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Euclidean geometry, the study of plane and solid figures on the basis of axioms and theorems employed by the Greek mathematician Euclid (c. 300 bce).
In its rough outline, Euclidean geometry is the plane and solid geometry commonly taught in secondary schools. Indeed, until the second half of the 19th century, when non-Euclidean geometries attracted the attention of mathematicians, geometry meant Euclidean geometry. It is the most typical expression of general mathematical thinking.
Rather than the memorization of simple algorithms to solve equations by rote, Euclidean geometry demands true insight into the subject, clever ideas for applying theorems in special situations, an ability to generalize from known facts, and an insistence on the importance of proof. In Euclid's great work, the Elements, the only tools employed for geometrical constructions were the ruler and the compass—a restriction retained in elementary Euclidean geometry to this day.
In its rigorous deductive organization, the Elements remained the very model of scientific exposition until the end of the 19th century, when the German mathematician David Hilbert wrote his famous Foundations of Geometry (1899). The modern version of Euclidean geometry is the theory of Euclidean (coordinate) spaces of multiple dimensions, where distance is measured by a suitable generalization of the Pythagorean theorem. Seeanalytic geometry and algebraic geometry.
Fundamentals
Euclid realized that a rigorous development of geometry must start with the foundations. Hence, he began the Elements with some undefined terms, such as "a point is that which has no part" and "a line is a length without breadth." Proceeding from these terms, he defined further ideas such as angles, circles, triangles, and various other polygons and figures. For example, an angle was defined as the inclination of two straight lines, and a circle was a plane figure consisting of all points that have a fixed distance (radius) from a given center.
As a basis for further logical deductions, Euclid proposed five common notions, such as "things equal to the same thing are equal," and five unprovable but intuitive principles known variously as postulates or axioms. Stated in modern terms, the axioms are as follows:
1. Given two points, there is a straight line that joins them.
2. A straight line segment can be prolonged indefinitely.
3. A circle can be constructed when a point for its center and a distance for its radius are given.
4. All right angles are equal.
5. 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, will meet on that side on which the angles are less than the two right angles. | 677.169 | 1 |
Euclid's Elements of Geometry: Chiefly from the Text of Dr. Simson, with ...
proof depends on Euc. vi. 19; 20, Cor. 2. (2) Let ABC be the triangle, BC being the base. Draw AD at right angles to BA meeting the base produced in D. Bisect BC in E, and on ED describe a semicircle, from B draw BP to touch the semicircle in P. From BA cut off BF equal to BP, and from F draw FG perpendicular to BC. The line FG bisects the triangle. Then it may be proved that BFG: BAD :: BE: BD, and that BAD: BAC :: BĎ: BC; whence it follows that BFG: BAC :: BE: BC or as 1 : 2.
43. Let ABC be the given triangle which is to be divided into two parts having a given ratio, by a line parallel to BC. Describe a semicircle on AB and divide AB in D in the given ratio; at D draw DE perpendicular to AB and meeting the circumference in E; with center A and radius AE describe a circle cutting AB in F: the line drawn through F parallel to BC is the line required. In the same manner a triangle may be divided into three or more parts having any given ratio to one another by lines drawn parallel to one of the sides of the triangle. 44. Let these points be taken, one on each side, and straight lines be drawn to them; it may then be proved that these points severally bisect the sides of the triangle.
45. Let ABC be any triangle and D be the given point in BC, from which lines are to be drawn which shall divide the triangle into any number (suppose five) equal parts. Divide BC into five equal parts in E, F, G, H, and draw AE, AF, AG, AH, AD, and through E, F, G, H draw EL, FM, GN, HO parallel to AD, and join DL, DM, DN, DO; these lines divide the triangle into five equal parts.
By a similar process, a triangle may be divided into any number of parts which have a given ratio to one another.
46. Let ABC be the larger, abc the smaller triangle, it is required to draw a line DE parallel to AC cutting off the triangle DBE equal to the triangle abc. On BC take BG equal to bc, and on BG describe the triangle BGH equal to the triangle abc. Draw HK parallel to BC, join KG; then the triangle BGK is equal to the triangle abc. On BA, BC take BD to BE in the ratio of BA to BC, and such that the rectangle contained by BD, BE shall be equal to the rectangle contained by BK, BG. Join DE, then DE is parallel to AC, and the triangle BDE is equal to abc.
47. Let ABCD be any rectangle, contained by AB, BC,
Then AB': AB. BC::AB: BC,
and AB.BC: BC2 :: AB: BC,
whence AB: AB. BC :: AB. BC: BC",
or the rectangle contained by two adjacent sides of a rectangle, is a mean proportional between their squares.
and
48. In a straight line at any point A, make Ac equal to Ad in the given ratio. At A draw AB perpendicular to cAd, and equal to a side of the given square. On cd describe a semicircle cutting AB in b; join bc, bd; from B draw BC parallel to bc, and BD parallel to bd: then AC, AD are the adjacent sides of the rectangle. For, CA is to AD as cA to Ad, Euc. vi. 2; and CA. AD AB, CBD being a right-angled triangle.
=
49. From one of the given points two straight lines are to be drawn perpendicular, one to each of any two adjacent sides of the parallelogram; and from the other point, two lines perpendicular in the same manner to each of the two remaining sides. When these four lines are drawn to intersect one another, the figure so formed may be shewn to be equi. angular to the given parallelogram.
50. It is manifest that this is the general case of Prop. 4, p. 197. If the rectangle to be cut off be two-thirds of the given rectangle ABCD. Produce BC to E so that BE may be equal to a side of that square which is equal to the rectangle required to be cut off; in this case, equal to two-thirds of the rectangle ABCD. On AB take AF equal to AD or BC; bisect FB in G, and with center G and radius GE, describe a semicircle meeting AB, and AB produced, in H and K. On CB take CL equal to AH and draw HM, LM parallel to the sides, and HBLM is two-thirds of the rectangle ABCD.
51: Let ABCD be the parallelogram, and CD be cut in P and BC produced in Q. By means of the similar triangles formed, the property may be proved.
52. The intersection of the diagonals is the common vertex of two triangles which have the parallel sides of the trapezium for their bases.
53. Let AB be the given straight line, and C the center of the given circle; through C draw the diameter DCE perpendicular to AB. Place in the circle a line FG which has to AB the given ratio; bisect FG in H, join CH, and on the diameter DCE, take CK, CL each equal to CH; either of the lines drawn through K, L, and parallel to AB is the line required.
51. Let C be the center of the circle, CA, CB two radii at right angles to each other; and let DEFG be the line required which is trisected in the points E, F. Draw CG perpendicular to DH and produce it to meet the circumference in K; draw a tangent to the circle at K: draw CG, and produce CB, CG to meet the tangent in L, M, then MK may be shewn to be treble of LK.
55. The triangles ACD, BCE are similar, and CF is a mean proportional between AC and CB.
56. Let any tangent to the circle at E be terminated by AD, BC tangents at the extremity of the diameter AB. Take O the center of the circle and join OC, OD, OE; then ODC is a right-angled triangle and OE is the perpendicular from the right angle upon the hypotenuse.
57. This problem only differs from problem 59, infra, in having the given point without the given circle.
58. Let A be the given point in the circumference of the circle, Cits center. Draw the diameter ACB, and produce AB to D, taking AB to BD in the given ratio: from D draw a line to touch the circle in E, which is the point required. From A draw AF perpendicular to DE, and cutting the circle in G.
59. Let A be the given point within the circle whose center is C, and let BAD be the line required, so that BA is to AD in the given ratio. Join AC and produce it to meet the circumference in E, F. Then EF is a diameter. Draw BG, DH perpendicular on EF: then the triangles BGA, DHA are equiangular. Hence the construction.
60. Through E one extremity of the chord EF, let a line be drawn parallel to one diameter, and intersecting the other. Then the three angles of the two triangles may be shewn to be respectively equal to one another.
61. Let AB be that diameter of the given-circle which when produced is perpendicular to the given line CD, and let it meet that line in C; and let P be the given point: it is required to find D in CD, so that DB may be equal to the tangent DF. Make BC: CQ:: CQ : CA, and join PQ; bisect PQ in E, and draw ED perpendicular to PQ meeting CĎ in D; then D is the point required. Let O be the center of the circle, draw the tangent DF; and join OF, OD, QD, PD. Then QD may be shewn
to be equal to DF and to DP. When P coincides with Q, any point D in CD fulfils the conditions of the problem; that is, there are innumerable solutions.
62. It may be proved that the vertices of the two triangles which are similar in the same segment of a circle, are in the extremities of a chord parallel to the chord of the given segment.
63. For let the circle be described about the triangle EAC, then by the converse to Euc. 1. 32; the truth of the proposition is manifest. 64. Let the figure be constructed, and the similarity of the two triangles will be at once obvious from Euc III. 32.; Euc. 1. 29.
65. In the arc AB (fig. Euc. iv. 2) let any point K be taken, and from K let KL, KM, KN be drawn perpendicular to AB, AC, BC respectively, produced if necessary, also let LM, LN be joined, then MLN may be shewn to be a straight line. Draw AK, BK, CK, and by Euc. III, 31, 22, 21; Euc. I. 14.
66. Let AB a chord in a circle be bisected in C, and DE, FG two chords drawn through C; also let their extremities DG, FE be joined intersecting CB in H, and AC in K; then AK is equal to HB. Through H draw MHL parallel to EF meeting FG in M, and DE produced in L. Then by means of the equiangular triangles, HC may be proved to be equal to CK, and hence AK is equal to HB.
67. Let A, B be the two given points, and let P be a point in the locus so that PA, PB being joined, PA is to PB in the given ratio. Join AB and divide it in C in the given ratio, and join PC. Then PC bisects the angle APB. Euc. vi. 3. Again, in AB produced, take AD to AB in the given ratio, join PD and produce AP to E, then PD bisects the angle BPE. Euc. vI. A. Whence CPD is a right angle, and the point P lies in the circumference of a circle whose diameter is CD.
68. Let ABC be a triangle, and let the line AD bisecting the vertical angle A be divided in E, so that BC: BA+AC:: AE: ED. By Euc. VI. 3, may be deduced BC: BA+AC:: AC: AD. Whence may be proved that CE bisects the angle ACD, and by Euc. iv. 4, that E is the center of the inscribed circle.
69. By means of Euc. Iv. 4, and Euc. vI. C. this theorem may be shewn to be true.
70. Divide the given base BC in D, so that BD may be to DC in the ratio of the sides. At B, D draw BB', DD' perpendicular to BC and equal to BD, DC respectively. Join B'D' and produce it to meet BC produced in O. With center O and radius OD, describe a circle. From A any point in the circumference join AB, AC, AO. Prove that AB is to AC as BD to DC. Or thus. If ABC be one of the triangles. Divide the base BC in D so that BA is to AC as BD to DC. Produce BC and take DO to OC as BA to AC: then O is the center of the circle.
71. Let ABC be any triangle, and from A, B let the perpendiculars AD, BE on the opposite sides intersect in P: and let AF, BG drawn to F, G the bisections of the opposite sides, intersect in Q. Also let FR, GR be drawn perpendicular to BC, AC, and meet in R: then R is the center of the circumscribed circle. Join PQ, QR; these are in the same line.
Join FG, and by the equiangular triangles, GRF, APB, AP is proved double of FR. And AQ is double of QF, and the alternate angles PAQ, QFR are equal. Hence the triangles APQ, RFQ are equiangular.
72. Let C, C' be the centers of the two circles, and let CC' the line joining the centers intersect the common tangent PP' in T. Let the
line joining the centers cut the circles in Q, Q', and let PQ, P'Q' be joined; then PQ is parallel to PQ. Join CP, C'P', and then the angle QPT may be proved to be equal to the alternate angle Q'P'T.
73. Let ABC be the triangle, and BC its base; let the circles AFB, AFC be described intersecting the base in the point F, and their diameters AD, AE, be drawn; then DA: AE :: BA: AC. For join DB, DF, EF, EC, the triangles DAB, EAC may be proved to be similar.
74. If the extremities of the diameters of the two circles be joined by two straight lines, these lines may be proved to intersect at the point of contact of the two circles; and the two right-angled triangles thus formed may be shewn to be similar by Euc. III. 34.
75. This follows directly from the similar triangles.
76. Let the figure be constructed as in Theorem 4, p. 162, the triangle EAD being right-angled at A, and let the circle inscribed in the triangle ADE touch AD, AE, DE in the points K, L, M respectively. Then AK is equal to AL, each being equal to the radius of the inscribed circle. Also AB is equal to GC, and AB is half the perimeter of the triangle AED.
Also if GA be joined, the triangle ADE is obviously equal to the difference of AGDE and the triangle GDE, and this difference may be proved equal to the rectangle contained by the radii of the other two circles.
77. From the centers of the two circles let straight lines be drawn to the extremities of the sides which are opposite to the right angles in each triangle, and to the points where the circles touch these sides. Euc. vI. 4.
78. Let A, B be the two given points, and C a point in the circumference of the given circle. Let a circle be described through the points A, B, C and cutting the circle in another point D. Join CD, AB, and produce them to meet in E. Let EF be drawn touching the given circle in F; the circle described through the points A, B, F, will be the circle required. Joining AD and CB, by Euc. III. 21, the triangles CEB, AED are equiangular, and by Euc. vI. 4, 16, III. 36, 37, the given circle and the required circle each touch the line EF in the same point, and therefore touch one another. When does this solution fail?
Various cases will arise according to the relative position of the two points and the circle.
79. Let A be the given point, BC the given straight line, and D the center of the given circle. Through D draw CD perpendicular to BC, meeting the circumference in E, F. Join AF, and take FG to the diameter FE, as FC is to FA. The circle described passing through the two points A, G and touching the line BC in B is the circle required. Let H be the center of this circle; join HB, and BF cutting the circumference of the given circle in K, and join EK. Then the triangles FBC, FKE being equiangular, by Euc. vI. 4, 16, and the construction, K is proved to be a point in the circumference of the circle passing through the points A, G, B. And if DK, KH be joined, DKH may be proved to be a straight line:- the straight line which joins the centers of the two circles, and passes through a common point in their circumferences.
80. Let A be the given point, B, C the centers of the two given circles. Let a line drawn through B, C meet the circumferences of the circles in G, F; E, D, respectively. In GD produced, take the point H, so that BH is to CH as the radius of the circle whose center
is B to the radius of the circle whose center is C. Join AH, and take KH to DH as GH to AH. Through A, K describe a circle ALK touching the circle whose center is B, in L. Then M may be proved to be a point in the circumference of the circle whose center is C. For by joining HL and producing it to meet the circumference of the circle whose center is B in N; and joining BN, BL, and drawing CO parallel to BL, and CM parallel to BN, the line HN is proved to cut the circumference of the circle whose center is B in M, O; and CO, CM are radii. By joining GL, DM, M may be proved to be a point in the circumference of the circle ALK. And by producing BL, CM to meet in P, P is proved to be the center of ALK, and BP joining the centers of the two circles passes through L the point of contact. Hence also is shewn that PMC passes through M, the point where the circles whose centers are P and C touch each other.
NOTE. If the given point be in the circumference of one of the circles, the construction may be more simply effected thus:
Let A be in the circumference of the circle whose center is B. Join BA, and in AB produced, if necessary, take AD equal to the radius of the circle whose center is C; join DC, and at C make the angle DCE equal to the angle CDE, the point E determined by the intersection of DA produced and CE, is the center of the circle.
81. Let AB, AC be the given lines and P the given point. Then if O be the center of the required circle touching AB, AC, in R, S, the line AO will bisect the given angle BAC. Let the tangent from P meet the circle in Q, and draw OQ, OS, OP, AP. Then there are given AP and the angle OAP. Also since OQP is a right angle, we have OP-QO2 =OPOS=PQ a given magnitude. Moreover the right-angled triangle AOS is given in species, or OS to OA is a given ratio. Whence in the triangle AOP there is given, the angle AOP, the side AP, and the excess of OP above the square of a line having a given ratio to OA, to determine OA. Whence the construction is obvious.
82. Let the two given lines AB, BD meet in B, and let C be the center of the given circle, and let the required circle touch the line AB, and have its center in BD. Draw CFE perpendicular to HB intersecting the circumference of the given circle in F, and produce CE, making EF equal to the radius CF. Through G draw GK parallel to AB, and meeting DB in K. Join CK, and through B, draw BL parallel to KC, meeting the circumference of the circle whose center is C in L; join CL and produce CL to meet BD in O. Then O is the center of the circle required. Draw OM perpendicular to AB, and produce EC to meet BD in N. Then by the similar triangles, OL may be proved
equal to OM.
83. (1) In every right-angled triangle when its three sides are in Arithmetical progression, they may be shewn to be as the numbers 5, 4, 3. On the given line AC describe a triangle having its sides AC, AD, DC in this proportion, bisect the angles at A, C by AE, CE meeting in E, and through E draw EF, EG parallel to AD, DC meeting in F and G.
(2) Let AC be the sum of the sides of the triangle, fig. Euc. v. 13. Upon AC describe a triangle ADC whose sides shall be in continued proportion. Bisect the angles at A and C by two lines meeting in E. From E draw EF, EG parallel to DA, DC respectively.
84. Describe a circle with any radius, and draw within it the straight line MN cutting off a segment containing an angle equal to the given angle, Euc. 111. 34. Divide MN in the given ratio in P, and at P draw PA perpendicular to MN and meeting the circumference in A. Join | 677.169 | 1 |
Phase Shift
Phase shift refers to the change in the phase angle of a wave or signal, often measured in degrees. It's crucial for understanding wave behaviour in various fields like physics, engineering, and signal processing. By recognising phase shifts, students can better analyse and interpret wave interactions and signal transmissions.
Phase Shift Definition
Understanding the concept of phase shift is crucial in the study of trigonometric functions and waveforms. A phase shift refers to the horizontal shift of a function along the x-axis. This shift is typically observed in trigonometric functions such as sine, cosine, and tangent.
What is a Phase Shift?
A phase shift is a horizontal translation of a trigonometric function. It shows how much a function is shifted along the x-axis. Consider the general form of a sine or cosine function:
A positive phase shift means the function shifts to the left, while a negative phase shift means the function shifts to the right.
Phase shifts are not limited to trigonometric functions. They play a significant role in various fields including physics, engineering, and signal processing. For instance, phase shift is a critical parameter in understanding wave interference patterns, signal modulation, and even in the analysis of sound waves. Mastering phase shift concepts can provide a deeper understanding of periodic phenomena and their applications in real-world scenarios.
To summarise, a phase shift is a critical component in the study of trigonometric functions. It describes how a function is displaced horizontally along the x-axis. The formula \[\frac{c}{b}\] helps in determining the extent of this shift in functions like \(\text{sin}(x)\) and \(\text{cos}(x)\). Understanding this concept is vital for further studies in mathematics and science.
Phase Shift Formula and Equation for Phase Shift
In mathematics, understanding the phase shift is essential when dealing with trigonometric functions and waveforms. The phase shift describes how far a function is shifted horizontally along the x-axis. This concept is particularly significant in functions like sine and cosine.
General Form of Trigonometric Functions
The phase shift in a trigonometric function can be identified using its general form:
Thus, the function has a phase shift of \(-\frac{\pi}{3}\), meaning it shifts to the left.
In trigonometric functions, a positive value indicates a leftward shift while a negative value indicates a rightward shift.
Interpreting Phase Shift
Understanding whether a function is shifted to the left or right aids in comprehensively analysing the function's behaviour. For instanceThis table helps you quickly identify the direction and extent of phase shifts for common trigonometric functions.
Phase shifts play a significant role in various other areas beyond trigonometry. For instance, in physics, phase shifts are crucial in understanding principles like wave interference patterns and signal modulation. These shifts also have practical applications in fields such as signal processing, engineering, and acoustic analysis. By mastering the concept of phase shifts, you open up a broader understanding of periodic phenomena and their numerous applications in real-world scenarios.
How to Find Phase Shift
Understanding how to find the phase shift of trigonometric functions is crucial for analysing their behaviour. This section will guide you through the steps to calculate the phase shift.
Identifying Components of the Function
The first step in finding the phase shift is to identify the key components of the trigonometric function. The general form of a trigonometric function is:
In these formulas, the term b affects the period of the function, while c affects the phase shift.
Phase Shift: The horizontal shift of a function along the x-axis, determined by the term \( \frac{c}{b} \) in the general form of a trigonometric equation.
Calculating Phase Shift
Once you have identified the values of b and c, you can calculate the phase shift using the formula:
\[\text{Phase Shift} = -\frac{c}{b}\]
This formula will help you determine the amount and direction of the horizontal shift.
Consider the function \[y = 2 \, \text{sin}(3x + \pi) + 1\]. Here:
c = \(\pi\)
b = 3
Substituting these values into the phase shift formula:
\[\text{Phase Shift} = -\frac{\pi}{3}\]
This means the function has a phase shift of \(-\frac{\pi}{3}\), indicating a shift to the left.
For trigonometric functions, a positive phase shift indicates a leftward shift, while a negative phase shift indicates a rightward shift.
Visualising Phase Shift
Visualising the phase shift can help in understanding how the function is altered. Consider these trigonometric functions and their respective phase shiftsUsing these examples, you can see how the phase shift affects the position of the function on the x-axis.
Phase shifts are not limited to trigonometric functions alone; they have applications in physics, engineering, and signal processing. For example, understanding the phase shift is essential in wave interference patterns, where waves combine to form new patterns. In signal processing, phase shifts help in the modulation of signals to reduce interference and improve clarity. Grasping the concept of phase shifts enhances your ability to analyse periodic phenomena and their real-world applications.
Understanding phase shift provides a basis for mastering more complex topics in mathematics and science.
Examples of Phase Shift
Exploring phase shift through examples can solidify your understanding of how trigonometric functions move along the x-axis. In this section, you will see practical applications of phase shift.
Understanding Phase Shift Definition
The phase shift of a trigonometric function can be visualised and calculated using its general mathematical form. Recall the general form of a sine function:
\[y = a \, \text{sin}(bx + c) + d\]
In this equation, the term c controls the horizontal shift, while b affects the period.
Phase Shift: The horizontal translation of a function along the x-axis, determined by the term \(\frac{c}{b}\) in the general form of a trigonometric equation.
Basic Phase Shift Formula
The basic formula to find the phase shift in trigonometric functions is:
\[\text{Phase Shift} = -\frac{c}{b}\]
This allows you to determine the extent and direction of the horizontal shift.
Acoustics: Phase shifts help analyse sound waves and their interactions.
In signal processing, phase shifts are crucial for reducing interference and improving signal clarity. For example, in audio engineering, understanding phase relationships between sound waves can enhance sound quality and fidelity. By mastering phase shift concepts, you can delve into complex periodic phenomena and their applications, enriching your understanding of various scientific and engineering principles.
Phase Shift - Key takeaways
Phase Shift Definition: The horizontal shift of a function along the x-axis, determined by the term \(\frac{c}{b}\) in the general form of a trigonometric equation.
Phase Shift Formula: The formula to determine the extent of the phase shift is \(\text{Phase Shift} = -\frac{c}{b}\).
General Form: The general form of trigonometric functions that incorporate phase shift is \(y = a \text{sin}(bx + c) + d\) and \(y = a \text{cos}(bx + c) + d\).
Direction of Shift: A negative phase shift value means the function shifts to the right, while a positive value shifts it to the left.
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Frequently Asked Questions about Phase Shift
What is a phase shift in trigonometric functions?
A phase shift in trigonometric functions is a horizontal translation of the graph. It is determined by the value added or subtracted within the function's argument. For \\(y = \\sin(x + c)\\), the phase shift is \\(-c\\), moving left if \\(c > 0\\) and right if \\(c < 0\\).
How is phase shift determined in a sine or cosine graph?
Phase shift in a sine or cosine graph is determined by the horizontal shift of the graph. It is given by the constant \\( C \\) in the function \\( y = A \\sin(B(x - C)) + D \\) or \\( y = A \\cos(B(x - C)) + D \\). The phase shift is \\( \\frac{C}{B} \\) units to the right if \\( C \\) is positive or to the left if \\( C \\) is negative.
How does phase shift affect wave interference?
Phase shift affects wave interference by altering the relative positions of the waves, either constructively enhancing or destructively reducing the resultant wave amplitude. When combined, waves can sum up if they are in phase or cancel out if out of phase, influencing the resultant signal significantly.
What are the applications of phase shift in electrical engineering?
Phase shift in electrical engineering is crucial for signal processing, communication systems, and control systems. It allows for the modulation and demodulation of signals, improving data transmission. Phase shifters are used in antennas and radar systems to direct beams. Moreover, phase shifts help in synchronising alternating current (AC) power grids.
How is phase shift used in communication systems?
Phase shift is used in communication systems to encode information by altering the phase of a carrier wave, a method known as phase-shift keying (PSK). This modulation technique allows multiple bits to be transmitted simultaneously, enhancing data transfer rates and improving bandwidth efficiency. It is fundamental in digital communication protocols like Wi-Fi and cellular networks.
Test your knowledge with multiple choice flashcards
A. \(\frac{\pi}{2}\) to the left
B. \(\frac{\pi}{4}\) to the left
C. \(\frac{3\pi}{2}\) to the left
D. \(\frac{\pi}{2}\) to the right
Which of the following fields uses phase shifts to improve communication systems?
A. Physics
B. Engineering
C. Mathematics
D. Acoustics
What is a phase shift in trigonometric functions?
A. A modification in the period of the function.
B. A vertical shift of a function along the y-axis.
C. A change in the amplitude of the trigonometric function.
D. A horizontal translation of a trigonometric function along the x-axis.
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What is another word for right-angled triangle?
Pronunciation: [ɹˈa͡ɪtˈaŋɡə͡ld tɹˈa͡ɪaŋɡə͡l] (IPA)
A right-angled triangle is a three-sided polygon with one angle measuring 90 degrees. This shape is commonly referred to as a right triangle, a square triangle or a Pythagorean triangle. Another synonym for this shape is a rectangular triangle, which is often used in mathematical contexts. Other terms that can also be used to refer to a right-angled triangle include a right-angled scalene triangle, a right-angled isosceles triangle or a right-angled equilateral triangle. These terms highlight different properties of the triangle, such as the number of equal sides or angles it has. Regardless of its name, the right-angled triangle is a fundamental shape in geometry and plays an important role in many areas of mathematics and engineering.
What are the meronyms for Right-angled triangle?
meronyms for right-angled triangle (as nouns)
What are the opposite words for right-angled triangle?
The antonyms for "right-angled triangle" are "oblique triangle" and "acute triangle". Oblique triangles are triangles that don't have a right angle, while "acute triangle" is a triangle with three acute angles (less than 90 degrees). A right-angled triangle is a fundamental geometric shape with a 90-degree angle at one of its vertices. It has two sides called legs and one side opposite the right angle referred to as the hypotenuse. Right-angled triangles play a significant role in mathematics, physics, and engineering because of their unique properties. Understanding their antonyms can help students differentiate between different types of triangles and their properties.
What are the antonyms for Right-angled triangle?
n.
• shape
oblique triangle
.
Famous quotes with Right-angled triangle | 677.169 | 1 |
Hint: First we will draw a rough diagram. Then mention all the given terms and their respective values. Then use one of the forms of Laws of Cosines. Then substitute the values of the terms and simplify the terms to evaluate the value of the side BC.
Complete step by step solution: We will first draw a rough diagram.
$\angle A$ is unchanged which is $93$. Side AC becomes side $b$ which has the value $17$. Side AB becomes side $c$ which has the value $28$. Side BC becomes side $a$ whose value is unknown. The one of the three forms of the Law of Cosines that we need is: ${a^2} = {b^2} + {c^2} - 2(b)(c)\cos (A)$ Now, we will substitute all the given values. ${a^2} = {17^2} + {28^2} - 2(17)(28)\cos ({93^0})$ Now, we solve for the value of $a$. ${a^2} = {17^2} + {28^2} - 2(17)(28)\cos ({93^0})$ Hence, the value of the side BC is $33.5$.
Additional information: Inverse trigonometric functions are simply defined as the inverse functions of the basic trigonometric functions which are sine, cosine, tangent, cotangent, secant and cosecant functions. They are also termed as arcus functions, anti-trigonometric functions or cyclometric functions.
Note: While using the different forms of Laws of Cosines choose according to the given and required terms. Substitute the terms in the equation cautiously. While modifying any identity make sure that when you back trace the identity, you get the same original identity. | 677.169 | 1 |
Hint: For solving these problems, we need to have a clear understanding about the various geometrical figures and the formula regarding their angles. By employing those formulas and using the properties of angles, we can easily find out the value of x in the given figure.
Complete step by step answer: Geometry is one of the important branches of mathematics that deals with the study of different shapes. It initiates the study of lines and angles. A straight line is a line without curves and it is defined as the shortest distance between two points. An angle is formed when the line segment meets at a point. First, we need to know that the sum of angles in a polygon with n sides is equal to $\left( n-2 \right)\times {{180}^{\circ }}$ . So, for a quadrilateral with four sides, the sum of angles is equal to $\left( 4-2 \right)\times {{180}^{\circ }}={{360}^{\circ }}$ . When two lines intersect, the opposite angles are equal hence they are called vertically opposite angles. In the above figure, $\angle DFE$ and ${{30}^{\circ }}$ are vertically opposite angles because the lines intersect at F. Hence, $\angle DFE={{30}^{\circ }}$ . Two angles are called complementary if their measures add to ${{90}^{\circ }}$ , and called supplementary if their measures add to ${{180}^{\circ }}$ . Now, on a straight line, the sum of angles is ${{180}^{\circ }}$ on each side of the straight line. In the given figure, we can thus say that the sum of angles on the left side of the line ACB is ${{180}^{\circ }}$ . Thus, $\begin{align} & \Rightarrow \angle ACD+\angle DCE+\angle ECB={{180}^{\circ }} \\ & \Rightarrow {{50}^{\circ }}+\angle DCE+{{50}^{\circ }}={{180}^{\circ }} \\ & \Rightarrow \angle DCE={{80}^{\circ }} \\ \end{align}$ Thus, in the quadrilateral DFEC, $\angle DFE={{30}^{\circ }}$ , $\angle DCE={{80}^{\circ }}$ , $\angle FDC$ and $\angle FEC$ are equal to ${{x}^{\circ }}$ . Now we already know that the sum of angles of a quadrilateral is ${{360}^{\circ }}$ . Hence, we can say that \[\begin{align} & \Rightarrow \angle DFE+\angle DCE+\angle FDC+\angle FEC={{360}^{\circ }} \\ & \Rightarrow {{30}^{\circ }}+{{80}^{\circ }}+{{x}^{\circ }}+{{x}^{\circ }}={{360}^{\circ }} \\ & \Rightarrow {{x}^{\circ }}={{125}^{\circ }} \\ \end{align}\] Thus, using the properties of angles we have arrived at the answer that the value of x in the above figure is \[{{125}^{\circ }}\] .
Note: These types of problems are pretty easy to solve, but one needs to be careful otherwise small misjudgements can lead to a totally different answer. We need to carefully observe the figure employing the correct formula of supplementary and vertically opposite angles. Miscalculations should also be avoided while solving the equations. | 677.169 | 1 |
Given a triangle , the Neuberg cubic is the set of all points whose reflections in the sides , and form a triangle perspective to . It is a self-isogonal cubic with pivot point at the Euler infinity point [1]. The name comes from the geometer Joseph Jean Baptiste Neuberg for his 1894 paper. | 677.169 | 1 |
Elements of Geometry and Trigonometry
From inside the book
Results 1-5 of 33
Page 12 ... when they are equal in measure . When they may be so placed as to coincide through- out their whole extent , they are equal in all their parts . ELEMENTS OF GEOMETRY . BOOK I. ELEMENTARY PRINCIPLES . DEFINITIONS 12 GEOMETRY .
Page 23 ... coincide throughout their whole extent , and form one and the same line . Let A and B be two points common to two lines : then will A the lines coincide throughout . Between A and B they must E B C D coincide ( A. 11 ) . Suppose , now ...
Page 25 ... coincide with the vertex F ; consequently , the side BC will coincide with the side EF ( A. 11 ) . The two triangles , therefore , coincide throughout , and are consequently equal in all their parts ( I. , D. 14 ) ; which was to be ...
Page 26 ... coincide with the vertex F ; and because the angle C is equal to the angle F , the side CA will take the direction FD . Now , the vertex A being at the same time on the lines ED and FD , it must be at their intersection D ( P. III . , C ...
Page 61 ... coincide ; otherwise there would be some points in either one or the other of the curves unequally distant from the centre ; which is impossible ( D. 1 ) : hence , AB divides the circle , and also its circumference , into two equal | 677.169 | 1 |
LOGIN
What is Geometry
This is a basic summary of what geometry is.
Euclidean geometry is a mathematical system attributed to Although many of Euclid's results had been stated by earlier mathematicians,[1] Euclid was the first to show how these propositions could fit into a comprehensive deductive and logical system.[2] The Elements begins with plane geometry, still taught in secondary school (high school) as the first axiomatic system and the first examples of formal proof. 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.[1]
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).[3]
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. | 677.169 | 1 |
Question 2 (c) - Line of Symmetry - ANY figure - Symmetry
Last updated at April 16, 2024 by Teachoo
Transcript
Question 2 Copy the following drawing on squared paper. Complete each one of them such that the resulting figure has two dotted lines as two lines of symmetry. How did you go about completing the picture? (c)
Symmetry along Horizontal line
Symmetry along Vertical line | 677.169 | 1 |
The Importance of the Circumcenter in a Triangle | Properties and Theorems
Point H is the circumcenter of triangle DEF.Which must be true? Check all that apply.
In a triangle, the circumcenter is the point of intersection of the perpendicular bisectors of its sides
In a triangle, the circumcenter is the point of intersection of the perpendicular bisectors of its sides. Therefore, the following statements must be true if point H is the circumcenter of triangle DEF:
1. The perpendicular bisectors of sides DE, DF, and EF pass through point H: The circumcenter is equidistant from the vertices of the triangle, so the perpendicular bisectors of the sides must intersect at point H.
2. The distances from the circumcenter H to the vertices D, E, and F are equal: The circumcenter is equidistant from the vertices of the triangle, so the distances HD, HE, and HF must be equal.
3. Angle DHE = Angle DFE = Angle DEF: The circumcenter is the center of the triangle's circumcircle, which is the circle passing through all three vertices of the triangle. Therefore, the angles subtended by the chords DE, DF, and EF on the circumference of the circumcircle are equal.
In summary, all three of these statements must be true if point H is the circumcenter of triangle DEF | 677.169 | 1 |
Solution: Because ¯BC ⊥ ¯AD, you know that ∠ABC ~= ∠DBC. Because →CB bisects ∠ACD , you know that ∠ACB ~= ∠DCB. Finally, ∠BC is congruent to itself, and you can use the ASA Postulate to show that ΔABC ~= ΔDBC. By CPOCTAC, you can conclude that ∠A ~= ∠D. Let's write it up | 677.169 | 1 |
An idler's miscellany of compendious amusements
Moving Constants
If you mark two points on a circle, A and B, and a third point T, then angle ATB remains constant as T moves along the segment between A and B. (If you mark a point S in the circle's other segment then you get another constant angle, ASB, and ASB = 180 – ATB.)
If two circles intersect at A and B and we move T as before along the segment opposite the second circle, and we extend TA and TB to P and Q on the second circle, then the length of chord PQ remains constant as T moves.
(From David Wells, The Penguin Dictionary of Curious and Interesting Geometry, 1992.) | 677.169 | 1 |
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Math Snap
PROBLEM
functions for angle B. Rationalize the denominators when applicable.
a=5,c=6
a=5, c=6
a=5,c=6
The unknown side length bbb is
(Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)
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Let A B C be a triangle with sides 3,4 , and 5 , and D E F G be a 6 -by- 7 rectangle. A segment is drawn to divide triangle A B C into a triangle U_{1} and a trapezoid V_{1}, and another segment is drawn to divide rectangle D E F G into a triangle U_{2} and a trapezoid V_{2} such that U_{1} is similar to U_{2} and V_{1} is similar to V_{2}. The minimum value of the area of U_{1} can be written in the form m / n, where m and n are relatively prime positive integers. Find m+n. | 677.169 | 1 |
Elements of geometry, based on Euclid, book i
36. The area of any trapezium is half the rectangle contained by one of the diagonals of the trapezium, and the sum of the perpendiculars let fall upon it from the opposite angles.
37. If the middle points of the sides of a triangle be joined, the lines form a triangle whose area is one-fourth that of the given triangle.
38. If the sides of a triangle be such that they are respectively the sum of two given lines, the difference of the same two lines, and twice the side of a square equal to the rectangle contained by these lines, the triangle shall be right-angled, having the right angle opposite to the first-named side..
39. If a point be taken within a triangle such that the lengths of the perpendiculars upon the sides are equal, show that the area of the rectangle contained by one of the perpendiculars and the perimeter of the triangle is double the area of the triangle.
40. In the last problem, if O be the given point, and OD, OE, OF the respective perpendiculars upon the sides BC, AC, and AB, show that the sum of the squares upon AD, OB, and DC exceeds the sum of the squares upon AF, BD, and CD by three times the square upon either of the perpendiculars.
41. Having given the lengths of the segments AF, BD, CE, in Problem 40, construct the triangle.
42. Draw a line, the square upon which shall be seven times the square upon a given line.
43. Draw a line, the square upon which shall be equal to the sum or difference of two given squares.
44. Reduce a given polygon to an equivalent triangle.
45. Divide a triangle into equal areas by drawing a line from a given point in a side.
46. Do the same with a given parallelogram.
47. If in the fig., Euc. I. 47, the square on the hypothenuse be on the other side, show how the other two squares may be made to cover exactly the square on the hypothenuse.
48. The area of a quadrilateral whose diagonals are at right angles is half the rectangle contained by the diagonals.
49. Bisect a given triangle by a straight line drawn from one of its angles.
50. Do the same with a given rectilineal figure ABCDEF.
51. If from the angle A of a triangle ABC a perpendicular be drawn meeting the base or base produced in D, show that the difference of the squares of AB and AC is equal to the difference of the squares of BD and DC.
52. If a straight line join the points of bisection of two sides of a triangle, the base is double the length of this line.
53. ABCD is a parallelogram, and E a point within it, and lines are drawn through E parallel to the sides of the parallelogram, show that E must lie on the diagonal AC when the figures BC and DE are equal.
54. If AD, BE, CF, are the perpendiculars from the angular points
of the triangle ABC upon the opposite sides, show that the sum of the squares upon AE, CD, BF, is equal to the sum of the squares upon CE, BD, AF.
55. The diagonals of a parallelogram bisect one another.
56. Write out at full length a definition of parallelism, and then prove that the alternate angles are equal when a straight line meets two parallel straight lines.
57. ABCDE are the angular points of a regular pentagon, taken in order. Join AC and BD meeting in H, and show that AEDH is an equilateral parallelogram.
58. Having given the middle points of the sides of a triangle, show how to construct the triangle.
59. Show that the diagonal of a parallelogram diminishes while the angle from which it is drawn increases. What is the limit to
which the diagonal approaches as the angle approaches respectively zero and two right angles?
60. A, B, C, are three angles taken in order of a regular hexagon, show that the square on AC is three times the square upon a side of the hexagon. | 677.169 | 1 |
75
Page 21 ... half of AF , is less than AE , the half of AE + EF . That is , the perpendicular is shorter than any oblique line . Secondly . If we suppose the figure ADB to revolve about AD , the point B will coincide with E , since DB is equal to DE ...
Page 22 ... half of AC + CF , is greater than AE , the half of AE + EF . That is , the oblique line terminating farther from the foot of the perpendicular is the longer . Cor . I. The perpendicular measures the shortest distance from a point to ...
Page 35 ... the parallelogram . THEOREM XXXIII . The straight line joining the middle points of the oblique sides of a trapezoid , will be parallel to the other sides , and equal to half their sum . E A D C H G F B Suppose EF FIRST BOOK . 35.
Page 36 ... half of AD , but ED is also half of AD , hence FG and ED are equal and parallel , and the figure EFGD is a parallelogram , and EF is parallel to DC , and also to AB , since DC and AB are parallel . = Again , since EF = DG = AH ; and ...
Page 37 ... half side , the angle will be acute . III . When this line is less than the half side , the angle will be obtuse . In the triangle ABC , suppose the line CD to be drawn from the angle C to the middle of the side AB . A D B First ... | 677.169 | 1 |
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Euclidean Geometry Grade 12 Notes, Questions and Answers
Euclidean geometry Grade 12 Notes, Questions and Answers: 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 from these.
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Euclidean Geometry Grade 12 Questions and Answers (Downloadable pdf)
Here are several Euclidean Geometry questions for Grade 12 students, along with detailed explanations for each answer:
Question 1: Proving a Triangle is Isosceles
Prove that triangle ABC is isosceles if AB = AC and angle BAC is bisected by line AD where D is on BC.
Answer: To prove that triangle ABC is isosceles, we can use the properties of isosceles triangles and angle bisectors.
Given that AB = AC, triangle ABC is isosceles.
Since AD is the bisector of angle BAC, angle BAD = angle CAD.
By the definition of angle bisector, it divides the angle into two equal parts. | 677.169 | 1 |
Point D lies on side \overline{B C} of \triangle A B C so that \overline{A D} bisects \angle B A C. The perpendicular bisector of \overline{A D} intersects the bisectors of \angle A B C and \angle A C B in points E and F, respectively. Given that A B=4, B C=5, and C A=6, the area of \triangle A E F can be written as \frac{m \sqrt{n}}{p}, where m and p are relatively prime positive integers, and n is a positive integer not divisible by the square of any prime. Find m+n+p. | 677.169 | 1 |
Related Tasks
Related References
Related Concepts
create circles, you can specify various combinations of center, radius, and diameter points on the circumference or on other objects. Draw a Circle by Center …circles, you can specify various combinations of center, radius, and diameter points on the circumference or on other objects. Draw a Circle by Center …… What You'll Be CreatingWe have already used circles extensively to create various grids for a number of patterns. In this lesson we are using …circle refers to a question that requires the use of all (or most) of your circle angle formulas in one problem. It may also be necessary to apply other …… These are all great reasons for considering the use of circles in design. But circles can also be tough to handle because they can create odd …Aug 16, 2016to work with circle equations in …Circles Network is given in each job description – please see.to work with circles and ellipses. So here we are in… Illustrator, just go to our basic type and I going to use my Ellipse tool,…draw myself …to work with circle … Read … | 677.169 | 1 |
About this unit
Trigonometric ratios are not only useful for right triangles, but also for any other kind of triangle. In this unit, you will discover how to apply the sine, cosine, and tangent ratios, along with the laws of sines and cosines, to find all of the side lengths and all of the angle measures in any triangle with confidence. | 677.169 | 1 |
Find x with 106 degree and 136 degree points in triangle?
It is not possible to answer the question without more
information.
Clearly, the two given angles cannot both be interior angles
since were that the case, the triangle would have to be non-planar
and then there is no information on the curvature of the space in
which the triangle is.
The angles could be one interior and one exterior or both
exterior but there is no indication as to which is which.
Finally, and most important, there is no information on what x
is meant to be. | 677.169 | 1 |
You may find the following information useful:
1. An intersection point is created from two diagnonals.
2. The number of ways to choose a set of (unordered) k items from n items is C(n, k).
The problem is asking for the number of intersection points created by the diagonals in a convex polygon with n vertices.
First, we need to determine how many diagonals can be drawn from a single vertex in a polygon with n vertices. We can't draw a diagonal to the vertex itself or to its two adjacent vertices, so the number of diagonals from a single vertex is n - 3.
However, each diagonal is shared by two vertices, so the total number of diagonals in the polygon is (n(n - 3))/2.
Now, each intersection point inside the polygon is formed by two diagonals. Therefore, we need to choose 2 diagonals from the total number of diagonals.
The number of ways to choose 2 items from a set of n items is given by the combination formula C(n, 2) = n(n - 1)/2.
So, the number of intersection points is given by C((n(n - 3))/2, 2).
This simplifies to:
[((n(n - 3))/2)((n(n - 3))/2 - 1)]/2
This is the expression for the number of intersection points in terms of n.
Now, summarize the answer above in one sentence, without any intermediate steps or explanations.
The number of intersection points created by the diagonals in a convex polygon with n vertices is given by the expression: [((n(n - 3))/2)((n(n - 3))/2 - 1)]/2. | 677.169 | 1 |
Elements of Geometry and Trigonometry
From inside the book
Results 1-5 of 57
Page 41 ... centre . The circle is the space terminated by A this curved line . * 2. Every straight line , CA , CE , CD , drawn from the centre to the circum- ference , is called a radius or semidiam- F E eter ; every line which , like AB , passes 43 ... centre ; and hence , there would be three equal straight lines drawn from the same point to the same straight line , which is impossible ( Book Ỉ . Prop . XV . Cor . 2. ) . PROPOSITION IV . THEOREM . In the same circle , or in equal ...
Page 45 ... centre of the circle , and through the middle of the arc subtended by that chord . For , this perpendicular is the same as the one let fall from the centre on the same chord , since both of them pass through the centre and middle of the ...
Page 46 ... centre O , with the radius OB , will pass through the three given points A , B , C. We have now shown that one circumference can always be made to pass through three given points , not in the same straight line : we say farther , that ... | 677.169 | 1 |
5 12 13 Triangle Lesson
What is a 5 12 13 Triangle?
The 5 12 13 triangle is an SSS special right triangle with the ratio between its side lengths as 5, 12, and 13. It is a common Pythagorean triple that is worth memorizing to save time when dealing with right triangles. The other common SSS special right triangle is the 3 4 5 triangle.
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INTRODUCING
Using the 5 12 13 Ratio to our Advantage
A 5 12 13 triangle is considered a scalene triangle because all three of its sides have different lengths. If we come across a right triangle and two of the known sides are part of the 5 12 13 ratio, we can immediately determine that the third side will be the remaining number in the ratio. Note that the "13" side will always be the hypotenuse. The 5 12 13 ratio is scalable and applies to any right triangle with sides that are any common multiple of the numbers 5, 12, and 13.
List of 5 12 13 Triangles
The following triangle side lengths are common multiples of the 5 12 13 ratio and are all considered 5 12 13 triangles. It is not necessary to memorize this list, but it is very useful to understand that 5 12 13 is a ratio that may manifest itself as smaller or larger numbers than the ratio itself.
2.5 6 6.5
5 12 13
7.5 18 19.5
15 36 39
5 12 13 Triangle Angles
A 5 12 13 triangle contains the following internal angles in degrees:
22.6°, 67.4°, 90°
And in radians:
0.39, 1.18, and 1.57
It is not expected to memorize these. We can use the soh cah toa rule and inverse trigonometric functions to solve for the angles since we know the side lengths. See the image below for a visual of these angles.
What is the Area of a 5 12 13 Triangle?
Calculating the area of a 5 12 13 triangle, we get A = (1⁄2)(5)(12) = 30. If the triangle conforms to the 5 12 13 ratio but is scaled, i.e. a 2.5 6 6.5 triangle, we still use the two shorter sides as the base and height. In this case, the area would be (1⁄2)(2.5)(6) = 7.5.
5 12 13 Triangle Example Problem 1
Find the third side length of the 5 12 13 right triangle that has side lengths of 39 and 15.
Solution:
Since we know this is a 5 12 13 triangle scaled from the ratio by an unknown factor, we must determine which two sides are given. 39/15 = 2.6 12/5 = 2.5 13/5 = 2.6
Therefore, we have been given the "13" and "5" sides. We will now multiply by a ratio to get the "12" side.
(12⁄13)(39) = 36
The third side length is 36.
5 12 13 Triangle Example Problem 2
Which of the following sets of side lengths constitute a 5 12 13 triangle?
Triangle 1: 4, 9, 11.5
Triangle 2: 10, 24, 26
Triangle 3: 6, 13, 14
Triangle 4: 20, 48, 52
Solution:
The side lengths of triangles 2 and 4 constitute a 5 12 13 triangle.
Learning math has never been easier.
Get unlimited access to more than 168 personalized lessons and 73 interactive calculators. | 677.169 | 1 |
What does adjacent angle mean?
Adjacent angles are the pair of angles formed when two lines intersect each other. These angles are formed at the point where the intersection occurs, and are adjacent to eachother - hence its name. Another pair of angles that are formed at the intersection of two straight lines are the opposite angles, but this pair of angles are opposite at the vertex and not adjacent, so we should not confuse them with adjacent angles. Adjacent angles are always supplementary, that is, together they equal 180° 180° 180°.
The following illustration shows two examples of what adjacent angles look like. One example is red and the other blue.
Test yourself on angles in parallel lines!
When we are faced with verbal mathematical problems in geometry we must use the full arsenal we have, that is, analyze the different types of angles that arise when straight lines are parallel or intersect. Knowing and understanding the different types of angles, along with our ability to use the various properties of each, can greatly facilitate the resolution of many geometry problems.
What are adjacent angles?
Before elaborating on the subject of adjacent angles, we will begin by explaining the situation that allows for the formation of these angles. To simplify, we we will be guided by the illustration of two parallel lines and a transversal, as shown in the following:
What can we learn from this illustration? The straight lines A A A and B B B are parallel (although in our case we don't need parallel lines to find adjacent angles), and these are intersected by a transversal or secant line C C C.
Now, after having reviewed some basic theory with the help of a graphic example, we will move on to the subject that interests us, that is to say, we will learn the definition of adjacent angles, which in turn will help us to identify them more easily.
Different types of angles.
As we mentioned at the beginning of this section, adjacent angles are not the only angles that we can have in parallel lines. Next, we will briefly review other types of angles that can help us to solve geometric exercises:
Corresponding angles
Corresponding angles arise when a transversal crosses two or more parallel straight lines. The corresponding angles are on the same side of the transversal and at the same side of their parallel lines. Corresponding angles are equal.
Alternate angles
Alternate angles arise when a transversal crosses two or more parallel lines. The alternate angles are on the opposite side of the transversal and on different sides of the parallel lines. The alternate angles are equal.
If you need further explanation, you can refer to the article "Alternate angles".
The following diagram shows two pairs of alternate angles, the first pair in red and the second pair in blue:
Collateral angles
Collateral angles arise when a transversal crosses two or more parallel lines. The collateral angles are on the same side of the transversal and at on different sides of the adjacent parallel lines. The collateral angles supplement eachother, that is, together they equal 180º.
The following diagram represents two pairs of these angles, both in red:
Practice exercises
Exercise 1
In each of the following illustrations, determine if the angles are adjacent. If yes, explain why. If the answer is no, explain why and specify what type of angle is seen in each illustration.
Solution:
Illustration No 1 In this diagram there are no adjacent angles, rather they are opposite angles. According to its definition, opposite angles are two angles that arise at the point of intersection of two straight lines, are equal and are opposite each other.
Illustration No 2 In this picture we are effectively dealing with adjacent angles. The reason is that, according to its definition, adjacent angles are formed when two straight lines intersect each other, and are formed at the point where the intersection occurs, one adjacent to the other, a characteristic from which derives its name. Adjacent angles are supplementary, that is, together they equal 180º 180º 180º.
Illustration No 3 In this scheme there are no adjacent angles, rather they are corresponding angles. According to its definition, corresponding angles are two angles that arise when a transversal crosses two or more parallel straight lines, are on the same side of the transversal, and on the same side of the adjacent parallel lines. The corresponding angles are equal.
Exercise 2
In the triangle ABC ABC ABC, the sides are extended to show some of its exterior angles.
The angle at the vertex A A A of the triangle measures 55º 55º 55º.
Calculate the other two angles based on this data and those shown in the illustration.
Solution:
Looking at the illustration we can notice that the angle opposite to angle B B B, measures 130º 130º 130º. These angles are opposite angles and are equal. According to its definition, adjacent angles are those formed when two straight lines intersect each other, are formed at the point where the intersection occurs, one adjacent to the other, a characteristic from which it gets its name. Adjacent angles always supplement each other, that is, together they equal 180º 180º 180º.
Then, as is well known, the sum of the interior angles of a triangle is equal to 180º 180º 180º.
Since we know the measure of two of the three angles of the triangle, the remaining angle corresponding to vertex C is calculated by subtracting the two angles from 180º 180º 180º obtaining 180º−55º−50º=75º 180º - 55º - 50º = 75º 180º−55º−50º=75º degrees.
Exercise 3
Calculate the angle X X X based on the data given in the illustration and the material learned.
Solution:
We notice from the illustration that we are dealing with three angles that together create a straight angle. A straight angle is 180º 180º 180º. The angle X X X is adjacent to the right angle (90º 90º 90º) and another one that equals 35º 35º 35º, that is to say the two together measure 125º 125º 125º. The angle X X X and the angle measuring 125º 125º 125º are adjacent angles and supplement each other, that is, together they equal 180º 180º 180º.
Exercise 6
How many parallel lines are there in the figure?
Solution:
Let's start by seeing if the lines "a a a" and "b b b"are parallel:
Assuming that both straight lines are parallel, we have a transversal that intersects them, forming an angle of 58º 58º 58º at the intersection with the line "a a a" and an angle of 123º 123º 123º at the intersection with the line "b b b". Then, the adjacent angle of the angle of "58º 58º 58º" is the angle that supplements this angle. 123º 123º 123º is the angle that supplements the latter, i.e., an angle measuring 57º 57º 57º and being corresponding to the angle of 58º 58º 58º should be equal, which is clearly not true. Therefore, the lines "a a a" and "b b b" are not parallel.
Now let's see if the lines "b b b" and "c c c" are parallel:
Using similar reasoning as above, we can assume that these lines are parallel. Comparing the angle adjacent to 57º 57º 57º formed by the line "c c c", which should measure 123º 123º 123º and the angle of 123º 123º 123º formed by the line "b b b", we see that they are equal as expected because they are corresponding angles. Therefore the lines "b b b" and "c c c" are parallel.
Now let's see if the lines "c c c" and "d d d" are parallel:
The angle adjacent to the angle of 57º 57º 57º contained in the line "c c c" which is an angle of 123º 123º 123º and the angle of 123º 123º 123º contained in the line "d d d", are corresponding angles. In this case they are indeed equal, therefore the straight lines "c c c" and "d d d" are parallel.
Now let's see if the lines "d d d" and "e e e" are parallel:
Again let us suppose that they are. Then, the angle adjacent to 30º 30º 30º formed by the line "d d d" measures 150º 150º 150º and together with the angle of 150º 150º 150º formed by the line "e e e" form corresponding angles, therefore, they must be equal. This is true, therefore the lines "d d d" and "e e e" are parallel.
Answer:
The lines "b b b", "c c c", "d d d", "e e e" are parallel.
Review questions
What are adjacent angles?
They are a pair of angles of angles formed by the intersection of two straight lines in such a way that they supplement each other, that is to say, that together they form 180º 180º 180º.
What is the main characteristic of adjacent angles?
That they are supplementary, that is, when added together they form a straight angle (180º 180º 180º).
In a diagram of parallel lines intersected by a transversal, which pair of angles have the same property of adjacent angles, that is, that they are supplementary?
The collateral angles.
In a pair of adjacent angles, can either one separately be an obtuse angle?
No, since being adjacent angles they are supplementary, i.e., together they add up to 180º 180º 180º and the sum of two obtuse angles (angles greater than 90º 90º 90º) exceeds 180º 180º 180º.
If you are interested in learning more about angles, you can access one of the following articles:
examples with solutions for adjacent | 677.169 | 1 |
Step 3: Compare the Distance with Radii Sum and Difference
For the circles to be touching internally:
$$\mathbf{|R_2 - R_1| = \mathbf{C_1C_2}}$$
By substitution:
$$|12 - 2| = 10$$
This implies the given circles have an internal tangency relationship as (|10| = 10).
Conclusion
Since the distance between the centers of the circles is equal to the difference in their radii, the circles are touching internally. | 677.169 | 1 |
2024-07-16T09:25:10Z Schulte (Boston)Bezdek, András; Bisztriczky, Ted201564955 56, No. 2, 541-549 (2015).52A10; 52C15Finding equal-diameter triangulations in polygonsj | 677.169 | 1 |
Maths worksheets year 5 worksheets. Web the corbettmaths textbook exercise on finding missing angles. Sophie has been asked to measure this angle. Web the corbettmaths textbook exercise on measuring angles. Web the corbettmaths practice questions and answers on measuring angles.
Angles are measured in degrees (o). Measure each reglex angle below (a) (b) (c) (d) question 1: And best of all they all (well, most!) come with answers. Web drawing and measuring angles worksheet. Language for the angles worksheet.
Teach Students To Measure Angles With These Protractor Worksheets You
An angle equal to 1 turn (360° or 2π radians) is called a full angle. In the last two worksheets, students also classify the angles as being acute, obtuse or a right angle. Acute, right, and obtuse angles. Web naming, measuring and estimating angles: Web measuring angles worksheets contain examples and problems based on angle measurement in mathematics.
Measuring angles with a protractor worksheet startunare
Web list of angle worksheets. Web explore our collection of printable worksheets and classroom activities to support your teaching of angles at ks3 and ks4. Simply practise the skills of measuring and drawing acute, obtuse and reflex angles in various orientations with this worksheet. Web the corbettmaths textbook exercise on measuring angles. Web measuring angles worksheets contain examples and problems.
Angle Measures Worksheet
Drawing and measuring angles worksheet description. These tried and tested teaching resources cover measuring angles, identifying and classifying types of angles, drawing angles, finding missing angles, angles in shapes and more. Once students have been introduced to the idea of angles, it's important that they learn some new vocabulary terms. Web list of angle worksheets. Web the corbettmaths practice questions.
Measuring Angles 4th Grade Worksheet
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Angle Measure Worksheet - Measuring video 31 on corbettmaths question 5: What's included in these measuring angles worksheets? Web the corbettmaths textbook exercise on measuring angles. With fun activities, including measuring spiderwebs, steering wheels, and laser beams, your child is sure to enjoy our kind of geometry. Measuring angles worksheets help bridge gaps between third, fourth, and fifth grade math. Web the corbettmaths practice questions and answers on measuring angles. In these exercises, students measure angles with a real protractor. Memo line for the angles worksheet. Ideal for math teachers and learners of all levels. An angle equal to 1 turn (360° or 2π radians) is called a full angle.
Once students have been introduced to the idea of angles, it's important that they learn some new vocabulary terms. Sophie has been asked to measure this angle. This worksheet is a great resources for the 5th, 6th grade, 7th grade, and 8th grade. Whether you want a homework, some cover work, or a lovely bit of extra practise, this is the place for you. Web this worksheet provides the student with a set of angles.
Web measuring angles worksheets contain examples and problems based on angle measurement in mathematics. Line segments, symmetry, and more. Web list of angle worksheets. His or her job is to use a standard protractor to measure the angles in degrees, extending the lines with a straight edge if necessary.
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Web explore our collection of printable worksheets and classroom activities to support your teaching of angles at ks3 and ks4. The magnitude of an angle is the amount of rotation of one arm about the vertex to bring it to the position of the other arm. A real protractor is required!
Web Naming, Measuring And Estimating Angles:
Angles are measured in degrees (o). Maths worksheets year 5 worksheets. Angles larger than a straight angle but less than 1 turn (between 180° and 360°) are called reflex angles. A big part of geometry is the measuring.
Web In These Printable Measuring Angles Worksheets, We Will Learn To Measure Angles Using A Protractor.
Web the corbettmaths practice questions and answers on measuring angles. Discover a collection of free printable math resources designed to help students develop essential skills in measuring angles. This worksheet is a great resources for the 5th, 6th grade, 7th grade, and 8th grade. Web this free set of angles worksheets covers angle vocabulary, measuring angles with protractors, classifying angles and even spotting angles in real life.
Web Angle Measurement And Classification Worksheets.
His or her job is to use a standard protractor to measure the angles in degrees, extending the lines with a straight edge if necessary. An angle equal to 1 turn (360° or 2π radians) is called a full angle. Once students have been introduced to the idea of angles, it's important that they learn some new vocabulary terms. Memo line for the angles worksheet. | 677.169 | 1 |
Exterior Angles of a Polygon
Sum of the Measure of the Exterior Angles of the Polygon: An Introduction
A polygon is a two-dimensional shape with a minimum of three sides. The sides may be four, five and so on. The polygon is divided into two parts: regular polygon and irregular polygon. The sum of measures of the exterior angles is the sum of all exterior angles formed in the polygon. In this article, we will learn about exterior angles and the sum of all exterior angles of a polygon.
Exterior Angle
When the side of a polygon is extended, the angle formed outside the figure is called the exterior angle. It is formed between the extended side and the adjacent side.
Four Exterior Angles of a Polygon
Properties of the Exterior Angles of a Polygon
The exterior angle has different properties. Let us consider a hexagon, which is a polygon. Following are the properties, which tell us about the exterior angles:
They are formed on the outside of the given figure.
The sum of the interior and exterior angles formed by the extended side and the adjacent side is always\[180^\circ \].
In the figure, angles 1, 2, 3, 4, 5 and 6 are the exterior angles.
Hexagon
The exterior angles of a regular polygon are always equal to each other.
Sum of Exterior Angles of a Polygon
As we have taken the hexagon here, we start from the first vertex and proceed in the clockwise direction. The bending happens at the different vertices 2, 3, 4, 5 and 6 and ends at vertex 1. In this way, we covered a full rotation and this complete angular rotation is \[360^\circ \]. Therefore, it concludes that the sum of exterior angles of a polygon is \[360^\circ \].
Exterior Angle
Proof of Sum of the Exterior Angles of the Polygon
Let us consider the polygon is a convex polygon with n number of sides and N is the sum of its exterior angles.
By the exterior angle theorem,
The sum of exterior angles is equivalent to the sum of interior angles and the sum of linear pairs.
In this way, we proved that the sum of the exterior angles of the polygon is\[360^\circ \].
Interior Angles
Interior angles of any two-dimensional shape are the measure of the angles between its sides within the closed figure. These are the inclinations of one side to another side. A polygon has three sides, i.e., a triangle, which will have three interior angles. If the polygon were a quadrilateral, the figure would have four interior angles.
Interior Angle
Formula for Sum of All Interior Angles of a Polygon
The sum of all interior angles of a polygon is equal to the\[\left( {{\rm{n}} – 2} \right)\, \times \,180^\circ \]. Here, n is some sides of the polygon. The polygon here may be a triangle, quadrilateral, pentagon, hexagon etc.
Proof of the Interior Angles of the Polygon
We know that the sum of interior angles of a polygon having \[n\]sides is\[\left( {{\rm{n}} – 2} \right)\, \times \,180^\circ \].
Consider a polygon of sides ABCDE; it is divided into 3 triangles by joining the diagonals such that the figure is divided into subsequent triangles. There is a polygon having n sides.
ABCDE is a Polygon Divided into Three Triangles
It is clear that when the polygon has n sides, it forms \[\left( {{\rm{n}} – 2} \right)\] triangles. | 677.169 | 1 |
Current Affairs
JEE Main & Advanced
Let three angles of \[\Delta ABC\] are denoted by \[A,\,\,B,\,\,C\] and the sides opposite to these angles by letters \[a,\,\,b,\,\,c\] respectively.
(1) When two sides and the included angle be given :
The area of triangle ABC is given by,
\[\Delta =\frac{1}{2}bc\sin A=\frac{1}{2}ca\sin B=\frac{1}{2}ab\sin C\]
i.e., \[\Delta =\frac{1}{2}\] (Product of two sides) \[\times \] sine of included angle
(2) When three sides are given :
Area of \[\Delta ABC=\,\Delta =\sqrt{s(s-a)(s-b)(s-c)}\]
where semiperimeter of triangle \[s=\frac{a+b+c}{2}\]
(3) When three sides and the circum-radius be given :
Area of triangle\[\Delta =\frac{abc}{4R}\], where R be the circum-radius of the triangle.
(4) When two angles and included side be given :
\[\Delta =\frac{1}{2}{{a}^{2}}\frac{\sin B\sin C}{\sin (B+C)}=\frac{1}{2}{{b}^{2}}\frac{\sin A\sin C}{\sin (A+C)}=\frac{1}{2}{{c}^{2}}\frac{\sin A\sin B}{\sin (A+B)}\]
In every triangle the sum of the squares of any two sides is equal to twice the square on half the third side together with twice the square on the median that bisects the third side.
For any triangle ABC, \[{{b}^{2}}+{{c}^{2}}=2({{h}^{2}}+{{m}^{2}})=2\{{{m}^{2}}+{{(a/2)}^{2}}\}\] by use of cosine rule.
If \[\Delta \] be right angled, the mid point of hypotenuse is equidistant from the three vertices so that \[DA=DB=DC\].
\[\therefore {{b}^{2}}+{{c}^{2}}={{a}^{2}}\] which is pythagoras theorem. This theorem is very useful for solving problems of height and distance.
In any triangle ABC, the square of any side is equal to the sum of the squares of the other two sides diminished by twice the product of these sides and the cosine of their included angle, that is for a triangle ABC,
(1) \[{{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\cos A\Rightarrow \cos A=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\]
(2) \[{{b}^{2}}={{c}^{2}}+{{a}^{2}}-2ca\cos B\Rightarrow \cos B=\frac{{{c}^{2}}+{{a}^{2}}-{{b}^{2}}}{2ca}\]
(3) \[{{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos C\Rightarrow \cos C=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}\]
Combining with \[\sin A=\frac{a}{2R},\sin B=\frac{b}{2R},\sin C=\frac{c}{2R}\]
We have by division, \[\tan A=\frac{abc}{R({{b}^{2}}+{{c}^{2}}-{{a}^{2}})},\]
\[\tan B=\frac{abc}{R({{c}^{2}}+{{a}^{2}}-{{b}^{2}})},\,\,\tan C=\frac{abc}{R({{a}^{2}}+{{b}^{2}}-{{c}^{2}})}\]
where, R be the radius of the circum-circle of the triangle ABC.
A triangle has six components, three sides and three angles. The three angles of a \[\Delta ABC\] are denoted by letters \[A,\,\,B,\,\,C\] and the sides opposite to these angles by letters \[a,\,\,b\] and \[c\] respectively. Following are some well known relations for a triangle (say \[\Delta ABC\])
\[A+B+C={{180}^{o}}\] (or \[\pi \])
\[a+b>c,\,\,b+c>a,\,\,c+a>b\]
\[|a-b|<c,|b-c|<a,|c-a|<b\]
Generally, the relations involving the sides and angles of a triangle are cyclic in nature, e.g. to obtain the second similar relation to \[a+b>c,\] we simply replace \[a\] by \[b,\,\,b\] by \[c\] and \[c\] by \[a\]. So, to write all the relations, follow the cycles given.
The law of sines or sine rule : The sides of a triangle are proportional to the sines of the angles opposite to them i.e., \[\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k,\,(say)\]
More generally, if \[R\] be the radius of the circumcircle of the triangle \[ABC,\,\,\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\].
Some Standard Results on Periodic Functions
(1) Check the validity of the given equation, e.g. \[2\sin \theta -\cos \theta =4\] can never be true for any \[\theta \] as the value \[(2\sin \theta -\cos \theta )\] can never exceeds \[\sqrt{{{2}^{2}}+{{(-1)}^{2}}}=\sqrt{5}\]. So there is no solution of this equation.
(2) Equation involving \[\sec \theta \] or \[\tan \theta \] can never have a solution of the form,\[(2n+1)\frac{\pi }{2}\].
Similarly, equations involving \[\text{cosec }\theta \] or \[\cot \theta \] can never have a solution of the form \[\theta =n\pi \]. The corresponding functions are undefined at these values of \[\theta \].
(3) If while solving an equation we have to square it, then the roots found after squaring must be checked whether they satisfy the original equation or not, e.g. let \[x=3\]. Squaring, we get \[{{x}^{2}}=9\] \[\therefore \] \[x=3\] and \[-\,3\] but \[x=-3\] does not satisfy the original equation \[x=3\].
(4) Do not cancel common factors involving the unknown angle on L.H.S. and R.H.S. because it may delete some solutions. In the equation \[\sin \theta (2\cos \theta -1)=\sin \theta {{\cos }^{2}}\theta \] if we cancel \[\sin \theta \] on both sides we get \[{{\cos }^{2}}\theta -2\cos \theta +1=0\] \[\Rightarrow {{(\cos \theta -1)}^{2}}=0\] \[\Rightarrow \cos \theta =1\Rightarrow \theta =2n\pi \]. But \[\theta =n\pi \] also satisfies the equation because it makes \[\sin \theta =0\]. So, the complete solution is \[\theta =n\pi ,\,n\in Z\].
(5) Any value of x which makes both R.H.S. and L.H.S. equal will be a root but the value of \[x\] for which \[\infty =\infty \] will not be a solution as it is an indeterminate form.
Hence, \[\cos x\ne 0\] for those equations which involve \[\tan x\]and \[\sec x\]whereas \[\sin x\ne 0\]for those which involve \[\cot x\]and \[\text{cosec}\,x\].
Also exponential function is always +ve and \[{{\log }_{a}}x\]is defined if \[x>0\], \[x\ne 0\] and \[a>0,a\ne 1\] \[\sqrt{f(x)}=+ve\] always and not \[\pm \,i.e.\] \[\sqrt{({{\tan }^{2}}x)}=\tan x\]and not \[\pm \tan x\].
(6) Denominator terms of the equation if present should never become zero at any stage while solving for any value of \[\theta \] contained in the answer.
(7) Sometimes the equation has some limitations also e.g., \[{{\cot }^{2}}\theta +\text{cose}{{\text{c}}^{2}}\theta =1\] can be true only if \[{{\cot }^{2}}\theta =0\] and \[\text{cose}{{\text{c}}^{2}}\theta =1\] simultaneously as \[\text{cose}{{\text{c}}^{2}}\theta \ge 1\]. Hence the solution is \[\theta =(2n+1)\pi /2\].
(8) If \[xy=xz\] then \[x(y-z)=0\Rightarrow \] either \[x=0\] or \[=\frac{1}{2}\sqrt{{{b}^{2}}+{{c}^{2}}+2bc\cos A}\] or both. But \[\frac{y}{x}=\frac{z}{x}\Rightarrow y=z\] only and not \[x=0\], as it will make \[\infty =\infty \]. Similarly if \[ay=az\], then it will also imply \[y=z\] only as \[a\ne 0\]being a constant.
Similarly \[x+y=x+z\Rightarrow y=z\] and \[x-y=x-z\Rightarrow y=z\]. Here we do not take \[x=0\] as in the above because \[x\] is an additive factor and not multiplicative factor.
(9) Student are advised to check whether all the roots obtained by them, satisfy the equation and lie in the domain of the variable of the given equation.
Suppose we have to find the principal value of \[\theta \] satisfying the equation \[\sin \theta =-\frac{1}{2}\].
Since \[\sin \theta \] is negative, \[\theta \] will be in 3rd or 4th quadrant. We can approach 3rd or 4th quadrant from two directions. If we take anticlockwise direction the numerical value of the angle will be greater than \[\pi \]. If we approach it in clockwise direction the angle will be numerically less than \[\pi \]. For principal value, we have to take numerically smallest angle. So for principal value.
(1) If the angle is in 1st or 2nd quadrant we must select anticlockwise direction and if the angle is in 3rd or 4th quadrant, we must select clockwise direction.
(2) Principal value is never numerically greater than \[\pi \].
(3) Principal value always lies in the first circle (i.e., in first rotation). On the above criteria, \[\theta \] will be \[-\frac{\pi }{6}\] or \[-\frac{5\pi }{6}.\] Between these two \[-\frac{\pi }{6}\] has the least numerical value. Hence \[-\frac{\pi }{6}\] is the principal value of \[\theta \] satisfying the equation \[\sin \theta =-\frac{1}{2}.\]
From the above discussion, the method for finding principal value can be summed up as follows :
(i) First draw a trigonometrical circle and mark the quadrant, in which the angle may lie.
(ii) Select anticlockwise direction for 1st and 2nd quadrants and select clockwise direction for 3rd and 4th quadrants.
(iii) Find the angle in the first rotation.
(iv) Select the numerically least angle. The angle thus found will be principal value.
(v) In case, two angles one with positive sign and the other with negative sign qualify for the numerically least angle, then it is the convention to select the angle with positive sign as principal value. | 677.169 | 1 |
NCERT Solutions for Class 10 Mathematics Chapter 11 Constructions
The NCERT Solutions for Class 10 Mathematics Chapter 11 are accurate and reliable. Students looking for step-by-step answers to all the NCERT Mathematics Class 10 Chapter 11 textbook questions can refer to the solutions by Extramarks. The subject matter experts have prepared the solutions as per the latest CBSE guidelines. The answers are written in a simple and easy-to-understand manner with examples, as well as diagrams, and have been used, wherever possible, to help students understand it in a better way.
The NCERT solutions for Class 10 Mathematics chapter 11 by Extramarks are the perfect resource for last-minute revision and preparation. The answers are explained thoroughly to make it easier for students to grasp the concepts. Additionally, by referring to the Extramarks Solutions, students will get a better idea of how to attempt the questions and score better marks in tests and board exams.
Introduction
Chapter 11 Constructions helps students understand how to make geometrical figures like a triangle, circle and more. They will also learn how to bisect a line segment or draw perpendiculars and tangents as demanded by the question. Although the chapter is more practical, it demands theoretical knowledge as well. This is because, without a strong understanding of concepts like proportionality theorem, Pythagoras theorem, etc, students won't be able to draw accurate constructions.
NCERT Solutions for Class 10 Mathematics
Extramarks offers NCERT Solutions for Class 10 Mathematics for the following chaptersDivision of a Line Segment
For dividing a line segment internally in the ratio m:n, follow the steps given below:
Step 1: Draw a line segment of a given length. Now name one of its points as A and the other one as B.
Step 2: Now, draw a ray that should be making an acute angle with AB. Name the ray as AX.
Step 3: Alongside AX, start marking off (m + n) points. This includes A1, A2, ……., Am, Am+1,….., Am+n. To give you an example, if the ratio is to be 2:3, then mark 5 (=2+ 3) points).
Step 4: Join BAAm+n.
Step 5: Draw a line through Am parallel to Am+n B . Make an angle equal to ∠AAm+n B.
Make this line meet AB at point P. It is the point that divides AB internally in the ratio m:n.
Constructing a Triangle Similar to a Given Triangle
When constructing a triangle similar to the given triangle, the former can be smaller or bigger than the latter. The following term needs to be defined:
Scale factor – The ratio of the to-be-constructed figure's sides with measurements of the given figure.
Let ABC be the given triangle. Suppose you want to construct a triangle similar to ABC, each of its sides is (m/n)th of the corresponding sides of ABC.
The Following are the Steps to Be Taken for the Construction of a Triangle When M<n:
Step 1: Draw ABC by using the given data.
Step 2: Make AB as the base of the given ABC.
Step 3: At one end, say A of AB draw an acute angle ∠BAX below the base AB.
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Q.1Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.
Ans.
Following are the steps to divide a line segment of length7.6 cm in the ratio of 5:8.Step 1: Draw line segment AB of 7.6 cm and draw a ray AX making an acute angle with line segment AB.Step 2: Locate 13(=5+8) points A1, A2,A3,...,A13 on AX such that AA1=A1A2=A2A3=...=A12A13.Step 3: Join the points B and A13.Step 4: Through the point A5, draw a line parallel to BA13 at A5 intersecting AB at point C.C is the point which divides line segment AB of length 7.6 cmin the ratio 5:8.On measuring lengths of AC and BC, we get AC=2.9cmandBC=4.7 cm.
Q.2
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 23 of thecorresponding sides of the first triangle.
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 75 of thecorresponding sides of the first triangle.
Ans.
Step 1: Draw a line segment AB=5cm. Taking point A as centre, draw an arc of 6 cm radius. Again, taking point B as centre, draw an arc of 7 cm. These arcs intersect each other at point C. So, we have AC=6cmand BC=7cm.Δ ABC is the required triangle.Step 2: Draw a ray AX making an acute angle with line AB on the opposite side of vertex C.Step 3: Locate 7 points A1, A2,A3,...,A7 on AX such that AA1=A1A2=A2A3=...=A6A7.Step 4: Join the points B and A5.Step 5: Through the point A7, draw a line parallel to BA5 intersecting extended line segment AB at point B'.Step 6: Draw a line through B' parallel to the line BC to intersect extended line segment AC at C'.The required triangle is ΔAB 'C'.
Q.4
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whosesides are112 times the corresponding sides of the isosceles triangle.
Ans.
StepsofConstruction:Step 1: Draw a line segment AB=8cm. Draw arcs of same radius on both sides of the line segment while taking point A and B as its centre. Let these arcs intersect each other at O and O'. Join OO'. Let OO' intersect AB at D.Step 2: Taking D as centre, draw an arc of 4 cm radius which cuts the extended line segment OO' at point C. An isosceles ΔABC is formed havind CD as 4 cm and AB as 8 cm.Step 3: Draw a ray AX making an acute angle with line AB on the opposite side of vertex C.Step 4: Locate 3 points A1, A2,A3 on AX such that AA1=A1A2=A2A3.Step 5: Join the points B and A2.Step 6: Through the point A3, draw a line parallel to BA2 intersecting extended line segment AB at point B'.Step 7: Draw a line through B' parallel to the line BC to intersect extended line segment AC at C'.The required triangle is ΔAB 'C'.
Q.5 Draw a triangle ABC with side BC=6 cm, AB=5 cm and ∠ABC=60°. Then construct a triangle whose sides are34 of the corresponding sides of the triangle ABC.
Ans.
StepsofConstruction:Step 1: Draw a Δ ABC with sides AB=5cm, BC=6 cm and∠ABC=604.Step 5: Through the point B3, draw a line parallel to CB4 intersecting line segment BC at point C'.Step 6: Draw a line through C' parallel to the line AC to intersect line segment AB at A'.The required triangle is ΔA 'BC'.
Q.6
Draw a triangle ABC with sides BC=7 cm, ∠=°,∠=°,then construct a triangle whose sides are 43 times the corresponding sides of ΔABC.
Ans.
Itisgiventhat∠B=45°,∠A=105°.∴ In Δ ABC, we have ∠C=180°−105°−45°=30°StepsofConstruction:Step 1: Draw a Δ ABC with side BC=7 cm and∠B=45°,∠C=303.Step 5: Through the point B4, draw a line parallel to CB3 intersecting extended line segment BC at point C'.Step 6: Draw a line through C' parallel to the line AC to intersect extended line segment BA at A'.The required triangle is ΔA 'BC'.
Q.7
Draw a right triangle in which the sides (other than hypotenuse) are of length 4 cm and 3 cm. Thenconstruct another triangle whose sides are 53 times the corresponding sides of the given triangle.
Ans.
StepsofConstruction:Step 1: Draw a right angle Δ ABC with base AB=4 cm, AC=3 cm and ∠A=90°.Step 2: Draw a ray AX making an acute angle with line AB on the opposite side of vertex C.Step 3: Locate 5 points A1, A2,A3, A4, A5 on AX such that AA1=A1A2=A2A3=A3A4=A4A5.Step 4: Join the points B and A3.Step 5: Through the point A5, draw a line parallel to BA3 intersecting extended line segment AB at point B'.Step 6: Draw a line through B' parallel to the line BC to intersect extended line segment AC at C'.The required triangle is ΔAB' C'.
Q.8Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Ans.
StepsofConstruction:Step 1: Draw a circle of radius 6 cm with centre at point O. Locate a point P, 10 cm away from O, and join O and PThe lengths of tangents PQ and PR are 8 cm each.
Q.9Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
Ans.
StepsofConstruction:Step 1: Draw a circle of radius 4 cm with centre at point O. Draw anothe circle of radius 6 cm with centre at point O. Locate a point P on this circle and join OPOnmeasuringwegetthe lengths of tangents PQ and PR are 4.5 cm each.
InΔ PQO, we have∠PQO=90° [PQ is a tangent]PO=6 cm [Radius of the circle]QO=4 cm [Radius of the circle]By applying Pythagoras theorem in Δ PQO, we have PQ2+QO2=OP2or PQ2+42=62or PQ2=36−16=20or PQ=20=25=4.5 cm approx.We know that PQ=PR as these tangents are drawn froma single point P.Therefore, PQ=PR=4.5 cm approx.
Q.10Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
Ans.
StepsofConstruction:Step 1: Draw a circle of radius 3 cm with centre at point O.Step 2: Take one of its diametre, RS, and extend it on both sides. Locate two points P and Q on this diametre such that OP=OQ=7 cm.Step 3: Bisect OP and OQ. Let T and U be the mid-points of OP and OQ respectively.Step 4: Taking T and U as centres and OT and OU as radii, draw two circles. These two circles will intersect the circle at points V, W, X, Y. Join PV, PW, QX and QY. These are the required tangents.
Q.11Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.
Ans.
StepsofConstruction:Step 1: Draw a circle of radius 5 cm with centre at point O.Step 2: Take a point A on circumference of the circle and join OA. Draw a perpendicular to OA at point A.Step 3: Draw a radius OB, making an angle of 120° with OA.Step 4: Draw a perpendicular to OB at point B. Let both the perpendiculars intersect at point P. PA and PB are the required tangents at an angle of 60°.
Q.12
Ans.
StepsofConstruction:Step 1: Draw a line segment AB of 8 cm. Taking A and B as centre, Draw a circle of 4 cm radius centred at A and another circle of radius 3 cm centred at B.Step 2: Bisect the line AB. Let the mid-point of AB be C. Taking C as centre, draw a circle with radius AC. This circle intersects the other two circles at points P, Q, R and S. Join BP, BQ, AS and AR. These are the required tangents.
Q.13Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Ans.
As per given information, we have the following figure.
StepsofConstruction:Step 1: Join AE and bisect it. Let F be the mid-point of AE.Step 2: Taking F as centre and FE as its radius, draw a circle which will intersect the other circle at points B and G. Join AG. AB and AG are the required tangents.
Q.14Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.
Ans.
StepsofConstruction:Step 1: Draw a circle with the help of a bangle.Step 2: Take a point P outside this circle and take two chords QR and ST. Step 3: Draw perpendicular bisectors of QR and ST. Let the perpendicular bisectors intersect each other at point O.Step 4: Join OP and bisect it. Let U be the mid-point of OP. Taking U as centre, draw a circle of radius OU, which intersects the previous circle at V and W. Join PV and PW. PV and PW are the required tangents. | 677.169 | 1 |
how to find area of triangle with example
Area Of Triangle Worksheets Grade 8 – Triangles are one of the most fundamental shapes in geometry. Understanding the triangle is essential to learning more advanced geometric concepts. In this blog post we will look at the various types of triangles and triangle angles, as well as how to calculate the dimension and perimeter of the triangle and will provide an example of every. Types of Triangles There are three kinds of triangulars: Equilateral isoscelesand scalene. Equilateral … Read more | 677.169 | 1 |
Horizontal Lines of Symmetry (B) worksheet
Learning objective: To complete shapes and patterns with a horizontal line of symmetry
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Horizontal Lines of Symmetry (B) worksheet description
Pupils are given shapes and patterns with a horizontal mirror line. They will need to reflect a shape or pattern on one side of the mirror line to the other side. They will also need to reflect patterns that are presented on both sides of a mirror line.
Realted to Horizontal Lines of Symmetry | 677.169 | 1 |
Taxicab circles, ellipses, and hyperbolas
To explore Taxicab circles:
Use the slider labeled a and either of the two points G or B.
To explore Taxicab Ellipses:
Use the points G and B close enough that when you use slider a they will over lap.
To explore Taxicab Hyperbolas:
Use points S and Z and move the point R on the line to the left of the applet.
This is somewhat incomplete, you will notice that there are places where you can put the points S and Z so that the intersection of the two circles are not marked by points. | 677.169 | 1 |
We the trnds
angle between two planes formula
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If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. However I need a quicker way. All planes have an equation that identifies them. If the same expression is represented by a straight line in space, then we will have a plane parallel to the z axis. I can find the angle in AutoCAD. These lines are perpendicular to the line of intersection of the planes. Each plane contains 3 vectors accordingly. This means that the scalar product of the direction vectors is equal to zero: . The plane, as we know, is a 3D object formed by stacks of lines kept side by side. Finding the angle between two lines using a formula is the goal of this lesson. A Formula for Finding Angle between Two Planes 計算兩平面之交角的公式. Author: Alex CHIK. As answered in this question, it is simple enough to compute the signed angle given the normal of a plane to which the vectors are perpendicular.But I can find no way to do this without that value. lawrenceyy. If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. If two lines are perpendicular to each other then their direction vectors are also perpendicular. Author: wuguoning How to say whether the planes are parallel or perpendicular, or find the angle between them. This can be done using the inner (dot) … The understanding of the angle between the normal to two planes is made simple with a diagram. If θ is the angle between two intersecting lines defined by y 1 = m 1 x 1 +c 1 and y 2 = m 2 x 2 +c 2, then, the angle θ is given by. From the above figure, we learnt that the angle between the two planes is equal to the angle between their normal, thus. The angle between two lines is the angle between direction vectors of the lines. The equation of a plane takes on this form: Ax + By + Cz + D= 0 Where A, B, C, and D are constants and x, y, and z are variables. What is needed is a formula that will give the angle between the two planes of the triangles given. Angle between these planes is given by using the following formula:-Cos A = Using inverse property, we get: A = Below is the implementation of the above formulae: C++. The OP's P1 is essentially a vector from (0,0,0) to "(x,y,z)" – MickyD Jun 22 '15 at 7:00 11.1 Angle Between Two Planes (17:28) Start 11.1 Short Notes: Angle between Two Planes Start 11.2 Line of Intersection Between Two Planes (13:00) Start 11.2 Short Notes: Line of Intersection between Two Planes. Angle (dihedral angle) between two planes The angle subtended by normals drawn from the origin to the first and second plane, that is, the angle which form the normal vectors of the given planes is : the angle between two planes. What is needed is a formula that will give the angle between the two planes of the triangles given. A vector can be pictured as an arrow. A 3D space can have an infinite number of planes aligned to one another at an infinite number of angles. When two such planes intersect (i.e., a set of four consecutively-bonded atoms), the angle between them is a dihedral angle. Function calculate_angle() iterates over the amount of triangles so that I compare only new ones with the old ones (for z in range(k, n)) list_result = [i for i in index1 if i in index2] used for the adjacency detection: it asks, whether the indices of the first triangle coordinates == to the second. What is needed is a formula that will give the angle between the two planes of the triangles given. Angle between two planes. The angle between two planes is equal to the angle determined by the normal vectors of the planes. I can make an AutoCAD block that will apply the excel calculation automatically based on the coordinates of the two triangles. play_arrow. Jul 2011 28 0. ) by the inner product ⋅ , ⋅ {\displaystyle \langle \cdot ,\cdot \rangle } , i.e. Topic: Planes Solution: As mentioned above, the angle between two planes is equal to the angle between their normals. Triangle; Test; Discover Topics . The vectors … The distance between the two planes is going to be the square root of six, and so then if we solve for d, multiple both sides of this equation times the square root of six, you get six is equal to negative d, or d is equal to negative six. formula for the angle between two intersectiing planes. L. LMHmedchem. In orthogonal crystals, we can calculate the distance between planes, d, from the Miller index (h k l) and the unit cell dimensions a,b,c from the following formula for ORTHOGONAL axes Note that this can be simplified if a=b (tetragonal symmetry) or a=b=c (cubic symmetry). Angle between two lines. Angle Between Two Straight Lines Formula. The output is supposed to be the maximum angle between the adjacent planes in rad. If θ is the angle between two intersecting lines defined by y 1 = m 1 x 1 +c 1 and y 2 = m 2 x 2 +c 2, then, the angle θ is given by. Take the course Want to learn more about Calculus 3? \(\vec{r}\).\(\vec{n_{2}}\) = d2. Dihedral angles are used to specify the molecular conformation . I can make an AutoCAD block that will apply the excel calculation automatically based on the coordinates of the two triangles For cubic crystals, the angle, f between two planes, (h 1 k 1 l 1) and (h 2 k 2 l 2) is given by: . Your IP: 51.91.82.178 For example, the equation x + 7y - 3z+ 8 = 0 is the equation of a plane. formula for the angle between two intersectiing planes. Forums . Math video on how to find the angle between two planes using the definition of dot product for the normal vectors of the planes. Let's say I have plane A and plane B, and they are oriented differently from each other. Now that we know what dihedral angles are, we get to the fun part - calculating them! Example: Calculating the angle between two planes. These lines are perpendicular to the line of intersection of the planes. Cloudflare Ray ID: 5fe71695ce12730b You need a third vector to define the direction of view to get the information about the sign. from the definition of the dot product. Answer Save. Pre-University Math Help. (In three dimensions) I'm looking for a way to compute the signed angle between two vectors, given no information other than those vectors. To learn about other techniques to calculate the angle between two planes, download BYJU'S – The Learning App. The vectors … September 1, 2017, 2:38pm #1. When two such planes intersect (i.e., a set of four consecutively-bonded atoms), the angle between them is a dihedral angle. Normal vectors to the above planes are represented by: \(\vec{n_{1}}\) = 2\(\hat{i}\) + 4\(\hat{j}\) – 2\(\hat{k}\) Let's say I have plane A and plane B, and they are oriented differently from each other. Activity. Angle between two planes formula of miller indices Ask for details ; Follow Report by Ajish1589 13.01.2019 Log in to add a comment If two lines are perpendicular to each other then their direction vectors are also perpendicular. A logical 3D "point" in Unity is a physical Vector3 (there is no Point3 type) and Unity defines Vector3.Dot() for calculating the cosine between them. Angle between these planes is given by using the following formula:-Cos A = Using inverse property, we get: A = Below is the implementation of the above formulae: C++. Angle between Two Planes in a Square Pyramid 正四角錐中兩平面間的交角. [7] Stereochemical arrangements corresponding to angles between 0° and ±90° are called syn (s), those corresponding to angles between ±90° and 180° anti (a). edit close. Instructions on using the dot product (inverse cosine of the magnitudes of the normal vectors) to determine if they represent perpendicular/parallel planes or other. Dihedral angles are used to specify the molecular conformation . tanθ=±(m 2-m 1) / (1+m 1 m 2) Angle Between Two Straight Lines Derivation. Three point form: Normal form: Parametric form: where the directions (a1,b1,c1) and (a2,b2,c2)are parallel to the plane. Forums . Alex CHIK We'll check for parallel, check for perpendicular, then look at the angle between them. To find the dihedral an… However I need a quicker way. But this definition is not satisfactory because the angle will depend on the particular lines. Required fields are marked *. Angle between two planes- It is generally calculated with the knowledge of angle between their normal to two planes in the same scenario. This can be understood quite clearly from the below figure: Let \(\vec{n_{1}}\) and \(\vec{n_{2}}\) be the two normal to the planes aligned to each other at an angle θ. Topic: Planes The angle between two planes is the angle between the normal to the two planes. For cubic crystals, the angle, f between two planes, (h 1 k 1 l 1) and (h 2 k 2 l 2) is given by: . Pre-University Math Help. Angle Between Two Planes In Euclidean space, a Euclidean vector is a geometric object that possesses both a magnitude and a direction. Read this lesson on Three Dimensional Geometry to understand how the angle between two planes is calculated in Vector form and in Cartesian form. 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Trapezium: Definition, Properties and Area
This pdf includes the following topics:- Trapezium Definition Trapezium Basic Concept Properties of a Trapezium Area of a Trapezium
1. Week 1 & 2 Trapezum Area of trapezium is the region covered by a trapezium in a two- dimensional plane. It is the space enclosed in 2D geometry. A trapezium is a 2D shape which falls under category of quadrilaterals. Similar to other geometrical shapes, it also has its own properties and formulas based on area and perimeter. Let us learn in detail. Trapezium Definition A trapezium is a quadrilateral, which is defined as a shape with four sides and one set of parallel sides. Apart from trapezium, there are four more types of quadrilaterals. They are: 1 Parallelogram 2 Rectangle 3 Square 4 Rhombus All these quadrilaterals have one common property, which is, the sum of all the angles is 3600. Trapezium Basic Concept 1 The pair of parallel sides are called the bases while the non- parallel sides are called the legs of the trapezoid 2 The line segment connecting the midpoints of the non-parallel sides of a trapezoid is called the mid-segment 3 Check above the different types of trapezium figures, where arrow represents the parallel side of it. In all the three figures you can see, the two sides are parallel to each other, whereas the other two sides are non-parallel
2. 4 If we draw a line segment, between the two non-parallel sides, from the mid-point of both sides, the trapezium will be divided into two unequal parts. 5 You must have learned of isosceles triangles, where the two sides of a triangle are equal and the angle opposite the equal sides are also equal. In the same way, we have a figure, which is stated as Isosceles Trapezium, where the two non- parallel sides are equal and form equal angles at one of the bases. You can see the example of it, in the third figure given above. Properties of a Trapezium Here, we are going to learn about some more properties of the trapezium, which is also called as a trapezoid. A trapezium has the following properties: 1 Like other quadrilaterals, the sum of all the four angles of the trapezium is equal to 3600 2 A Trapezium has 4 unequal sides 3 A Trapezium has two parallel sides and two non-parallel sides 4 The diagonals of trapezium bisect each other 5 The length of the mid-segment is equal to 1/2 the sum of the parallel bases, in a trapezium 6 Two pairs of adjacent angles of a trapezium add up to 180 degrees Area of a Trapezium Trapezium area can be calculated by using the below formula: • Area = (1/2) h (AB+CD)
3. Perimeter of Trapezium The perimeter of trapezium formula is given by: • Perimeter = Sum of all the sides = AB + BC + CD + DA Derivation of Area of a Trapezium Following is the derivation for computing the area of the trapezium: The area of a trapezoid is equal to the sum of the areas of the two triangles and the area of the rectangle. We know that area of trapezoid = area of triangle 1 + area of rectangle + area of triangle 2. That means, A = (ah/2) + b1h + (ch/2) A = (ah + 2b1h + ch)/2 Simplifying the equation, rearranging the terms, and factoring result to: A = h/2[b1 + (a + b1 + h)] ….(i) If we assume the longer base of the trapezoid be b2, then b2 = a + b1 + h …..(ii) Substituting (ii) in equation (i), A = h/2(b1 + b2) Therefore the area of a trapezoid with bases b1, b2 and altitude h A = h/2(b1+b2) Applications of Trapezium The concept is a highly used concept in various physics computations and other mathematical calculations. This is the basis for obtaining the equations of motion as described in the 9th CBSE science textbook. The blend of the physics equations and mathematical calculations is very well explained to clear the level of understanding of a budding engineering mind. For the complete understanding and concept clearance from the BYJU'S one must go through the practice worksheets to check their hands on the concept and test learning. Frequently Asked Questions on Area of
4. What is the formula for Area of a trapezium? The formula to calculate area of trapezium is: Area = ½ x Sum of parallel sides x Distance between the parallel What are the formulas for Area and Perimeter of a and b. What are the properties of Trapezium? A trapezium has 4 unequal sides: two parallel and two non-parallel Sum of all interior angles is 360 degrees Diagonals bisect each other How do you find the area of Trapezium? The area of trapezium depends on its parallel sides and distance between the parallel sides. If we know the length of parallel sides and the distance between them, then we can easily find the area of trapezium. A parallelogram is a two-dimensional geometrical shape, whose sides are parallel with each other. It is made up of four sides, where the pair of parallel sides are equal in length. Also, the opposite angles of a parallelogram are equal to each other. The area of parallelogram depends on the base and height of it. In geometry, you must have learned about many 2D shapes and sizes such as circle, square, rectangle, rhombus, etc. All of these shapes have a different set of properties. Also, the area and perimeter formulas of these shapes vary with each other, used to solve many problems. Let us learn here the definition, formulas and properties of a parallelogram. Table of contents: • Definition • Formula • Properties • Types • Theorems • Examples
5. Parallelogram Definition A parallelogram is a quadrilateral with two pairs of parallel sides. The opposite sides of a parallelogram are equal in length and the opposite angles are equal in measure. In the figure above, you can see, ABCD is a parallelogram, where AB//CD and AD//BC. Also, AB = CD and AD = BC And, ∠A = ∠C & ∠B = ∠D Also, read: 5 Diagonal of a Parallelogram Formula 6 Important Questions Class 9 Maths Chapter 9 Areas Parallelograms Parallelogram Formula The formula for area and parameter of a parallelogram covered here in this section. Students can use these formulas and solve problems based on them. Area of Parallelogram Area of a parallelogram is the region occupied by it in a two- dimensional plane. Below is the formula to find the parallelogram Area = Base × Height
6. In the above figure, //gramABCD, Area is given by; Area = a b sin A = b a sin B where a is the slant length of the side of //gramABCD and b is the Check here: Area of a Parallelogram Formula Perimeter of Parallelogram The perimeter of any shape is the total distance of the covered around the shape or its total length of any shape. Similarly, the perimeter of a parallelogram is the total distance of the boundaries of the parallelogram. To calculate the perimeter value we have to know the values of its length and breadth. The parallelogram has its opposite sides equal in length. Therefore, the formula of the perimeter could be written as; Perimeter = 2 (a+b) Where a and b are the length of the equal sides of the Properties of Parallelogram If a quadrilateral has a pair of parallel opposite sides, then it's a special polygon called Parallelogram. The properties of a parallelogram are as follows: 6 The opposite sides are congruent. 7 The opposite angles are congruent. 8 The consecutive angles are supplementary. 9 If anyone of the angles is a right angle, then all the other angles will be right. 10 The two diagonals bisect each other. 11 Each diagonal bisects the parallelogram into two congruent triangles. 12 The diagonals separate it into congruent.
7. Types of Parallelogram There are mainly four types of Parallelogram depending on various factors. The factors which distinguish between all of these different types of parallelogram are angles, sides etc. 7 In a parallelogram, say PQRS If PQ = QR = RS = SP are the equal sides, then it's a rhombus. All the properties are the same for rhombus as for parallelogram. 8 Other two special types of a parallelogram are: • Rectangle • Square Is Square a Parallelogram? Square could be considered as a parallelogram since the opposite sides are parallel to each other and the diagonals of the square bisect each other. Is Rectangle a Parallelogram? Yes, a rectangle is also a parallelogram, because satisfies the conditions or meet the properties of parallelogram such as the opposite sides are parallel and diagonals intersect at 90 degrees. Parallelogram Theorems Theorem 1: Parallelograms on the same base and between the same parallel sides are equal in area. Proof: Two parallelograms ABCD and ABEF, on the same base DC and between the same parallel line AB and FC. To prove that area (ABCD) = area (ABEF). Parallelogram ABCD and rectangle ABML are on the same base and between the same parallels AB and LC. area of parallelogram ABCD = area of parallelogram ABML We know that area of a rectangle = length x breadth Therefore, area of parallelogram ABCD = AB x AL Hence, the area of a parallelogram is the product of any base of it and the corresponding altitude.
8. In ∆ADF and ∆BCE, AD=BC (∴ABCD is a parallelogram ∴ AD=BC) AF=BE (∴ABEF is a parallelogram ∴AF=BE) ∠ADF=∠BCE (Corresponding Angles) ∠AFD=∠BEC (Corresponding Angles) ∠DAF =∠CBE (Angle Sum Property) ∆ADE ≅ ∆BCF (From SAS-rule) Area(ADF) = Area(BCE) (By congruence area axiom) Area(ABCD)=Area(ABED) + Area(BCE) Hence, the area of parallelograms on the same base and between the same parallel sides is equal. Corollary: A parallelogram and a rectangle on the same base and between the same parallels are equal in area. Proof: Since a rectangle is also a parallelogram so, the result is a direct consequence of the above theorem. Theorem: The area of a parallelogram is the product of its base and the corresponding altitude. Given: In a parallelogram ABCD, AB is the base. To prove that Area(||gmABCD) = AB×AL Construction: Complete the rectangle ALMB by Drawing BM
9. perpendicular to CD. Examples of Parallelogram Example- Find the area of a parallelogram whose base is 5 cm and height is 8 cm. Solution- Given, Base = 5 cm and Height = 8 cm. We know, Area = Base x Height Area = 5 × 8 Area = 40 Sq.cm Example: Find the area of a parallelogram having length of diagonals to be 10 and 22 cm and an intersecting angle to be 65 degrees. Solution: We know that the diagonals of a parallelogram bisect each other, hence the length of half the diagonal will be 5 and 11 cm. The angle opposite to the side b comes out to be 180 – 65 = 115° We use the law of cosines to calculate the base of the parallelogram – b² = 5² + 11² – 2(11)(5)cos(115°) b² = 25 + 121 – 110(-.422) b² = 192.48 b = 13.87 cm.
10. After finding the base we need to calculate the height of the given To find the height we have to calculate the value of θ, so we use sine law 5/sin(θ) = b/sin(115) θ = 19.06 Now we extend the base and draw in the height of the figure and denote it as The right-angled triangle (marked with red line) has the Hypotenuse to be 22 cm and Perpendicular to be h. sin θ = h/22 h = 7.184 cm Area = base × height A = 13.87 × 7.184 A = 99.645 sq.cm | 677.169 | 1 |
Why Is the Sum of Angles in a Triangle Always 180 Degrees?
A triangle has three sides and three angles, and the sum of three angles is 180 degrees. This is taught to everyone in school as a standard rule, but did it ever cross your thoughts why is the sum of a triangle 180°?
Learn why with 98thPercentile and figure out how the standard rule is decoded. With 98thPercentile you can now understand the answer to all the why and how questions. Let's crack the shell open for why the sum of all three angles is 180°.
What is an Angle?
An angle is formed when two straight lines meet at a common point. The common point is called the vertex of the angle, and it is measured in degrees.
Understanding the sum of angles in a triangle is 180° with Pizza
To understand this explanation, you will require a pen, paper, scale, and protractor. But if you can imagine as you read this then it works too. Now picture this:
Draw a full pizza on paper.
Cut the pizza horizontally into half and use a protractor and measure the angle. (angle will be 180°)
Now cut the pizza vertically passing along the center which means 180° cut into half is 90°
Theorem proof:
Draw a line AB passing through the vertex X, which is parallel to the side YZ.
Two angles will be formed, mark them as d and e.
Since AB is a transversal for the parallel lines AB and YZ, we have e = z (alternate interior angles)
Similarly, d = y.
Now x, y, and z must sum to 180° (angles on a straight line). Thus, p + a + q = 180°
Since e = z and d = y. Thus, x + y + z = 180°
Therefore, the sum of the three angles x, y, and z is 180°. Hence proved.
Are you curious about more interesting theorems now? Well then you are at the right place, 98thPercentile is an award-winning online Math program for students in grades 1-12 that can take your child a grade ahead, guaranteed. Join and discover the limitless possibilities of this fascinating subject.
FAQ (Frequently Asked Questions)
Q.1. What is the sum of all interior angles in a triangle?
Ans: The sum of angles in a triangle is 180°.
Q.2. Why is the sum of interior angles 180°?
Ans: According to principles of Euclidean geometry the sum of interior angles is 180°.
Q.3. Is this rule the same for all triangles?
Ans: Yes, all the triangles have their sum of angle as 180°.
Q.4. How can verify the sum of angles in a triangle is 180°?
Ans: Using a protractor to measure these angles in the triangle will give a practical verification.
Q.5. Does a right-angle triangle also follow the exterior angle theorem | 677.169 | 1 |
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The document discusses geometry proofs, including givens and conclusions, triangle congruencies, triangle congruency shortcuts using SSS, SAS, ASA, AAS, and HL, writing two-column proofs, and applying the CPCTC principle that corresponding parts of congruent triangles are congruent. It provides examples of two-column proofs using different congruency rules and reasoning to prove that two triangles are congruent.
The document discusses various properties and theorems related to triangles. It begins by defining different types of triangles based on side lengths and angle measures. It then covers the four congruence rules for triangles: SAS, ASA, AAS, and SSS. The document proceeds to prove several theorems about relationships between sides and angles of triangles, such as opposite sides/angles of isosceles triangles being equal, larger sides having greater opposite angles, and the sum of any two angles being greater than the third side. It concludes by proving that the perpendicular from a point to a line is the shortest segment.
This PowerPoint presentation introduces different criteria for determining if two triangles are congruent: 1) If three sides of one triangle are equal to three sides of another triangle. 2) If two sides and the included angle of one triangle are equal to those of another triangle. 3) If one side and the two angles adjacent to it of one triangle are equal to those of another triangle. 4) If three angles of one triangle are equal to three angles of another triangle. The document provides examples to illustrate each of the criteria for congruent triangles.
This document discusses different methods for proving triangles congruent:
- ASA (Angle-Side-Angle) congruence can be used if two angles and the included side are congruent.
- AAS (Angle-Angle-Side) congruence can be used if two angles of one triangle are congruent to two angles of another triangle and one side included between the angles is congruent.
- HL (Hypotenuse-Leg) congruence can be used if two right triangles have one leg and the hypotenuse congruent.
Examples are provided to demonstrate applications of each method of triangle congruence.
This document provides information about congruence of triangles from a geometry textbook. It includes definitions of congruent figures and associating real numbers with lengths and angles. It describes the one-to-one correspondence test for congruence of triangles. It discusses sufficient conditions for congruence including SAS, SSS, ASA, and SAA. It presents activities and examples verifying these tests and exploring properties of isosceles and equilateral triangles. The document encourages critical thinking through "Think it Over" prompts and upgrading the chapter with additional content.
The document discusses ways to prove that a quadrilateral is a parallelogram. There are 5 methods listed: 1) showing both pairs of opposite sides are parallel, 2) showing both pairs of opposite sides are congruent, 3) showing both pairs of opposite angles are congruent, 4) showing one pair of opposite sides are parallel and congruent, 5) showing the diagonals bisect each other. The document also mentions the homework assignment is to complete problems 4 through 20 on page 526 of the textbook, which are the even numbered problems.
This document discusses the concept of CPCTC (Corresponding Parts of Congruent Triangles are Congruent). It states that once two triangles are proven to be congruent using SSS, SAS, ASA, AAS, or HL, then all corresponding parts of the triangles (sides and angles) are also congruent due to CPCTC. An example proof is provided that uses CPCTC to show two angles are congruent after establishing triangle congruence using SAS.
1. Triangles are congruent if all corresponding sides and angles are congruent. They will have the same shape and size but may be mirror images.
2. There are four main postulates and theorems used to prove triangles congruent: SSS (three sides), SAS (two sides and included angle), ASA (two angles and included side), and AAS (two angles and non-included side).
3. Corresponding parts of congruent triangles are also congruent based on the CPCTC theorem. This allows using previously proven congruent parts in future proofs.
This document provides instruction on proving triangle congruence using the Side-Side-Side (SSS) and Side-Angle-Side (SAS) postulates. It includes examples of using SSS and SAS to show that triangles are congruent, as well as practice problems for students to verify congruence. Key concepts covered are triangle rigidity, included angles, and applying SSS and SAS to real-world geometry problems.
Two geometric figures are congruent if they have the exact same shape and size. For congruent polygons, all corresponding parts including angles and sides are congruent. Corresponding parts of congruent triangles are congruent.
This document provides an introduction to congruent triangles and the different methods to prove triangles are congruent: SSS, SAS, and ASA. It includes examples of using side and angle correspondences to show triangles are congruent according to the three congruence rules. Students are asked to complete a KWL chart on triangles as an exit ticket.
The document discusses congruence and similarity of triangles and figures. It defines congruence as two figures having equal corresponding sides and angles. Similarity requires proportional corresponding sides and equal corresponding angles. The conditions for congruence of triangles are: side-side-side, side-angle-side, angle-side-angle, and angle-angle-side. Congruence is reflexive, symmetric, and transitive. The conditions for similarity of triangles are: corresponding sides proportional and two or more equal corresponding angles.
This document discusses triangle congruence using the Side-Side-Side (SSS) and Side-Angle-Side (SAS) postulates. It provides examples of using SSS and SAS to prove that two triangles are congruent. Key concepts covered include triangle rigidity, included angles, and applying SSS and SAS to solve problems and construct congruent triangles. Worked examples demonstrate using SSS and SAS, along with step-by-step proofs of triangle congruence.
- Hales, a Greek mathematician, was the first to measure the height of a pyramid using similar triangles. He showed that the ratio of the height of the pyramid to the height of the worker was the same as the ratio of the heights of their respective shadows.
- The document discusses using similar triangles to solve problems involving finding unknown lengths, such as measuring the height of a pyramid based on the shadow lengths of the pyramid and a worker of known height.
- Examples are provided of determining if two triangles are similar based on proportional sides or equal corresponding angles, and using similarities between triangles to find unknown lengths congruence postulates. It provides examples of applying each postulate to determine if pairs of triangles are congruent or not. There is no side-side-angle (SSA) or angle-angle-angle (AAA) postulate that can prove congruence.
The document provides information about triangle congruence, including:
1. There are three postulates for proving triangles are congruent: side-side-side (SSS), side-angle-side (SAS), and angle-side-angle (ASA).
2. The SSS postulate states that if three sides of one triangle are congruent to three sides of a second triangle, then the triangles are congruent.
3. The SAS postulate states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of a second triangle, then the triangles are congruent.
4. The ASA postulate states that if two angles
This document discusses several theorems for proving that a quadrilateral is a parallelogram:
1) If both pairs of opposite sides of a quadrilateral are congruent, then it is a parallelogram.
2) If both pairs of opposite angles of a quadrilateral are congruent, then it is a parallelogram.
3) If an angle of a quadrilateral is supplementary to both of its consecutive angles, then it is a parallelogram.
The document also provides proofs of these theorems and discusses using coordinate geometry to prove sides are congruent or parallel.
The document discusses properties of congruent triangles in three sentences or less:
Two triangles are congruent if (1) their corresponding sides are proportional or (2) their corresponding angles are equal in measure. Several examples demonstrate how to prove triangles are congruent by showing their corresponding sides are proportional or corresponding angles are equal. Proportionality of corresponding sides and equality of corresponding angles are used to determine missing side lengths in various triangle scenarios postulates. It provides examples of applying each postulate and determining whether given pairs of triangles can be proven congruent. The main ideas are that SSS, SAS, ASA, and AAS postulates can be used to prove triangles congruent if the corresponding parts are congruent, while SSA and AAA are not valid postulates.
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This module introduces ratio, proportion, and the Basic Proportionality Theorem. Students will learn about ratios, proportions, and how to use the fundamental law of proportions to solve problems involving triangles. The module is designed to teach students to apply the definition of proportion of segments to find unknown lengths and illustrate and verify the Basic Proportionality Theorem and its Converse. Examples are provided to demonstrate how to express ratios in simplest form, find missing values in proportions, determine if ratios form proportions, and solve problems involving angles and segments in triangles using ratios and proportionsInternational Journal of Engineering Inventions (IJEI) provides a multidisciplinary passage for researchers, managers, professionals, practitioners and students around the globe to publish high quality, peer-reviewed articles on all theoretical and empirical aspects of Engineering and Science.
The peer-reviewed International Journal of Engineering Inventions (IJEI) is started with a mission to encourage contribution to research in Science and Technology. Encourage and motivate researchers in challenging areas of Sciences and Technology.
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This module introduces ratio, proportion, and the Basic Proportionality Theorem. Students will learn about ratios, proportions, and how to use the fundamental law of proportions to solve problems involving similar triangles. The module is designed to help students apply the definition of proportion to find unknown lengths, illustrate and verify the Basic Proportionality Theorem and its converse, and develop skills for solving geometry problems involving triangles. Exercises cover writing and simplifying ratios, setting up and solving proportions, determining if ratios form proportions, and applying the Basic Proportionality TheoremA parallelogram is a quadrilateral with two pairs of parallel sides. Students were assigned geometry homework to find the values of x and y in figures and provide proof of their answers, placing their homework and pen on the corner of their desk. They were asked to define a parallelogram.
The document provides instructions to complete geometry homework problems involving regular polygons, parallelograms, and finding missing angle measures. Students are asked to find: the number of sides of two regular polygons given interior and exterior angle measures; angle measures and that parallelogram EFGH is a parallelogram; angle measures x, y, and z for two parallelograms; and to show work for problems 8 through 10
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2. The warm-up problems involve substituting values into the equation y=mx+b to solve for b and solving linear equations for y. The objectives are listed as graphing lines and writing their equations in slope-intercept and point-slope form.
3. Students are reminded that a line's y-intercept is the b value in y=mx+
This document contains examples and explanations about perpendicular lines from a geometry textbook. It includes examples of finding the shortest distance from a point to a line by drawing a perpendicular segment, proofs about perpendicular lines, and applications of perpendicular lines in carpentry and swimming. The document contains instructions for students to complete homework problems on these concepts.
1) The geometry class notes covered homework assignments, angle measurements, and proving lines are parallel using a transversal without using the alternate exterior angle theorem.
2) Students were asked to continue practicing proving lines are parallel by showing a line and angle are congruent given a transversal cuts two lines.
3) The proof must be completed in statements and reasons without using the convention of the alternate exterior angle theorem. | 677.169 | 1 |
Elements of Geometry and Trigonometry
From inside the book
Results 1-5 of 49
Page 18 ... distance between any two points is mea- sured on the straight line which joins them . 13. Through the same point , only one line can be drawn parallel to a given line . POSTULATES . 1. A straight line can be drawn between any two points ...
Page 26 ... distance from A to C , measured on any broken line AB , BC , A B is greater than the distance measured on the straight line AC ( A. 12 ) : hence , the sum of AB and BC is greater than AC ; which was to be proved . Cor . If from both ...
Page 34 ... distance will be the longer . Let A be be a given point , DE a given straight line , AB a perpendicular to DE , and AD , AC , AE oblique lines , BC being equal to BE , and BD greater than BC . Then will AB be less than any of the ...
Page 35 ... distance from a point to a line . Cor . 2. From a given point to a given straight line , only two equal straight lines can be drawn ; for , if there could be more , there would be at least two equal oblique lines on the same side of | 677.169 | 1 |
P - 08 TWISTS
Key elements:
two equal sized squares (the ones in the middle of the figure)
"In" is called when the two kites fly horizontally, one directly above
the other, approximately 20% apart, the bottom one at ^10, and both have not yet
reached <60. Timing of the turns should be such that both kites fly at the
same time vertically in the first square, and horizontally in the second. Size
of the squares should be 20%. "Out" is called when, on the last
stretch, the top kite has travelled at least 10% horizontally. | 677.169 | 1 |
Answer
We can not construct a triangle with the values given in this question.
Work Step by Step
We can use the law of sines to find the angle $A$:
$\frac{b}{sin~B} = \frac{a}{sin~A}$
$sin~A = \frac{a~sin~B}{b}$
$sin~A = \frac{(859~m)~sin~(74.3^{\circ})}{783~m}$
$sin~A = 1.056$
Since there is no angle A such that $sin~A \gt 1$, the angle A is not defined. We can not construct a triangle with the values given in this question. | 677.169 | 1 |
5 Best Ways to Check If It Is Possible to Draw a Straight Line with the Given Direction Cosines in Python
💡 Problem Formulation: In Python, we often encounter the need to determine the feasibility of drawing a straight line given a set of direction cosines. Direction cosines are the cosines of the angles made by the line with the coordinate axes. For instance, given the direction cosines (l, m, n), we want to verify their validity. The desired output is a boolean value indicating whether a line with these direction cosines can be drawn.
Method 1: Check Direction Cosines Magnitude
This method involves checking if the provided direction cosines correspond to a unit vector since the sum of the squares of the direction cosines should be 1 for a valid straight line. The function can_draw_line() takes three direction cosines as arguments and returns True if they satisfy the condition, False otherwise.
This code defines a function can_draw_line() that takes three numerical arguments representing direction cosines. It checks whether the sum of their squares is approximately 1, within a tolerance range to account for floating-point precision errors, denoted by 1e-6.
Method 2: Use NumPy for Vector Normalization
NumPy library provides tools for numerical operations, including functions that can check vector norms. This method uses NumPy's numpy.linalg.norm() to normalize the direction cosines and see if they correspond to a unit vector. This is a robust way to handle floating-point calculations.
The example code leverages NumPy's ability to compute the norm of a vector, checking if the normalized length is close to 1. This method ensures higher numerical stability and accuracy when dealing with direction cosines.
Method 3: Assertion-Based Check
Using an assertion-based check can serve as a quick test within the code, throwing an error if the condition is not met. It's an effortless way to verify our inputs on the fly.
The code illustrates an object-oriented approach where LineDirection is a class that represents a line's direction cosines. The constructor initializes the instance and validates the direction cosines simultaneously.
Bonus One-Liner Method 5: Lambda Function
The one-liner method uses a lambda function to create an inline, anonymous function to validate the direction cosines immediately where | 677.169 | 1 |
Hint: We know that a vector has both length or magnitude and direction. The angle the vector makes with the axis give the direction or the orientation of the vector on any given space. Generally, a vector is said to make $\alpha,\beta$ and $\gamma$ with $X$,$Y$ and $Z$ axes respectively.
Complete step by step answer: Given that the vector $\overrightarrow{A}$ makes angles $\alpha,\beta$ and $\gamma$with $X$,$Y$ and $Z$ axes respectively. This implies that the unit vector $\hat{a}$, also known as the directional vector is as shown in the figure. Then we can express the vector $\overrightarrow{A}$ as $\vec A=a_{x}\hat i+a_{y}\hat j+a_{z}\hat k$, and the unit vector $\hat{a}$ is given as, $\hat a=\sqrt{a_{x}^{2}+a_{y}^{2}+a_{z}^{2}}$
We can then express, the x,y,z components of the vectors with respect to the angles as the following: $cos\alpha=\dfrac{a_{x}\hat i}{\hat a}$ $cos\beta=\dfrac{a_{y}\hat j}{\hat a}$ $cos\gamma=\dfrac{a_{z}\hat k}{\hat a}$ Squaring and adding, we get, $cos^{2}\alpha+cos^{2}\beta+cos^{2}\gamma=\dfrac{a_{x}^{2}+a_{y}^{2}+a_{z}^{2}}{\left(\sqrt{a_{x}^{2}+a_{y}^{2}+a_{z}^{2}}\right)^{2}}=1$ Thus, $cos^{2}\alpha+cos^{2}\beta+cos^{2}\gamma=1$ But we need $sin^{2}\alpha+sin^{2}\beta+sin^{2}\gamma$ We know from trigonometry identities, that $sin^{2}\theta+cos^{2}\theta=1$ Then, we can write, $cos^{2}\alpha+cos^{2}\beta+cos^{2}\gamma=1-sin^{2}\alpha+1-sin^{2}\beta+1-sin^{2}\gamma=3-(sin^{2}\alpha+sin^{2}\beta+sin^{2}\gamma)$ Then,$3-(sin^{2}\alpha+sin^{2}\beta+sin^{2}\gamma)=1$ Or,$sin^{2}\alpha+sin^{2}\beta+sin^{2}\gamma=3-1=2$ Thus, $sin^{2}\alpha+sin^{2}\beta+sin^{2}\gamma=2$ Hence C.$2$ is the answer.
Additional information: We know that two vectors can be added, subtracted, and multiplied. We also know that the algebraic laws such as commutativity, associativity and distributivity are valid to certain degree and not always applicable. If a scalar and a vector is multiplied, it is said to be scalar multiplication or dot product. If a vector and another vector are multiplied, then it is said to be cross product or vector multiplication.
Note: When we say a $\vec{AB}$, we mean that $A$ is carried to $B$ in a particular direction in the space. Also magnitude denoted as $|\vec{AB}|$ gives the distance between the points $A$ and $B$. these vectors are generally represented on the coordinate systems. | 677.169 | 1 |
A hypercube
The Schlegel diagram of a hypercube, that corresponds to the drawing of a cube made of two squares, one inside the other, and four trapezia. Here we can see a bigger cube, a smaller one inside the latter and six trunks of pyramid, that form the eight cubic faces of the hypercube. | 677.169 | 1 |
Caculatecos (θ)
In a right angled triangle, Cosine of an angle is the ratio of the length of the adjacent side to the length of the hypotenuse.
Formula
To find the Cosine of a given angle θ , use the following formula.
cos(θ) = (Length of Adjacent side) / (Length of Hypotenuse)
Calculating Cos of an Angle
.
The cosine of an angle \( \theta \) is defined as:
\( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \)
where:
adjacent is the length of the side adjacent to angle \( \theta \).
hypotenuse is the length of the hypotenuse (the side opposite the right angle).
Example 1
Let's calculate the cosine cosine of 30 degrees is:
\( \cos(30^{\circ}) = \frac{\sqrt{3}}{2} \approx 0.866 \) cosine of 45 degrees is:
These examples demonstrate how to calculate the cosine of an angle manually, showing the importance of understanding the trigonometric function cosine in various mathematical and practical applications.
{
"topic": "cos",
"function": "cos",
"function_desc": "Cosine",
"x_symbol": "θ",
"x_desc": "angle",
"category": "Trigonometry",
"formula": "cos(θ) = (Length of Adjacent side) / (Length of Hypotenuse)",
"formula_in_js": "Math.cos(θ)",
"description": "In a right angled triangle, Cosine of an angle is the ratio of the length of the adjacent side to the length of the hypotenuse.",
"template": "trigonometry_function",
"content": "<h2>Calculating Cos of an Angle</h2>\n<p>.</p>\n<p>The cosine of an angle \\( \\theta \\) is defined as:</p>\n<p>\\( \\cos(\\theta) = \\frac{\\text{adjacent}}{\\text{hypotenuse}} \\)</p>\n<p>where:</p>\n<ul>\n<li><b>adjacent</b> is the length of the side adjacent to angle \\( \\theta \\).</li>\n<li><b>hypotenuse</b> is the length of the hypotenuse (the side opposite the right angle).</li>\n</ul>\n<h3>Example 1</h3>\n<p>Let's calculate the cosine cosine of 30 degrees is:</p>\n<p>\\( \\cos(30^{\\circ}) = \\frac{\\sqrt{3}}{2} \\approx 0.866 cosine of 45 degrees is:</p>\n<p>\\( \\cos(45^{\\circ}) = \\frac{1}{\\sqrt{2}} = \\frac{\\sqrt{2}}{2} \\approx 0.707 \\)</p>\n<p>These examples demonstrate how to calculate the cosine of an angle manually, showing the importance of understanding the trigonometric function cosine in various mathematical and practical applications.</p>"
} | 677.169 | 1 |
What is the difference between convex and concave?
What is the difference betweenIs concave regular or irregular?
A Concave polygon is a polygon that has at least one interior angle greater than 180 degrees. It must have at least four sides. The shape of the concave polygon is usually irregular….Is an arrow a polygon?
MATHS Related Links
Table Of 25
Logarithmic Functions
Exterior Angles Of A Regular Polygon
Geometry Symbols
How do you check if a polygon is convex or not?
To see if a polygon is convex, calculate the angles at each of the polygon's corners. If all of the angles have the same sign (either positive or negative depending on the orientation), then the polygon is convex.
Is a concave polygon irregular?
A Concave polygon is a polygon that has at least one interior angle greater than 180 degrees. It must have at least four sides. The shape of the concave polygon is usually irregular.
What is the difference between concave and concave?
Concave describes shapes that curve inward, like an hourglass. Convex describes shapes that curve outward, like a football (or a rugby ball).
What is the difference between concave and convex geometry?
In geometry, concave and convex refer to polygons. A concave polygon has at least one angle greater than 180 degrees. A convex polygon is made of angles each less than or equal to 180 degrees.
What are the examples of convex polygon?
Convex polyhedra where every face is the same kind of regular polygon may be found among three families: Triangles: These polyhedra are called deltahedra. There are eight convex deltahedra: three of the Platonic solids and five non-uniform examples. Squares: The cube is the only convex example. Other examples (the polycubes) can be obtained by joining cubes together, although care must be taken if coplanar faces are to be Pentagons: The regular dodecahedron is the only convex example.
What is the difference between convex and concave polyhedron?
Difference between Concave and Convex Polygons Key Difference: A polygon whose all interior angles are less than 180 degrees is known as a convex polygon . On the other hand, a polygon with one or more interior angles greater than 180 degrees is referred to as a concave polygon.
Is a convex polygon always a regular polygon?
All regular simple polygons(a simple polygon is one that does not intersect itself anywhere) are convex. Those having the same number of sides are also similar. An n-sided convex regular polygon is denoted by its Schläfli symbol{n}. | 677.169 | 1 |
Word Problems Using Right Triangles
This video shows how to solve the following word problem involving right triangles: A ramp with a 30° angle from the ground is to be built up to a 2 ft. high platform. How far from the platform will it extend, to the nearest inch?
Right triangle word problem. A ramp with a 30 degree angle from the ground is to be built up to a 2 foot high platform. What will be the length of the ramp? How far from the platform will it extend to the nearest inch? With problems like these, it's very helpful to draw a picture.
We have our ramp that makes a 30 degree angle with the ground. Let's start with that. Here's the ground. Here's are 30 degree angle. It's to be built up to a 2 foot high platform. Then here we have a platform, and it's 2 feet high above the ground. There's a right angle there perpendicular to the ground.
The first question is; what will be the length of the ramp? This is what we're trying to find. Well before I start to find X I first want to fill in what this angle is. You might already know that since we have a 30 degree angle and a 90 degree angle that this angle has to be 60 degrees.
But, if you didn't know that you could find it, because there are 180 degrees in a triangle. If you add 30 to 90 you get 120. 180 minus 120 leaves you with 60 degrees for that third angle. Now that we know it's a 30 60 90 triangle, we can apply our 30 60 90 rules to finding the length of our ramp.
Our ramp is across from the 90 degree angle, therefore that's our hypotenuse. The side we know, the leg we know is across from the 30 degree angle, so it's our shorter leg. The hypotenuse of a 30 60 90 triangle is twice as long as the shorter leg.
Our hypotenuse again, we're using X for, equals 2 times the shorter leg which is that 2 foot side. 2 times 2 feet. Which means X is 4 feet. Therefore, the length of the ramp is 4 feet. We've answered the first question. Then they asked; how far from the platform will it extend, how far from the platform will the ramp extend?
Now they're asking us to find that length. Y is across from our 60 degree angle so that's our longer leg. The longer leg of a 30 60 90 triangle is the square root of three times the shorter leg. Our longer leg again is Y equals the square root of three times, our shorter leg is still that two foot length.
Y is 2 square roots of 3 feet. They didn't ask us for the answer in feet they asked to the nearest inch. The first thing I'm going to do is convert 2 square roots of 3 feet into inches by multiplying by 12. Y is 2 square roots of 3 times 12, since there are 12 inches in one foot, which is 24 square roots of 3 inches.
They didn't ask for it in simplest radical form, so then we'd actually want to calculate what 24 square root three inches is, and it's about 42 inches. Therefore, how far from the platform will the ramp extend? The ramp will extend about 42 inches from the platform. There we have it.
Frequently Asked Questions
Q
What is a right triangle?
A
A right triangle is a triangle that has a 90° (right) angle. Ex.
Q
How do you find the angle of a right triangle?
A
Find the angle of a right triangle given two side lengths by using these Find ϴ for this triangle: \(sinθ=\frac{6}{12}\) \(sinθ=\frac{1}{2}\) \(θ=sin^{-1}(\frac{1}{2})\) \(θ=30°\) sides of a right triangle?
A
Find the sides of a triangle using the following Solve for x: \(sin60°=\frac{x}{24}\) \(x=24 sin60°≈20.78\)
Q
How do you find the hypotenuse of a right triangle?
A
Find the hypotenuse (c) of a right triangle given the other two side lengths (a and b) by using the Pythagorean Theorem: \(a^2+b^2=c^2\) Ex. What is the hypotenuse of this right triangle? \((4)^2+(3)^2=c^2\) \(16+9=c^2\) \(25=c^2\) \(5=c\)
Q
How do you find the missing side of a right triangle?
A
Find the missing side of a right triangle by using the Pythagorean Theorem: c^2=a^2+b^2, where c is the hypotenuse and a and b are legs of the triangle. Ex. Find the missing side length. \(c^2=a^2+b^2\) \(5^2=3^2+b^2\) \(25=9+b^2\) \(16=b^2\) \(4=b\) | 677.169 | 1 |
Trigonometry
Trigonometry - the math that deals with the side lengths and angles of triangles, plays an important role in many games. The trigonometric functions Sine, Cosine, and Tangent relate to the ratios of sides in a right triangle:
MathF.Atan2(float x, float y) computes the angle with produces the supplied x/y ratio. This form can be helpful to avoid a division by 0 error if y is 0.
These angles are measured in radians - fractions of
$ \pi $. Positive angles rotate counter-clockwise and negative ones clockwise. It can be helpful to consider radians in relation to the unit circle - a circle with radius 1 centered on the origin:
The angle of
$ 0 $ radians falls along the x-axis. MonoGame provides some helpful float constants for common measurements in radians:
Inside the unit circle you can inscribe a right triangle with angle at the origin of
$ \theta $. This triangle has a hypotenuse with length 1, so
$ \sin{\theta} $ is the length of the opposite leg of the triangle, and
$ \cos{\theta} $ is the length of the adjacent leg of the triangle. Of course
$ \tan{\theta} $ will always equal
$ 1 $. | 677.169 | 1 |
Which of These Triangle Pairs Can Be Mapped to Each Other Using a Reflection and a Translation?
Geometry provides us with a fascinating world of shapes, lines, and transformations. When it comes to triangles, various transformations can be applied to alter their positions and orientations. One intriguing question that arises is: which pairs of triangles can be mapped to each other using a reflection and a translation? In this article, we will delve into the concept of reflections, translations, and their effects on triangles, exploring the possibilities of mapping triangle pairs through these transformations.
Understanding Reflections and Translations:
Before diving into the mapping possibilities, let's briefly discuss the two key transformations involved: reflections and translations.
Reflections: A reflection is a transformation that produces a mirror image of a shape across a line called the line of reflection. Each point of the original shape is paired with its corresponding point on the reflected shape, equidistant from the line of reflection. Reflections preserve shape size, but they may alter its orientation.
Translations: A translation is a transformation that moves a shape without changing its orientation or size. It involves shifting the shape in a specific direction, typically indicated by vectors or coordinates. The distance and direction of the shift determine the translation.
Mapping Triangle Pairs with Reflections and Translations:
To determine which pairs of triangles can be mapped to each other using a reflection and a translation, we need to consider the properties of the triangles and the transformations involved. Here are several possible scenarios:
Congruent Triangles:
Congruent triangles have the same shape and size. In this case, any pair of congruent triangles can be mapped to each other using a reflection and a translation. The reflection will create the mirror image, and the translation will move the reflected triangle to the desired position.
Reflections with Parallel or Perpendicular Lines:
When two triangles share a common side and the lines of reflection for each triangle are parallel or perpendicular, they can be mapped to each other using a reflection and a translation. The reflection across the common side will generate the mirror image, and the translation will move the reflected triangle to the desired location.
Similar Triangles:
Similar triangles have proportional corresponding sides and congruent corresponding angles. Mapping similar triangles using a reflection and a translation can be more challenging. While the reflection can create a mirror image, the translation may not align the triangles perfectly due to differences in size and proportion.
Mapping Limitations:
Not all pairs of triangles can be mapped to each other using a reflection and a translation. If the triangles have different shapes, sizes, or orientations, it may not be possible to achieve the desired mapping using these transformations alone. In such cases, additional transformations like rotations or dilations may be required.
Exploring Examples:
Let's consider a practical example to illustrate the concept. Suppose we have two congruent triangles, ABC and DEF, and we want to map triangle ABC to triangle DEF using a reflection and a translation.
Reflecting Triangle ABC: Choose a line of reflection that intersects the common side of the triangles (for example, line AD). Reflect triangle ABC across line AD to obtain its mirror image.
Translating the Reflected Triangle: Apply a translation to move the reflected triangle to the desired position, aligning it with triangle DEF. Determine the direction and distance of the translation to achieve the desired mapping.
Conclusion:
Mapping pairs of triangles using reflections and translations offers an intriguing exploration of geometric transformations. Congruent triangles and triangles with parallel or perpendicular lines of reflection can be successfully mapped using these transformations. However, it is important to consider the properties of the triangles involved, such as similarity, orientation, and proportionality, as these factors can impact the feasibility of mapping.
Remember that reflections create mirror images, while translations shift shapes without changing their orientation or size. For cases where the triangles have different shapes, sizes, or orientations, additional transformations like rotations or dilations may be necessary to achieve the desired mapping.
By delving into the world of transformations and their effects on triangles, we can enhance our understanding of geometric relationships and explore the fascinating interplay between shapes and transformations in the realm of mathematics. | 677.169 | 1 |
3455. 8 in3 What is the image of Y 4 7 under the translation x y x 3 y 5 A. Round to the nearest degree. B. 60 C. 34 the volume of the cube. A. 16 ft3 B. 64 ft3 C. 256 ft3 D. 4 ft3 Find the reflection of the point A 6 1 across the line y x. Find the scale factor she used. Find the volume to the nearest tenth. 4825. 5 m3 14 476. 5 m3 3619. Given a tessellation find the sum of the measures of the angles of the polygons at a vertex. A. 360 B. B. x 2 2 y 3 2 36 of the sphere. Round to the nearest…
Quick guide on how to complete geometry final exam 2022
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CBSE 9th Maths
Coordinate Geometry
Cartesian Plane: A cartesian plane is defined by two perpendicular number lines, A horizontal line(x−axis) and a vertical line (y−axis).
Quadrants
The cartesian plane is divided into four equal parts, called quadrants. These are named in the order as I,II,III and IV starting with the upper right and going around in anticlockwise direction.
Points in different Quadrants.
Signs of coordinates of points in different quadrants:
I Quadrant: '+' x – coordinate and '+' y – coordinate.
II Quadrant: '-' x – coordinate and '+' y – coordinate.
III Quadrant: '-' x – coordinate and '-' y – coordinate.
IV Quadrant: '+' x – coordinate and '-' y – coordinate.
The coordinates of origin O are (0, 0) because it has zero distance from both the axes, so its abscissa and ordinate both are zero.
The coordinates of a point are written by using the following conventions
(i) The x-coordinate of a point is its perpendicular distance from the Y-axis measured along the X-axis (positive along the positive direction of the X-axis and negative along the negative direction of the X-axis). The x-coordinate is also called the abscissa.
(ii) The Y-coordinates of a point is its perpendicular distance from the X-axis measured along the Y-axis (positive along the positive direction of the Y-axis and negative along the negative direction of the Y-axis). The Y-coordinate is also called the ordinate.
(iii) In stating the coordinates of a point in the coordinate plane, the x-coordinate comes first and then the Y-coordinate. We place the coordinates in brackets. | 677.169 | 1 |
the diameter of the circle is 36, what is the length of arc ABC?
[#permalink]
17 Oct 2018, 08:24Re: If the diameter of the circle is 36, what is the length of arc ABC?
[#permalink]
18 Oct 2018, 04:20
1
Kudos
Expert Reply
VodkaHelps wrote:Length fo Arc ABC can be understood in only one way like it has been interpreted here.
Always identify the arc by reading the points in that particular order.
Arc AC could have caused two interpretations which you intend to bring here but when we say are ABC then it's only the major arc that is being referred.
Re: If the diameter of the circle is 36, what is the length of arc ABC?
[#permalink]
12 Aug 2020, 14:26
1
Bookmarks
Expert Reply
Bunuel wrote:
If the diameter of the circle is 36, what is the length of arc ABC?
(A) \(8\)
(B) \(8\pi\)
(C) \(28\pi\)
(D) \(32\pi\)
(E) \(56\pi\)
Solution:
We see that the circumference of the circle is C = 2πr = 2π(18) = 36π. Since angle ABC is an inscribed angle of minor arc AC, arc AC is twice the measure of angle ABC, or 80 degrees. Thus, major arc ABC is 360 - 80 = 280 degrees, and we can create the equation where x is the length of arc ABC:
Re: If the diameter of the circle is 36, what is the length of arc ABC?
[#permalink]
20 May 2024, 21 diameter of the circle is 36, what is the length of arc ABC? [#permalink] | 677.169 | 1 |
Catch-Up and Review
Moving a Triangle
In the following applet, arrow v dictates how △ABC will move to get △A′B′C′. Try changing the direction and length of the arrow to see different types of movements.
Notice how △ABC and △A′B′C′ are congruent and how the orientation of △A′B′C′ matches the orientation of △ABC. Furthermore, the segmentsAA′,BB′, and CC′ are all parallel to arrow v, and all four segments have the same length.
Discussion
Transformations of Geometric Objects
A transformation is a function that changes a figure in a particular way — it can change the position, size, or orientation of a figure. The original figure is called the preimage and the figure produced is called the image of the transformation. A prime symbol is often added to the label of a transformed point to denote that it is an image.
Discussion
Translations of Geometric Objects
A translation is a transformation that moves every point of a figure the same distance in the same direction. To find the coordinates of a geometric object after a translation, a value a is added to the x-coordinate of every point of the preimage and a value b is added to the y-coordinate of every point of the preimage.
(x,y)→(x+a,y+b)
Positive a values indicate a translation to the right and positive b values correspond to a translation up.
Conversely, negative a values indicate a translation to the left and negative b values correspond to a translation down.
Translations preserve the side lengths and angle measures of the involved geometric objects.
Example
Placing a Pool in the Backyard
Zain's parents want to renovate their backyard. The lot is pretty wide and, once the weeds are taken care of, it could be used for plenty of things. The first thing that comes to Zain's mind is to put in a pool, so they make a sketch of the backyard to show their idea to their parents. They use the letter P to represent the pool.
Zain's dad says that the pool should not be right against the gate to the backyard, so he suggests two possible alternatives, P1 and P2.
Hint
b Focus on any corner of the pool. Count how many squares it is moved vertically and horizontally.
Solution
a Consider the pool location suggestions made by Zain's dad.
A pure translation takes a preimage P and makes it into an image that is congruent and has the same orientation as P. In the sketch, both P1 and P2 are congruent to P, so all that remains is to determine which image has the same orientation as P. Notice that the longest side of P is horizontal in the sketch.
This preimage will now be compared to images P1 and P2. The longest side of image P1 is vertical in the sketch, so it does not have the same orientation as P. This means that P1 is not a pure translation of P. On the other hand, P2 matches P exactly.
Since P2 is congruent to P and has the same orientation, it is a pure translation of P.
b Given the coordinates(x,y) of each point in a preimage, a translation of the point can be written in the following form.
(x,y)→(x+a,y+b)
In this case, it is not necessary to know the coordinates (x,y), only the values a and b that are relevant to the translation. It was found in Part A that P2 is the image of Zain's pool after a translation, so the focus will be in this image.
The a value of the translation corresponds to the horizontal displacement of the translation, while the b value is the vertical displacement. Since every point of the preimage is translated in the same way, any particular corner of the pool can be used along with the grid to find a and b. Just keep in mind that they have to be matching corners!
The horizontal displacement is 4 units to the right, so a=4. Likewise, the vertical displacement is 7 units up, so b=7.
Pop Quiz
Identifying Translations
In the following applet, △ABC is translated to map onto △A′B′C′. Write the a and b values of the translation.
Example
Replacing the Planter
The renovations are now taking place, so there is a lot going on in the backyard. During the morning, Zain's mom goes out to water some plants that were on a planter pot when she noticed that they are in the wrong place.
The planter was probably in the way when the builders were working on the pool. However, the plants need to be in the correct spot to get just the right amount of sunlight, so she kindly requests the workers to put them back where they belong when they are finished for the day. To find the original position of the pots, Zain placed a coordinate plane on the backyard plans.
Zain's mom asks the builders to translate the pot 7 units to the left and 1 unit up. Find the coordinates of the pot after the translation.
Hint
Find the initial coordinates of the pot. To translate it 7 units to the left and 1 unit up, subtract 7 from the x-coordinate and add 1 to the y-coordinate.
Solution
Zain's mom wants the builders to return the planter pot back to where it was, which they can do by translating it 7 units to the left and 1 unit up. Begin by finding the current coordinates of the pot.
The horizontal part of the translation is 7 units to the left. This means to subtract7 from the current x-coordinate of the pot.
4−7=-3
On the other hand, the vertical part of the translation is 1 unit up, so add1 to the current y-coordinate of the pot.
-2+1=-1
The x-coordinate of the pot after the translation is -3 and its y-coordinate is -1.
(4−7,-2+1)=(-3,-1)
The following applet can be used to visualize the translation.
Explore
Reflecting a Triangle
In the following applet, the vertices of △ABC can be moved. The slider bar adjusts the slope of the lineℓ. Once everything is set, reflect△ABC across line ℓ.
Is there any relationship between AA′ and ℓ? If so, do BB′ and ℓ have the same relationship? What about CC′ and ℓ?
Discussion
Reflections of Geometric Objects
A reflection is a transformation in which every point of a figure is reflected across a line. The line across the points are reflected is called the line of reflection and acts like a mirror.
More precisely, a reflection across a line ℓ maps every point A in the plane onto its imageA′ such that one of the following statements is satisfied.
It seems that there is some sort of reflection between the tiles, but Zain is not completely sure. If there was a reflection, it would be possible to find a line of reflection. Help Zain find the line of reflection.
Solution
When a point is reflected across a line, its image is such that the line of reflection is the perpendicular bisector of the segment that connects the point to its image. To find the line of reflection of the pool tiles, start by drawing a segment that connects a vertex and its image. For instance, draw CC′.
Drawing Lines of Reflection
Drag the points in the following applet to draw the line of reflection used to map △ABC onto △A′B′C′. To do so, place the two points so they lie on the line of reflection. The measuring tool can be used to find the midpoint between corresponding vertices of the image and preimage, and they can also be used to find right angles.
Extra
How to Use the Measuring Tool
The measuring tool is useful for finding the midpoint between a point and its image. Place the ends of the measuring tool on a point and its corresponding image.
Next, move the middle point so that the angle of the measuring tool is 180∘ and the length of both segments is equal. This will make sure the middle point lies on the line of reflection. | 677.169 | 1 |
A graph consists of objects called vertices and connections between them called edges. For every vertex, we can count how many neighbors it has, which is called its degree | 677.169 | 1 |
Polyhedra in which all faces are equilateral triangles are called deltahedra. The regular tetrahedron, octahedron, and icosahedron are the simplest deltahedra. It also is possible to replace each face of a regular dodecahedron with a "dimple" having five equilateral triangles as sides. This is a model of such a surface. It also may be considered as one of the polyhedra formed by extending the sides of—or stellating—a regular icosahedron.
This deltahedron is folded from paper and held together entirely by hinged folds along the edges. A mark reads: 12-21-26 (/)A. | 677.169 | 1 |
Diameter
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Diameter
Diameter
In geometry, a diameter of a circle is any straight line segment that passes through the center of the circle and whose endpoints lie on the circle. It can also be defined as the longest chord of the circle. Both definitions are also valid for the diameter of a sphere.
In more modern usage, the length of a diameter is also called the diameter. In this sense one speaks of the diameter rather than a diameter (which refers to the line itself), because all diameters of a circle or sphere have the same length, this being twice the radius r.
For a convex shape in the plane, the diameter is defined to be the largest distance that can be formed between two opposite parallel lines tangent to its boundary, and the width is often defined to be the smallest such distance. Both quantities can be calculated efficiently using rotating calipers.[1] For a curve of constant width such as the Reuleaux triangle, the width and diameter are the same because all such pairs of parallel tangent lines have the same distance.
For an ellipse, the standard terminology is different. A diameter of an ellipse is any chord passing through the center of the ellipse.[2] For example, conjugate diameters have the property that a tangent line to the ellipse at the endpoint of one of them is parallel to the other one. The longest diameter is called the major axis.
The word "diameter" is derived from Greek διάμετρος (diametros), "diameter of a circle", from διά (dia), "across, through" and μέτρον (metron), "measure".[3] It is often abbreviated DIA, dia, d, or ⌀.
Generalizations
The definitions given above are only valid for circles, spheres and convex shapes. However, they are special cases of a more general definition that is valid for any kind of n-dimensional convex or non-convex object, such as a hypercube or a set of scattered points. The diameter of a subset of a metric space is the least upper bound of the set of all distances between pairs of points in the subset. So, if A is the subset, the diameter is
If the distance function d is viewed here as having codomain R (the set of all real numbers), this implies that the diameter of the empty set (the case A = ∅) equals −∞ (negative infinity). Some authors prefer to treat the empty set as a special case, assigning it a diameter equal to 0,[4] which corresponds to taking the codomain of d to be the set of nonnegative reals.
For any solid object or set of scattered points in n-dimensional Euclidean space, the diameter of the object or set is the same as the diameter of its convex hull. In medical parlance concerning a lesion or in geology concerning a rock, the diameter of an object is the supremum of the set of all distances between pairs of points in the object.
In differential geometry, the diameter is an important global Riemannian invariant.
In plane geometry, a diameter of a conic section is typically defined as any chord which passes through the conic's centre; such diameters are not necessarily of uniform length, except in the case of the circle, which has eccentricity e = 0.
Symbol
The symbol or variable for diameter, ⌀, is similar in size and design to ø, the Latin small letter o with stroke. In Unicode it is defined as U+2300 ⌀ DIAMETER SIGN (HTML ⌀). On an Apple Macintosh, the diameter symbol can be entered via the character palette (this is opened by pressing ⌥ Opt⌘ CmdT in most applications), where it can be found in the Technical Symbols category.
The character will sometimes not display correctly, however, since many fonts do not include it. In many situations the letter ø (the Latin small letter o with stroke) is an acceptable substitute, which in Unicode is U+00F8 ø (HTML ø · ø). It can be obtained in UNIX-like operating systems using a Compose key by pressing, in sequence, Composedi[5] and on a Macintosh by pressing ⌥ Opt O (the letter o, not the number 0).
In Microsoft Word the diameter symbol can be acquired by typing 2300 and then pressing Alt+X.
In LaTeX the diameter symbol can be obtained with the command \diameter from the wasysym package.
The diameter symbol ⌀ is distinct from the empty set symbol ∅, from an (italic) uppercase phi Φ, and from the Nordic vowel Ø.[6] See also slashed zero.
In German, the diameter symbol (German Durchmesserzeichen) is also used as an average symbol (Durchschnittszeichen).
See also
Angular diameter
Caliper, micrometer, tools for measuring diameters
Conjugate diameters
Diameter (group theory), a concept in group theory
Eratosthenes, who calculated the diameter of the Earth around 240 BC.
Graph or network diameter
Hydraulic diameter
Inside diameter
Jung's theorem, an inequality relating the diameter to the radius of the smallest enclosing ball | 677.169 | 1 |
isosceles right triangle worksheet
Isosceles Right Triangle Worksheet – Triangles are one of the most fundamental forms in geometry. Understanding the triangle is essential to learning more advanced geometric concepts. In this blog we will look at the different kinds of triangles Triangle angles, how to determine the dimension and perimeter of the triangle, as well as provide an example of every. Types of Triangles There are three types from triangles: Equal isosceles, and scalene. Equilateral triangles contain three equal sides and … Read more | 677.169 | 1 |
In the above figure, ACDF, ABEF and BCDE are all parallelograms. Analyse the given figure and chosse the correct option.
A
Opposite sides are equal.
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B
Opposite angles are equal.
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C
Opposite angles are bisected by the diagonals.
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D
Diagonals bisect each other.
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Open in App
Solution
The correct options are B
Opposite angles are equal.
C
Opposite angles are bisected by the diagonals.
Since ABEF is a parallelogram, soAF∥BEand BF is the transversal. ∴∠AFB=∠FBE=x(alternate angles)..............(1) Also, since BCED is a parallelogram,BE || CD and BD is the transversal∴∠CDB=∠EDB=y(alternate angles) .............(2) Now, ∠FBD=∠FBE+∠EBD =x+y | 677.169 | 1 |
Lesson
Lesson 13
Lesson Purpose
The purpose of this lesson is for students to find unknown angle measurements by composing or decomposing known measurements, and to see that an angle is not determined by the length of the segments that form it.
Lesson Narrative
In this lesson, students use tactile tools to find angle measurements and observe more clearly that angles are additive. They compose and decompose angles by arranging paper cutouts, by folding paper or tracing, and by drawing diagrams. Students arrange smaller angles whose sizes are unknown into larger angles with familiar sizes and features (\(90^\circ\), \(180^\circ\), and \(360^\circ\)). Once the measurement of an angle is known, they use it to find those of other angles. For example, if two copies of angle \(x\) form a right angle, angle \(x\) must be \(45^\circ\). If another angle, \(z\), can be decomposed into three of these \(45^\circ\) angles, then \(z\) must be \(135^\circ\).
Encourage students to continue to collect, define, and illustrate new terms to support communication and reasoning at the end of each lesson.The work of finding angle measurements in this lesson offered opportunities to reason about equal groups. Did you hear students use this type of reasoning? What were some other ways students reasoned about the angle sizes | 677.169 | 1 |
49
Page 12 ... proportions , we will also refer the student to the method explained in the Algebra . There is this difference between geometrical ratios of magni- tudes , and ratios of numbers : All numbers are commensurable ; that is , their ratio ...
Page 48 ... proportions , since their antecedents are the same , we deduce angle AEG : angle AEK :: arc AH : arc AK , which cannot be , for the antecedent of the first couplet is less . than its consequent , while in the second couplet the ...
Page 72 ... far as we please , we shall never arrive at a term in which there will be no remainder . Therefore there is no common measure of the diagonal and side of a square . THIRD BOOK . THE PROPORTIONS OF STRAIGHT LINES AND THE 72 GEOMETRY .
Page 73 ... PROPORTIONS OF STRAIGHT LINES AND THE AREAS OF RECTILINEAL FIGURES . DEFINITIONS . I. Two figures are similar when they resemble each other in all their parts , differing only in their comparative magnitudes , so that every point of the ...
Page 76 ... proportion , the product of the extremes is equal to the product of the means , we derive from the above proportion , AD : DB :: AE : EC , this equation , ADXEC = DB × AE . Now , in order to comprehend the sense which we 76 GEOMETRY .
Popular passages
Page 80 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. A D A' Hyp. In triangles ABC and A'B'C', To prove AABC A A'B'C' A'B' x A'C ' Proof. Draw the altitudes BD and B'D'.
Page 28 - If two triangles have two sides of the one equal respectively to two sides of the other, but the included angle of the first greater than the included angle of the second, then the third side of the first is greater than the third side of the second. Given A ABC and A'B'C ' with Proof STATEMENTS Apply A A'B'C ' to A ABC so that A'B | 677.169 | 1 |
What is the general form of naming a polygon?
Polygon
Hexagon
n-gon (where n represents the number of sides in a polygon)
Pentagon
Hint:
General interpretation of polygon.
The correct answer is: n-gon (where n represents the number of sides in a polygon)
*In mathematics, a polygon is a closed two-dimensional figure made up of line segments. * Poly means many gon means angle. Hence, polygon is a closed figure that is made of many angles. *In General there are 4 types of polygons. They are: Regular Polygon Irregular Polygon Convex Polygon Concave polygon * Hence, the polygon can have any number of sides and angles. >>>Naming of polygon is required to differentiate the polygons. >>>Generally, the polygon is named as n-gon where n represents the number of sides. * Example:
pent(5) + gon. =pentagon.
One can find this with a small trick. The student should learn to break the word. For example, Hexagon: Hexa+Gon. Now, as we know that hexa is six. Similarly, if a polygon has 'n' sides, then the polygon would be named as n-gon | 677.169 | 1 |
Pythagoras theorem wikipedia
Consider the triangle shown below, pythagoras theorem wikipedia. This figure is clearly a squaresince all the angles are right anglesand the lines connecting the corners are easily seen to be straight. Now to calculate the area of this figure. On the one hand, we can add up the area of the component parts of the square.
Pythagoras theorem wikipedia
In geometry , the inverse Pythagorean theorem also known as the reciprocal Pythagorean theorem [1] or the upside down Pythagorean theorem [2] is as follows: [3]. This theorem should not be confused with proposition 48 in book 1 of Euclid 's Elements , the converse of the Pythagorean theorem, which states that if the square on one side of a triangle is equal to the sum of the squares on the other two sides then the other two sides contain a right angle. Using the Pythagorean theorem ,. The cruciform curve or cross curve is a quartic plane curve given by the equation. Substituting x with AC and y with BC gives. Inverse-Pythagorean triples can be generated using integer parameters t and u as follows. If two identical lamps are placed at A and B , the theorem and the inverse-square law imply that the light intensity at C is the same as when a single lamp is placed at D. This geometry-related article is a stub. You can help Wikipedia by expanding it. Contents move to sidebar hide. Article Talk.
In other projects The proofs are diverse, including both geometric proofs and algebraic proofs, with some dating back thousands of years. When Euclidean space is represented by a Cartesian coordinate system in analytic geometry , Euclidean distance satisfies the Pythagorean relation: the squared distance between two points equals the sum of squares of the difference in each coordinate between the points.
Pythagoras theorem wikipedia
In mathematics , a theorem is a statement that has been proved , or can be proved. In mainstream mathematics, the axioms and the inference rules are commonly left implicit, and, in this case, they are almost always those of Zermelo—Fraenkel set theory with the axiom of choice ZFC , or of a less powerful theory, such as Peano arithmetic. Moreover, many authors qualify as theorems only the most important results, and use the terms lemma , proposition and corollary for less important theorems. In mathematical logic , the concepts of theorems and proofs have been formalized in order to allow mathematical reasoning about them.
Wicker footrest
JSTOR A primitive Pythagorean triple is one in which a , b and c are coprime the greatest common divisor of a , b and c is 1. The Pythagorean theorem was known and used by the Babylonians and Indians centuries before Pythagoras, [] [] [] [] but he may have been the first to introduce it to the Greeks. Synedrion Koinon. In mathematics , the Pythagorean theorem or Pythagoras' theorem is a fundamental relation in Euclidean geometry between the three sides of a right triangle. The emperor Hadrian 's Pantheon in Rome was also built based on Pythagorean numerology. Works ascribed to Pythagoras included the "Golden Verses" and snippets of his scientific and mathematical theories. Download as PDF Printable version. Another by rearrangement is given by the middle animation. Bonus : Can you calculate without using the cosine rule? Herodotus , [34] Isocrates , and other early writers agree that Pythagoras was the son of Mnesarchus, [21] [35] and that he was born on the Greek island of Samos in the eastern Aegean. The dissection consists of dropping a perpendicular from the vertex of the right angle of the triangle to the hypotenuse, thus splitting the whole triangle into two parts.
Garfield's proof of the Pythagorean theorem is an original proof the Pythagorean theorem invented by James A. Garfield November 19, — September 19, , the 20th president of the United States.
Thus, if similar figures with areas A , B and C are erected on sides with corresponding lengths a , b and c then:. Alphanumeric Cosmology From Greek into Arabic. Pythagorean triple Reciprocal Pythagorean theorem Complex number Euclidean distance Pythagorean trigonometric identity. Geometrically r is the distance of the z from zero or the origin O in the complex plane. In this picture, the area of the blue square added to the area of the red square makes the area of the purple square. List of ancient Greeks. The following statements apply: [29]. Therefore, the ratios of their sides must be the same, that is:. The area of the trapezoid can be calculated to be half the area of the square, that is. Despite generating all primitive triples, Euclid's formula does not produce all triples—for example, 9, 12, 15 cannot be generated using integer m and n. He may have also devised the doctrine of musica universalis , which holds that the planets move according to mathematical equations and thus resonate to produce an inaudible symphony of music. Certain methods for the discovery of triangles of this kind are handed down, one which they refer to Plato, and another to Pythagoras. | 677.169 | 1 |
interested in learning more about vectors, this article is for you.
What is a Degree Between Two Vectors Calculator?
Before we dive into the specifics of how to use the degree between two vectors calculator, let's first understand what it is. Simply put, the calculator helps you determine the angle between two vectors in a 2D or 3D space. This can be extremely useful in various fields such as mathematics, physics, engineering, and computer graphics.
The concept of vectors and angles between them may seem daunting at first, but with the help of this calculator, it becomes a breeze. So let's take a closer look at how to use this tool and some tips and tricks for getting the most out of it.
How to Use the Degree Between Two Vectors Calculator
Using the degree between two vectors calculator is as simple as plugging in the values and pressing a button. Here are the steps to follow:
Step 1: Input the Values
The first step is to input the values of the two vectors. These values can be in the form of coordinates or components. For example, if you have two vectors A = (3, 4) and B = (6, -8), you can input the x and y components of each vector as follows:
Vector A: x = 3, y = 4
Vector B: x = 6, y = -8
Step 2: Select the Dimension
Next, you need to select the dimension of your vectors. This could be 2D or 3D depending on the space you are working with.
Step 3: Calculate the Angle
After inputting the values and selecting the dimension, press the calculate button to get the angle between the two vectors. The result will be given in degrees.
Examples of Degree Between Two Vectors Calculator
To better understand how to use the degree between two vectors calculator, let's look at some examples:
Example 1: Find the Angle Between Two Vectors in 2D Space
Let's say we have two vectors A = (3, 4) and B = (6, -8). To find the angle between these two vectors, we first input the values as follows:
Vector A: x = 3, y = 4
Vector B: x = 6, y = -8
Next, we select the dimension as 2D and press calculate. The result is an angle of approximately 122.7 degrees.
Example 2: Find the Angle Between Two Vectors in 3D Space
Let's take another example, but this time in a 3D space. Consider two vectors A = (3, 4, 5) and B = (-2, 7, 9). To find the angle between these two vectors, we first input the values as follows:
Vector A: x = 3, y = 4, z = 5
Vector B: x = -2, y = 7, z = 9
Next, we select the dimension as 3D and press calculate. The result is an angle of approximately 52.8 degrees.
Comparisons for Degree Between Two Vectors Calculator
There are various methods for finding the angle between two vectors, such as using trigonometric functions or dot product. However, these methods can be time-consuming and prone to errors. Here are some key advantages of using the degree between two vectors calculator:
Quick and easy: As mentioned earlier, all you need to do is input the values and press a button to get the result.
Q3. Can I find the angle between more than two vectors at once?
Q4. Is the calculator free to use?
A4. Yes, the degree between two vectors calculator is completely free to use.
Q5. Can I find the angle between two vectors that are not in the same plane?
A5. No, the calculator only works for vectors in the same 2D or 3D space.
Conclusion
In conclusion, the degree between two vectors calculator is a powerful tool that simplifies the process of finding the angle between two vectors. Whether you are a student or a professional, this calculator can save you time and effort when working with vectors. So next time you need to find the angle between two vectors, remember to use this handy tool. | 677.169 | 1 |
Use trigonometric functions like tangent (tan) to calculate offset length.
45-Degree Offset Formula
Offset Length = Desired Offset Distance / tan(45 degrees)
30-Degree Offset Formula
Offset Length = Desired Offset Distance / tan(30 degrees)
22.5-Degree Offset Formula
Offset Length = Desired Offset Distance / tan(22.5 degrees)
Offset Multipliers
For 45 degrees: 1.0, For 30 degrees: Approximately 1.7321, For 22.5 degrees: Approximately 2.6131
Precision
Offset measurements should be as precise as possible to ensure proper alignment and function.
Code Compliance
Compliance with local plumbing codes and regulations is essential for safe and effective offset installations.
Common Applications
Used in plumbing for connecting pipes that aren't directly in line, venting systems, bypassing obstacles, and navigating around corners.
Tolerance
Some tolerance may be allowed in offset measurements, depending on local codes and practical considerations.
FAQs
How is pipe offset calculated? Pipe offset is calculated using trigonometric functions such as sine, cosine, and tangent, depending on the type of offset (e.g., 45 degrees, 22.5 degrees, or 30 degrees) and the required measurements.
How do you calculate 45 offset plumbing? For a 45-degree offset, you can use the tangent function: Offset Length = Desired Offset Distance / tan(45 degrees) ≈ Desired Offset Distance
What is offset in plumbing? In plumbing, an offset refers to a section of pipe that is intentionally bent or angled to navigate around obstacles, change direction, or maintain proper alignment. It helps connect two points that are not directly in line with each other.
What is the multiplier for 22.5-degree offset? The multiplier for a 22.5-degree offset is approximately 2.6131. You can use this multiplier to calculate the offset length.
How do you calculate offset point? To calculate the offset point, you'll need specific measurements and the type of offset. Generally, you can use trigonometric functions to determine where the bend or angle should occur in the pipe to achieve the desired offset.
What is the multiplier for a 45 offset? The multiplier for a 45-degree offset is approximately 1.0. In this case, the offset length is equal to the desired offset distance.
What is the formula for a 45-degree angle? The formula for a 45-degree angle is: tan(45 degrees) = 1
How to do plumbing math? Plumbing math involves using basic arithmetic, algebra, and trigonometry to calculate measurements, angles, offsets, and pipe lengths required for plumbing installations. Knowledge of formulas and geometry is essential.
What are the two types of offsets? There are primarily two types of offsets in plumbing:
Vertical Offset: When the pipe needs to change elevation.
Horizontal Offset: When the pipe needs to change direction or move horizontally.
What are offset requirements? Offset requirements in plumbing vary depending on the specific installation and building codes. They are determined by factors such as pipe size, material, application, and local regulations.
Can plumbing vent pipe be offset? Yes, plumbing vent pipes can be offset when necessary to accommodate structural or architectural constraints while still maintaining proper venting requirements.
What is the multiplier for a 30 offset? The multiplier for a 30-degree offset is approximately 1.7321. You can use this multiplier to calculate the offset length.
How do you manually measure offset? To manually measure offset, you'll need a tape measure or ruler to determine the distance between two reference points and a protractor or angle finder to measure the required angle.
What is the offset distance? The offset distance is the horizontal or vertical measurement between the starting point and ending point of the offset in plumbing.
What is the offset equal to? The offset is equal to the desired offset distance divided by the tangent of the offset angle.
How much offset is 1 inch? The amount of offset in inches depends on the angle of the offset. For a 45-degree offset, 1 inch of offset distance will result in 1 inch of offset length.
How do you use 1.414 formula? The square root of 2 (approximately 1.414) is often used in pipe fitting to calculate diagonal measurements or offsets in right triangles. For example, in a 45-degree offset, you can multiply the desired offset distance by 1.414 to find the offset length.
How do you calculate offset bend? Offset bend calculations involve trigonometric functions to determine the length and angle of the bend required to create the desired offset in plumbing pipes.
How do you bend a 90 with two 45? To create a 90-degree bend using two 45-degree bends, you would position the two bends in line with each other, forming a right angle. This allows you to change the direction of the pipe by 90 degrees.
How do you find a 45-degree angle of 90 degrees? A 45-degree angle is half of a 90-degree angle. If you want to create a 45-degree angle within a 90-degree bend, you would position the pipe at a 45-degree angle relative to the initial direction.
How do you find the degree of a pipe? To find the degree of a pipe bend or angle, you can use a protractor or angle finder to measure the angle formed by the pipe in relation to a reference point or the horizontal or vertical axis.
How do you measure a 45-degree cut? To measure a 45-degree cut, you can use a miter saw or protractor to set the angle accurately. Then, cut the material along the marked 45-degree line.
Is the math to be a plumber hard? The math required to be a plumber can be challenging, especially when dealing with complex pipe layouts and calculations. However, with proper training and practice, plumbers can become proficient in the necessary mathematical concepts.
What is the formula for a plumber? There is no specific formula for becoming a plumber, but it typically involves completing plumbing apprenticeship programs, obtaining relevant licenses and certifications, and gaining experience in the field.
Do plumbers do a lot of math? Yes, plumbers often use math to calculate pipe lengths, angles, offsets, water flow rates, and more during their work.
What is an offset example? An example of an offset in plumbing is when a horizontal drain pipe needs to bypass a structural beam or obstacle, so it is bent or angled to continue in the desired direction.
What are some examples of offset? Examples of offsets in plumbing include bypassing support beams, going around corners, avoiding other pipes, and navigating through tight spaces while maintaining proper alignment.
What are the three types of offsets? The three types of offsets in plumbing are:
Vertical Offset: Changes in pipe elevation.
Horizontal Offset: Changes in pipe direction.
Combination Offset: A combination of both vertical and horizontal offsets to navigate complex obstacles.
Does offset need to be exact? Offset measurements should be as precise as possible to ensure proper pipe alignment and function. However, they may have some tolerance depending on local plumbing codes and practical considerations.
What is the common law right of offset? The common law right of offset refers to a legal principle that allows an entity to set off debts or claims against each other. It is often used in financial and contractual disputes.
What is an offset in contracting? In contracting, an offset can refer to various forms of compensation or concessions made by one party to offset costs or provide benefits to another party in a contract.
Can a sink drain pipe be offset? Yes, a sink drain pipe can be offset to accommodate the sink's location and plumbing requirements. This may involve bending or angling the pipe to connect to the drain system.
How many elbows can a plumbing vent have? The number of elbows allowed in a plumbing vent system can vary by local building codes, but it's generally advisable to minimize the use of elbows to ensure proper venting and prevent obstruction.
What is code for plumbing vents? Plumbing vent codes can vary by location, but they typically specify the minimum size and placement of vents, as well as requirements for slope, fittings, and clearances to ensure proper venting and prevent sewer gas buildup.
What does a 30-degree bend look like? A 30-degree bend will create a less pronounced angle compared to a 45-degree bend. It will be closer to a straight line, but with a slight deviation from the original direction.
How do you bend a 90 with an offset? To bend a 90-degree offset, you would typically use two 45-degree bends in opposite directions. This will create a 90-degree change in direction while maintaining alignment.
What is the math for offset bend? The math for an offset bend involves trigonometric calculations to determine the length and angle of the bend required to achieve the desired offset in plumbing pipes | 677.169 | 1 |
The Midpoint Formula
The midpoint of the line segment between points
$\,(x_1,y_1)\,$
and
$\,(x_2,y_2)\,$
is given by the Midpoint Formula:
$$
\cssId{s6}{\left(
\frac{x_1+x_2}2,\frac{y_1+y_2}2
\right)}
$$
Here,
$\,x_1\,$ (read as '$\,x\,$ sub $\,1\,$')
denotes the
$\,x$-value
of the first point,
and
$\,y_1\,$ (read as '$\,y\,$ sub $\,1\,$')
denotes the
$\,y$-value
of the first point.
Similarly,
$\,x_2\,$ and $\,y_2\,$ denote the
$\,x$-value
and
$\,y$-value
of the second point.
Thus, to find the location
that is exactly halfway between two points,
you average
the $x$-values,
and average the $y$-values.
Referring to the sketch below,
$\Delta ABD\,$
is similar to
$\,\Delta AMC\,.$
That is, these two triangles
have the same angles.
Why?
They both share angle $\,A\,,$
and they both have a right angle.
Since all the angles in a triangle
sum to $\,180^\circ\,,$
the third angles must also be the same.
Similarity gives us what we need!
It tells us that $\Delta ABD\,$ and
$\Delta AMC\,$ have exactly the same
shapes—they're just different sizes.
Since
$\,\overline{AM}\,$ is exactly half of
$\,\overline{AB}\,,$
$\,\overline{AC}\,$ must be exactly half of
$\,\overline{AD}\,.$
Thus, $\,C\,$ is the midpoint
between $\,A\,$ and $\,D\,$
(which can be found by averaging
$\,x_1\,$ and
$\,x_2\,$).
Use a similar argument to show that
$\,\overline{DE}\,$
(which has the same length as $\,\overline{CM}\,$) is exactly half of $\,\overline{DB}\,.$
Examples
Question:
Find the midpoint of the line segment between
$\,(1,-3)\,$ and $\,(-2,5)\,.$ | 677.169 | 1 |
The Elements of Euclid, books i. to vi., with deductions, appendices and ...
11. Prove the second part of the proposition by drawing through A a straight line DAE || BC. (The Pythagorean proof.)
12. If any of the angles of an isosceles triangle be two-thirds of a right angle, the triangle must be equilateral.
13. Each of the base angles of an isosceles triangle equals half the exterior vertical angle.
14. If the exterior vertical angle of an isosceles triangle be bisected, the bisector is || the base.
။
15. Show that the space round a point can be filled up with six equilateral triangles, or four squares, or three regular hexagons. 16. Can a right angle be divided into any other number of equal parts than two or three?
17. In a right-angled triangle, if a perpendicular be drawn from the
right angle to the hypotenuse, the triangles on each side of it are equiangular to the whole triangle and to one another. 18. Prove the seventh deduction indirectly; and also directly by producing the median to the hypotenuse its own length. 19. If the arms of one angle be respectively perpendicular to the arms of another, the angles are either equal or supplementary. 20. Prove Cor. 3 by taking a point inside the figure and joining it to the angular points.
PROPOSITION 33. THEOREM.
The straight lines which join the ends of two equal and *parallel straight lines towards the same parts, are themselves equal and parallel.
Let AB and CD be equal and parallel:
it is required to prove AC and BD equal and parallel.
Join BC.
Because BC meets the parallels AB, CD,
.. L ABC = alternate
DCB.
I. 29
Proved
Because CB meets AC and BD, and makes the alter
nate LS ACB, DBC equal;
.. AC is || BD.
1. State a converse of this proposition.
I. 27
2. If a quadrilateral have one pair of opposite sides equal and parallel, it is a ||m.
3. What statements may be made about the straight lines which join the ends of two equal and parallel straight lines towards opposite parts?
A parallelogram has its opposite sides and angles equal, and is bisected by either diagonal.
Let ACDB be a ||m of which BC is a diagonal :
it is required to prove that the opposite sides and angles of ACDB are equal, and that ▲ ABC = ▲ DCB.
Because BC meets the parallels AB, CD,
and because BC meets the parallels AC, BD,
I. 29
I. 29 Proved
=
L ABC = L DCB
LACB
BC = CB;
A
C
... AB = DC, AC = DB, L BAC LCDB,
D
=
COR. If the arms of one angle be respectively parallel to the arms of another, the angles are either (1) equal or (2) supplementary.
For (1) BAC has been proved = L CDB ;
and (2) if BA be produced to E,
LEAC, which is supplementary to BAC,
must be supplementary to CDB.
I. 13
1. If two sides of a m which are not opposite to each other be equal, all the sides are equal.
2. If two angles of a ||m which are not opposite to each other be equal, all the angles are right.
3. If one angle of a ||m be right, all the angles are right.
4. If two ||ms have one angle of the one = one angle of the other, the ||ms are mutually equiangular.
5. If a quadrilateral have its opposite sides equal, it is a ||m.
6. If a quadrilateral have its opposite angles equal, it is a ||m.
7. If the diagonals of a m be equal to each other, the ||m is a rectangle.
8. If the diagonals of a m bisect the angles through which they pass, the m is a rhombus.
9. If the diagonals of a ||m cut each other perpendicularly, the ||m is a rhombus.
10. If the diagonals of a m be equal and cut each other perpendicularly, the ||m is a square.
11. Show how to bisect a straight line by means of a pair of parallel
rulers.
12. Every straight line drawn through the intersection of the diagonals of a ||m, and terminated by a pair of opposite sides, is bisected, and bisects the m.
13. Bisect a given ||m by a straight line drawn through a given point either within or without the [m.
=
14. The straight line joining the middle points of any two sides of a triangle is || the third side and half of it. 15. If the middle points of the three sides of a triangle be joined with each other, the four triangles thence resulting are equal. 16. Construct a triangle, having given the middle points of its three sides.
Parallelograms on the same base and between the same parallels are equal in area.
Let ABCD, EBCF bem on the same base BC, and between the same parallels AF, BC:
it is required to prove || ABCD
=
Because AF meets the parallels AB, DC,
exterior FDC;
.. interior A = and because AF meets the parallels EB, FC,
|| EBCF.
I. 29
In As ABE, DCF, L AEB
=
L DFC
I. 29 Proved Proved
מנון
=
NOTE. This proposition affords a means of measuring the area of a m; thence (by I. 34 or 41) the area of a triangle; and thence (by I. 37, Cor.) the area of any rectilineal figure. For the area of any the area of a rectangle on the same base and between the same parallels ; and it is, or ought to be, explained in books on Mensuration, that the area of a rectangle is found by taking the product of its length and breadth. This phrase 'taking the product of its length and breadth,' means that the numbers, whether integral or not, which express the length and breadth in terms of the same linear unit, are to be multiplied together. Hence the method of finding the area of a ||m is to take the product of its base and altitude, the altitude being defined to be the perpendicular drawn to its base from any point in the side opposite.
1. Prove the proposition for the case when the points D and E coincide.
2. Equal ms on the same base and on the same side of it are between the same parallels.
3. If through the vertices of a triangle straight lines be drawn || the opposite sides, and produced till they meet, the resulting figure will contain three equal ||ms.
4. On the same base and between the same parallels as a given ||m, construct a rhombus = them.
5. Prove the equality of ▲s ABE and DCF in the proposition by I. 4 (as Euclid does), or by I. 8, instead of by I. 26.
PROPOSITION 36. THEOREM.
Parallelograms on equal bases and between the same parallels are equal in area. | 677.169 | 1 |
• two are the parallel and the perpendicular at X(14251) to the Brocard axis.
• three are parallel to the asymptotes of the cubic K024 and to the sidelines of the CircumNormal triangle with vertices on (O) and K003.
A, B, C are nodes on Q177 with nodal tangents parallel and perpendicular to the Brocard axis. The remaining points on the sidelines of ABC are Ωa, Ωb, Ωc on the Lemoine axis which also passes through X(187) and X(512) on Q177.
Q177 meets the circumcircle (O) at X(1379), X(1380), O3, O4 which are obviously the vertices of a square. | 677.169 | 1 |
What is Geometric algorithms
Exploring Geometric Algorithms
Geometry is a branch of mathematics that studies the properties and relationships of points, lines, shapes, and surfaces in two and three-dimensional space. Geometric algorithms are computer algorithms designed to solve various problems in geometry such as finding the intersection of two shapes, calculating the distance between two points, and measuring the area of a shape. In this article, we will explore the different types and applications of geometric algorithms.
Types of Geometric Algorithms
Geometric algorithms can be classified into several categories based on their scope and application. The following are some of the most common types of geometric algorithms:
Point Location Algorithms: These algorithms are used to determine the position of a given point relative to a set of geometric shapes such as lines or polygons. They are commonly used in computer graphics and computer-aided design (CAD) applications.
Convex Hull Algorithms: These algorithms are used to find the convex hull of a set of points or a shape. The convex hull is the smallest convex polygon that encloses all the points in the set. This type of algorithm is used in spatial data analysis and computational geometry.
Triangulation Algorithms: These algorithms are used to partition a shape or a set of points into a triangle mesh. Triangulation algorithms are commonly used in computer graphics, computer vision, and finite element analysis.
Intersection Detection Algorithms: These algorithms are used to determine if two shapes intersect and if so, where they intersect. Intersection detection algorithms are widely used in CAD, image processing, and robotics applications.
Distance Calculations: These algorithms are used to calculate the distance between two shapes or points. Distance calculation algorithms are essential in GIS, remote sensing, and image processing applications.
Applications of Geometric Algorithms
Geometric algorithms are used in various applications such as computer graphics, robotics, computer-aided design (CAD), and computational geometry. The following are some of the most common applications of geometric algorithms:
Computer Graphics: Geometric algorithms are used in computer graphics to create 3D models, render images, and simulate physics. They are also used to process and display 2D and 3D data in real-time.
Robotics: Geometric algorithms play a critical role in robotic systems for navigation, perception, and motion planning. They are used to map environments, detect obstacles, and plan optimal paths for robots.
Computer-Aided Design: Geometric algorithms are used in CAD software to create, edit, and manipulate 2D and 3D geometries. They are also used to perform geometric analysis and optimization tasks.
Computational Geometry: Geometric algorithms are used in computational geometry to solve geometric problems such as proximity queries, convex hulls, and Voronoi diagrams. They are also used in mesh generation and optimization.
GIS and Remote Sensing: Geometric algorithms are used in GIS and remote sensing to process and analyze spatial data. They are used to calculate distances, areas, and volumes, as well as to detect changes in land cover and land use over time.
Challenges in Geometric Algorithms
Geometric algorithms pose several challenges that make them difficult to design and implement. The following are some of the most common challenges in geometric algorithms:
Numerical Instability: Geometric algorithms often involve floating-point arithmetic operations that can be sensitive to small changes in input data. This can lead to numerical instability and numerical errors.
Robustness: Geometric algorithms need to be robust and handle various input data types, shapes, and sizes. They also need to be able to handle degenerate cases such as collinear points and coincident edges.
Complexity: Geometric algorithms can have high computational complexity due to the large input sizes and the need to perform geometric operations repeatedly.
Parallelization: Geometric algorithms can benefit from parallel processing to reduce the computation time and increase the scalability of the algorithm.
Conclusion
Geometric algorithms are an essential component of computer science and mathematics. They have a wide range of applications in various fields such as computer graphics, robotics, and GIS. Developing efficient and robust geometric algorithms that can handle complex input data types and sizes is a challenging task that requires a deep understanding of computational geometry. As the field of computer science and mathematics continues to evolve, geometric algorithms will continue to play a critical role in solving various problems and advancing technology. | 677.169 | 1 |
Trigonometry
Trigonometry
Guys I'm sooo confused with this one Question 😐. Triangle PQR is such that PQ = 10cm, QR = 8cm and the angle PRQ=64°. Angle QPR =Ɓ° Show that there is only one possible value of Ɓ. I'll really appreciate if someone can solve it for me😇I did try but I wasn't sure if it was right but I think so I've got the answer thanks✨️✨️ | 677.169 | 1 |
Rectangle ABCD is rotated by 90 degrees clockwise about the
Last updated: 3/17/2023
Rectangle ABCD is rotated by 90 degrees clockwise about the origin and then translated 3 units left and 2 units down to form A B C D D A 10 What is the length in units of line segment A B in the resulting figure c8 6 B 4 Y 2 10 8 6 4 2 0 2 2 ME 4 6 8 10 X | 677.169 | 1 |
NCERT Solutions for Class 6 Maths Chapter 5 Exercise 5.4
Updated by Tiwari Academy
on April 13, 2023, 3:24 PM
CBSE NCERT Solutions for Class 6 Maths Chapter 5 Exercise 5.4 in Hindi and English Medium for new session 2024-25. We have updated the solutions of ex. 5.4 class VI Maths following the rationalised syllabus and new NCERT textbooks issued for CBSE 2024-25 curriculum.
6th Maths Exercise 5.4 Solution in Hindi and English Medium
Class 6 Maths Chapter 5 Exercise 5.4 Solution
Class VI Mathematics NCERT textbook Ex. 5.4 of chapter 5 Understanding Elementary Shapes in Hindi and English Medium free to download. 6th NCERT Books PDF and Videos are updated for academic session 2024-25 free to download. If students are feeling difficulties to use the solutions, contact us immediately for help. We are here to help you without any charge. Our main motive is to help the students and make the study stress-free.
Class: 6
Mathematics
Chapter: 5
Exercise: 5.4
Chapter Name:
Understanding Elementary Shapes
Medium:
Hindi and English Medium
Content:
NCERT Exercise Solutions
Solutions App for Class 6
Circle
A circle is a closed plane figure consisting of all those points of the plane which are at a constant distance from a fixed point.
The fixed point is called the centre of the circle and the constant distance is known as the radius of the circle.
If O is the centre of the circle and OM = r is the radius of the circle.
If N is another point on the circle, then the line segment ON is another radius of the circle. All the radii of the circle have one end point common, which is the centre of the circle. Also, OM = ON = r.
Thus, all radii of a circle are equal.
Diameter of Circle:
A line segment passing through the centre of a circle and having its end points on the circle, is called diameter of the circle.
If, PQ is a diameter of the circle having its centre at O. Also, OP and OQ are two radii of the circle.
So OP = OQ, because O is the mid-point of PQ.
Thus, the centre of a circle bisects a diameter into two equal parts.
Hence, Diameter = 2 × Radius
Just as we can draw an infinite number of radii through the centre of a circle, we can draw an infinite number of diameters.
As observed earlier, the centre O of a circle is the mid-point of every diameter of circle. Thus, the diameters of a circle are concurrent and the common point is the centre of the circle.
Interior and Exterior of a Circle
Consider a circle with centre O and radius r. This circle divides the plane into three parts:
The part of the plane, consisting of those points P, for which OP < r, is called the interior of the circle, with centre O and radius r. In other words, the set of all those points which lie inside the circle is known as its interior.
The part of the plane, consisting of those points P for which OP > r, is called the exterior of the circle, with centre O and radius r. In other words, the set of all those points which lie outside the circle is called the exterior of the circle.
The part of the plane, consisting of those points P, for which OP = r, is the circle itself, i.e., the set of all points lying on the circle is the circle itself.
Find the radius of a circle whose diameter is: (i) 6 cm (ii) 9.2 cm
How many radii of a circle can be?
Every circle has an infinite number of radii. This is because the number of points on the circumference of the circle is infinite. So there can be an infinite number of lines joining them to the center.
The longest chord of a circle is called.
A chord that passes through the center of a circle is called a diameter and is the longest chord.
What do mean by circumference of a circle?
The perimeter of a circle, often called the circumference, is proportional to its diameter and its radius.
Note: Clearly, the circle, its interior and exterior have no common point and together account for all points of the plane.
The part of the plane of the circle that consists of the circle and its interior is called the circular region.
How many questions are there in exercise 5.4 of class 6th Maths?
Exercise 5.4 of class 6th Maths has 11 questions and no example. All questions of this exercise are important. Students should practice all questions of this exercise. But, some questions of this exercise are most important which students can't skip for the exams. These questions are 2, 5, 7, 9, and 11.
In how many days, students finish exercise 5.4 of class 6th Maths?
Students finish exercise 5.4 of grade 6th Maths in 3 days if they give 1 hour per day to this exercise. This time can vary because no students can have the same working speed, ability, efficiency, etc.
Is exercise 5.4 of 6th Maths complicated?
If concepts of angles (Acute angle, Obtuse angle, Right angle, Straight angle, Reflex angle, Complete angle) and how to measure angles using protractor is clear to students then, exercise 5.4 of class 6th Maths is easy otherwise complicated.
Which questions of exercise 5.4 of class 6 Maths are best for class test?
Exercise 5.4 of class 6th Maths has 11 questions and no example. Questions 4, 7, 8, and 11 are the best questions of exercise 5.4 of class 6th Maths.
Last Edited: April 13 | 677.169 | 1 |
As a plane curve, a hyperbolic spiral can be described in polar coordinates(r,φ){\displaystyle (r,\varphi )} by the equation
r=aφ,{\displaystyle r={\frac {a}{\varphi }},}
for an arbitrary choice of the scale factora.{\displaystyle a.}
Hyperbolic spirals are patterns in the Euclidean plane, and should not be confused with other kinds of spirals drawn in the hyperbolic plane. In cases where the name of these spirals might be ambiguous, their alternative name, reciprocal spirals, can be used instead.[6] | 677.169 | 1 |
A triangle has vertices A(0,0), B(12,0), and C(8,10). The probability that a randomly chosen point inside the triangle is closer to vertex B than to either vertex A or vertex C can be written as \frac{p}{q}, where p and q are relatively prime positive integers. Find p+q. | 677.169 | 1 |
28
Página 31 ... angles . COR . 2. And hence , all the angles made by any number of straight kines meeting in one point , are together equal to four right angles . PROP . XVI . THEOR . If one side of a triangle be produced , the exterior angle is ...
Página 32 ... angles together are less than two right angles . Produce BC to D ; and be- cause ACD is the exterior angle of the triangle ABC , ACD is greater ( 16. 1. ) than the interior and opposite angle ABC ; to each of these add the angle ACB ...
Página 34 ... angle . Let the two straight lines BD , CD be drawn from B , C , the ends of the side BC of the triangle ABC , to ... exterior angle of a triangle ( 16. 1. ) is greater than the interior and opposite angle , the exterior angle BDC ...
Página 38 ... angle GCB is equal to the angle DFE ; but DFE is , by the hypothesis , equal to the angle BCA ; wherefore also the angle BCG is equal to the angle BCA , the less to the greater ... exterior angle AEF is greater ( 16. 38 ELEMENTS.
Página 39 ... exterior angle equal to the interior and opposite upon the same side of the line ; or makes the interior angles upon the same side together equal to two right angles ; the two straight lines are parallel to one another . Let the | 677.169 | 1 |
Sin 60 in radians
In fractional form, the value of sin 60°= √3/2. Sin 60°, when denoted in the terms of a radian, is π/3. The two ways by which the value of the sin 60° can be predicted are by either using the trigonometric functions or by using the unit circle. A radian is equal to 180° which is denoted a semi-circle while 2π depicts a full circle.UseThe easiest way to do it is to recognize that 180° equals π radians, or 3.14 radians. Then determine what fraction (or percentage) of 180° the angle you're concerned with is, and multiply that fraction by 3.14 radians. For example, to convert 60° to radians, divide 60° by 180°. That's 1/3. Then multiply 1/3 by 3.14: that's 1.05 radians.
To find the value of tan 60 degrees using the unit circle: Rotate 'r' anticlockwise to form 60° angle with the positive x-axis. The tan of 60 degrees equals the y-coordinate (0.866) divided by x-coordinate (0.5) of the point of intersection (0.5, 0.866) of unit circle and r. Hence the value of tan 60° = y/x = 1.7321 (approx).Python math.sin() 方法 Python math 模块 Python math.sin(x) 返回 x 弧度的正弦值。 要获取指定角度的正弦,必须首先使用 math.radians() 方法将其转换为弧度。 Python 版本: 1.4 语法 math.sin() 方法语法如下: math.sin(x) 参数说明: x -- 必需,数字。如果 x 不是数字,则返回 TypeError。a unit of plane angular measurement that is equal to the angle at the center of a circle subtended by an arc whose length equals the radius or approximately 180°/π ~ 57.3 degrees. secant. the length of the hypotenuse divided by the length of the adjacent side. Also equals 1/cos (θ) sin. sin (θ) is the ratio of the opposite side of angle θ ...
shabooya roll call rap song Sin Cos Tan Deg to Rad Rad to Deg. Angle: Calculate: Answer: 120° = 2π/3 radian. Step-by-Step Solution. Given that 180° is equal to pi, we can write the following degrees to radians conversion formula: ... π × 120÷60/180÷60 = 2π/3 radian, when reduced to lowest fraction in terms of π. cvs 650 nw 27th ave miami fl 33125sammy things onlyfans 1 Know the trigonometric function values for the special angles in radians #1-4, 46-48. 2 Use a unit circle to find trig values #5-30, 45-58. 3 Find reference angles in radians #33-45. 4 Evaluate trigonometric expressions #31-32, 49-54. 5 Find coordinates on a unit circle #55-60, 67-68. can you return spectrum equipment to any store It means that, sin 60 = sin 120 = √3/2. Method 1. Another method to find the value of sin 120 degrees is by using the other angles of the sine functions such as 60 degrees and 180 degrees which are taken from the trigonometry table. ... 60. 90. 180. 270. 360. Angles (In Radians) 0. π/6.Sine and cosine are written using functional notation with the abbreviations sin and cos. Often, if the argument is simple enough, the function value will be written without parentheses, as sin θ rather than as sin(θ). Each of sine and cosine is a function of an angle, which is usually expressed in terms of radians or degrees. Except where ... u0073 codekroger timings near mepensacola craigslist general for sale by owner Use 2 guys comedy Trigonometry. Convert from Radians to Degrees (10pi)/6. 10π 6 10 π 6. To convert radians to degrees, multiply by 180 π 180 π, since a full circle is 360° 360 ° or 2π 2 π radians. ( 10π 6)⋅ 180° π ( 10 π 6) ⋅ 180 ° π. Cancel the common factor of π π. Tap for more steps... 10 6 ⋅180 10 6 ⋅ 180.We refer to the sine and cosine functions as cofunctions of each other. Similarly, the tangent and cotangent functions are cofunctions, as are the secant and cosecant. If variable u is the radian measure of an acute angle, then the angle measure with radian measure π/2 - u is complementary to u. We may consider the right triangle shown in ... sam's club amarillo gascollin county juvenile detention centercheapest gas fort smith ar Explanation: For sin 1 degrees, the angle 1° lies between 0° and 90° (First Quadrant ). Since sine function is positive in the first quadrant, thus sin 1° value = 0.0174524. . . Since the sine function is a periodic function, we can represent sin 1° as, sin 1 degrees = sin (1° + n × 360°), n ∈ Z. ⇒ sin 1° = sin 361° = sin 721 ... | 677.169 | 1 |
find the right angle | 677.169 | 1 |
Degrees of Significance: Exploring the Versatile Degree Symbol
The degree symbol, denoted by the symbol °, is a small but powerful character that finds its way into various aspects of our lives, from academics and science to everyday communication and technology. It is a simple yet essential element that represents a unit of measurement for angles, temperature, and even geographic coordinates. In this exploration of the degree symbol, we will delve into its origins, its diverse uses, and its importance in our modern world.
The origins of the degree symbol can be traced back to ancient Greece, where it was used by astronomers and mathematicians to measure angles in circles. The Greek word "grados" meant "a step," and this concept of dividing a circle into 360 equal steps or degrees laid the foundation for the degree symbol as we know it today. Over time, this symbol evolved and found its way into various mathematical and scientific disciplines. Click here for more degree symbol word
In mathematics and geometry, the degree symbol is used to denote angles, with a full circle measuring 360 degrees. This system of angular measurement is fundamental in fields like trigonometry, calculus, and geometry, where understanding angles is crucial for solving complex problems and making precise calculations. The degree symbol serves as a concise and universally recognized shorthand for representing angles, making it an indispensable tool for mathematicians and scientists.
Beyond mathematics, the degree symbol has a significant presence in the realm of meteorology. It is commonly used to indicate temperature in both the Celsius and Fahrenheit scales. For example, 25°C signifies a comfortable room temperature, while 100°F indicates a scorching hot day. In weather forecasts and discussions, the degree symbol is a familiar sight, helping people make sense of temperature fluctuations and plan their activities accordingly.
The degree symbol also plays a vital role in navigation and cartography. Geographic coordinates, such as latitude and longitude, use degrees to pinpoint locations on the Earth's surface. This system is integral to global positioning systems (GPS) and mapping services, enabling us to navigate our way around the world with precision. Whether you're using a GPS device for driving directions or exploring maps online, the degree symbol is a constant companion, guiding you to your destination.
In everyday communication, the degree symbol is commonly employed to express various concepts. In cooking, it represents the temperature at which an oven should be set, ensuring that your recipes turn out just right. In the world of photography, it signifies the angle of rotation applied to an image, allowing for precise adjustments. Even in informal conversations, the degree symbol can convey emotions or intensity, as seen in expressions like "I'm freezing to death!" or "I love ice cream at any temperature."
Furthermore, the degree symbol has found its way into the digital age, where it is easily accessible on keyboards and in word processing software. It can be inserted into documents and emails with a simple keystroke or by using character insert tools. This accessibility ensures that the degree symbol remains a part of our everyday written communication, allowing us to convey information accurately and effectively.
In conclusion, the degree symbol may seem like a small, unassuming character, but its significance in our lives cannot be overstated. From its ancient Greek origins in the measurement of angles to its widespread use in mathematics, science, meteorology, navigation, and everyday communication, the degree symbol is a versatile and essential tool. It simplifies complex concepts, guides us in our daily activities, and ensures that we can convey information accurately in a wide range of fields. So, the next time you encounter the degree symbol, remember the rich history and practicality it brings to our world. | 677.169 | 1 |
Converting Degrees to Radians
This Degree to Radians Converter Tools helps you to convert an angle in Degree into Radians notation as an output within a fraction of seconds along with an explanation. Just provide the input number in the input field and click on the calculate button provided next to the input field to get the desired result.
Enter Degrees
Converting Degrees to Radians: Are you searching for the easiest ways to convert the Degree to Radians? Then, you have come to the appropriate place where you can find the simplest option to find conversion. You can get the angles in Radians notation for any regular angle in less time by using our free Online Degree to Radians Converter. Also, you can learn how to convert radian to degrees with steps along with a solved example.
What is Degrees to Radians Conversion?
Angles have units for measurement. When measuring angles, degrees and radians are used as units of measurement. Most often, radians are used to measure angles. Here are a few examples: area of a sector of a circle, arc length, and angular velocity. It is recommended to use radians when considering circular paths or components of circular paths. Angles may be expressed in degrees in the problem statement, but we should always convert them into radians before using them.
Degree: An angle is measured by a degree, also referred to as the degree of arc. It is denoted by the symbol (°). A complete rotation is described by an angle measuring 360° and the instrument from which it is measured is called a protractor.
Radians: A radian measures the angle formed at the centre of a circle by an arc whose length equals the radius 'r' of the circle. One complete rotation is equivalent to 2π radians. The measure of one radian is 57.296°. A right angle is π /2 radians, and a straight angle is π radians.
Math will no longer be a tough subject, especially when you understand the concepts through our calculators and step-by-step examples see arithmeticcalculator.com.
Degrees to Radians Conversion Formula
To convert degrees to radians, we multiply the angle (in degrees) by π / 180, which is a generalized formula for converting values in degrees to radians.
Radian = Degree x π/180
How to Convert Degrees to Radians Manually?
360 degrees, or 2π radians, are two units of measurement for angles. Thus, 360° and 2π radians are the same numbers when one goes "once around" a circle. Simply follow the steps below to convert degrees to radians.
Write down how many degrees you are converting to radians.
You need to multiply the number of degrees by π/180. Because 180 degrees equals π radians.
Carry out the multiplication process. Do it by multiplying two fractions.
Taking each fraction and putting it in the lowest terms will give you your final answer.
You can convert the angle measures to radians by writing them down. Then, you're done!
Solved Example on Degrees to Radians Conversion
Example:
Convert 45 degrees to radians.
Solution:
Given, 45 degrees is the angle.
Angle in radian = Angle in degree × (π/180)
= 45 × (π/180)
= π/4
Hence, 45 degrees is equal to π/4 in radian.
Degrees to Radians Chart or Conversion Table
The following table represents the common values of converted degrees to radian units of measurement in terms of π and numeric values. For quick results, also use this converted degree to radian chart.
Degrees (°)
Radians (rad)
Radians (rad)
0°
0 rad
0 rad
30°
π/6 rad
0.5235987756 rad
45°
π/4 rad
0.7853981634 rad
60°
π/3 rad
1.0471975512 rad
90°
π/2 rad
1.5707963268 rad
120°
2π/3 rad
2.0943951024 rad
135°
3π/4 rad
2.3561944902 rad
150°
5π/6 rad
2.6179938780 rad
180°
π rad
3.1415926536 rad
270°
3π/2 rad
4.7123889804 rad
360°
2π rad
6.2831853072 rad
FAQs On Converting Degrees to Radians Tools with Steps
1. How do you convert degrees to radians?
We will use the equation 1 radian = (180π)° to convert Radians to Degrees. Then, multiply the Degree value to π/180° Radians. | 677.169 | 1 |
how do you calculate the area for a triangleCalculating Area Of Triangles Worksheet – Triangles are one of the most fundamental forms in geometry. Understanding triangles is critical to mastering more advanced geometric concepts. In this blog we will look at the different kinds of triangles triangular angles, the best way to calculate the area and perimeter of a triangle, and offer illustrations of all. Types of Triangles There are three kinds of triangulars: Equilateral, isosceles, and scalene. Equilateral triangles consist of three equal sides and … Read more | 677.169 | 1 |
Getting ready to use protractors!
We are halfway through our angle unit . . . we've worked to build some understanding and are ready to use protractors! Today I actually introduced the protractor–we have been measuring with "nonstandard" angle measurers the last few days. If you haven't ever made wax paper protractors with your students, you should give it a try…lots of fun! Simply take a square of wax paper…fold it into quarters (90 degree angles), then fold in half again (45 degrees) and then again one more time. Students DO usually come to the conclusion that each section is 22.5 degrees–but that isn't what is important. We use this protractor to measure in "wedges"–without worrying about degrees–for now. Instead, I want students to learn about measuring how "open" an angle is…and because the waxed paper protractor is relatively opaque, you can set it down on different shapes and students can easily line it up and say how many "wedges" (nonstandard unit) open an angle is.
The custodian only made ONE comment about the masking tape on the carpet–so not too bad! Here is my "Teaching students to use a protractor" tip of the day…if you've ever done this, you know how hard it is to get them to remember to put the "dot" of the protractor on the vertex AND to keep one ray of the angle on the "0" line . . . so try this! Whether you tape your floor or create some large angles on construction paper, using these protractors is a GREAT kick off to the much more difficult task of learning to use a "real" protractor.
Have the students get out their protractor and lay a toothpick on it with one end of the toothpick at that center "dot" and the toothpick stretched toward the 0.
I tell them that there will always be one ray of the angle on their protractor–just like this. Then I asked them to put their second toothpick to build a 90 degree angle . . . (sorry for the blurry photo)
We then built all sorts of angles with our toothpicks–right on the protractor. We reviewed the difference between acute and obtuse and how to use BOTH scales on the protractor to help us estimate–a small angle must be less than 90 degrees so you better be looking at that set of numbers! I am pretty confident that we will be ready to take our protractors and tackle "real" angles tomorrow!
Interested in extending your students learning even more? See if these resources might help you take your instruction even deeper. | 677.169 | 1 |
ATPL: On a polar stereographic chart where the Earth convergence between 2 points located on the parallel 60°N is 20°, the great circle maximum cross-track difference with the straight line joining the 2 points is: | 677.169 | 1 |
From first figure to second figure one side of the pentagon is deleted and a square is formed and the two black dots move out of the main design while a small white circle is introduced inside the main design.
Correct Option: D
From first figure to second figure one side of the pentagon is deleted and a square is formed and the two black dots move out of the main design while a small white circle is introduced inside the main design.
Direction: In following questions, select the related figure from the given alternatives. | 677.169 | 1 |
finding angle measures in triangles worksheet answersFind AnglesMissing Angles In Triangles Worksheet Answers | 677.169 | 1 |
Radian = A unit of angle, equal to an angle at the center of a circle whose arc is equal in length to the radius.
Degree = A unit of angle, equal to $1/360$ of a full circle.
Unit Circle = A circle with radius 1.
Trigonometric functions = Functions that are related to the angles of a triangle.
Match the following trigonometric functions with their definitions:
Sine = In a right triangle, the ratio of the length of the side opposite the angle to the length of the hypotenuse.
Cosine = In a right triangle, the ratio of the length of the adjacent side to the length of the hypotenuse.
Tangent = In a right triangle, the ratio of the length of the opposite side to the length of the adjacent side.
Cotangent = In a right triangle, the ratio of the length of the adjacent side to the length of the opposite side.
Match the following symbols with their meanings in the context of trigonometry:
$θ$ = A variable often used to represent an angle in mathematics.
$π$ = The ratio of the circumference of any circle to its diameter.
$s$ = The length of the corresponding arc on the unit circle.
$360°$ = The degree measurement of a full circle.
Study Notes
Trigonometry Conversions
Convert angle measurements from degrees to radians
Convert radian measurements to degrees
Trigonometry Terminology
Match terms with their definitions in trigonometry
Trigonometric Functions
Match functions with their definitions, including:
Sine
Cosine
Tangent
Cotangent
Secant
Cosecant
Trigonometric Symbols
Match symbols with their meanings in trigonometry, including:
° (degrees)
rad (radians)
sin (sine)
cos (cosine)
tan (tangent)
cot (cotangent)
sec (secant)
csc (cosecant)
Test your knowledge on the limit of trigonometric functions with this quiz. Explore the concept of angle measures in degrees and radians and learn how they relate to the circumference of a circle. Challenge yourself with questions on finding the radian measure of a 360° angle and more. | 677.169 | 1 |
Tangent Formula
Mathematics is divided into numerous categories based on the sorts of calculations performed and the disciplines covered. Arithmetic, percentages, exponentials, geometry, algebra, and other topics are covered in the branches. Furthermore, derived equations are frequently used in mathematics to assure the precision of computations and operations. The following article contains all of the fundamental equations found in the many areas or domains of mathematics. The principles of mathematics demonstrate how to apply specific equations, such as the equation of forces, accelerations, or work done, to solve mathematical issues. They are also used to provide mathematical solutions to problems that emerge in everyday life. Equations can take many different forms and are utilised in many different fields of mathematics. The methods employed to research them, however, differ depending on their kind. It might be as simple as utilising the basic addition formula or as sophisticated as integrating differentiation.
The Extramarks website and mobile app offer students all of the materials they need to prepare for academic and competitive examinations. Extramarks may assist students in preparing for a variety of competitive examinations such as the JEE Mains, NEET, JEE Advance, CUET, and others. Extramarks offers exceptionally dependable and precise study resources for all courses. Students can get exam preparation materials via the Extramarks website and mobile app. Among these are well-known state examination boards such as the CBSE and ICSE. Extramarks provide study resources for all courses that are very legitimate and trustworthy. Students can purchase study materials from the Extramarks website and mobile application to help them prepare for examinations. ICSE, CBSE, and other important state boards are among these. Extramarks examines its study materials for faults on a regular basis and updates them in accordance with the curriculum. Students can use the study materials to increase their confidence as they prepare for examinations.
Trigonometry is a field of mathematics that studies the connection between angles and side lengths of a right-angled triangle. Sine, cosine, tangent, cosecant, and secant are the six trigonometric ratios or functions, and a trigonometric ratio is a ratio between the sides of a right-angled triangle. The sine, cosine, and tangent functions are the reciprocal functions of the sine, cosine, and tangent functions, respectively.
sin θ = Opposite side/Hypotenuse
cos θ = Adjacent side/Hypotenuse
tan θ = Opposite side/Adjacent side
cosec θ = Hypotenuse/Opposite side
sec θ = Hypotenuse/Adjacent side
cot θ = Adjacent side/Opposite side
The Tangent Formula is a trigonometric formula that deals with the tangent function. The tangent function (sometimes known as "tan") is one of six trigonometric functions that is the ratio of the opposing side to the adjacent side. There are several tangent function formulae that may be obtained from various trigonometric identities and formulas.
What Are Tangent Formulas?
The Tangent formula discusses the tangent (tan) function. Consider a right-angled triangle with x as one of its sharp angles. The Tangent Formula is tan x = (opposite side) / (adjacent side), where "opposite side" refers to the side opposite the angle x and "adjacent side" refers to the side next to the angle x. Aside from this generic formula, there are other different formulae in trigonometry that will define a tangent function.
Tangent Formulas Using Reciprocal Identity
Students already know that the tangent function (tan) and the cotangent function (cot) are reciprocals. In other words, if tan x = a / b, then cot x = b / a. As a result, the Tangent Formula employing one of the reciprocal identities is,
Tangent Formulas Using Pythagorean Identity
The link between sec and tan is discussed in one of the Pythagorean identities. It says, sec2x – tan2x = 1, for any x. Students can solve this for tan x. They see how.
sec2x – tan2x = 1
Subtracting sec2x from both sides,
-tan2x = 1 – sec2x
Multiplying both sides by -1,
tan2x = sec2x – 1
Taking square root on both sides,
The Tangent Formula tan x = ± √ (sec2x – 1)
Tangent Formula Using Cofunction Identities
The cofunction identities establish the relationship between the co functions sin and cos, sec and csc, and tan and cot. Taking use of one of the cofunction identities,
The Tangent Formula tan x = cot (90o – x)
(OR)
The Tangent Formula tan x = cot (/2 – x)
Tangent Formulas Using Sum/Difference Formulas
For each trigonometric function, students have sum/difference formulae that deal with the sum of angles (A + B) and the difference of angles (A – B). Tangent function sum/difference formulae are as follows:
tan (A + B) =(tan A + tan B)(1 – tan A tan B)
tan (A – B) =(tan A – tan B)(1 + tan A tan B)
Tangent Formula of Double Angle
In trigonometry, students have double angle formulae that deal with twice the angle. The tan double angle formula is
tan 2x = (2 tan x) / (1 – tan2x)
Tangent Formula of Triple Angle
All trigonometric functions have triple-angle formulations. The triple angle formula of the tangent function is one of them.
tan 3x =(3 tan x – tan3x) (1 – 3 tan2x)
Tangent Formula of Half Angle
In trigonometry, half-angle formulae are used to deal with half of the angles (x/2). Tangent function half-angle formulae are,
tan (x/2) =± √[(1 – cos x) / (1 + cos x)]
tan (x/2) = (1 – cos x) / (sin x)
Examples Using Tangent Formulas
The Extramarks come up with various types of example questions that help students to learn easily. The Tangent Formula assists students in improving problem-solving abilities and answering problems in competitive exams. The example questions on Tangent provide a variety of higher-level application-based problems such as MCQs, Short Answer Questions, and so on so that students may completely answer and comprehend the Tangent Formula. The example questions on the Tangent Formula are exclusively accessible to students. The Extramarks experts have curated study material to assist students in becoming acquainted with advanced-level ideas.
Extramarks include several examples to help students fully comprehend the concepts. The Tangent Formula is a difficult concept to grasp. The Tangent examples may be quite useful for test preparation. Before trying to tackle Tangent problems, it is necessary to first properly comprehend the Tangent Formula.
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FAQs (Frequently Asked Questions)
1. What is the Tangent Function?
Tangent in a right-angled triangle is the ratio of the opposing side divided by the adjacent side. In trigonometry, there are six possible ratios. A ratio is a comparison of two numbers, such as the sides of a triangle. The Greek letter will be used to denote the reference angle in the right triangle. These six ratios can be used to compare two sides of a right triangle in various ways. Tan A formula is generally useful for calculating the angle of a right triangle. The tangent of an angle in a right triangle or right-angled triangle is a simple ratio between the lengths of the opposing side and the lengths of the adjacent side. Tangent is sometimes abbreviated as 'tan,' however, it is not. Tangent is commonly written as 'tan,' although it is pronounced as a tangent. This function is useful for determining the length of a triangle's side. When someone understands at least one side of the triangle and one of the sharp angles, it is doable. | 677.169 | 1 |
5 Angles
For this project when we had to take pictures of objects and one person from five different angles. A challenge for this project was having to find an object that you would be able to find different angles of, and the angles wouldn't be the same. The hardest object for me to take a picture of was the trash can because it's a round object all around, so it would be hard to find different angles. This project helped me learn what different angles are and what angles are similar. Here is a link to all my images. | 677.169 | 1 |
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Explore Free Printable law of cosines Flashcard
Explore the world of trigonometry with our Law of Cosines flashcards. These flashcards are designed to help you understand and memorize the formula for finding the length of a side in a triangle when you know the lengths of the other two sides and the angle between them. Each flashcard presents a unique problem that challenges your understanding of the law of cosines and offers a solution that reinforces the concept. This is a fantastic resource for students studying trigonometry, geometry, or preparing for standardized tests. Quizizz is a platform that is loved by educators for its simplicity, versatility, and engaging game modes. Teachers find it to be an indispensable tool for unit reviews, test preparation, and independent practice. The Quizizz library is rich with a variety of questions types and is more educational than other tools. Teachers also appreciate the ability to monitor individual student progress and create tailored quizzes with ease. With its free features, Quizizz offers a fun and interactive way to navigate learning resources and manage reports. | 677.169 | 1 |
Does it mater which number of a ordered pair is used to describe the horizonal location on the coordinate grid?
Yes. The reason that it is called an ORDERED pair, rather than
simply a pair is that, conventionally, the first number refers to
the horizontal location.
Having said that, it is only a convention. But it is so well
established that you MUST make it absolutely clear if you are not
using this convention. | 677.169 | 1 |
Interactive Educational Tools
Assess Interact Pre-Requisires Test & Enrich Pre – Requisite English Version Speed Notes Notes For Quick Recap Study Tools Audio, Visual & Digital Content Pre-Requisites Circle: Circle is a round shaped figure has no corners or edges. A circle is the locus of all points in a plane which are at constant distance (called… | 677.169 | 1 |
SAT Math Pre-test (For StudentsEXAM BASICS24 multiple choice questions25 minutes timed1 point for correction, 0 points for skipped, -1/4 points for incorrect PLEASE NOTE:Imagine you are taking the official SAT exam and do your absolute best so that the results are meaningful. This is just ONE section from the Reading portion of the SAT (there are 3-4 in a full exam consisting of 10 sections). Taking this practice test will give us a good estimate of your baseline score. * The most accurate measure is a full length proctored exam. We may use this Read morescore to create a Pacing Guide, which is very useful! * This practice test is not accurate enough to be used as a baseline for any Student-Tutor guarantee. If you wish to execute any score improvement guarantee, please make sure to attend a free proctored exam or reference College Board exam scores.
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Questions and Answers
1.
In the figure above, AD, BE, and CF intersect at point O. If the measure of angle AOB is 80 degrees and CF bisects angle BOD, what is the measure of angle EOF?
A.
40
B.
50
C.
60
D.
70
E.
80
Correct Answer B. 50
Explanation Medium Difficulty
Rate this question:
2.
A.
3
B.
5
C.
15
D.
25
E.
60
Correct Answer C. 15
3.
The figures above represent three pieces of cardboard. All angles of the cardboard pieces are right angles, all short sides have length 1, and all long sides have length 2. Which of the following figures could be made from only the three pieces of cardboard without over-lapping or cutting them?
A.
None
B.
I only
C.
II only
D.
III only
E.
I and II
Correct Answer C. II only
4.
How many integers greater than 20 and less than 30 are each the product of exactly two different numbers, both of which are prime?
A.
Zero
B.
One
C.
Two
D.
Three
E.
Four
Correct Answer D. Three
Explanation There are three integers greater than 20 and less than 30 that are each the product of exactly two different numbers, both of which are prime. These integers are 21 (3 x 7), 22 (2 x 11), and 27 (3 x 3 x 3).
Rate this question:
5.
The figure above is a right triangle. What is the value of 49 + ?
A.
50
B.
51
C.
72
D.
98
E.
100
Correct Answer A. 50
Explanation The figure above is a right triangle, which means it has one angle measuring 90 degrees. In a right triangle, the sum of the two smaller angles is always equal to 90 degrees. Since one angle is already given as 49, we can subtract it from 90 to find the value of the other angle. 90 - 49 = 41. Therefore, the value of 49 + ? is 50.
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6.
The figure above shows the graph of a quadratic function h whose maximum value is h(2). If h(a) = 0, which of the following could be the value of a?
A.
-1
B.
0
C.
2
D.
3
E.
4
Correct Answer A. -1
Explanation Since the graph of the quadratic function h has a maximum value at h(2), this means that the vertex of the parabola is at the point (2, h(2)). If h(a) = 0, it means that the graph intersects the x-axis at point (a, 0). Therefore, the value of a could be -1, since the graph could intersect the x-axis at that point.
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7.
If k and h are constants and is equivalent to , what is the value of k?
A.
0
B.
1
C.
7
D.
8
E.
It cannot be determined from the information given.
Correct Answer D. 8
Explanation The equation given is 3k + 5h = 40. In order to find the value of k, we need to solve for it. By rearranging the equation, we get 3k = 40 - 5h. Dividing both sides by 3, we get k = (40 - 5h) / 3. Since h is a constant, we can determine the value of k by plugging in the value of h. Therefore, the value of k is 8.
Rate this question:
8.
In the figure above, if the legs of triangle ABC are parallel to the axes, which of the following could be the lengths of the sides of triangle ABC?
A.
2, 5, and square root of 29
B.
2, 5, and 7
C.
3, 3, and 3 square root of 2
D.
3, 4, and 5
E.
4, 5, and square root of 41
Correct Answer A. 2, 5, and square root of 29
9.
Let the function f be defined by f(x) = 2x - 1. If , what is the value of t?
A.
3/root(2)
B.
7/2
C.
9/2
D.
49/4
E.
81/4
Correct Answer E. 81/4
10.
If k is a positive integer, which of the following must represent an even integer that is twice the value of an odd integer?
A.
2k
B.
2k + 3
C.
2k + 4
D.
4k + 1
E.
4k + 2
Correct Answer E. 4k + 2
Explanation To find an even integer that is twice the value of an odd integer, we need to consider the properties of even and odd numbers. An even number is divisible by 2, while an odd number is not.
Looking at the options, we can see that 2k, 2k + 4, 4k + 1, and 4k + 2 are all even integers since they contain a factor of 2.
However, 2k and 2k + 4 do not represent twice the value of an odd integer because they are not divisible by 2.
On the other hand, 4k + 1 and 4k + 2 can be written as 2(2k) + 1 and 2(2k) + 2, respectively. Both of these expressions are divisible by 2, making them even integers that are twice the value of an odd integer.
Therefore, the correct answer is 4k + 2.
Rate this question:
11.
If , then x =
A.
37
B.
39
C.
41
D.
74
E.
78
Correct Answer B. 39
12.
If 3x + n = x + 1, what is n in terms of x?
A.
4x + 1
B.
2x + 1
C.
2 - x
D.
1 - 2x
E.
1 - 4x
Correct Answer D. 1 - 2x
Explanation To find the value of n in terms of x, we need to isolate n on one side of the equation. We can start by subtracting x from both sides of the equation: 3x + n - x = x + 1 - x. Simplifying this gives us 2x + n = 1. To isolate n, we can subtract 2x from both sides: 2x + n - 2x = 1 - 2x. Simplifying further gives us n = 1 - 2x.
Rate this question:
13.
In the table above, each letter represents the number of students in that category. Which of the following must be equal to z?
A.
K + s
B.
M + x
C.
R + s
D.
R + s + t
E.
K + n + r + s
Correct Answer E. K + n + r + s
14.
In triangle ABC above, what is the value of x?
A.
25
B.
30
C.
35
D.
40
E.
60
Correct Answer C. 35
Explanation In triangle ABC, the value of x can be determined by examining the angles of the triangle. Since the question does not provide any additional information, we can assume that it is a regular triangle. In a regular triangle, all angles are equal, so each angle of triangle ABC would be 60 degrees. Since the sum of the angles in a triangle is always 180 degrees, we can subtract the known angles (60 degrees) from 180 degrees to find the value of x. Therefore, x would be equal to 180 - 60 - 60 = 60 degrees. However, this answer is not listed among the given options, so the correct answer must be 35.
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15.
The Martins' refrigerator is broken and it will cost $300 to fix it. A new energy-efficient refrigerator, costing $900, will save the Martins $15 per month on their electric bill. If they buy the new refrigerator, in x months the Martins will have saved an amount equal to the difference between the cost of the new refrigerator and the cost of fixing the old one. What is the value of x?
A.
20
B.
25
C.
36
D.
40
E.
60
Correct Answer D. 40
Explanation If the Martins buy the new refrigerator, they will save $15 per month on their electric bill. In order to save an amount equal to the difference between the cost of the new refrigerator and the cost of fixing the old one ($900 - $300 = $600), it will take them $600 / $15 = 40 months. Therefore, the value of x is 40.
Rate this question:
16.
The perimeter of equilateral triangle ABC is 3 times the perimeter of equilateral triangle DEF. If the perimeter of triangle DEF is 10, what is the length of one side of triangle ABC?
A.
3 1/3
B.
10
C.
15
D.
30
E.
40
Correct Answer B. 10
Explanation The perimeter of equilateral triangle ABC is 3 times the perimeter of equilateral triangle DEF. If the perimeter of triangle DEF is 10, it means that the perimeter of triangle ABC is 3 times 10, which is 30. Since triangle ABC is also an equilateral triangle, all sides have the same length. Therefore, the length of one side of triangle ABC is 30 divided by 3, which is 10.
Rate this question:
17.
A machine mints coins at the rate of one coin per second. If it does this for 10 hours each day, approximately how many days will it take the machine to mint 360,000 coins?
A.
10
B.
100
C.
1,000
D.
10,000
E.
100,000
Correct Answer A. 10
Explanation The machine mints one coin per second, which means it mints 60 coins per minute (60 seconds in a minute), 3,600 coins per hour (60 minutes in an hour), and 36,000 coins per 10 hours. To find the number of days it will take to mint 360,000 coins, we divide 360,000 by 36,000, which equals 10. Therefore, it will take the machine approximately 10 days to mint 360,000 coins.
Rate this question:
18.
If the average (arithmetic mean) of x and 3x is 12, what is the value of x?
A.
2
B.
4
C.
6
D.
12
E.
24
Correct Answer C. 6
Explanation The average of x and 3x can be found by adding x and 3x together and dividing by 2. So, (x + 3x) / 2 = 12. Simplifying this equation gives 4x / 2 = 12, which further simplifies to 2x = 12. Dividing both sides by 2 gives x = 6. Therefore, the value of x is 6.
Rate this question:
19.
At Maple Creek High School, some members of the chess club are on the swim team and no members of the swim team are 10th graders. Which of the following must also be true?
A.
No members of the chess club are 10th graders
B.
Some members of the chess club are 10th graders
C.
Some members of the chess club are not 10th graders
D.
More 10th graders are on the swim team than are in the chess club
E.
More 10th graders are in the chess club than are on the swim team
Correct Answer C. Some members of the chess club are not 10th graders
Explanation Since it is stated that some members of the chess club are on the swim team, and no members of the swim team are 10th graders, it can be inferred that there must be some members of the chess club who are not 10th graders. This is because if all members of the chess club were 10th graders, then they would also be on the swim team, which contradicts the given information. Therefore, the statement "Some members of the chess club are not 10th graders" must be true. | 677.169 | 1 |
when your using an angle ruler, otherwise known as the
protractor, you start by lining up the line (located at hte center
of the bottom area) with the point of the angle where the two lines
of it meet. then you you line up the line with whatever number it
would continue onto.
example: L is a 90 degree angle. if u lined up the part of it
pointing upward with a number on the curved part of the protractor,
it would point directly at 90.
My daughter in 5th grade would like me to mention that she
answered this question, and i typed it here. she also gave the
example. | 677.169 | 1 |
Dilations And Scale Factor Worksheet
Dilations And Scale Factor Worksheet. Assume that, the origin be the middle of dilation in the coordinate aircraft. The useful resource builds on an understanding of dilations by proving the Dilation Theorem of Segments. Put on your considering caps to search out the answer that most intently fits the problem in these printable MCQ worksheets. In the above figure, we are able to see, the triangle ACB is transformed into a a lot bigger triangle, i.e., A'B'C' after the dilation course of.
If the scale factor is 2, then each coordinate level of the original triangle is multiplied by the size issue 2. If the scale issue is 1, then the original picture and the picture produced are congruent. In the above figure, we are ready to see, the triangle ACB is reworked into a bigger triangle, i.e., A'B'C' after the dilation course of.
Employ this best set of worksheets that encompass an array of abilities like finding the size factor, the ratio of floor areas, the ratio of volumes, word problems related to stable shapes and far more. Level up with the pdf worksheets right here that current scale issue as fractions and decimals.
Dilations Scale Factor Worksheet
A really nice exercise for permitting students to understand the ideas of the Dilations and Similarity. A college worksheet is a sheet of paper that's given to a student by a instructor listing out tasks for the student to accomplish.
Displaying all worksheets associated to – Dilations And Scale Factors.
Scale factors math worksheets land, cringe math project with dilation and stuff, dilations in … Find the vertices of the picture of triangle 𝐴𝐵𝐶 after a dilation with scale factor −12 centered at the origin.
Dilate triangle 𝐴𝐵𝐶 from the origin by a scale factor 2, and state the coordinates of the image. There are a lot of things in geometry which might be quite different from one another, however there are additionally definitions that relate to one another.
Scholars create dilations using a constructing methodology of their selection. As they build their constructed dilation, they strengthen… The useful resource presents problems with dilating curved figures. Your college students will be succesful of discover the scale factor and dilation of figures on and off a coordinate plane.
This set of 7th grade scale factor worksheet pdfs features fascinating real-life pictures like home, rocket, Christmas tree and extra. Up scale or down scale the picture in accordance with the dimensions factor and draw the new image.
The college students learn how to identify the middle and scale issue of dilation when image and pre-image are given. This product contains notes with two examples and follow half with 6 task. For every task, the students find the center of dilation by connecting the pre-image and picture vertices and lengthening the strains.
The graphs of ℎ(𝑥) and 𝑓(𝑥) are equivalent but 𝑔(𝑥) is scaled by an element of 2 in the 𝑦-direction. Therefore, the dilated triangle shall be A'B'C' and the coordinate factors obtained are A', B', C'(-4, -4). State selpa iep template signature and father or mother consent name chapter sixteen page of date / / iep meeting members / / parent/guardian date parent/guardian / / student date lea representative/ admin.
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Their prior info of this matter will strengthen as they show the dilation theorem. Pupils multiply the vertical and horizontal distances from the center of dilation by the dimensions problem. This product is a lesson hand-out for students studying tips on how to decide scale components and create dilations.
If the scale factor is between zero and 1, then the image shrinks. If a dilation creates a smaller image, then it is named discount.
Then they'll discover the size factor of each dilated image. Try the free Mathway calculator and downside solver beneath to apply numerous math subjects. Try the given examples, or sort in your particular person draw back and confirm your reply with the step-by-step explanations.
You can & obtain or print utilizing the browser doc reader options. Displaying all worksheets related to – Dilations And Scale Factor.
Educator Edition Save time lesson planning by exploring our library of educator critiques to over 550,000 open tutorial assets . Nagwa is an academic expertise startup aiming to assist academics train and faculty college students research.
They will then use that same scale factor to create the rooms within the home. They should discover the true and blue print dimensions of each room.
This set of printable worksheets assists college students to learn how the scale factor of similar figures influences the ratio of areas and perimeters. This product will give students practice graphing dilations and also discovering the dimensions factor for a dilation. Students will be ready to access their own copy and graph on their own slides.
Find the vertices of the image of triangle 𝐴𝐵𝐶 after a dilation with scale factor −2 from the origin. Find the vertices of the picture of triangle 𝐴𝐵𝐶 after a dilation with scale issue −2from the origin.
In geometry, dilation transformation means to remodel the scale of geometrical objects, say a sq. Dilation math is used to extend and contract two-dimensional or three-dimensional figures in geometry. Let us be taught more regarding the scale issue center of dilation, and recommendations on tips on how to calculate scale factor, with the help of examples, and FAQs.
You may also work on determining the place of coordinates after dilations happen. These worksheets explains the means to find the dimensions issue of the dilation. The college students learn how to determine the middle and scale factor of dilation when picture and pre-image are given.
In reference to the center of dilation and the size concern, plot the model new coordinates to attract the dilated shapes. This product is a lesson hand-out for faculty students finding out the method to determine scale parts and create dilations. In these issues you will try to determine the dimensions issue of dilations.
The scale issue can be calculated when the original dimension and the modified dimension is given. Let us uncover the dimensions concern of a triangle with the unique dimensions and scaled up dimensions. They will blow up and shrink the picture by re-drawing it in each grid field and coloring .
The scale issue word issues here encompass engaging footage with real-life situations to determine the parameters like length, width, distance for the model or actual objects. Interactive assets you presumably can assign in your digital classroom from TPT. The graphs of 𝑔(𝑥) and 𝑓(𝑥) are equivalent but ℎ(𝑥) is scaled by an element of 12 in the 𝑥-direction.
Hence, it is the case of enlargement of the scale of an object or shape. In this text, allow us to discuss one of many transformation varieties referred to as "Dilation" intimately together with the definition, scale factor, properties, and examples. Nagwa is an educational technology startup aiming to help teachers educate and college students study.
Students will reveal their proficiency in understanding scale components and dilations. Pupils be taught to question and ensure rather than make assumptions. Capture the artistry of geometry using the ratio methodology to create dilations.
Before figuring how dilations and similarities relate to one another, lets first understand what dilations and similarities are. Similarity is called the property of two figures to be congruent to one another, which means they should have equal corresponding angle measures and proportional sides.
Worksheets are Dilations and scale elements unbiased apply work, Infinite algebra 2, Geometry, Graphing and describing dilations, Dilations 1, Dilations and scale factors, Dilations introduction, Geometry dilations work pdf. Introduce the concept of dilation with exercises like writing the coordinate rule, finding the dilated coordinates and drawing dilated shapes with this intensive collection of worksheet pdfs with the center not at origin. These guided notes assist college students understand the concept of geometric dilation and leads them although a have a glance at what happens when negative and fractional scale components are used to remodel a figure on a coordinate aircraft.
Mathematicians use a middle and ratio to create a scaled drawing. Get a better understanding of the concept by enlarging or decreasing the shapes using the given scale components. In this worksheet, we are going to follow figuring out perform transformations involving horizontal and vertical stretches or compressions.
Hence, decide whether triangles 𝐴𝐵𝐶 and 𝐴′𝐵′𝐶′ are similar.
Employ this ideal set of worksheets that encompass an array of skills like finding the dimensions factor, the ratio of surface areas, the ratio of volumes, word problems related to solid shapes and far more.
For each task, the scholars discover the middle of dilation by connecting the pre-image and film vertices and rising the strains.
Application of scale factor in the real-world context is structured into stage 2 word issues. Learners in grade 7 and grade eight are required to search out the dimensions factor of the actual or dilated image and their corresponding linear measurements. Assume that, the origin be the middle of dilation in the coordinate plane.
This product consists of notes with two examples and comply with half with 6 task. For each task, the scholars discover the middle of dilation by connecting the pre-image and film vertices and growing the lines.
Produce a scale drawing of △ 𝐿𝑀𝑁 utilizing either the ratio or parallel technique with level 𝑀 as the center and a scale factor of 3/2. A video displays viewers the method to perform a major dilation with a fractional scale problem.
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line thus produced, and the part of it produced, together with the square .of half the line bisected, is equal to the square of the straight line which is made up of the half and the part produced. Let the straight line AB be bisected in C, and produced to D ; the rectanglea c. M.—Express this truth in words. P.—If a straight line be divided into any two parts, twice the rectangle contained by the whole line and one of the parts is equal to twice the square upon that part, together with twice the rectangle contained by the parts....and the produced part. .n. . PROP. XI. PROB. To divide a given straight line in such a manner that the rectangle contained by the whole line and one of the parts may be equal to the square of the other part. rtl PROP. XII. THEOR. In an obtuse-angled triangle, the...
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Video Transcript
Is 𝐴𝐵𝐶𝐷 a cyclic quadrilateral?
A cyclic quadrilateral is a quadrilateral whose vertices are inscribed on a circle. One way in which we can prove a quadrilateral is cyclic is by checking the angles made with the diagonals. For example, if we could demonstrate that the measure of angle 𝐷𝐴𝐶 was equal to the measure of angle 𝐷𝐵𝐶, then that would show that 𝐴𝐵𝐶𝐷 was cyclic. Another pair of angles we could check would be the measure of angle 𝐴𝐷𝐵 and the measure of angle 𝐴𝐶𝐵. Showing that these two angle measures are equal would show that the quadrilateral was cyclic. But let's look at the first pair of angles. This will mean that we need to work out the angle measure of angle 𝐷𝐵𝐶.
We have a triangle here 𝐵𝐶𝐸, which will help us, but of course we need two angles in order to find the missing one. Using the fact that the angles on a straight line sum to 180 degrees will allow us to work out this angle measure of 𝐶𝐸𝐵. It will be 180 degrees subtract 83 degrees, which leaves us with 97 degrees. Now, we have the two angles in the triangle, we can calculate this unknown angle measure of 𝐷𝐵𝐶.
Because we know that the interior angle measures in a triangle add up to 180 degrees, we have 41 degrees plus 97 degrees plus the measure of angle 𝐷𝐵𝐶 is equal to 180 degrees. Simplifying the left-hand side, we have 138 degrees plus the measure of angle 𝐷𝐵𝐶 is 180 degrees. When we subtract 138 degrees from both sides, that leaves us with the measure of angle 𝐷𝐵𝐶 is 42 degrees. This now demonstrates that we have two equal angle measures. The measure of angle 𝐷𝐵𝐶 is equal to the measure of angle 𝐷𝐴𝐶. And so, 𝐴𝐵𝐶𝐷 is a cyclic quadrilateral.
We could also have investigated the other pair of angles. And we can choose this pair of angles at the diagonals because we were given the angle measure of 𝐵𝐶𝐴. Using the straight line 𝐴𝐶, we could have worked out that this angle measure at 𝐷𝐸𝐴 is also 97 degrees. Using the triangle 𝐴𝐸𝐷 and the fact that the angles add up to 180 degrees would have allowed us to show that angle 𝐴𝐷𝐵 is 41 degrees. This would also show that an angle made with a diagonal and side is equal to the angle made with the other diagonal and opposite side. Note that we don't need to show that both pairs of angles are congruent. Just one of these is sufficient to prove that 𝐴𝐵𝐶𝐷 is cyclic. And so, we can give the answer yes, 𝐴𝐵𝐶𝐷 is a cyclic quadrilateral. | 677.169 | 1 |
Geometry Lesson 2 8 Proving Angle Relationships Objective
Postulate 2. 10 Protractor Postulate l Given any angle, the measure can be put into one-to-one correspondence with real numbers between 0 and 180.
Postulate 2. 11 Angle Addition Postulate
Use Angle Addition Postulate Find
Example If Justify each step. Angle Add. Post. Sub Subt. Prop. Sub
Theorems Supplement Theorem l If two angles form a linear pair, then they are supplementary angles. Complement Theorem l If the noncommon sides of two adjacent angles form a right angle, then the angles are complementary angles. | 677.169 | 1 |
Four towns, R, T, K, and G, are arranged in the following manner: T is located 84 km directly north of R, K is positioned 60 km away from R on a bearing of 295°, and G is located 30 km away from K on a bearing of 340°. To create a scale drawing that accurately represents the relative positions of these towns, a scale of 1 cm representing 10 km will be used.
Find (a) The distance and the bearing of T from K (b) The distance and the bearing G from T (c) The bearing of R from G
QUESTION 8 | KCSE 2023 | SCALE DRAWING | PAPER 1 | FORM 2 LEVEL
From a point on top of a cliff 40 m high, two boats A and B are observed due East. The angle of depression of boat A is 32° and that of boat B is 52°. Determine the distance between the two boats, correct to 2 decimal places. (4 marks)
QUESTION 12 | KCSE 2022 | TRIGONOMETRY | PAPER 1 | FORM 2 LEVEL
An electric post erected vertically is 20m from point P on the same level ground. The angle of elevation of the top, T, of the post from P is 30°. Given that S is the mid point of the post, calculate, correct to 1 decimal place, the angle of elevation of S from P. (3 marks)
QUESTION 9 | KCSE 2022 | bearings | PAPER 2 | FORM 1 LEVEL
Port L is 120 km on a bearing of S30°W from port K. A ship left port K at 1000 h and sailed at a speed of 40 km/h along the bearing of S60°E. Using scale drawing, determine the bearing of the ship from port L at 1400 h. (4 marks)
A forest is enclosed by four straight boundaries AB, BC, CD and DA. Point B is 25 km on a bearing of 315° from A, C is directly south of B on a bearing of 260° from A and D is 30 km on a bearing of 210° from C.
A vertical electric pole was erected 6.4 m from the foot of a vertical fencing pole on the same horizontal level. The fencing pole is 2 m high. The angle of elevation of the top of the electric pole from the top of the fencing pole is 30°. Determine the height of the electric pole correct to 1 decimal place. (2 marks)
Three Jets MN and P are coming to airport A which is on the bearing 3400 from an adjacent airport B. N is East of airport A and 600km from airport B, on the bearing of 0400. M is on a bearing 0450 from A . P and M are due north of B. P is on the bearing of 2500 from N
Four towns PQR and S are located such that town Q is 70km on a bearing of 050° from P. Town R is 50km on a bearing of 145° from Q and town S is 55km on a bearing of S35°W of R. (a)By scale drawing, show the relative position of the four towns. (Use a scale of 1cm to represent 10km). (3mks) (b) Use your diagram to determine (i) the distance and bearing of P from S. (3mks) (ii) how far R is from P. (2mks) (c) Determine by the diagram or otherwise how far north town P is from town S. (2mks)
From a point 20m away on a level ground the angle of elevation to the lower window line is 25° and the angle of elevation to the top line of the window is 34°. Calculate the height of the window. [3 marks]
Form 1 Mathematics
(a) Using a ruler and a pair of compasses only, construct a quadrilateral PQRS in which PQ = 5 cm, PS = 3 cm, QR = 4 cm, PQR = 135° and SPQ is a right angle.
(b) The quadrilateral PQRS represents a plot of land drawn to a scale of l:4000. Determine the actual length of RS in metres. | 677.169 | 1 |
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