(O\) to point \(A\) is \(\begin{aligned} \overrightarrow{OA}&=x_1\utilde{i}+y_1\utilde{j} \end{aligned}\). \(\begin{aligned} \overrightarrow{OA} \end{aligned}\) is known as a position vector.
\(\blacksquare\) The magnitude for a vector
\(\utilde{r}=x\utilde{i}+y\utilde{j}\) is
\(|r|=\sqrt{x^2+y^2}\).
\(\blacksquare\) The unit vector in the direction of \(\utilde{r}\) is | 677.169 | 1 |
Enlargement, sometimes called scaling, is a kind of transformation that changes the size of an object.
The image created is similar to the object. Despite the name enlargement, it includes making objects smaller.
For every enlargement, a scale factor must be specified. The scale factor is how many times larger than the object the image is. Length of side in image = length of side in object × scale factor
For any enlargement, there must be a point called the center of enlargement. Distance from center of enlargement to point on image = Distance from Centre of enlargement to point on object X scale factor
The Centre of enlargement can be anywhere, but it has to exist.
This process of obtaining triangle A' B 'C' from triangle A B C is called enlargement.
Translation vector moves every point of an object by the same amount in the given vector direction.
It can be simply be defined as the addition of a constant vector to every point.
Translations and vectors: The translation at the left shows a vector translating the top triangle 4 units to the right and 9 units downward. The notation for such vector movement may be written as:
Vectors such as those used in translations are what is known as free vectors. Any two vectors of the same length and parallel to each other are considered identical. They need not have the same initial and terminal points. | 677.169 | 1 |
The Triangle and its Properties
The Triangle and its Properties: Angle sum property (with notions of proof & verification through paper folding, proofs using property of parallel lines, difference between proof and verification.), Exterior angle property. Sum of two sides of a it's third side. Pythagoras Theorem. | 677.169 | 1 |
Law of Sines
The Law of Sines is one of the most useful triangle equations we can
find in the world of math. If you've already tackled the Pythagorean
theorem, you're already aware of how interesting triangles can be.
The Law of Sines takes this one step further, allowing us to make
all kinds of calculations based on the angles of triangles
rather than their sides. But how do we use the Law of Sines? Let's
find out.
What is the Law of Sines?
Also known simply as the "Sine Rule," the Law of Sines is
represented by the following equation:
asinA=bsinB=csinC
Remember, the lower-case letters represent the sides of a
triangle, whereas the upper-case letter represents the
angles of a triangle. A is always opposite to a, B is always
opposite to b, and C is always opposite to c.
But what about the phrase sin?
"Sin" represents "sine." Alone with cosine and tangent, these values
represent the ratio of a side in a right-angled triangle. Sine is
written like this:
Flashcards covering the Law of Sines
Practice tests covering the Law of Sines
Set your student up with a qualified math tutor
Whether your student is struggling with the Law of Sines or they're
ready to steam ahead toward greater challenges, a math tutor can be
an excellent ally. When you contact Varsity Tutors, we will find
your student a math tutor whose skills match their unique needs.
Speak with one of our Educational Directors today to discuss your
options and get your student started with a private math tutor. | 677.169 | 1 |
Finding cos Value from Trigonometric Table
We know the values of the trigonometric ratios of some
standard angles, viz, 0°, 30°, 45°, 60° and 90°. While applying the concept of
trigonometric ratios in solving the problems of heights and distances, we may
also require to use the values of trigonometric ratios of nonstandard angles,
for example, sin 54°, sin 63° 45′, cos 72°,
cos 46° 45′ and tan 48°. The approximate values,
correct up to 4 decimal places, of natural sines, natural cosines and natural
tangents of all angles lying between 0° and 90°, are available in trigonometric
tables.
Reading Trigonometric Tables
Trigonometric tables consist of three parts.
(i) On the extreme left, there is a column containing 0 to 90 (in degrees).
(ii) The degree column is followed by ten columns with the headings
0′, 6′, 12′, 18′, 24′, 30′, 36′, 42′, 48′ and 54′ or
0.0°, 0.1°, 0.2°, 0.3°, 0.4°, 0.5°, 0.6°, 0.7°, 0.8° and 0.9°
(iii) After that, on the right, there are five columns known as mean difference columns with the headings 1′, 2′, 3′, 4′ and 5′.
Note: 60′ = 60 minutes = 1°.
1.
Reading the values of cos 67°
To
locate the value of cos 67°, look at the extreme left column. Start from the
top and move downwards till we reach 67.
We
want the value of cos 67°, i.e., cos 67° 0′. Now, move to the right in the row
of 67 and reach the column of 0′.
We
find 0.3907
Therefore,
cos 67° = 0.3907.
2.
Reading the values of cos 67° 48′
To
locate the value of cos 67° 48′, look at the extreme left column. Start from
the top and move downwards till you reach 67.
Now,
move to the right in the row of 67 and reach the column of 48′.
We
find 3778 i.e., 0. 3778
Therefore,
cos 67° 48′ = 0. 3778.
3.
Reading the values of cos 67° 41′
To
locate the value of cos 67° 41′, look at the extreme left column. Start from
the top and move downwards till you reach 67.
Now,
move to the right in the row of 67 and reach the column of 36′.
We
find 3811 i.e., 0.3811
So,
cos 67° 41′ = 0.3811 - mean difference for 5′
= 0.3811
-
14 [Subtraction, because cos 67° 41′ < cos 67° 36′]
0.3797
Therefore,
cos 67° 41′ = 0.3797.
Conversely,
if cos θ = 0.1097 then θ = cos 83° 42′ because in the table, the value 0.1097 corresponds
to the column of 42′ in the row of 83, i.e., 83°.
Tangram is a traditional Chinese geometrical puzzle with 7 pieces (1 parallelogram, 1 square and 5 triangles) that can be arranged to match any particular design. In the given figure, it consists of o… | 677.169 | 1 |
...circle two angles at the centre are in the same ratio as their intercepted arcs. (bpolygon, and less than the perimeter of any regular circumscribed polygon. § 361. (2) The ratio of the circumference of a circle to its diameter is the same for all circles. C = ird = 2 irr. § 362. (3) The area of a circle is equal to one-half the product of its circumference...
...cuts orthogonallv any circle through a pair of harmonic conjugates to A, B. TRIGONOMETRY. 7. Assuming that the ratio of the circumference of a circle to its diameter is a constant, -n say, show that the circular measure of two right angles is ir radians, and prove that...
...could be made, the quotient in each case would be the same and equal to 3.14159 + , ie the ratio of the circumference of a circle to its diameter is the same for all circles. The ratio is denoted by the Greek letter TT (pi). The value of TT found in geometry is 3. 14159+ ....
...circle two angles at the centre are in the same ratio as their intercepted arcs. (6circle two angles at the centre are in the same ratio as their intercepted arcs. (6) The ratio of. a circumference of a circle to its diameter is the same for all circles. [See Art. 9 (6).] For the proof of (a), reference may be made to any plane geometry ; for instance,...
...explanation, gives out that gives approximate values for the square roots of several large numbers, and proves that the ratio of the circumference of a circle to its diameter is less than 37- but greater than 3^, (4) that the first approach to the rapidity with which the decimal...
...the work will not be accepted unless the constructions are made accurately with ruler and compass. 1. Divide a line externally in extreme and mean ratio...(without proofs) the method of determining this value. GEOMETRY (A), JUNE, 1901 1. State and prove a theorem relating to the squares of the sides of an obtuse-angled...
...17. Assuming that a circle may be treated as a regular polygon with an infinite number of sides, show that the ratio of the circumference of a circle to its diameter is constant. What is the circular measure of the least angle whose sine is J, and what is the measure...
...of circle ABCD in terms of its diameter; ie to compute the value of TT. ARGUMENT 1. The ratio of the circumference of a circle to its diameter is the same for all circles. 2. Since the diameter of the given circle is ' unity, the side of an inscribed square will be V2. REASONS... | 677.169 | 1 |
The Sin A Sin B Formula: Exploring its Applications and Significance
Trigonometry, a branch of mathematics that deals with the relationships between the angles and sides of triangles, has numerous formulas that aid in solving complex problems. One such formula is the sin A sin B formula, which plays a crucial role in various trigonometric calculations. In this article, we will delve into the details of this formula, its applications, and its significance in solving real-world problems.
Understanding the Sin A Sin B Formula
The sin A sin B formula is derived from the trigonometric identity known as the product-to-sum formula. This identity states that the product of two trigonometric functions can be expressed as the sum or difference of two other trigonometric functions. Specifically, the sin A sin B formula is derived from the product-to-sum formula for sine functions:
sin A sin B = (1/2) * [cos(A – B) – cos(A + B)]
This formula allows us to express the product of two sine functions in terms of cosine functions. By utilizing this formula, we can simplify trigonometric expressions and solve various trigonometric equations.
Applications of the Sin A Sin B Formula
The sin A sin B formula finds applications in a wide range of fields, including physics, engineering, and navigation. Let's explore some of its key applications:
1. Harmonic Analysis
In the field of signal processing, the sin A sin B formula is used in harmonic analysis. Harmonic analysis involves decomposing complex signals into simpler sinusoidal components. By utilizing the sin A sin B formula, the product of two sinusoidal signals can be expressed as the sum or difference of two other sinusoidal signals. This simplification aids in analyzing and manipulating signals in various applications, such as audio processing and image compression.
2. Electrical Engineering
In electrical engineering, the sin A sin B formula is employed in the analysis of alternating current (AC) circuits. AC circuits involve sinusoidal voltages and currents, and the sin A sin B formula allows engineers to calculate the power dissipated in such circuits. By expressing the product of voltage and current as the sum or difference of cosine functions, engineers can determine the average power consumed by electrical devices and optimize circuit designs.
3. Navigation and Astronomy
The sin A sin B formula is also utilized in navigation and astronomy to solve problems related to celestial navigation. By considering the angles between celestial bodies, such as the sun, moon, and stars, navigators and astronomers can determine their positions and make accurate calculations. The sin A sin B formula aids in these calculations by allowing the conversion of angular measurements into trigonometric functions, enabling precise celestial navigation.
Significance of the Sin A Sin B Formula
The sin A sin B formula holds significant importance in trigonometry and its applications. Here are some key reasons why this formula is of great significance:
1. Simplification of Trigonometric Expressions
The sin A sin B formula allows for the simplification of complex trigonometric expressions. By expressing the product of two sine functions in terms of cosine functions, trigonometric equations can be simplified and solved more efficiently. This simplification aids in various mathematical calculations and reduces the complexity of trigonometric problems.
2. Versatility in Problem Solving
The sin A sin B formula provides a versatile tool for problem-solving in various fields. Its applications in harmonic analysis, electrical engineering, and navigation demonstrate its wide-ranging utility. By utilizing this formula, professionals in different domains can tackle complex problems and derive accurate solutions.
3. Connection to Trigonometric Identities
The sin A sin B formula is derived from the product-to-sum formula, which is a fundamental trigonometric identity. By understanding and utilizing this formula, individuals can gain a deeper understanding of the relationships between trigonometric functions. This knowledge can be further applied to derive and utilize other trigonometric identities, expanding the scope of trigonometry as a whole.
Summary
The sin A sin B formula is a powerful tool in trigonometry, enabling the simplification of trigonometric expressions and solving various real-world problems. Its applications in harmonic analysis, electrical engineering, and navigation highlight its significance in different fields. By understanding and utilizing this formula, individuals can enhance their problem-solving abilities and gain a deeper understanding of trigonometry as a whole.
Q&A
1. What is the sin A sin B formula?
The sin A sin B formula is derived from the product-to-sum formula for sine functions. It states that the product of two sine functions can be expressed as the sum or difference of two cosine functions.
3. Why is the sin A sin B formula significant?
The sin A sin B formula is significant because it simplifies trigonometric expressions, provides a versatile problem-solving tool, and connects to fundamental trigonometric identities. It enhances mathematical calculations, aids in various fields, and expands the understanding of trigonometry.
4. How does the sin A sin B formula simplify trigonometric expressions?
The sin A sin B formula expresses the product of two sine functions in terms of cosine functions. This simplification reduces the complexity of trigonometric expressions and allows for more efficient solving of trigonometric equations.
5. Can the sin A sin B formula be applied to other trigonometric functions?
No, the sin A sin B formula is specific to sine functions. However, similar product-to-sum formulas exist for other trigonometric functions, such as cosine and tangent | 677.169 | 1 |
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1992 IMO Problems/Problem 4
Contents
Problem
In the plane let be a circle, a line tangent to the circle , and a point on . Find the locus of all points with the following property: there exists two points , on such that is the midpoint of and is the inscribed circle of triangle .
Video Solution
Solution
Note: This is an alternate method to what it is shown on the video. This alternate method is too long and too intensive in solving algebraic equations. A lot of steps have been shortened in this solution. The solution in the video provides a much faster solution,
Let be the radius of the circle .
We define a cartesian coordinate system in two dimensions with the circle center at and circle equation to be
We define the line by the equation , with point at a distance from the tangent and cartesian coordinates
Let be the distance from point to point such that the coordinates for are and thus the coordinates for are
Let points , , and be the points where lines , , and are tangent to circle respectively.
First we get the coordinates for points and .
Since the circle is the incenter we know the following properties:
and
Therefore, to solve for the coordinates for point by calculating the intersection of and as follows:
Solving for we get:
Solving for we get:
Now we need to find the limit of and as approaches infinity:
This means that the locus of starts at point on the circle but that point is not included in the locus as that is the limit.
If we assume that the locus is a ray that starts at let's calculate the slope of such ray:
Since the calculated slope of such locus at any point is not dependent on and solely dependent on fixed and , then this proves the slope is fixed and thus the locus is a ray that starts at excluding that point and with a slope of in the cartesian coordinate system moving upwards to infinity.
We can also write the equation of the locus as: and
~Tomas Diaz. [email protected]
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page. | 677.169 | 1 |
geometry glossary
acute
The word 'acute' comes from the Latin acus for 'needle' (which also forms
the root for acid, acupuncture, and acumen). In mathematics, an acute angle is one that is less than 90°. An acute triangle is one in which all three interior angles are acute.
area
central angle
The central angle is the angle subtended at the center of a circle by an arc or a chord;
in other words, it is the angle between two radii.
centroid
For a triangle, the centroid is the point of intersection
of the medians (i.e., those lines drawn from
each angle to the midpoint of the facing angle). For any other shape, the
centroid is the point whose coordinates are the average of the coordinates
of the shape's vertices. The centroid is
the center of mass of a figure.
circumcenter
The circumcenter is the center of a circle that passes through the vertices of a given polygon,
usually a triangle. For a triangle, it
is the same as the point of intersection of the perpendicular bisectors of the three sides. The point is equidistant from the triangle's
vertices and hence the center of the circle that may be circumscribed about
it (see circumscription).
circumcircle
The circumcircle is the circle that passes through all three vertices of a given triangle.
The circumcircle is said to circumscribe the triangle.
circumference
The circumference is the curve which incloses a circle, ellipse,
or other closed, plane figure. In figures bounded by straight lines, such
as the triangle, square, or other type of polygon, the term perimeter is used to designate the sum of all the bounding lines taken together.
The length of the circumference depends partly on the nature of the curve;
thus, that of the circle = 2πr = 2πd; and that of
the ellipse = 2π[1 – (1/2)2e2/1 – (1 ×
3/2 × 4)2e 4/3 – (1 × 3 × 5/2
× 4 × 6)2e 6/5 – ...], where a is the semi-major axis, and e the eccentricity.
circumscription
In plane geometry, circumscription is the construction of a circle such that all vertices of a particular polygon lie on its circumference, the polygon
then being said to be inscribed within the circle. All regular polygons and all triangles may be circumscribed.
In three dimensions circumscription implies the construction of a sphere such that on its surface lie all the vertices of a polyhedron.
concave
congruent
In the case of geometric figures, congruent means having exactly the same shape and size,
i.e., differing only in their positions in space. Two congruent figures
will coincide exactly when superimposed. In contrast, two figures are said
to be similar if the figures cannot be superimposed
without being scaled or reflected.
coplanar
Coplanar means lying in the same plane. It is
possible to construct a plane through any set of three points,
but this is true of sets of four points only in special cases. Any intersecting
or parallel straight lines are coplanar.
diagonal
a diagonal is a line that joins any two vertices of a polygon, if the vertices are not
next to each other; or a line that joins two vertices of a polyhedron that are not on the same face.
diameter
A diameter is a line joining any two points on the surface
of a geometrical figure and passing through its center. The term is most
often used in connection with the circle and sphere, all of whose infinitely many
diameters have the same length.
isosceles
A shape that is isosceles has two sides of the same length, as in the case of an isosceles triangle.
An isosceles trapezoid in the United States
is the equivalent of a trapezium in Britain. 'Isosceles' comes from the
Greek iso (same or equal) and skelos (legs).
perimeter
The perimeter is the distance around a two-dimensional shape, especially one bounded by straight
lines, such as a triangle, square or polygon. In the case of a curved shape,
such as a circle or ellipse, the term circumference is more commonly used.
perpendicular
plane
straight
Something that is straight has no deviations. A straight line is usually simply
called a line. A straight angle,
or flat angle, is exactly 180°.
vertex
A vertex is a point where two sides of a triangle (or other polygon), or two sides of an angle, intersect, or at which three or more sides of a pyramid or other polyhedron intersect; otherwise known as a corner. A cube, for example, has eight vertices. The point of a cone is also known as a vertex. | 677.169 | 1 |
If the extremities of a line segment of length l moves in two fixed perpendicular straight lines, then the locus of the point which divides this line segment in the ratio 1 : 2 is-
A
a parabola
B
an ellipse
C
a hyperbola
D
None of these
Video Solution
Text Solution
Verified by Experts
The correct Answer is:2
Let the two fixed perpendicular straight lines be the co-ordinate axis let P(h, k) be the point whose locus is required let PA : PB = 1 : 2 Then PA=l3 and PB = 2l3 k=l3sinθor3k=lsinθ.....(i) and h=2l3cosθor3h2=lcosθ....(ii) Sqauring and adding (i) and (ii) we get 9k2+9h24=l2 ∴ locus 9x2+36y2=4l2 Which is an eclipse | 677.169 | 1 |
hbrproductions
Help please!!!!!Carlotta is constructing an equilateral triangle. She has already constructed the li...
4 months ago
Q:
help please!!!!!Carlotta is constructing an equilateral triangle. She has already constructed the line segment and arc shown. What should Carlotta do for her next step?Place the point of the compass on point Q and draw an arc that intersects the first arc, using the same width for the opening of the compass as the first arc. Use the straightedge to draw a line that passes through point P and intersects the arc. Use the straightedge to draw a line that passes through point Q and intersects the arc. Place the point of the compass on point Q and draw an arc that intersects the first arc, using a width for the opening of the compass that is less than 1/2PQ.
Accepted Solution
A:
a. Place the point of the compass on point Q and draw an arc that intersects the first arc, using the same width for the opening of the compass as the first arc. | 677.169 | 1 |
Explore unit 8 trigonometry homework answers.
Welcome to the Warren Institute blog! In this article, we will delve into unit 8 right triangles and trigonometry homework 3 answers key. Understanding these fundamental concepts is crucial for mastering the principles of trigonometry. By providing the answers key, we aim to help you reinforce your learning and gain confidence in tackling similar problems. Join us as we explore the right triangles and their application in real-world scenarios. Get ready to sharpen your skills and deepen your understanding of trigonometry!
The Pythagorean Theorem and Right Triangles
The Pythagorean Theorem is a fundamental concept in trigonometry and right triangle geometry. It states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. This theorem is crucial for solving problems involving right triangles and is often used in trigonometric calculations. Understanding the Pythagorean Theorem is essential for mastering unit 8 right triangles and trigonometry.
Trigonometric Ratios and Solving Right Triangles
In this section, students learn about the trigonometric ratios: sine, cosine, and tangent. These ratios are used to relate the angles of a right triangle to the lengths of its sides. Students also learn how to use these ratios to solve for missing side lengths and angle measures in right triangles. Mastering these concepts is important for developing a strong foundation in trigonometry and understanding the relationships between angles and sides in right triangles.
Applications of Trigonometry in Real-Life Scenarios
Understanding how trigonometry applies to real-world situations is an essential part of learning unit 8 right triangles and trigonometry. Students explore applications of trigonometry in areas such as architecture, engineering, physics, and navigation. By solving practical problems using trigonometric principles, students gain a deeper appreciation for the relevance and utility of trigonometry in various fields.
Challenging Problems and Critical Thinking Exercises
To reinforce their understanding of right triangles and trigonometry, students engage in challenging problems and critical thinking exercises. These tasks require students to apply their knowledge of trigonometric concepts to solve complex problems, enhancing their problem-solving skills and critical thinking abilities. By tackling these challenging exercises, students develop a deeper understanding of the principles covered in unit 8 and build confidence in their trigonometry skills.
frequently asked questions
How can I access the answer key for Unit 8 Right Triangles and Trigonometry Homework 3?
You can access the answer key for Unit 8 Right Triangles and Trigonometry Homework 3 by asking your teacher or checking the online resources provided by your textbook publisher.
What strategies or resources can I use to check my work for Unit 8 Right Triangles and Trigonometry Homework 3?
You can use online calculators, geometry software, or mathematics textbooks as resources to check your work for Unit 8 Right Triangles and Trigonometry Homework 3.
Are there any online platforms or communities where I can discuss and compare answers for Unit 8 Right Triangles and Trigonometry Homework 3?
Yes, you can join online forums such as Reddit's r/matheducation or the Math Help Forum to discuss and compare answers for Unit 8 Right Triangles and Trigonometry Homework 3.
Can the teacher provide additional guidance on Unit 8 Right Triangles and Trigonometry Homework 3 answers?
How can I ensure that I understand the concepts covered in Unit 8 Right Triangles and Trigonometry Homework 3 before checking the answers?
To ensure understanding of the concepts covered in Unit 8 Right Triangles and Trigonometry Homework 3 before checking the answers, review the textbook explanations and examples thoroughly, practice solving similar problems on your own, and seek help from your teacher or classmates if needed.
In conclusion, mastering the concepts covered in Unit 8: Right Triangles and Trigonometry is crucial for building a strong foundation in mathematics. The homework 3 answers key serves as a valuable tool for students to assess their understanding and reinforce their learning. By diligently practicing and comprehending these fundamental principles, students can enhance their problem-solving skills and | 677.169 | 1 |
Michael Lugo
Overview
Like most things mathematical, there is a nice Wikipedia article describing the Hyperplane Separation Theorem. An immediate consequence of the Hyperplane Separation Theorem is the Separating Axis Theorem (SAT): two closed convex sets are disjoint if and only if there exists a hyperplane between the two. This theorem is extremely useful in game programming to detect object collisions. Although this theorem works for any dimension, I will focus on the easiest to draw: the second dimension.
The Second Dimension
First, lets consider how to interpret the components of the SAT in my favorite space, \(\mathbb{R}^2\). If you are familiar with hyperplanes in the second dimension, then you can safely skip this section. In any dimension, a hyperplane is the set of all vectors \(\vec{u}\) that satisfy the equation \(\vec{v} \cdot \vec{u} = c\) for a fixed vector \(\vec{v}\) and constant \(c\). When we specialize to \(\mathbb{R}^2\), then we can expand this equation in terms of components. Let \(\vec{v} = \langle v_1, v_2 \rangle\) and \(\vec{u} = \langle x,y \rangle\).
$$
\vec{v} \cdot \vec{u} = c
\Rightarrow
v_1x + v_2y = c
$$
This form should look familiar, since it is the standard form for a line in \(\mathbb{R}^2\)! So hyperlanes in \(\mathbb{R}^2\) are straight lines. As I continuously remind my students, the only information you need to define a line is a point and a slope. If we simply reinterpret "slope" as "direction", then we recover that any hyperplane can be defined by a point and a direction. Only now, our direction is normal to the line we want. Note that this avoids the slope issue of a vertical line. Use the canvas below to drag the base point of the line and rotate the normal direction.
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Convex Polygons
Now that we have the first ingredient for the SAT (a 2D definition of hyperplane), we focus on the hypothesis of a closed convex set. A set is simply a collection of objects, which is not very enlightening. In this setting, we specifically want a set of points in \(\mathbb{R}^2\), which is a little more enlightening since this has a geometric interpretation as some points on the plane. A rough definition of a closed set is a set of points that contains its boundary. So if you were to draw it, there would be no dashed lines. A convex set of points satisfies the following condition: if \(x\) and \(y\) are points of your set, then any point \(c\) lying on the line segment between \(x\) and \(y\) is also in the set. For example, a circle is closed, but it is not convex. A disk without its boundary circle is convex, but it is not closed. A disk with its boundary is closed and convex.
The class of closed convex sets in \(\mathbb{R}^2\) is quite large, and many such sets are computationally expensive to verify that they are indeed convex and closed. That is why here we focus on a familiar and easy to use class of sets: polygons. If you have a filled polygon (you include all the points within the polygon) then the set of those points is automatically closed. The only thing we would need to check quickly and efficiently is if the set is also convex. For our most famous regular polygons, this is automatically true. However, this is not true for all polygons. To quickly test for convexity, a programmer may assume the polygon's information is presented in a certain way. In my canvas below, I have assumed that the points of my polygon are given clockwise, with the edges formed sequentially along pairs of vertices. With information like this, a simple cross product (imagining the vectors nested in \(\mathbb{R}^3\)) reveals the orientation between adjacent edges, and a consistent orientation signals the presence of a convex polygon.
Use the canvas below to test this with a polygon containing . Click and drag points to reposition them. If you break the convexity of the polygon (with non-intersecting lines), the background will turn red.
Canvas not supported
Projections onto Normals
We now have all the pieces necessary to discuss the SAT in the case of two convex polygons: two convex polygons are disjoint if and only if there exists some line that can be drawn between them. The question now becomes, "How do we find such a line?" or "Where can this line be?". Once again, our choice to use polygons aids us. If you stare at two disjoint polygons, you will find that you can always draw a line betwen them, and this line can be chosen to be parallel to an edge. It's a fact that two disjoint convex polygons will have a separating line whose normal direction is a normal direction to an edge of one of the polygons. Thus our question simplifies to, "How can we tell when there is a gap between two convex polygons along in a fixed direction?". The answer is to use projections.
When we fix a direction vector \(\vec{v}\), then we can project any other vector \(\vec{p}\) onto \(\vec{v}\) using the equation
$$
\operatorname{proj}\,_{\vec{v}}(\vec{p}) = \frac{\vec{v}\cdot\vec{p}}{\vec{v}\cdot \vec{v}}\,\vec{v}
$$
If we really only cared about the direction of our vector \(\vec{v}\), then we might as well use a unit vector for \(\vec{v}\), which simplifies the above to \(\operatorname{proj}\,_{\vec{v}}(\vec{p}) = (\vec{v} \cdot \vec{p})\vec{v}\). Now, if we think of \(\vec{v}\) as defining the direction of an axis, then all the information we need is the magnitude of the projection, which is called the component of projection. Still using a unit vector \(\vec{v}\), this simplifies to
$$
\operatorname{comp}\,_{\vec{v}}(\vec{p})
= |\operatorname{proj}\,_{\vec{v}}(\vec{p})|
= \vec{v} \cdot \vec{p}
$$
So now we can quickly calculate the component of projection in any fixed direction. If we do that for every point of a polygon, we get a list of numbers. Now, suppose I told you to project the entire polygon onto the axis in the direction of unit vector \(\vec{v}\)? We would geometrically think of casting a shadow of the polygon onto the axis, which would result in an interval along this axis. To find this interval, we simply need to find the maximum and minimum components of projection for each point of the polygon, which is again an incredibly quick operation! Then to determine if there is a gap between the convex polygons in this fixed direction, we simply check for a gap in the projected intervals.
Use the canvas below to rotate the axis. The projection of these shapes onto the axis are shown in red and green for square and triangle respectively. Note that by rotating our axis, we can make the intervals intersect. However, this (clearly) doesn't imply the shapes intersect. To show intersection, we would need to show that for EVERY axis normal to an edge, the projection intervals intersect.
Canvas not supported
Putting It Together
Here you can see a complete example. The shape you formed above and a box are places on a canvas, and you can drag around your shape. The program projects each shape against the normal axis for each edge. The box's projections are green, while your shape's projections are blue. If the shapes are colliding, then the you will see each normal axis will have intersecting projection intervals. Otherwise, there will be at least one axis where the intervals are disjoint. (If the canvas is blank, then your shape above is not convex.)
Canvas not supported
Conclusion
The Separating Axis Theorem is used to determine collisions. Using convex polygons, this is quickly computed using projections onto normal axes. This enables simple and efficient collision detection between two convex polygons.
P.S. Try extending this idea to collisions between a disk and a convex polygon. | 677.169 | 1 |
1.5 Angle Addition Postulate Worksheet Answers
Points d h and p are coplanar.
1.5 angle addition postulate worksheet answers. Angle addition postulate the angle addition postulate states that if d is in the interior of abc then abd cbd abc the following diagram gives an example of the angle addition postulate. G and p are collinear. Proof in the figure at the right point b is the midpoint of.
Angle addition postulate worksheets these angles worksheets are great for practicing the angle addition postulate. Draw a picture to help. These worksheets will produce 9 problems per page.
T is collinear with between m and n. S is collinear with between d and p 2. 20 write a segment addition problem using three points like question 11 that asks the student to solve for x but has a solution x.
The postulate that can be used to show each statement is true. Free geometry worksheets created with infinite geometry. Scroll down the page for more examples and solutions.
F 32j0 l1 s1x okeu xtha2 3sworfht hwdadr we9 ulzl 5ci. Ab x bc ac 2 create your own worksheets like this one with infinite geometry. These angles worksheets are great for practicing identifying if a point is interior exterior or on the | 677.169 | 1 |
...triangle parallel to the third side,'it divides the two sides proportionally. Prop. 76. // a line divides two sides of a triangle proportionally, it is parallel to the third side. Prop. 79. Two triangles are similar when they are mutually equiangular. Prop. 79, Cor. I. Two triangles...
...a triangle parallel to the third side, it divides these sides proportionally. (6f) The segments cut off on two transversals by a series of...
...corresponding segment. Prove this, using Theorems XL III and XL IV. THEOREM XLV 191. // a line divides two sides of a triangle proportionally it is parallel to the third side. Given A ABC having line PQ cutting AB at P and p> r> f>f) BC at Q, so that To prove PQ || AC. Proof....
...807. In a given line, AB, to find a point, C, so that PROPOSITION XVI. THEOREM 300. If a line divides two sides of a triangle proportionally, it is parallel to the third side. Given in A AEC, AB : BC = AD: DM. To prove DB parallel to EC. Proof. Through C, draw CE' parallel toc) The segments cut off on two transversals by a series of...
...L, PARALLEL FROM PROPORTIONAL SEGMENTS 195 Proposition XII. Theorem 202. // a straight line divides two sides of a triangle proportionally, it is parallel to the third side. Hyp.: Given the triangle ABC with EF so placed that CE : EA = CF : FB. Det. : To prove that EF is parallel...
...it divides those sides proportionally. Proof for the commensurable case only. b. If a line divides two sides of a triangle proportionally, it is parallel to the third side. c. The segments cut off on two transversals by a series of parallels are proportional. d. The bisector...
...triangle parallel to the third side it divides these sides proportionally. [57, cd] (b) If a line divides two sides of a triangle proportionally it is parallel to the third side. (Proofs for commensurable cases only.) [58*, cd*] (c) The segments cut off on two transversals by a... | 677.169 | 1 |
Section3.1Cartesian Coordinates
Objectives:PCC Course Content and Outcome Guide
When we model a relationship between two variables visually, we use the Cartesian coordinate system. This section covers the basic vocabulary and ideas that come with the Cartesian coordinate system.
Figure3.1.1.Alternative Video Lesson
René Descartes.
Several ideas and conventions used in mathematics are attributed to (or at least named after) René Descartes 1 . The Cartesian coordinate system is one of these.
The Cartesian coordinate system identifies the location of every point in a plane. Basically, the system gives every point in a plane its own "address" in relation to a starting point. We'll use a street grid as an analogy. Here is a map with Carl's home at the center. The map also shows some nearby businesses. Assume each unit in the grid represents one city block.
Figure3.1.2.Carl's neighborhood
If Carl has an out-of-town guest who asks him how to get to the restaurant, Carl could say:
"First go \(2\) blocks east (to the right on the map), then go \(3\) blocks north (up on the map)."
Two numbers are used to locate the restaurant. In the Cartesian coordinate system, these numbers are called coordinates and they are written as the ordered pair\((2,3)\text{.}\) The first coordinate, \(2\text{,}\) represents distance traveled from Carl's house to the east (or to the right horizontally on the graph). The second coordinate, \(3\text{,}\) represents distance to the north (up vertically on the graph).
Figure3.1.3.Carl's path to the restaurant
To travel from Carl's home to the pet shop, he would go \(3\) blocks west, and then \(2\) blocks north.
In the Cartesian coordinate system, the positive directions are to the right horizontally and up vertically. The negative directions are to the left horizontally and down vertically. So the pet shop's Cartesian coordinates are \((-3,2)\text{.}\)
Figure3.1.4.Carl's path to the pet shop
Remark3.1.5.
It's important to know that the order of Cartesian coordinates is (horizontal, vertical). This idea of communicating horizontal information before vertical information is consistent throughout most of mathematics.
Checkpoint3.1.6.
Use Figure 2 to answer the following questions.
What are the coordinates of the bar?
What are the coordinates of the gas station?
What are the coordinates of Carl's house?
Warning3.1.7.Notation Issue: Coordinates or Interval?
Unfortunately, the notation for an ordered pair looks exactly like interval notation for an open interval. Context will help you understand if \((1,3)\) indicates the point \(1\) unit right of the origin and \(3\) units up, or if \((1,3)\) indicates the interval of all real numbers between \(1\) and \(3\text{.}\)
Traditionally, the variable \(x\) represents numbers on the horizontal axis, so it is called the \(x\)-axis. The variable \(y\) represents numbers on the vertical axis, so it is called the \(y\)-axis. The axes meet at the point \((0,0)\text{,}\) which is called the origin. Every point in the plane is represented by an ordered pair, \((x,y)\text{.}\)
In a Cartesian coordinate system, the map of Carl's neighborhood would look like this:
Figure3.1.8.Carl's Neighborhood in a Cartesian Coordinate System
Definition3.1.9.Cartesian Coordinate System.
The Cartesian coordinate system 2 is a coordinate system that specifies each point uniquely in a plane by a pair of numerical coordinates, which are the signed (positive/negative) distances to the point from two fixed perpendicular directed lines, measured in the same unit of length. Those two reference lines are called the horizontal axis and vertical axis, and the point where they meet is the origin. The horizontal and vertical axes are often called the \(x\)-axis and \(y\)-axis.
The plane based on the \(x\)-axis and \(y\)-axis is called a coordinate plane. The ordered pair used to locate a point is called the point's coordinates, which consists of an \(x\)-coordinate and a \(y\)-coordinate. For example, the point \((1,2)\text{,}\) has \(x\)-coordinate \(1\text{,}\) and \(y\)-coordinate \(2\text{.}\) The origin has coordinates \((0,0)\text{.}\)
A Cartesian coordinate system is divided into four quadrants, as shown in Figure 10. The quadrants are traditionally labeled with Roman numerals.
Figure3.1.10.A Cartesian grid with four quadrants marked
Example3.1.11.
On paper, sketch a Cartesian coordinate system with units, and then plot the following points: \((3,2),(-5,-1),(0,-3),(4,0)\text{.}\)
Explanation.
Reading QuestionsReading Questions
1.
What are the coordinates of the gas station in the map of Carl's neighborhood?
2.
A Cartesian coordinate system has seven "places" within that are worth noting. What are they? (For example, one of them is Quadrant I.)
ExercisesExercises
Identifying Coordinates.
12Creating Sketches of Graphs.
3.
Sketch the points \((8,2)\text{,}\)\((5,5)\text{,}\)\((-3,0)\text{,}\) and \((2,-6)\) on a Cartesian plane.
4.
Sketch the points \((1,-4)\text{,}\)\((-3,5)\text{,}\)\((0,4)\text{,}\) and \((-2,-6)\) on a Cartesian plane.
5.
Sketch the points \((208,-50)\text{,}\)\((97,112)\text{,}\)\((-29,103)\text{,}\) and \((-80,-172)\) on a Cartesian plane.
6.
Sketch the points \((110,38)\text{,}\)\((-205,52)\text{,}\)\((-52,125)\text{,}\) and \((-172,-80)\) on a Cartesian plane.
7.
Sketch the points \((5.5,2.7)\text{,}\)\((-7.3,2.75)\text{,}\)\(\left(-\frac{10}{3},\frac{1}{2}\right)\text{,}\) and \(\left(-\frac{28}{5},-\frac{29}{4}\right)\) on a Cartesian plane.
8.
Sketch the points \((1.9,-3.3)\text{,}\)\((-5.2,-8.11)\text{,}\)\(\left(\frac{7}{11},\frac{15}{2}\right)\text{,}\) and \(\left(-\frac{16}{3},\frac{19}{5}\right)\) on a Cartesian plane.
9.
Sketch a Cartesian plane and shade the quadrants where the \(x\)-coordinate is negative.
10.
Sketch a Cartesian plane and shade the quadrants where the \(y\)-coordinate is positive.
11.
Sketch a Cartesian plane and shade the quadrants where the \(x\)-coordinate has the same sign as the \(y\)-coordinate.
12.
Sketch a Cartesian plane and shade the quadrants where the \(x\)-coordinate and the \(y\)-coordinate have opposite signs.
Cartesian Plots in Context.
13.
This graph gives the minimum estimates of the wolf population in Washington from 2008 through 2015.
What are the Cartesian coordinates for the point representing the year 2014?
Between 2014 and 2015, the wolf population grew by wolves.
List at least three ordered pairs in the graph.
14.
Here is a graph of the foreign-born US population (in millions) during Census years 1960 to 2010.
What are the Cartesian coordinates for the point representing the year 1960?
Between 1960 and 1980, the US population that is foreign-born increased by million people.
List at least three ordered pairs in the graph.
Regions in the Cartesian Plane.
15.
The point \({\left(-7,4\right)}\) is in Quadrant
I
II
III
IV
.
The point \({\left(1,-1\right)}\) is in Quadrant
I
II
III
IV
.
The point \({\left(-5,-9\right)}\) is in Quadrant
I
II
III
IV
.
The point \({\left(10,2\right)}\) is in Quadrant
I
II
III
IV
.
16.
The point \({\left(-4,-3\right)}\) is in Quadrant
I
II
III
IV
.
The point \({\left(10,-7\right)}\) is in Quadrant
I
II
III
IV
.
The point \({\left(-9,7\right)}\) is in Quadrant
I
II
III
IV
.
The point \({\left(10,7\right)}\) is in Quadrant
I
II
III
IV
.
17.
Assume the point \((x,y)\) is in Quadrant II18.
Assume the point \((x,y)\) is in Quadrant IV19.
Answer the following questions on the coordinate system:
For the point \((x,y)\text{,}\) if \(x>0>0yPlotting Points and Choosing a Scale.
20.
What would be the difficulty with trying to plot \((12,4)\text{,}\)\((13,5)\text{,}\) and \((310,208)\) all on the same graph?
21.
The points \((3,5)\text{,}\)\((5,6)\text{,}\)\((7,7)\text{,}\) and \((9,8)\) all lie on a straight line. What can go wrong if you make a plot of a Cartesian plane with these points marked, and you don't have tick marks that are evenly spaced apart? | 677.169 | 1 |
Quilt Making in Three Dimensions
Introduction: Quilt Making in Three Dimensions
About: The Lesley STEAM Learning Lab is a center designed to research new opportunities for learning through engagement and inquiry-based exploration.
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This project investigates the geometry embedded in visual art and quilt making, through the creation of a rhombic dodecahedron, which is a convex polyhedron or three-dimensional shape with flat polygonal (fractal) faces, straight edges and sharp corners or vertices. The project is inspired by the quilt-based art of Sanford Biggers and Adam Savage's "Tested" episode with mathematician Matt Parker:
In the "Tested" tutorial, Savage and Parker work with laser cut acrylic rhombus shapes and create their own rhombic dodecahedron. Savage and his team also add other elements to their objects (see optional steps).
Sanford Biggers' paintings and sculptures are done directly on or made from pre-1900 antique quilts. Many of the quilts are created using geometric shapes. The tradition of quilt-making holds a significant place in American culture and has special resonance in African American communities as witnessed in the quilts by the Gee's Bend — a small, insulated community in Alabama — that has produced quilts from the 19th century to the present.
Students will learn about geometric shapes and polyhedra, find and identify the shapes in the quilts and art, and use found materials such as cardboard, old plastic CD cases, or anything that can be cut into shapes of the same size and thickness.
Step 1: Learn More About Shapes & Polyhedra
Explore the four shapes and five polyhedra in the photo. Use the "Illuminations" handout (attached) from the National Council of Teachers of Mathematics to determine the number of faces, edges, and vertices (corners) for each shape. Answer the following questions:
Find the sum of the number of faces and the number of vertices. How does this sum compare with the number of edges?
Do you think this may be a rule for the other shapes?
Add the number of faces and the number of corners for the other shapes. Compare the sum of faces and corners to the number of edges. What did you find out? Is there a rule for all of the shapes?
For this project we will focus on the 2D rhombus and 3D dodecahedron shapes.
Step 2: Find the Shapes in Quilts and Art
Quilts are often pieced in a patchwork pattern of basic shapes or blocks, with the blocks being made individually and assembled into the finished work. The fundamental geometries of the quilts include single repeating patches: triangles, squares, diamonds, and hexagons. See if you can identify the shapes in the quilt on the left (courtesy of Eli Leon). How do these shapes relate to the previous step?
Artist Sanford Biggers remixes these quilts by extruding or expanding the 2D shapes to create the flat polygonal faces. The polygons are covered with real quilt patterns. Many of the quilts Biggers uses come from the African American quilting tradition that is linked to West and Central African textile design that often juxtaposes unlike motifs or interrupts orderly repetition by shifts in texture, direction, pattern or scale.
Can you identify shapes and polyhedra in Biggers' artwork?
Use the Decoding the Cypher handout (PDF) to explore this question.
The Bronx Museum of the Arts catalog includes a comic by John Jennings with David Brame and Esperanza Bey.
Step 3: Collect Your Materials
For this project you will need at least 12 rhombuses cut from cardboard or other stiff materials. All of the rhombuses should be the same size and thickness.
Position two of the shape together like in the photo and tape both sides. Make sure you can flex or bend the taped joints to 120 degrees.
SVG files are included if you have access to a laser cutter (to cut out the shapes).
Step 4: Connect the Shapes and Joints
Keep connecting the shapes and joints using the tape until you complete the 3D object (see photo).
You can also follow along with Adam Savage and Matt Parker in the introduction video.
Step 5: Optional: Add Mirror Film and LEDs to the Shapes
As an additional step you can add infinity mirror film to the inside shapes before you tape them together, to create additional effects. In the "Tested" video they add LED strips and black tape to the edges of the object.
The LED strips can be attached to the inside or outside edges. Black tape is used on the outside edges to hide this work.
Step 6: Optional: Create Special Effects
In this example, two opposite sides of the rhombic dodecahedron were left clear and the object was placed against a computer screen displaying a sound-generated animation such as what is created by Magic Music Visuals, a music visualizer and VJ software program. To simulate the style of improvisational quilts, a kaleidoscope effect was applied to a GLSL shader, which was created using OpenGL Shading Language (code).
You can add colorful beads or other tiny objects (colored tissue paper) to the inside of the mirrored object to create effects.
You can also use a projector to map and project images inside of the 3D object.
This work is made possible by support from STAR, a Biogen Foundation Initiative. The team at Lesley supporting this initiative includes faculty and staff in the Lesley STEAM Learning Lab, Science in Education, the Center for Mathematics Achievement, and other related Lesley University departments and programs. | 677.169 | 1 |
Suppose you choose three random points in the unit square. What is the probability that the triangle determined by the three points is acute? (All angles less than 90°?)
(Write a program that uses a Monte Carlo procedure to estimate this probability.)
Take a structure point defined by
typedef struct{
double x;
double y;
} point;
which represents a point in the plane. Define the structure segment by
typedef struct{
point p1;
point p2;
} segment;
which represents a (closed) line segment in the plane starting at point p1 and ending at point p2. It may happen that the segment degenerates to a point and p1=p2.
Define the structure circle by
typedef struct{
point center;
double radius;
} cicle;
which represents a (closed) circle in the plane centered at the point center and with radius radius.
Write a program that asks the user to type in the data for a line segment and for a circle. The program determines whether the segment is inside the circle, intersects the circle but is neither inside nor outside the circle, or is outside the circle. The program should also state the number of points of intersection between the
segment and the circle. For example, the segment from (1,2) to (4,3) lies inside the circle with center (0,0) of radius 5 and touches it at exactly one point, the point (4,3). Whereas, the segment from (-3,-399) to (-3,541) neither is inside nor outside the same circle but touches it at exactly two points, the points (-3,-4) and (-3,4). | 677.169 | 1 |
Related Puzzles
Physical Education (staying healthy)
Antibiotic Awareness Week 2022
Summer Vacation
Canada
Poppy
QUESTIONS LIST: isosceles : _ trapezoid - two sides are parallel and base angles are equal, non parallel sides are equal length, parallelogram : sides are parallel two by two, trapezoid : two sides are parallel. side length and angles are not equal, square : four sides of equal length, four right angles, rectangle : four angles are right, kite : which one of the diagonals is an axis of symmetry, rhombus : a parallelogram whose side are equal, especially when it is not a square, quadrilateral : four-sided polygon | 677.169 | 1 |
Andhra Pradesh State Board STD 9th Maths Chapter 12 Circles Ex 12.1 Books Solutions with Answers are prepared and published by the Andhra Pradesh Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of AP Board Class 9th Maths Chapter 12 Circles Ex 12.1 Books Answers Solutions, then you are in the right place. Here is a complete hub of Andhra Pradesh State Board Class 9th Maths Chapter 12 Circles Ex 12.1 solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Andhra Pradesh Board STD 9th Maths Chapter 12 Circles Ex 12.1 Textbooks. These Andhra Pradesh State Board Class 9th Maths Chapter 12 Circles Ex 12.1 Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Andhra Pradesh Board.
Question 2. State true or false. i) A circle divides the plane on which it lies into three parts. ( ) ii) The area enclosed by a chord and the minor arc is minor segment. ( ) iii) The area enclosed by a chord and the major arc is major segment. ( ) iv) A diameter divides the circle into two unequal parts. ( ) v) A sector is the area enclosed by two radii and a chord. ( ) vi) The longest of all chords of a circle is called a diameter. ( ) vii) The mid point of any diameter of a circle is the centre. ( ) Solution: i) True ii) True iii) True iv) False v) False vi)True vii) True | 677.169 | 1 |
The length of tangent from a point A at a distance of 5 cm from the
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Class 8-9-10, JEE & NEET
The length of tangent from a point A at a distance of 5 cm from the centre of the circle is 4 cm. What is the radius of the circle?
Solution:
Let us first draw whatever is given for a better understanding of the problem.
Let O be the center of the circle and B be the point of contact.
We know that the radius of the circle will be perpendicular to the tangent at the point of contact. Therefore, we have as a right triangle and we have to apply Pythagoras theorem to find the radius of the triangle. | 677.169 | 1 |
Which Set Of Angles Can Form A Triangle
In geometry, a triangle is a closed figure with three straight sides and three angles. The sum of the interior angles of a triangle is always 180 degrees. Therefore, not all sets of angles can form a triangle. For a set of angles to form a triangle, they must satisfy the triangle inequality theorem, which states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. We will discuss which set of angles can form a triangle.
Let's assume that we have three angles, A, B, and C. The sum of these angles is 180 degrees. Therefore, we can write the following equation:
A + B + C = 180 degrees
We know that for a set of angles to form a triangle, the sum of any two angles must be greater than the third angle. Therefore, we can write the following three inequalities:
A + B > C B + C > A A + C > B
Let's consider the three angles A, B, and C separately and see what ranges of values they can take to satisfy the triangle inequality theorem.
Angle A:
Let's assume that angle A is the largest angle among the three. In this case, the other two angles must be less than A. Therefore, we can write the following inequality:
A > B, A > C
Now, let's consider the two smaller angles, B and C. For them to form a triangle with A, their sum must be greater than A. Therefore, we can write the following inequality:
B + C > A
By combining these two inequalities, we get:
A > B, A > C, B + C > A
Therefore, any value of angle A that satisfies these three inequalities can form a triangle.
Angle B:
Now let's assume that angle B is the largest angle among the three. In this case, the other two angles must be less than B. Therefore, we can write the following inequality:
B > A, B > C
Now, let's consider the two smaller angles, A and C. For them to form a triangle with B, their sum must be greater than B. Therefore, we can write the following inequality:
A + C > B
By combining these two inequalities, we get:
B > A, B > C, A + C > B
Therefore, any value of angle B that satisfies these three inequalities can form a triangle.
Angle C:
Finally, let's assume that angle C is the largest angle among the three. In this case, the other two angles must be less than C. Therefore, we can write the following inequality:
C > A, C > B
Now, let's consider the two smaller angles, A and B. For them to form a triangle with C, their sum must be greater than C. Therefore, we can write the following inequality:
A + B > C
By combining these two inequalities, we get:
C > A, C > B, A + B > C
Therefore, any value of angle C that satisfies these three inequalities can form a triangle.
Conclusion
For a set of angles to form a triangle, the sum of any two angles must be greater than the third angle. Therefore, we can use the triangle inequality theorem to determine which set of angles can form a triangle. Specifically, for any given set of three angles A, B, and C, we can determine which angle is the largest and then use the corresponding inequality to determine the range of values for that angle that can form a triangle. If all three inequalities are satisfied, then the set of angles can form a triangle | 677.169 | 1 |
RIGHTANGLE
Right angle
In geometry and trigonometry, a right angle is an angle that bisects the angle formed by two halves of a straight line. More precisely, if a ray is placed so that its endpoint is on a line and the adjacent angles are equal, then they are right angles. As a rotation, a right angle corresponds to a quarter turn | 677.169 | 1 |
...them both within it, dr touches the one without and the other within ? PROB. xxxin. Given the three lines, drawn from the angles of a triangle to the middle of each of the opposite sides ; to determine the triangle. PROB. xxxu. The base of a triangle, the sum...
...being identical, it follows, that the 'three perpendiculars intersect in the same point. 62. PROP. 3. The lines drawn from the angles of a triangle to the middle points of the opposite sides, intersect in the same point. Let AM, BN, CP, (fig. 34.) be drawn to the...
...= 135. l 55"/ X I93X.84 = Kíi-.lí MATHEMATICS AND PHYSICS. 70» The intersection of three linee, drawn from the angles of a triangle, to the middle of the opposite sides, is its centre of gravity and percussion. If the lines bisect the angles, it is the centre of the inscribed...
...l-4th of the height. In the human body it is in the pelvis, between the hips. The intersection of three lines, drawn from the angles of a triangle, to the middle of the opposite sides, is its centre of gravity and percussion. If the lines bisect the angles, it is the centre of the inscribed...
...to the middle of the hase, being given ; to determine the triangle. PROR, xxxin.— Given the three lines, drawn from the angles of a triangle to the middle of each of the opposite sides; to determine the triangle. PROR, xxxiv. — Given the hase of в triangle,...
...line drawn I'rom the vertex to the middle of the base, to determine the triangle. 12. Given the three lines drawn from the angles of a triangle to the middle of each of the opposite sides, to determine the triangle. 13. Given the base of a triangle, and the sum...
...[9] gives this THEOREM. Sixteen times thf5.36 12.06 21.44 = 33.64 = 48.25 65.08 85.69 108.56 134.01 162.12 193. 0 The intersection of three lines, drawn from the angles of a triangle, to the "middle of the opposite sides, is its centre of gravity and percussion. If the lines bistct the angles, it is the centre of the inscribed...
...[9] gives this THEOREM. Sixteen times thelying towards the bisected side is half of the other. From this it readily follows that all the three lines drawn from the angles of a triangle to the middle of the opposite sides, pass through one and the same point. If we bisect the angle a of the triangle abc (pi. 3, fig. 37),... | 677.169 | 1 |
Advanced mathematics
Trig Identity
In this diagram $O$ is the centre of a unit circle (i.e. a circle with radius 1).
Use it to find two trigonometrical relationships between $A$ and $2A$.
Did you know ... ?
Trigonometrical expressions can often be derived algebraically or geometrically. In more advanced applications the functions of trigonometry are defined as infinite power series rather than geometric ratios | 677.169 | 1 |
A Course of Mathematics for the Use of Academies, as Well as Private Tuition
96. From the given equation of the curve put into fluxions, find the value of or y, which value substitute instead of it in the equation z+y; then the fluents, being taken, will give the value of z, or the length of the curve, in terms of the absciss or ordinate.
EXAMPLES.
EXAM. 1. To find the length of the arc of a circle, in terms of the sine, versed sine, tangent, and secant.
The equation of the circle may be expressed in terms of the radius, and either the sine, or the versed sine, or tangent, or secant, &c of an arc. Let therefore the radius of the circle be ca or ce=r, the versed sine ad (of the arc AE the right sine DE=y, the tangent TE―t, and the secant cт=s, then, by the nature of the circle, there arise these equations, viz.
Then, by means of the fluxions of these equations, with the general fluxional equation 22+y, are obtained the following fluxional forms, for the fluxion of the curve, the fluent of any one of which will be the curve itself; viz.
Hence the value of the curve, from the fluent of each of these, gives the four following forms, in series, viz. putting d2r the diameter, the curve is
2.3d 2.4.5d 2.4.6.7d3+ &c.)✓ dr,
z=(1+
=(1+
ta
+
+ &c.).y,
574
3
+
2.38.3
to
ts
+ &c.) t,
728 91.8
3(85 — r3) + &c.) r、
2.4.5.85
Now, it is evident that the simplest of these series, is the third in order, or that which is expressed in terms of the tangent. That form will therefore be the fittest to calculate an example by in numbers. And for this purpose it will be convenient to assume some arc whose tangent, or at least the it, is known to be some small simple number. Now, square the arc of 45 degrees, it is known, has its tangent equal to
the
the radius; and therefore, taking the radius r=1, and conse- quently the tangent of 45°, or t≈1 also, in this case the arc of 45° to the radius 1, or the arc of the quadrant to the dia- meter 1, will be equal to the infinitė series 1 — } + } } + } − + 13 &c.
But as this series converges very slowly, it will be proper to take some smaller arc, that the series may converge faster; such as the arc, of 30 degrees, the tangent of which is =√✓}}, or its square t2: which being substituted in the series, the length of the arc of 30° comes out
1
1
(1- +
3
1
1
+9.3+
&c.) ✓.
Hence, to com3.3 5.32 7.33 7.339.3+ pute these terms in decimal numbers, after the first, the succeeding terms will be found by dividing, always by 3, and these quotients again by the absolute number 3, 5, 7, 9, &c.; and lastly, adding every other term together, into two sums, the one the sum of the positive terms, and the other the sum of the negative ones: then lastly, the one sum taken from the other leaves the length of the arc of 30 degrees; which being the 12th part of the whole circumference when the radius is 1, or the 6th part when the diameter is 1, consequently 6 times that arc will be the length of the whole circumference to the diameter 1. Therefore multiplying the first term by 6, the product is 123-4641016; and hence the operation will be conveniently made as follows:
+Terms. 3.4641016
Terms.
0.3849002
183286
EXAM. 2. To find the length of a parabola.
EXAM. 3. To find the length of the semicubical parabola, whose equation is ax2=y3.
EXAM. 4. To find the length of an elliptical curve.
EXAM. 5. To find the length of an hyperbolic curve.
1
OF QUADRATURES; OR, FINDING THE AREAS OF CURVES.
97. The Quadrature of Curves, is the measuring their areas, or finding a square, or other right-lined space, equal to a proposed curvilineal one.
By art. 9 it appears, that any flowing quantity being drawn into the fluxion of the line along which it flows, or in the direction of its motion, there is produced the fluxion of the quantity generated by the flowing. That is, Dd X DE or yx is the fluxion of the area ADE. Hence this rule.
t
E C
A
X
RULE.
98. From the given equation of the curve, find the value either of or of y; which value substitute instead of it in the expression yx; then the fluent of that expression, being taken, will be the area of the curve sought.
EXAMPLES.
EXAM. 1. To find the area of the common parabola.
The equation of the parabola being ax = y2; where a is the parameter, x the absciss AD, or part of the axis, and y the ordinate DE.
From the equation of the curve is found Y ✔ax. This substituted in the general fluxion of the area y gives x ✔ ax or a2x2; the fluxion of the parabolic area; and the fluent of this, or ga2x2 = 3x / ax = 3xy, is the area of the parabola ADE, and which is therefore equal to of its circumscribing rectangle.
EXAM. 2. To square the circle, or find its area.
The equation of the circle being y2 ax x2, or y Vax-x2, where a is the diameter; by substitution, the
general
general fluxion of the area yz, becomes x ✔ax x2, for the But as the fluent of this cannot
fluxion of the circular area.
be found in finite terms, the
quantity axx is thrown
into a series, by extracting the root, and then the fluxion of the area becomes
1.3x3
1.3.5x4 2.4.6.8a4
- &c.);
&c.) ;
and then the fluent of every term being taken, it gives
2 1.x
x √ax × (
x2
3 5α 4.7a2 4.6.9a3
1.3.5.x4 4.6.8.11a4
for the general expression of the semisegment ADE.
And when the point o arrives at the extremity of the diameter, then the space becomes a semicircle, and x = a; and then the series above becomes barely
2 1 3 5 4.7
1.3.5.
4.6.8.11
for the area of the semicircle whose diameter is a.
&c.)
EXAM. 3. To find the area of any parabola, whose equation
is amzn=ym+n
EXAM. 4. To find the area of an ellipse.
EXAM. 5. To find the area of an hyperbola.
EXAM. 6. To find the area between the curve and asymp
tote of an hyperbola.
EXAM. 7. To find the like area in any other hyperbola whose general equation is "y"—am+n,
99. In the solid formed by the rotation of any curve about its axis, the surface may be considered as generated by the circumference of an expanding circle, moving perpendicularly along the axis, but the expanding circumference moving along the arc or curve of the solid. Therefore, as the fluxion of any generated quantity is produced by drawing the generating quantity into the fluxion of the line or direction in which it moves, the fluxion of the surface will be found by drawing the circumference of the generating circle into the fluxion of the curve. That is, the fluxion of the surface, BAE, is equal to AE drawn into the circumference BCEF, whose radius is the ordinate DE.
100. But if c be 3.1416, the circumference of a circle whose
4
whose diameter is 1, xAD the absciss, y=DE the ordi- nate, and z = AE the curve; then 2y the diameter BE, and 2cy the circumference BCEF; also, AE V+y2: therefore 2cyz or 2cy+y is the fluxion of the surface. And consequently if, from the given equation of the curve, the value of x or y be found, and substituted in this expression 2cy/x+y2, the fluent of the expression being then taken, will be the surface of the solid required.
EXAMPLES.
EXAM. 1. To find the surface of a sphere, or of any seg
ment.
In this case, AE is a circular arc, whose equation is y2 =ax -x2, or y=√αx—x3.
The fluxion of this gives y=
α-2x
This value of z, the fluxion of a circular arc, may be found more easily thus: In the fig. to art. 95, the two triangles EDc, Eae are equiangular, being each of them equiangular to the triangle ETC conseq. ED EC:: Ea: Ee, that is,
And the
The value z of being found, by substitution is obtained 2cy ż cacx for the fluxion of the spherical surface generated by the circular arc in revolving about the diameter Ad. fluent of this gives acx for the said surface of the spherical segment BAE.
But ac is equal to the whole circumference of the gene- rating circle; and therefore it follows, that the surface of any spherical segment, is equal to the same circumference of the generating circle, drawn into x or AD, the height of the seg-
ment.
Also when x or AD becomes equal to the whole diameter a, the expression acx becomes aca or ca2, or 4 times the area of the generating circle, for the surface of the whole sphere. And these agree with the rules before found in Mensuration of Solids.
EXAM. 2. To find the surface of a spheroid. EXAM. 3. To find the surface of a paraboloid. EXAM. 4. To find the surface of an hyperboloid. | 677.169 | 1 |
Does a parallelogram have two right angles?
A parallelogram is a quadrilateral with 2 pair of opposite sides parallel. A rectangle is a special parallelogram that has 4 right angles. However, a trapezoid could have one of the sides connecting the two parallel sides perpendicular to the parallel sides which would yield two right angles.
Can a parallelogram have 4 right angles?
A rectangle is a parallelogram with four right angles, so all rectangles are also parallelograms and quadrilaterals.
Is a parallelogram 360 degrees?
Explanation: Parallelograms have angles totalling 360 degrees, but also have matching pairs of angles at the ends of diagonals.
What angles do a parallelogram have?
Which quadrilateral can have exactly 2 right angles?
trapezoid
The quadrilateral that can have only two right angles is a trapezoid. Not all trapezoids have right angles, but we can construct one that does.
How do you find the angle of a right parallelogram?
There are six important properties of parallelograms to know:
Opposite sides are congruent (AB = DC).
Opposite angels are congruent (D = B).
Consecutive angles are supplementary (A + D = 180°).
If one angle is right, then all angles are right.
The diagonals of a parallelogram bisect each other.
What are some real life examples of parallelograms?
Real Life Examples Rectangle-shaped objects – Books, tabletops, mobile phones, and TV screens. Square-shaped objects – Chessboard, wall clock, and a slice of bread. Parallelogram-shaped objects – Street and traffic sign, the structures on the neck of a guitar, and the United States Postal Service logo.
How many angles does a parallelogram have?
Lesson Summary. You
Does parallelogram have four equal sides?
A parallelogram is a polygon with four sides (a quadrilateral). It has two pairs of parallel sides (sides which never meet) and four edges. The opposite sides of a parallelogram have the same length (they are equally long). The word "parallelogram" comes from the Greek word "parallelogrammon" (bounded by parallel lines).
What is the difference between a square and a parallelogram?
As nouns the difference between square and parallelogram. is that square is any simple object with four nearly straight and nearly equal sides meeting at nearly right angles while parallelogram is (geometry) a convex quadrilateral in which each pair of opposite edges are parallel and of equal | 677.169 | 1 |
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Trigonometry
GCSELevel 4-5Level 6-7Cambridge iGCSEEdexcel iGCSE
Trigonometry Revision
Trigonometry
Trigonometryis the study of the relationship between the angles of a triangle and its lengths. We have functions that can tell us these relationships, the primary ones we will be looking at are the sine function, thecosine function, and the tangent function. We abbreviate these to sin, cosand tan.
The trigonometric functionsare functions of angles.
SOHCAHTOA
SOHCAHTOA is a memory device for recalling how the trigonometric functions relate to the lengths of a right angle triangle.
Abbreviating \text{sin}(x) to \text{S}, \text{cos}(x) to \text{C}, \text{tan}(x) to \text{T}, Opposite to \text{O}, Adjacent to \text{A} and Hypotenuse to \text{H}, we get
\text{S} = \dfrac{O}{H}
\text{C} = \dfrac{A}{H}
\text{T} = \dfrac{O}{A}
Which is where SOHCAHTOA comes from.
Level 4-5GCSECambridge iGCSEEdexcel iGCSE
Angles of Elevation and Depression
When looking straight forward, your eyeline is parallel to the horizon. When you want to look at something above your eyeline, you look up. The angle at which you look up is called the angle of elevation.
Similarly, if you want to view something below your eyeline, you look down. The angle to which you look down is called theangle of depression.
We can use these angles to construct right angled triangles, on which we can do trigonometry. For example, if you were standing 308 meters from the base of the Shard, and you knew that your eye level was 1.6 meters from the ground, if your angle of elevation is 45\degree then you can use this to calculate the height of the Shard.
You have the angle 45\degree, and the adjacent length of 308 \text{ m}. So recalling SOHCAHTOA
\text{tan}(x) = \dfrac{\text{Opposite}}{\text{Adjacent}}
So, using our calculator for \text{tan}(45) = 1 we have
1 = \dfrac{\text{Opposite}}{308}
\text{Opposite} = 308 \text{ m}
We must remember that it is 308 plus our eye level, giving us a height for the shard of 309.6 \text{ m}.
The same process can be used with an angle of depression to find the depth of things below our eyeline.
Level 6-7GCSECambridge iGCSEEdexcel iGCSE
Shortest Distance
For a point P and a line L, the shortest line from P to L is perpendicularto L.
This must be the shortest length between P and L as any line connecting P and L at any other angle is the hypotenuse to a triangle with the perpendicular line. As the hypotenuse is always the largest side of a right angled triangle, the perpendicular line must be the shortest possible distance from P to L.
We may use this fact to calculate, using SOHCAHTOA, the distances between points and lines.
Example: If the line AB passes though points A and B whilst line PB = 8 \text{ cm} and \angle PBA = 40 \degree then we can work out the shortest distance from P to line AB.
As the shortest distance is the line perpendicular to AB from P to the line AB, we can construct a right angled triangle, with the angle 40 \degree and the hypotenuse of PB = 8 \text{ cm}. Which would leave us looking for the opposite side.
By SOHCAHTOA
\text{Opposite} = \text{sin}(x) \times \text{Hypotenuse}
with our values
\text{Shortest distance}=\text{sin}(40) \times 8
\text{Shortest distance}=5.1\text{ cm}
to 1 decimal place.
Level 6-7GCSECambridge iGCSE
Example 1: Finding a Side
Find the missing length y to 1 decimal place.
[2 marks]
First, we can see that we have the angle 37 \degree and it's adjacent length 5 \text{ cm}. We are looking for the length of the hypotenuse, so with SOHCAHTOA, we know
\text{cos}(x) = \dfrac{\text{adjacent}}{\text{hypotenuse}}
substituting in our values
\text{cos}(37) = \dfrac{5}{y}
y = \dfrac{5}{\text{cos}(37)}
Using our calculators we get
y = 6.260 \dots
So the length of the hypotenuse is 6.3 \text{ cm} to 1 decimal place.
Level 4-5GCSECambridge iGCSEEdexcel iGCSE
Example 2: Finding an Angle
Find the missing angle x to the nearest degree.
[2 marks]
We are looking for an angle x and we have it's adjacent length 1.3 \text{ cm} and it's opposite length 3.1 \text{ cm}. Using SOHCAHTOA we have
\text{tan}(x) = \dfrac{\text{opposite}}{\text{adjacent}}
Substituting our values in
\text{tan}(x) = \dfrac{3.1}{1.3} = \dfrac{31}{13}
We must use the inverse function of \text{tan}, this is the \text{tan}^{-1} function. You should be able to find this on a calculator. The inverse function is defined such that:
\text{tan}^{-1}(\text{tan}(x)) = x
So substituting in \text{tan}(x) = \dfrac{31}{13}
x =\text{tan}^{-1} \left( \dfrac{31}{13} \right) = 67.249 \dots
So x to the nearest degree is 67\degree.
Level 4-5GCSECambridge iGCSEEdexcel iGCSE
Example 3: Angle of Depression
Tower A is 200 \text{ m} away from tower B. From the peak of tower A, the angle of depression when looking down at the peak of tower B is 24 \degree, how much shorter is tower B than tower A, to the nearest metre?
[4 marks]
We can construct the right angled triangle to find the height difference.
Now we have a right angled triangle with an angle of 24 \degree, the adjacent length is 200 \text{ m} and the opposite length is what we are looking for. From SOHCAHTOA we have
\text{tan}(x) = \dfrac{\text{opposite}}{\text{adjacent}}
Substituting in the values we have gives us
\text{tan}(24) = \dfrac{\text{opposite}}{200}
\text{opposite} = 200 \times\text{tan}(24)
\text{opposite} = 89.045 \dots \text{ m}
So tower A is 89 \text{ m} taller than tower B.
Level 6-7GCSECambridge iGCSEEdexcel iGCSE
Trigonometry Example Questions
Question 1: Find the missing length y to 1 decimal place.
[2 marks]
Level 4-5GCSECambridge iGCSEEdexcel iGCSE
We have a known angle and the hypotenuse. We are looking for the adjacent side. So using SOHCAHTOA we know
\text{cos}(x) = \dfrac{\text{adjacent}}{\text{hypotenuse}}
So putting in our values
\text{cos}(64) = \dfrac{y}{5.2}
y = \text{cos}(64) \times 5.2 = 2.279 \dots
So the missing length is 2.3 \text{ mm} to 1 decimal place.
Question 2: Find the missing angle x to the nearest degree.
[2 marks]
Level 4-5GCSECambridge iGCSEEdexcel iGCSE
From the angle we are looking for we have its adjacent side and its opposite side. From SOHCAHTOA we have
\text{tan}(x) = \dfrac{\text{opposite}}{\text{adjacent}}
Substituting in our values
\text{tan}(x) = \dfrac{3}{3.3}
So using \text{tan}^{-1} to find x
x = \text{tan}^{-1} \left( \dfrac{3}{3.3} \right)
x = 42.27 \dots
So the missing angle is 42 \degree to the nearest degree.
Question 3: Nina is looking at her house. She knows that the house is 6.5 \text{ m} tall, and her eyeline is 1.5 \text{ m} high. If her angle of elevation to see the top of her house is 25 \degree, how far away, to 1 decimal place, is she standing from her house?
[4 marks]
Level 6-7GCSECambridge iGCSEEdexcel iGCSE
If Nina's house is 6.5 \text{ m} and her eyeline is 1.5 \text{ m} off the ground, then when she looks up at her house, she is only looking 5 \text{ m} up. Given this and the angle of elevation of 25 \degree, we can construct a right angled triangle.
Now to find out how far away from her house she is, we can use SOHCAHTOA. We have an angle and it's opposite side, while we are looking for the adjacent side.
\text{tan}(x) = \dfrac{\text{opposite}}{\text{adjacent}}
So using our values
\text{tan}(25) = \dfrac{5}{\text{adjacent}}
\text{adjacent} = \dfrac{5}{\text{tan}(25)} = 10.722 \dots
So Nina is standing 10.7 \text{ m} away from her house.
Question 4: From a hotel balcony, Jack can see the beach. He knows that from the ground floor to the beach is just 25 \text{ m} and when looking at the beach from the balcony he looks with an angle of depression 32 \degree. How high up is Jack, to 1 decimal place?
[4 marks]
Level 6-7GCSECambridge iGCSEEdexcel iGCSE
If the distance from jack to the beach is 25 \text{ m} and it has an angle of depression of 32 \degree then we can construct a right angled triangle.
Now we want to find the opposite side to the angle and we have the adjacent angle. Using SOHCAHTOA we know
\text{tan}(x) = \dfrac{\text{opposite}}{\text{adjacent}}
Substituting our values in
\text{tan}(32) = \dfrac{\text{opposite}}{25}
\text{opposite} = 25 \times \text{tan}(32) = 15.621 \dots
Jack is 15.6 \text{ m} high.
Question 5: For three points A, B and C where BC = 3.5 \text{ cm} and \angle ABC = 28 \degree, what is the shortest distance from C to the line AB?
Give you answer to 2 decimal places.
[4 marks]
Level 6-7GCSECambridge iGCSE
The shortest distance from C to AB is the length of the line from C to AB that is perpendicular to AB.
So we construct a right angled triangle where BC is the hypotenuse and the perpendicular line from C to AB is the opposite side to the angle \angle ABC.
Using SOHCAHTOA
\text{Opposite} = \text{sin}(x) \times \text{Hypotenuse}
With our values
\text{Shortest Distance}= \text{sin}(28) \times 3.5
\text{Shortest Distance}= 1.64 \text{ cm}
To 2 decimal place | 677.169 | 1 |
Class 8 Courses
Trigonometric Ratios of Compound Angle Revision Video | Class 11, JEEThrough this revision video you will be able to revise Trigonometric ratios of compound angles, an important Class 11 topic. This Video tutorial covering Compound angles in trigonometry, Compound angles based questions, Formula of Cos(A + B), Formula of Cos(A - B), Formula of Sin(A + B), Formula of Sin(A - B), Formula of tan(A + B), Formula of tan(A- B), Cos(A + B)= CosACosB – SinASinB, Cos(A - B)= CosACosB + SinASinB, trigonometric identities, common angles, and more. So it will help students to prepare for JEE and Other competitive exams like Comed-K, BITSAT etc. Mind Map is also shared in the revision video to learn all the important formulae. This revision technique will help you solve Compound angles based questions like a topper and get a better score. | 677.169 | 1 |
Assume the center to be X
if OT is t than PX should be t ...coz the lines r parallel
if PX is t then XS should be t, too
now we have a diagonal PS = 2t ...which is by adding the two t's
then you move to the right so PS + SR = 2t+ p
now assume the midpoint of the line XQ to be Y
to move vertically straight up u would be needing vector PY , which we don't have so your method isn't correct | 677.169 | 1 |
The Element of Geometry
Im Buch
Seite 16 ... PROB . To draw a straight line at right angles to a given straight line , from a given point in the same . Let AB be the given straight line , and C the given point in it ; it is required to draw a straight line from C , at right angles ...
Seite 17 ... PROB . To draw a straight line perpendicular to a given straight line of an un- limited length , from a given point without it . Let AB be the given straight line , which may be produced to any length both ways , and let C be a point ...
Seite 19 ... one nearer thereto . Therefore , If , & c . Q. E. D. COR . A perpendicular from a point to a straight line , measures the shortest distance between the point and the line . PROP . XIV . PROB . At a given point OF GEOMETRY . BOOK I. 19.
Seite 20 John Playfair. PROP . XIV . PROB . At a given point in a given straight line , to make an angle equal to a given angle . Let B be the given point in the given straight line AB , and let DEF be the given angle ; it is required to make at ...
Seite 25 ... PROB . To draw a straight line through a given point parallel to a given straight line . Let A be the given point , and BC the given straight line ; it is re- quired to draw a straight line through the point A parallel to the straight | 677.169 | 1 |
Page 50 - The third side is found by the proportion. As the sine of the given angle is to the sine of the angle opposite the required side, so is the side opposite the given angle to the required side.
Page 41 - Since, when an angle is acute its supplement is obtuse, it follows from the preceding proposition, that the sine and cosecant of an obtuse angle are positive, while its cosine, tangent, cotangent, and secant, are negative.
Bibliographic information
Title | 677.169 | 1 |
What is Plane geometry: Definition and 16 Discussions
Euclidean geometry is a mathematical system attributed to Alexandrian Greek mathematician Euclid, which he described in his textbook on geometry: the Elements. Euclid's method consists in assuming a small set of intuitively appealing axioms, and deducing many other propositions (theorems) from these. Although many of Euclid's results had been stated by earlier mathematicians, Euclid was the first to show how these propositions could fit into a comprehensive deductive and logical system. The Elements begins with plane geometry, still taught in secondary school (high school) as the first axiomatic system and the first examples of mathematical proofs. It goes on to the solid geometry of three dimensions. Much of the Elements states results of what are now called algebra and number theory, explained in geometrical language.For more than two thousand years, the adjective "Euclidean" was unnecessary because no other sort of geometry had been conceived. Euclid's axioms seemed so intuitively obvious (with the possible exception of the parallel postulate) that any theorem proved from them was deemed true in an absolute, often metaphysical, sense. Today, however, many other self-consistent non-Euclidean geometries are known, the first ones having been discovered in the early 19th century. An implication of Albert Einstein's theory of general relativity is that physical space itself is not Euclidean, and Euclidean space is a good approximation for it only over short distances (relative to the strength of the gravitational field).Euclidean geometry is an example of synthetic geometry, in that it proceeds logically from axioms describing basic properties of geometric objects such as points and lines, to propositions about those objects, all without the use of coordinates to specify those objects. This is in contrast to analytic geometry, which uses coordinates to translate geometric propositions into algebraic formulas.
Currently, as far as I know, the two main ways to express any given point on a plane is through either cartesian plane or polar coordinates. Both of which requires an ordered pair of two numbers to express a point. However, I wonder if there exists such a system that could express any given...
Before looking at the proof of basic theorems in Euclidean plane geometry, I feel that I should draw pictures or use other physical objects to have some idea why the theorem must be true. After all, I should not just plainly play the "game of logic". And, it is from such observations in real...
Mod note: Moved from a technical forum section, so missing the homework template.
@fab13 -- please post homework problems in the appropriate section under Homework & Coursework.
I have the following exercise to solve : I have to find all the points on the surface ##x^2+y^2+z^2=36## (so a sphere...
Hello,
I am totally bad at geometry , by geometry I mean plane euclidean geometry with similarities and circles. I sometimes feel totally lost with problems. For example:
The parallel sides of trapezoid ABCD are 3 cm and 9 cm(AB and DC).The non parallel sides are 4 cm and 6 cm(AD and BC).A...
Homework Statement
Problem 99 from "Kiselev's Geometry Book I - Planimetry":
Two isosceles triangles with a common vertex and congruent lateral sides cannot fit one inside the other.
Homework EquationsThe Attempt at a Solution
The statement is obviously true. If we visualize each isosceles...
Homework Statement
Problem 55 from Kiselevś Geometry - Book I. Planimetry: "Prove that each diagonal of a quadrilateral either lies entirely in its interior, or entirely in its exterior. Give an example of a pentagon for which this is false."
Homework EquationsThe Attempt at a Solution
The...
Hello,
After reading both How to Prove It: A Structured Approach - By Daniel J Velleman, and one of the Lost Feynman Lectures on Planetary Orbits, I'm wondering if anyone could suggest to me any good books they've read (or heard about) pertaining to logic (paired with analysis), or plane...
Homework Statement
Find the equation of plane which contains the line l:\left\{\begin{array}{l} x=t+2 \\y=2t-1\\z=3t+3 \end{array}\right., and makes the angle of \frac{2\pi}{3} with the plane \pi:x+3y-z+8=0.
The Attempt at a Solution
My attempt was to find the normal vector of plane which...
Consider a point P inside a triangle ABC. Angle PBC is 10 degrees, angle PCB is 20 degrees, and angle BAC is 100 degrees. Find angle PAC.
Question is that is this problem even solvable? I found it in an Olympiad training book...
Hi everyone,
Can anyone solve the followong by plane Euclidean geometry?
I got it by co-ordinate geometry, but couldn't get it by plane...
>In an acute - angled triangle PQR , angle P=\pi/6 , H is the orthocentre, and M is the midpoint of QR . On the line HM , take a point T such that...
Hey there
I do study a textbook at School called Plane Geometry, it is about 3d dimensional figures, it starts with explaining Planes, points and lines, then It starts with theories, then I have to prove something like a line perpendicular to a plane and things like that. I googled Plane...
This is the last problem on a geometry problem set that I can't seem to finish.
AB and BC are chords in a circle where AB > BC. D is the midpoint of minor arc ADBC. If DE is perpendicular to AB, prove that AE = EB + BC.
I would really appreciate just the proper way to approach this... | 677.169 | 1 |
Sam is making a repeating pattern for the rule "Triangle, Square, Circle, and Rectangle." The 32nd shape in the pattern is
Triangle
Square
Rectangle
Circle
Hint:
We are given a repeating pattern of four shapes made by Sam. The three shapes are triangle, square, circle and rectangle. The rule to write shapes is "Triangle, Square, Circle and rectangle". We are asked to find the shape at 32th place.
The correct answer is: Rectangle
The given pattern in words is Triangle, Square, Circle, Rectangle, Triangle, Square, Circle, Rectangle We are asked to find the shape at the 32th place. The rule used to write the pattern is " Triangle, Square, Circle, Rectangle". Four shapes are repeating. And, the fourth shape is a rectangle. So, all multiples of four will have rectangle in their position. We have to find the shape at 32th place. It is a multiple of 4. 4 × 8 = 32 The 32th place will have rectangle. So, the shape at the 32th place is a rectangle.
While solving the question, we have to be careful about the order of the sequence | 677.169 | 1 |
Use Pythagorean theorem to measure topography
In this article
Description
Students build measuring tools from cardboard or LEGO® bricks to create an initial transportation plan in Excel for an island national park. Next, they use the Pythagorean theorem to design their road and bring their national park to life by adding topographic elements in Paint 3D.
Teacher resources
Feedback
Coming soon: Throughout 2024 we will be phasing out GitHub Issues as the feedback mechanism for content and replacing it with a new feedback system. For more information see: | 677.169 | 1 |
What Is a Congruence Statement: Explained | Legal Topic
Unlocking the Mysteries of Congruence Statements
Have you ever wondered what exactly a congruence statement is? Fear not, for we are here to unravel the complexities and shed light on this intriguing topic. Congruence statements are a fundamental concept in geometry, and understanding them is crucial for a deeper comprehension of geometric principles. So, let's dive right and explore fascinating world congruence statements!
Defining Congruence Statements
Simply put, a congruence statement asserts that two geometric figures are congruent, meaning they have the same size and shape. In mathematical terms, if two triangles, for example, have the same angles and side lengths, they are considered congruent. This relationship denoted by symbol '≅' and essential various geometric proofs and constructions.
The Importance of Congruence Statements
Congruence statements play a vital role in geometry and have practical applications in various fields. From architecture and engineering to art and design, the concept of congruence underpins many aspects of our modern world. By understanding and utilizing congruence statements, we can accurately measure and replicate geometric shapes, ensuring precision and consistency in our creations.
Examples of Congruence Statements
Let's consider few examples illustrate concept congruence statements:
Example
Congruence Statement
1
Triangle ABC ≅ Triangle DEF
2
Rectangle PQRS ≅ Rectangle WXYZ
3
Circle O ≅ Circle P
Congruence statements are a cornerstone of geometric reasoning and have far-reaching implications in the world around us. By grasping the essence of congruence, we can unlock a deeper understanding of shapes and their relationships, paving the way for new discoveries and innovations. So, the next time you encounter a congruence statement, marvel at its significance and appreciate the beauty of its mathematical elegance.
Legal Contract: Understanding Congruence Statements
This contract is entered into on this day by and between the undersigned parties, for the purpose of defining and clarifying the concept of congruence statements.
Article 1 – Definition
For the purposes of this contract, congruence statement refers to a mathematical expression that asserts the equality of two geometric figures in terms of size and shape.
Article 2 – Legal Relevance
This contract holds legal relevance in accordance with the laws governing mathematical principles and educational curriculum to ensure the accurate understanding of congruence statements.
Article 3 – Obligations
It is the obligation of the involved parties to adhere to the definitions and principles outlined within this contract and the associated laws and regulations.
Article 4 – Duration
This contract shall remain in effect indefinitely, unless otherwise terminated or revised by mutual agreement of the involved parties.
Article 5 – Termination
In the event of termination, all parties agree to adhere to the provisions outlined in the termination clause of this contract.
Top 10 Legal Questions About Congruence Statements
Question
Answer
1. What is a congruence statement?
A congruence statement is a declaration that two geometric figures are congruent, meaning they have the same size and shape. It is often written in the form of "triangle ABC is congruent to triangle DEF" and is used to show that the corresponding parts of the two triangles are equal. It`s a powerful tool in geometry and helps in proving the equality of shapes.
2. Why are congruence statements important in legal matters?
Congruence statements are crucial in legal matters, especially in cases involving property disputes, construction contracts, and design patents. They provide a clear and formal way to establish the equality of geometric figures, which can be used as evidence in court. In legal matters, precision and accuracy are paramount, and congruence statements help in achieving that.
3. Can a congruence statement be used in real estate disputes?
Absolutely! Congruence statements are often used in real estate disputes to prove the equality of property boundaries, especially in cases where there are conflicting survey reports or conflicting claims to a piece of land. By using congruence statements, lawyers can demonstrate the exact match of property boundaries, which can be pivotal in resolving disputes.
4. How are congruence statements relevant in construction contracts?
Congruence statements play a crucial role in construction contracts, particularly in cases where there are disagreements about the dimensions and shapes of structural elements. By using congruence statements, contractors and architects can formally establish the equality of geometric figures, ensuring that the construction meets the specified requirements and standards. It`s a way to uphold the integrity of the build and avoid legal disputes down the line.
5. Can congruence statements be used in intellectual property law?
Yes, congruence statements are valuable in intellectual property law, especially in cases involving design patents. When seeking protection for a new design, being able to demonstrate its congruence with existing designs is crucial. Congruence statements provide a formal and precise way to establish the equality of shapes, which is essential in proving the uniqueness and originality of a design.
6. Are there any legal precedents involving the use of congruence statements?
Yes, there have been legal cases where congruence statements played a pivotal role in resolving disputes. One notable case involved a property boundary dispute, where the use of congruence statements helped in clearly establishing the equality of survey reports, leading to a fair and conclusive resolution. The legal precedents set by such cases underscore the importance of congruence statements in legal matters.
7. What are the potential legal implications of an incorrect congruence statement?
An incorrect congruence statement can have serious legal implications, especially in cases where property boundaries, construction dimensions, or design patents are involved. It can lead to disputes, claims of misrepresentation, and even financial losses. Therefore, it`s crucial for lawyers and professionals to ensure the accuracy and validity of congruence statements in legal matters.
8. How can lawyers effectively utilize congruence statements in their legal arguments?
Lawyers can effectively utilize congruence statements in their legal arguments by employing them as concrete evidence of equality in geometric figures. By presenting well-crafted congruence statements supported by accurate measurements and precise reasoning, lawyers can bolster their arguments and provide compelling visual evidence to support their client`s position. It`s a powerful tool in building a strong legal case.
9. Is there a specific format for writing congruence statements in legal documents?
While there is no strict legal format for writing congruence statements in legal documents, it`s important to ensure that they are clear, precise, and formally structured. Using standardized notation, such as the "triangle ABC is congruent to triangle DEF" format, can enhance the clarity and professionalism of the statement. Additionally, including relevant measurements and geometric reasoning can further strengthen the congruence statement in a legal context.
10. What role do expert witnesses play in establishing the validity of congruence statements in court?
Expert witnesses play a crucial role in establishing the validity of congruence statements in court by providing professional expertise and testimony regarding geometric congruence. Their knowledge and experience in geometry, surveying, or design can lend credibility to the congruence statement, especially in complex legal cases where precise measurements and technical understanding are paramount. Expert witnesses can help ensure that congruence statements are accurately interpreted and upheld in the legal proceedings. | 677.169 | 1 |
You have really impressed us with your newly acquired knowledge of the properties of 2d shapes. Over the next few days, we are going to look more closely at the properties of 3d shapes. Before you look at the videos and powerpoints, you will need to know the following terminology:
3d shapes have a vertex/vertices (corner points); edges and faces; and
a prism is a 3d shape that, when cut through its cross section, the faces at either end remain the same shape.
Maths is Fun websiteDiagrams and a more detailed explanation of the terminology: vertices, faces and edges | 677.169 | 1 |
Trigonometry with Bearings
Trigonometry with Bearings Revision Content
Understanding Bearings
Bearings are used in navigation to define the direction of one point relative to another. They are measured in degrees from the North line in a clockwise direction, typically expressed as a three-figure bearing.
Always remember that bearing measurements should be between 0 and 360 degrees.
Basic Principles
When reading or writing bearings, remember that a bearing must be a three-figure measurement. For example, "60°" would be written as "060°".
Be aware that bearings are always measured clockwise.
Calculating Bearings in Trig Problems
To find the bearing, you can use basic geometry (usually involving right-angled triangles) and trigonometry. The cosine rule, sine rule or Pythagoras' theorem might also be used, depending on the problem. | 677.169 | 1 |
Construction of a Quadrilateral when two adjacent sides and three angles are given
The minimum n...
Question
The minimum number of measurements that is required to construct a quadrilateral is .
A
5
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B
36
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Solution
The correct option is A 5 To construct a quadrilateral we need any one of the following measurements.
(i) 4 sides and any one diagonal
(ii) 3 sides and 2 diagonals
(iii) 3 sides and 2 included angles
(iv) 3 angles and 2 adjacent sides
(v) 4 sides and 1 diagonal
Thus, the minimum number of measurements we require to construct a quadrilateral is 5. | 677.169 | 1 |
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