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For each of the following, draw the terminal side of the indicated angle on a coordinate system and determine the values of the six trigonometric functions of that angle
The terminal side of the angle \(\alpha\) is in the first quadrant and \(\sin(\alpha) = \dfrac{1}{\sqrt{3}}\)
The terminal side of the angle \(\beta\) is in the second quadrant and \(\cos(\beta) = -\dfrac{2}{3}\)
The terminal side of the angle \(\gamma\) is in the second quadrant and \(\tan(\gamma) = -\dfrac{1}{2}\)
The terminal side of the angle \(\theta\) is in the second quadrant and \(\sin(\theta) = \dfrac{1}{3}\)
Answer
2. We first use the Pythagorean Identity and obtain \(\sin^{2}(\beta) = \dfrac{5}{9}\). Since the terminal side of \(\beta\) is in the second quadrant, \(\sin(\beta) = \dfrac{\sqrt{5}}{3}\). In addition,
\[\tan(\beta) = -\dfrac{\sqrt{5}}{2}\]
\[\cot(\beta) = -\dfrac{2}{\sqrt{5}}\]
\[\sec(\beta) = -\dfrac{3}{2}\]
\[\csc(\beta) = \dfrac{3}{\sqrt{5}}\]
Exercise \(\PageIndex{3}\)
For each of the following, determine an approximation for the angle \(\theta\) in degrees (to three decimal places) when \(0^\circ \leq \theta \leq 360^\circ\).
The point \((3, 5)\) is on the terminal side of \(\theta\).
The point \((2, -4)\) is on the terminal side of \(\theta\).
\(\sin(\theta) = \dfrac{2}{3}\) and the terminal side of \(\theta\) is in the second quadrant.
\(\sin(\theta) = -\dfrac{2}{3}\) and the terminal side of \(\theta\) is in the fourth quadrant.
\(\cos(\theta) = -\dfrac{1}{4}\) and the terminal side of \(\theta\) is in the second quadrant.
\(\cos(\theta) = -\dfrac{3}{4}\) and the terminal side of \(\theta\) is in the third quadrant.
Answer
3. Since the terminal side of \(\theta\) is in the second quadrant, \(\theta\) is not the inverse sine of \(\dfrac{2}{3}\). So we let \(\alpha = \arcsin(\dfrac{2}{3})\). Using \(\alpha\) as the reference angle, we then see that \[\theta = 180^\circ - \arcsin(\dfrac{2}{3}) \approx 138.190^\circ\]
For each of the following right triangles, determine the value of \(x\) correct to the nearest thousandth.
Answer
\(x = 6\tan(47^\circ) \approx 6.434\).
\(x = 3.1\cos(67^\circ) \approx 1.211\).
\(x = \tan^{-1}(\dfrac{7}{4.9}) \approx 55.008^\circ\).
\(x = \sin^{-1}(\dfrac{7}{9.5}) \approx 47.463^\circ\).
Exercise \(\PageIndex{7}\)
One angle in a right triangle is \(55^\circ\) and the side opposite that angle is 10 feet long. Determine the length of the other side, the length of the hypotenuse, and the measure of the other acute angle.
Exercise \(\PageIndex{8}\)
One angle in a right triangle is \(37.8^\circ\) and the length of the hypotenuse is 25 inches. Determine the length of the other two sides of the right triangle.
Exercise \(\PageIndex{9}\)
One angle in a right triangle is \(27^\circ 12'\) and the length of the side adjacent to this angle is 4 feet. Determine the other acute angle in the triangle, the length of the side opposite this angle, and the length of the hypotenuse. Note: The notation means that the angle is 27 degrees, 12 seconds. Remember that 1 second is \(\dfrac{1}{60}\) of a degree.
Answer
The other acute angle is \(64^\circ 48'\).
The side opposite the \(27^\circ 12'\) angle is \(4\tan(27^\circ 12') \approx 2.056\) feet.
Note that the Pythagorean Theorem can be used to check the results by showing that \(4^{2} + 2.056^{2} \approx 4.497^{2}\). The check will not be exact because the \(2.056\) and \(4.497\) are approximations of the exact values.
Exercise \(\PageIndex{10}\)
If we only know the measures of the three angles of a right triangle, ex- plain why it is not possible to determine the lengths of the sides of this right triangle.
Exercise \(\PageIndex{11}\)
Suppose that we know the measure \(\theta\) of one of the acute angles in a right triangle and we know the length \(x\) of the side opposite the angle \(\theta\). Explain how to determine the length of the side adjacent to the angle \(\theta\) and the length of the hypotenuse.
Exercise \(\PageIndex{12}\)
In the diagram to the right, determine the values of \(a\), \(b\), and \(h\) to the nearest thousandth.
The given values are:
Figure \(\PageIndex{1}\)
\[\alpha = 23^\circ\]
\[\beta = 140^\circ\]
\[c = 8\]
Answer
We first note that \(\theta = 40^\circ\). We use the following two equations to determine \(x\).
We can then use right triangles to obtain \(h \approx 6.872\space ft, a \approx 10.691\space ft, and b \approx 17.588\space ft\)
Exercise \(\PageIndex{13}\)
A tall evergreen tree has been damaged in a strong wind. The top of the tree is cracked and bent over, touching the ground as if the trunk were hinged. The tip of the tree touches the ground 20 feet 6 inches from the base of the tree (where the tree and the ground meet). The tip of the tree forms an angle of 17 degrees where it touches the ground. Determine the original height of the tree (before it broke) to the nearest tenth of a foot. Assume the base of the tree is perpendicular to the ground.
Exercise \(\PageIndex{14}\)
Suppose a person is standing on the top of a building and that she has an instrument that allows her to measure angles of depression. There are two points that are 100 feet apart and lie on a straight line that is perpendicular to the base of the building.
Now suppose that she measures the angle of depression to the closest point to be \(35.5^\circ\) and that she measures the angle of depression to the other point to be \(29.8^\circ\). Determine the height of the building.
Exercise \(\PageIndex{15}\)
A company has a 35 foot ladder that it uses for cleaning the windows in their building. For safety reasons, the ladder must never make an angle of more than \(50^\circ\) with the ground.
What is the greatest height that the ladder can reach on the building if the angle it makes with the ground is no more than \(50^\circ\).
Suppose the building is 40 feet high. Again, following the safety guidelines, what length of ladder is needed in order to have the ladder reach the top of the building?
Figure \(\PageIndex{2}\)
Note
For Exercises \(\PageIndex{16}\) through \(\PageIndex{19}\), use the Law of Sines.
Exercise \(\PageIndex{16}\)
Two angles of a triangle are \(42^\circ\) and \(73^\circ\). The side opposite the \(73^\circ\) angle is 6.5 feet long. Determine the third angle of the triangle and the lengths of the other two sides.
Answer
The third angle is \(65^\circ\). The side opposite the \(42^\circ\) angle is \(4.548\) feet long. The side opposite the \(65^\circ\) angle is \(6.160\) feet long.
Exercise \(\PageIndex{17}\)
A triangle has a side that is 4.5 meters long and this side is adjacent to an angle of \(110^\circ\). In addition, the side opposite the \(110^\circ\) angle is 8 meters long. Determine the other two angles of the triangle and the length of the third side.
Exercise \(\PageIndex{18}\)
A triangle has a side that is 5 inches long that is adjacent to an angle of \(61^\circ\). The side opposite the \(61^\circ\) angle is 4.5 inches long. Determine the other two angles of the triangle and the length of the third side.
Answer
There are two triangles that satisfy these conditions. The sine of the angle opposite the \(5\) inch side is approximately \(0.9717997\).
Exercise \(\PageIndex{19}\)
In a given triangle, the side opposite an angle of \(107^\circ\) is 18 inches long. One of the sides adjacent to the \(107^\circ\) angle is 15.5 inches long. Determine the other two angles of the triangle and the length of the third side.
Note
For Exercises \(\PageIndex{20}\) through \(\PageIndex{21}\), use the Law of Cosines.
Exercise \(\PageIndex{20}\)
The three sides of a triangle are 9 feet long, 5 feet long, and 7 feet long. Determine the three angles of the triangle.
Answer
The angle opposite the \(9\) foot long side is \(95.739^\circ\). The angle opposite the \(7\) foot long side is \(50.704^\circ\). The angle opposite the \(5\) foot long side is \(33.557^\circ\).
Exercise \(\PageIndex{21}\)
A triangle has two sides of lengths 8.5 meters and 6.8 meters. The angle formed by these two sides is \(102^\circ\). Determine the length of the third side and the other two angles of the triangle
Note
For the remaining exercises, use an appropriate method to solve the problem.
Exercise \(\PageIndex{22}\)
Two angles of a triangle are \(81.5^\circ\) and \(34^\circ\). The length of the side opposite the third angle is 8.8 feet. Determine the third angle and the lengths of the other two sides of the triangle.
Exercise \(\PageIndex{23}\)
In the diagram below, determine the value of \(\gamma\) (to the nearest hundredth of a degree) and determine the values of \(h\) and \(d\) (to the nearest thousandth) if it is given that \[a = 4\]\[b = 8\]\[c = 10\]\[\theta = 26^\circ\]
Figure \(\PageIndex{3}\)
Exercise \(\PageIndex{24}\)
In the diagram below, it is given that:
Figure \(\PageIndex{4}\)
The length of \(AC\) is \(2\).
The length of \(BC\) is \(2\).
\(\angle ACB = 40^\circ\)
\(\angle CAD = 20^\circ\)
\(\angle CBD = 45^\circ\)
Determine the lengths of \(AB\) and \(AD\) to the nearest thousandth.
Exercise \(\PageIndex{25}\)
A ski lift is to be built along the side of a mountain from point \(A\) to point \(B\) in the following diagram. We wish to determine the length of this ski lift.
Figure \(\PageIndex{5}\)
A surveyor determines the measurement of angle \(BAC\) to be \(155.6^\circ\) and then measures a distance of 800 ft from Point \(A\) to Point \(C\). Finally, she determines the measurement of angle \(BCA\) to be \(17.2^\circ\). What is the the length of the ski lift (from point \(A\) to point \(B\))?
Answer
The ski lift is about \(1887.50\) feet long.
Exercise \(\PageIndex{26}\)
A boat sails from Muskegon bound for Chicago, a sailing distance of 121 miles. The boat maintains a constant speed of 15 miles per hour. After encountering high cross winds the crew finds itself off course by \(20^\circ\) after 4 hours. A crude picture is shown in the following diagram, where \(\alpha = 20^\circ\).
Figure \(\PageIndex{6}\)
How far is the sailboat from Chicago at this time?
What is the degree measure of the angle \(\gamma\) (to the nearest tenth) in the diagram? Through what angle should the boat turn to correct its course and be heading straight to Chicago?
Assuming the boat maintains a speed of \(15\) miles per hour, how much time have they added to their trip by being off course?
Answer
The boat is about \(67.8\) miles from Chicage.
\(\gamma \approx 142.4^\circ\). So the boat should turn through an angle of about \(180^\circ - 142.4^\circ = 37.6^\circ\)
The direct trip from Muskegon to Chicago would take \(\dfrac{121}{15}\) hours or about \(8.07\) hours. By going off-course, the trip now will take \(\dfrac{127.8}{15}\) hours or about \(8.52\) hours.
Exercise \(\PageIndex{27}\)
Figure \(\PageIndex{7}\)
Two trees are on opposite sides of a river. It is known that the height taller of the shorter of the two trees is \(13\) meters. A person makes the following angle measurements:
The angle of elevation from the base of the shorter tree to the top of the taller tree is \(\alpha = 20^\circ\).
The angle of elevation from the top of the shorter tree to the top of the taller tree is \(\beta = 12^\circ\).
Determine the distance between the bases of the two trees and the height of the taller tree.
Exercise \(\PageIndex{28}\)
One of the original Seven Wonders of the World, the Great Pyramid of Giza (also known as the Pyramid of Khufu or the Pyramid of Cheops), was believed to have been built in a \(10\) to \(20\) year period concluding around \(2560\) B.C.E. It is also believed that the original height of the pyramid was \(146.5\) meters but that it is now shorter due to erosion and the loss of some topmost stones.
To determine its current height, the angle of elevation from a distance of \(30\) meters from the base of the pyramid was measured to be \(46.12^\circ\), and then the angle of elevation was measured to be \(40.33^\circ\) from a distance of \(60\) meters from the base of the pyramid as shown in the following diagram. Use this information to determine the height \(h\) of the pyramid. (\(138.8\) meters)
Figure \(\PageIndex{8}\)
Exercise \(\PageIndex{29}\)
Two sides of a triangle have length \(2.5\) meters and \(3.5\) meters, and the angle formed by these two sides has a measure of \(60^\circ\). Determine the area of the triangle. Note: This is the triangle in Progress Check 3.17 on page 200.
Figure \(\PageIndex{9}\)
Exercise \(\PageIndex{30}\)
A field has the shape of a quadrilateral that is not a parallelogram. As shown in the following diagram, three sides measure \(50\) yards, \(60\) yards, and \(70\) yards. Due to some wetland along the fourth side, the length of the fourth side could not be measured directly. The two angles shown in the diagram measure \(127^\circ\) and \(132^\circ\).
Figure \(\PageIndex{10}\)
Determine the length of the fourth side of the quadrilateral, the measures of the other two angles in the quadrilateral, and the area of the quadrilateral. Lengths must be accurate to the nearest hundredth of a yard, angle measures must be correct to the nearest hundredth of a degree, and the area must be correct to the nearest hundredth of a square yard.
Exercise \(\PageIndex{31}\)
In each of the following diagrams, one of the vectors \(\textbf{u}\), \(\textbf{v}\), and \(\textbf{u} + \textbf{v}\) is labeled. Label the vectors the other two vectors to make the diagram a valid representation of \(\textbf{u} + \textbf{v}\)
In the following diagram, \(|\textbf{a}| = 10\) and \(|\textbf{a}| + |\textbf{b}| = 14\). In addition, the angle \(\theta\) between the vectors \(\textbf{a}\) and \(\textbf{b}\) is \(30^\circ\). Determine the magnitude of the vector \(\textbf{b}\) and the angle between the vectors \(\textbf{a}\) and \(\textbf{a} + \textbf{b}\).
Figure \(\PageIndex{13}\)
Answer
The angle between the vectors \(\textbf{a}\) and \(\textbf{a} + \textbf{b}\) is approximately \(9.075^\circ\). In addition, \(|\textbf{b}| \approx 4.416\).
Exercise \(\PageIndex{34}\)
Suppose that vectors \(\textbf{a}\) and \(\textbf{b}\) have magnitudes of 125 and 180, respectively. Also assume that the angle between these two vectors is \(35^\circ\). Determine the magnitude of the vector \(\textbf{a} + \textbf{b}\) and the measure of the angle between the vectors \(\textbf{a}\) and \(\textbf{a} + \textbf{b}\).
Exercise \(\PageIndex{35}\)
A car that weighs \(3250\) pounds is on an inclined plane that makes an angle of \(4.5^\circ\) with the horizontal. Determine the magnitude of the force of the car on the inclined plane, and determine the magnitude of the force on the car down the plane due to gravity. What is the magnitude of the smallest force necessary to keep the car from rolling down the plane?
Exercise \(\PageIndex{36}\)
An experiment determined that a force of \(45\) pounds is necessary to keep a \(250\) pound object from sliding down an inclined plane. Determine the angle the inclined plane makes with the horizontal.
Exercise \(\PageIndex{37}\)
A cable that can withstand a force of \(4500\) pounds is used to pull an object up an inclined plane that makes an angle of 15 degrees with the horizontal. What is the heaviest object that can be pulled up this plane with the cable? (Assume that friction can be ignored.)
Exercise \(\PageIndex{38}\)
Determine the magnitude and the direction angle of each of the following vectors | 677.169 | 1 |
A student is standing with a banner at the top of a 100 m high college building. From a point on the ground, the angle of elevation of the top of the student is 60° and from the same point, the angle of elevation of the top of the tower is 45°. Find the height of the student.
A.
35 m
B.
73.2 m
C.
50 m
D.
75m
Detailed Solution for Test: Trigonometry- 1 - Question 1
Let BC be the height of the tower and DC be the height of the student.
In rt. ΔABC,
AB = BC cot 45° = 100 m
Two poles of equal height are standing opposite to each other on either side of a road which is 100 m wide. Find a point between them on road, angles of elevation of their tops are 30∘ and 60∘. The height of each pole in meter, is:
Anil looked up at the top of a lighthouse from his boat and found the angle of elevation to be 30 degrees. After sailing in a straight line 50 m towards the lighthouse, he found that the angle of elevation changed to 45 degrees. Find the height of the lighthouse.
A.
25
B.
25√3
C.
25(√3-1)
D.
25(√3+1)
Detailed Solution for Test: Trigonometry- 1 - Question 10
If we look at the above image, A is the previous position of the boat. The angle of elevation from this point to the top of the lighthouse is 30 degrees.
After sailing for 50 m, Anil reaches point D from where the angle of elevation is 45 degrees. C is the top of the lighthouse.
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Trigonometric Meaning In Urdu
Trigonometric Meaning in English to Urdu is مُثَلّثاتی تفاعَل, as written in Urdu and , as written in Roman Urdu. There are many synonyms of Trigonometric which include Algebraic, Algorithmic, Analytical, Arithmetical, Geometrical, Math, Measurable, Numerical, Scientific, Computative, etc.
Synonyms For Trigonometric , Similar to Trigonometric
Trigonometric Urdu Meaning - Find the correct meaning of Trigonometric in Urdu, it is important to understand the word properly when we translate it from English to Urdu. There are always several meanings of each word in Urdu, the correct meaning of Trigonometric in Urdu is مُثَلّثاتی تفاعَل, and in roman we write it .
The other meanings are , , , , Qatay and Mamaas. There are also several similar words to Trigonometric in our dictionary, which are Algebraic, Algorithmic, Analytical, Arithmetical, Geometrical, Math, Measurable, Numerical, Scientific and Computative. After English to Urdu translation of Trigonometric | 677.169 | 1 |
In this routine the quality of each element is determined. To this end the
ratio of the largest edge to the radius of the inscribed sphere is used. One can prove that the radius of the inscribed sphere of a
linear tetrahedral is three times the volume divided by the sum of the area of
its faces
[25]. Therefore, the quality for element can be written as:
(715)
The factor
is such that the quality of an equilateral
tetrahedron is 1. For all other tetrahedra it exceeds 1. The larger the value,
the worse the element. The cut-off of was introduced to avoid
dividing by zero or getting a negative value. | 677.169 | 1 |
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Similarity Example Problems | Similarity | Geometry | Khan Academy
Transcript:
In the first task we want to find segment length, segment CE. We have these two parallel lines. AB is parallel to DE. We have these two intersecting, of these two triangles. Let's see what we can do. The first thing that may come to mind is that this angle and this angle are opposite (vertex) angles. So, they will be equal. Another thing you can think of is that the angle CDE is named after the angle CBA. We have this intersection here and these will be cross angles – that is, they will be equal. If we continue this intersection, we will get the corresponding angle with CDE here. And this one is just the opposite. Either way, this angle and this angle will be equal. So, we found that we have two triangles. And they have two corresponding angles that are equal. That, in itself, suffice it to say that triangles are similar. In fact – In fact, we can say that this and that angle are also equal as cross angles. But it is not necessary. We already know that they are similar. In fact, we can say that at the cross angles, they will also be equal. But we already know enough to say they're similar, even before we do that. We have already met this triangle. I will try to color it so we have it identical corresponding vertices. It is very important to know which angles correspond to which countries so as not to confuse the proportions, and to know what corresponds to each other. Thus, we know that triangle ABC is similar to triangle .. And this peak A corresponds to peak E right here – is similar to peak E. A peak B here corresponds to vertex D, EDC. How does this help us? This tells us the proportions of the respective countries will be equal. They will be equal. They will be a constant. So, we have a corresponding country .. The proportion, for example, the corresponding side of the BC will be DC. We can see it. Just about the way we proved the similarity. If this is true, then BC is corresponding to the DC side We know that BC, the length of BC on DC, right here, will be equal to the length of .. First we have to find how much CE is, this is what interests us. I use BC and DC because I know their values. Thus, BC over DC will be equal to of the respective CE country. The relevant country here will be CA .. It will be equal to CA on CE corresponding countries. This is the last and the first, the last and the first ARE on CE. We know how many aircraft, The aircraft here is 5. We know how much DC is She is 3. We know how many CAs or ACs are here, CA is 4 And now we can calculate the CE. We can there are many ways you can calculate it. Multiply by the cross, which is actually a multiplication of the numerator by the denominator. We get 5 along the length of the CE, which is equal to 3 by 4, which in turn is equal to 12. that's how we get CE. CE is equal to 12 over 5. This is the same as 2 and 2 fifths. or 2.4 This will be 2 and 2 fifths We're ready. We used the similarity to get this country, as we simply knew that the proportions between the respective countries will be the same. Let's solve this problem here now. Let's look at this one here. I will draw a small line here. This is a different task. Here we want to find how much is DE. Again we have these two parallel lines. We know that the corresponding angles are equal. We know that this angle will be equal to this angle. Because this can be considered intersecting here. Also, we know this angle here will be equal to this angle here. Once again, we have corresponding angles for the intersections. In both triangles. I'm looking at CBD and CAE triangles. They share a common corner here. In fact, we showed once again that we can stop at two angles. We really showed that all three angles of these two triangles, all three corresponding angles, are equal to each other. Now we know .. The important thing we need to do is once again to make sure we get what you write, in the right order, when we prove the similarity. Now, we know that triangle CBD is similar, not equal, is similar to triangle CAE. This means that the relationship between the parties concerned will be a constant. We know, for example, that the relationship between CDs it will be the same such as the relationship between CB to CA. Let's write it. We know that CB on CA will be equal to the ratio between CD on CE. We know that CB is, CB here is 5. We know how much is CA? We have to be very careful now. Not 3. SA, this whole country will be 5 plus 3 This is equal to 8. We also know that CD is, CD will be 4. Now we can multiply crosswise again. We have 5 CE, 5 CE equals 8 to 4 8 by 4 is 32. So CE is equal to 32 over 5. Or we can present it in another way. 6 and 2 fifths. We are not ready because we are not looking for how much CE is. Only this part is wanted here. Search with is DE. We know the whole length, CE here. She is 6th and 2nd fifth. So, DE is here what we need to find. It will be equal to this length, 6 and 2 fifth minus, minus CD. This will be equal to 2 and 2 fifths. 6 and 2 fifths minus 4 is 2 and 2 fifths. We're ready! DE is 2 and 2 fifths | 677.169 | 1 |
(b) State and prove the theorem on the sum of the angles of any polygon.
2. If two circumferences intersect, the straight line joining their centers bisects their common chord at right angles.
State the corresponding theorem when the circumferences are tangent to each other.
3. A straight line parallel to one side of a tria: gle divides the other two proportionally.
4 When is a straight line divided in extreme and mean ratio?
Give and prove the construction for dividing a given straight line in extreme and mean ratio.
To the construction of what regular polygon does this construction apply?
5. What is the meaning of π in geometry?
Find the length of the side of an equilateral triangle whose aera equals that of a circle of radius R.
JUNE 1899. SOLID AMD SPHERICAL.
I. When is a straight line perpendicular to a plane? (a) All the perpendiculars to a given straight line at the same point lie in the plane perpendicular to that line at that point.
(b) The perpendicular frem a point to a plane is the shortest line that can be drawn from that point to the plane. 2. The sum of the face angles of any convex polyhedral angle is less than four right angles.
3. What is a cone of revolution?
State and prove the theorem regarding the lateral area of a cone of revolution.
4. Through two points on the surface of a sphere not the extremities of a diameter one and only one great circle may be drawn.
5. At what distance from the centre of a given sphere of radius R will one-quarter of the surface of that sphere become visible
SEPTEMBER 1899. PLANE.
1. What is the distinction between equal figures and equivalent figures?
State the various cases of two equal triangles, and prove any one of them.
2 When are two magnitudes commensurable? they incommensurable?
When are
In equal circles or in the same circle two angles at the center have the same ratio as their intercepted arcs, whether these arcs are commensurable or incommensurable.
3. If three or more straight lines drawn through a common point intersect two parallels, the corresponding segments of the parallels are porportional.
4. Show how to construct a square equivalent to a given polygon.
5. If the circumference of a circle be subdivided into three or more equal arcs:
(a) Their chords form a regular polygon, whose center is the center of the circle;
(b) The tangents at the points of division form a regular circumscribed polygon, whose center is the center of the circle.
SEPTEMBER, 1899. SOLID AND SPHERICAL.
1. Between two straight lines not in the same plane a common perpendicular can be drawn, and only one.
2. What is a rectangular parallelopiped?
The volume of a rectangular parallelopiped equals the product of its three dimensions, if the unit of volume is a cube whose edge is the linear unit.
3. The bases of a cylinder are equal.
4. The area of a zone of a sphere equals the altitude of the zone times the circumference of a great circle.
5. What must be the radius of a sphere if its surface and volume are numerically equal?
TRIGONOMETRY
JUNE. 1899.
1. Explain the two methods of measuring angles in ordinary use in trigonometry.
2, (a) What angles have the same sine as ÷3?
(b) What ones have the same cosine?
(c) What ones have the same tangent?
3. (a) Derive a formula for expressing 2 cos a sin b as a difference of sines,
(b) Derive one for expressing cos p-cos q as a double product of sines.
(4) Derive a formula for expressing tan 1⁄2a in terms of
cos a.
5. Transform the first member of the following identity into the second:
6. Solve the equation sec2 cosec2 + 2 cosec10 = 8. 7. (a) Write the formulae for solving a triangle when two sides, a and b, and the angle A,opposite a, are given, (b) How many solutions would there be in each of the following cases: | 677.169 | 1 |
Common Core Alignment
CCSS.Math.Content.2.G.1 - Recognize and draw shapes having specified attributes, such as a given number of angles or a given number of equal faces.5 Identify triangles, quadrilaterals, pentagons, hexagons, and cubes. | 677.169 | 1 |
Plane and Spherical Trigonometry, Surveying and Tables
From inside the book
Results 1-5 of 14
Page 30 George Albert Wentworth. 59. What is the angle of elevation of an inclined plane if it rises 1 foot in a horizontal distance of 40 feet ? 60. A ship is sailing due north - east with a velocity of 10 miles an hour . Find the rate at which ...
Page 74 ... angle ACB = 62 ¯ 31 ' , were measured . Find the distance from A to B. 17. Two inaccessible objects A and B are each ... hour , the other at the rate of 40 miles an hour . How far apart are the trains at the end of half an hour ...
Page 87 ... angle is 13 . 38. A privateer , 10 miles S.W. of a harbor , sees a ship sail from it in a direction S. 80 ¯ E. , at a rate of 9 miles an hour . In what direction , and at what rate , must the privateer sail in order to come up with the ...
Page 172 ... Hour Circles , or Circles of Declination , are great circles . passing ... angle which the ecliptic makes with the equinoctial is called the obliquity ... hour circle passing through the star , ZMDZ the vertical through the star | 677.169 | 1 |
A quick review of transformations in the coordinate plane. ("Isometry" is another term for "rigid transformation".)
Line Reflections
Remember that a
reflection is simply a flip. Under a reflection, the figure does not change size
(it
is a rigid transformation or isometry).
It is simply flipped over the line of
reflection. The orientation (lettering of the diagram) is reversed.
Reflection
in the x-axis:
(x,y) → (x,-y)
When you reflect a
point across the x-axis, the x-coordinate remains the
same, but the y-coordinate is transformed into its opposite.
or
When working with the graph of y = f (x),
replace y with -y.
Reflection
in the y-axis: (x,y) → (-x,y)
When you reflect a
point across the y-axis, the y-coordinate remains the
same, but the x-coordinate is transformed into its opposite.
or
When working with the graph of y = f (x),
replace x with -x.
Reflection
in y = x: (x,y) → (y,x)
When you reflect a
point across the line y = x, the x-coordinate
and the y-coordinate change places.
or
Reflection
iny = -x:
(x,y) → (-y,-x)
When you reflect a
point across the line y = -x, the x-coordinate
and the y-coordinate change places and are negated (the signs
are changed).
or
Point Reflections
A
point reflection exists when a figure is
built around a single point called the center of the
figure. For every
point in the figure, there is another point found directly opposite it on
the
other side of the center. The figure does not change size (it is
a rigid transformation or isometry).
Reflection in the Origin: (x,y) → (-x,-y)
While any point in the
coordinate plane may be used as a point of reflection, the most
commonly used point is the origin.
or
When
working with the graph of y = f (x),
replace x with -x
and y with -y.
Rotations
A rotation turns a figure through an angle about a fixed point called the center.
The center of rotation is assumed to be the origin, unless stated
otherwise. A
positive angle
of rotation turns the figure
counterclockwise, and a negative angle of rotation turns the figure
in a
clockwise direction. The figure does
not change size (it is a rigid transformation or isometry).
Counterclockwise (CCW):referred to as positive angles
Rotation of 90º:
Rotation of 180º:
(same as reflection in origin)
Rotation of 270º:
Clockwise (CW):referred to as negative angles
Rotation of 90º:
R0,90º (x,y) = (y,-x)
Rotation of 180º:
(same as reflection in origin)
Rotation of 270º:
R0,270º (x,y) = (-y,x)
Notice how a rotation of 90º CCW is the same as a rotation of 270º CW,
a rotation of 180º CCW is the same as a rotation of 180º CW,
and, a rotation of 270º CCW is the same as a rotation of 90º CW.
Translations
A translation "slides" an object a fixed
distance in a given direction. The original object and its translation have the same shape and size (rigid transformation or isometry), and they face in the same direction. The translation may be indicated by a translation vector.
Translation of h, k: (x,y) → (x + h, y + k)
vector: < h, k >
or vector < h, k >
Under the image of y = f (x) is y = f (x - h) + k..
If h > 0, the original graph is shifted h units to the
right.
If h < 0, the original graph is shifted | h | units to the left.
If k > 0, the original graph is shifted k units up.
If k < 0, the original graph is shifted | k | units down.
Dilations
A dilation is not a rigid transformation.
Adilation is a
transformation that produces an image that is the same shape as the
original, but is
a different size (the figures are similar). The description of a dilation includes the scale factor
and the center of the
dilation. A dilation "shrinks" or "stretches" a figure
(and is not a rigid transformation or isometry).
Dilation of scale factor k:
(x,y) → (kx,ky)
The center of a
dilation is most often the origin, O. It may however, be some other point in the coordinate plane which will be specified.
NOTE: There-posting of materials(in part or whole) from this site to the Internet
is copyright violation
and is not considered "fair use" for educators. Please read the "Terms of Use". | 677.169 | 1 |
Geometry: Symmetry
Symmetry
Geometry
Just what is symmetry? Symmetry is one of those things that you can recognize but cannot put into words. Many words or phrases have a meaning similar to symmetry; balanced and well-proportioned are two that immediately come to mind. However, those words don't help us when we are trying to come up with a mathematically precise definition of symmetry.
For example, take a look at the three triangles in Figure 25.7. The first triangle is a scalene triangle. The angles of the triangle are all distinct, as are the lengths of the sides. The second triangle is an isosceles triangle. It has more symmetry than the scalene triangle, because two of the sides and two of the angles are congruent. An isosceles triangle has more balance than the scalene triangle. The third triangle is an equilateral triangle. All three sides are congruent, as are all three angles. It is the most symmetrical triangle in the group, but why?
Figure 25.7Three triangles with different levels of symmetry.
Imagine that you are small enough to stand at a vertex of these triangles. Suppose you start by standing on the vertices of the scalene triangle. Think about the views if you stand on vertex A, vertex B, and vertex C. Your view will change depending on which of the three vertices you are standing on. The segments that you look out on have different lengths. The included angles that the sides make have different measures in each case. There isn't much in common with the three perspectives, and that can be disconcerting. Each view involves something different.
What if you do the same thing with the isosceles triangle? Your view from vertex B and your view from vertex C are the same. The segment that you look out on has the same length in both views, and the measures of the angles formed by the included sides are the same. So there's some familiarity in what you see from vertex B when compared to what you see from vertex C. Things look the same from vertex B and vertex C, and the symmetry of this triangle is greater than the symmetry of the scalene triangle.
Moving on to the equilateral triangle, imagine your view from each of the three vertices. Because all sides are congruent and all angles are congruent, the view is the same from all three vertices. Identical views go hand in hand with maximum symmetry.
Now look at things from a different perspective. Instead of shrinking down to the triangle, imagine that you are all-powerful and are in complete control of the triangle. You can rotate it and flip it. If you rotate the scalene triangle, it looks different until you have gone full circle. Any way that you flip the scalene triangle it looks different. If you rotate the isosceles triangle, it also looks different until you have gone full circle, but if you flip it (interchanging vertices B and C) it looks like you haven't done anything. Whenever you can rotate or flip a figure and it looks like you haven't done anything, the figure you are playing with has some symmetry. The more ways you can move a figure and have it look like nothing has happened, the more symmetry the figure has. You are getting closer to how a mathematician describes symmetry.
A symmetry of an object is an isometry that moves the object back onto itself. In a symmetry, once the movement is complete it looks like nothing has been done to it. All objects have at least one symmetry; the identity isometry can always be applied. But this is a boring symmetry, hardly worth mentioning. But because the identity isometry is an isometry, it must be mentioned. The symmetries of an object follow the isometries of an object, so the names of the symmetries will be similar to the names of the isometries. There is a translation symmetry, a reflection symmetry, a rotation symmetry, and a glide reflection symmetry.
Solid Facts
A symmetry of an object is an isometry that moves the object back onto itself.
Example 1: What are the symmetries of a square?
Solution: A square looks pretty symmetrical (from a nonmathematical perspective) so there are probably lots of symmetries around. Figure 25.8 will help sort out the symmetries of a square.
Figure 25.8The symmetries of a square.
First of all, look at reflections. If you reflect the square across l1, you'll end up with what you started with. The same holds true for reflection across l2, l3, and l4. So there are four reflection symmetries for a square.
Tangent Line
There are some obvious and not-so-obvious applications of symmetry. Symmetry has obvious applications in architecture and art. The not-so-obvious applications include golf ball design and chemistry. The arrangement of the dimples of a golf ball affects a variety of properties of the ball, and the symmetry of a molecule affects its chemical properties.
Next, consider the rotations. If the center of the square is the point O (where the diagonals intersect), you can rotate the square 90º, 180º, 270º, and 360º. Realize that the last rotation is just the identity rotation.
So there are eight symmetries: four reflection symmetries and four rotation symmetries. That's a lot of symmetries, but you shouldn't be surprised, because the square is pretty symmetrical.
This is just the tip of the iceberg. I could write volumes on symmetry alone! The applications to art, music, construction, science, and nature are endless! And the mathematics behind symmetry are as elegant as the symmetry of a snowflake. I hope this brief introduction inspires you to learn more | 677.169 | 1 |
Myriagon
In geometry, a myriagon is a polygon with 10000 sides or 10000-gon. Several philosophers have used the regular myriagon to illustrate issues regarding thought.
Regular myriagon
A regular myriagon is represented by Schläfli symbol {10000} and can be constructed as a truncated 5000-gon, t{5000}, or a twice-truncated 2500-gon, tt{2500}, or a thrice-truncaed 1250-gon, ttt{1250), or a four-fold-truncated 625-gon, tttt{625}.
The measure of each internal angle in a regular myriagon is 179.964°. The area of a regular myriagon with sides of length a is given by
Because 10000 = 24× 54, the number of sides is neither a product of distinct Fermat primes nor a power of two. Thus the regular myriagon is not a constructible polygon. Indeed, it is not even constructible with the use of neusis or an angle trisector, as the number of sides is neither a product of distinct Pierpont primes, nor a product of powers of two and three.
One interior angle in a regular triacontagon is 168°, meaning that one exterior angle would be 12°. The triacontagon is the largest regular polygon whose interior angle is the sum of the interior angles of smaller polygons: 168° is the sum of the interior angles of the equilateral triangle (60°) and the regular pentagon (108°).
WIN Party
It was founded by a group of publicans and bar-owners who objected to the government's ban on smoking in bars and restaurants, introduced in December 2004 its leader was Nav Karan Parmar who was an owner of a bar himself. WIN's slogan was "Freedom of Choice", and the party said that it was fighting a growing trend in which "the average Kiwi ... is being told more and more what they can and can't do". According to the party's leaders, opposition to the smoking ban was the party's primary campaign plank, but other related issues were also given attention.
WIN's leader was John van Buren, a publican from the Banks Peninsula area. Geoff Mulvihill, a publican from Timaru, was the deputy leader. Both van Buren and Mulvihill have been accused by the Ministry of Health of not enforcing the smoking ban, as required by law. Mulvihill is known for supporting freedom of choice for his patrons, having once lost his liquor licence for operating around the clock prior to the legislation of 24-hour trading. The party had been given official registration, but chose not to field candidates in the 2005 elections. Instead, it endorsed the larger United Future party.
We Win
We Win We're gonna shout loud, loud until the walls come down shout loud, loud until the walls come down loud until the walls come down We're gonna shout loud, loud until the walls come down shout loud, loud until the walls come down loud until the walls come down Yeah yeah yeah Because we've already won loud until the walls come down We're gonna shout loud, loud until the final sound shout loud, loud until the final sound loud until the final sound Because we've already won And You don't have a chance Yeah we've already won No you don't have a chance It's already done And you don't have a chance Because we've already won! Because we've already won And You don't have a chance Yeah we've already won No you don't have a chance It's already done And you don't have a chance Because we've already won! We have already won | 677.169 | 1 |
Plane Geometry: Harmonic Mean in Geometry. Visual Summary
In geometry, the harmonic mean is used to find a point on
a line segment that divides it into two parts in a specific
ratio. This technique is known as harmonic division and is
based on the properties of a harmonic progression, which is
a sequence of numbers that has a certain relationship
between its terms.
The harmonic mean is calculated
by taking the reciprocal of each number in a set, finding
the arithmetic mean of those reciprocals, and then taking
the reciprocal of that result. In the context of geometric
constructions, the harmonic mean can be used to solve
various problems involving the division of line segments,
the construction of circles and ellipses, and other
applications. | 677.169 | 1 |
Truncated Tetrahedron Calculator
Introduction
Overview of the Truncated Tetrahedron
The truncated tetrahedron is a polyhedron derived from a regular tetrahedron by truncating (cutting off) its four vertices. This results in a new shape with 8 faces, 18 edges, and 12 vertices. The faces consist of 4 equilateral triangles and 4 regular hexagons. The truncated tetrahedron is a type of Archimedean solid, known for its symmetrical and aesthetically pleasing structure. It finds applications in various fields, including chemistry, architecture, and art, where understanding its properties and measurements is crucial.
Importance of Geometric Calculators
Geometric calculators are essential tools for accurately computing various properties of complex shapes like the truncated tetrahedron. These calculators simplify the process of determining surface area, volume, edge lengths, and other critical parameters, saving time and reducing the risk of errors. They are invaluable for students, educators, engineers, and researchers who regularly work with geometric shapes, providing quick and precise results that facilitate deeper understanding and practical application of geometric principles.
Understanding the Truncated Tetrahedron
Definition and Properties
The truncated tetrahedron is a type of Archimedean solid created by truncating (cutting off) the vertices of a regular tetrahedron. This modification transforms the original tetrahedron into a polyhedron with 8 faces, 18 edges, and 12 vertices. The faces of a truncated tetrahedron consist of 4 equilateral triangles and 4 regular hexagons, making it both symmetrical and aesthetically appealing. Key properties of the truncated tetrahedron include its uniform face shapes and the fact that its vertices are all equivalent, contributing to its overall symmetry.
Historical Background
The concept of truncating polyhedra dates back to ancient Greek mathematicians, with Archimedes being one of the earliest scholars to study such shapes. The truncated tetrahedron is one of the thirteen Archimedean solids, which were thoroughly examined and described by Archimedes in his works on geometry. These shapes gained further prominence during the Renaissance when mathematicians and artists, fascinated by their symmetry and beauty, studied and utilized them in various designs and artworks.
Applications in Various Fields
The truncated tetrahedron has numerous applications across different fields due to its unique geometric properties. In chemistry, it is used to model the structure of certain molecular compounds and crystal lattices. In architecture, the shape is employed in designing aesthetically pleasing and structurally sound buildings and sculptures. Additionally, in art, the truncated tetrahedron serves as an inspiration for creating intricate and visually appealing patterns. Its study also finds relevance in fields such as materials science, gaming (for dice design), and education, where it helps in teaching complex geometric concepts in an intuitive manner.
Calculator Overview
Purpose and Benefits of Using a Calculator for Truncated Tetrahedrons
The primary purpose of using a calculator for truncated tetrahedrons is to facilitate quick and accurate computation of various geometric properties. Calculating parameters such as surface area, volume, edge lengths, and radii manually can be time-consuming and prone to errors, especially given the complexity of the formulas involved. A dedicated calculator streamlines this process, allowing users to input basic measurements and obtain precise results instantly. This tool is invaluable for students, educators, architects, engineers, and researchers who need reliable data for their work or studies.
The benefits of using a calculator for truncated tetrahedrons include:
Input Parameters
Explanation of Required Inputs
To accurately compute the properties of a truncated tetrahedron, the calculator requires certain input parameters. Each parameter corresponds to a specific geometric property of the shape. Understanding these inputs and how to measure them ensures precise calculations.
Edge Length (a)
The edge length (a) refers to the length of one of the edges of the truncated tetrahedron. It is a crucial input as it directly influences the calculations of other properties such as surface area and volume. To measure the edge length, use a ruler or caliper to measure the distance between two adjacent vertices on one of the hexagonal or triangular faces.
Edge Length of the Tetrahedron (a')
The edge length of the original tetrahedron (a') is the length of an edge before the vertices were truncated. This input helps to understand the transformation from the regular tetrahedron to the truncated version. It can be calculated or provided as a known value if the original shape is considered.
Surface Area (A)
The surface area (A) is the total area of all the faces of the truncated tetrahedron. This includes the areas of the four equilateral triangles and the four regular hexagons. Surface area can be measured by summing the areas of individual faces or derived using specific geometric formulas.
Volume (V)
The volume (V) is the total space enclosed by the truncated tetrahedron. It is a measure of the three-dimensional space occupied by the shape. The volume can be calculated using the edge length or other geometric properties.
Circumsphere Radius (rc)
The circumsphere radius (rc) is the radius of a sphere that passes through all the vertices of the truncated tetrahedron. It is a significant parameter in understanding the spatial properties of the shape. This radius can be computed based on the edge length and the geometric relationships of the vertices.
Midsphere Radius (rm)
The midsphere radius (rm) refers to the radius of a sphere that is tangent to all the edges of the truncated tetrahedron. This radius helps in analyzing the internal proportions of the shape and can be determined using specific geometric calculations.
Surface-to-Volume Ratio (A/V)
The surface-to-volume ratio (A/V) is a dimensionless value that compares the surface area to the volume of the truncated tetrahedron. It is an important parameter in various scientific and engineering applications, particularly in materials science where surface properties are critical. This ratio can be calculated directly once the surface area and volume are known.
How to Measure and Input Values
To use the truncated tetrahedron calculator effectively, follow these steps for measuring and inputting values:
Measure Accurately: Use appropriate tools like rulers, calipers, or geometric formulas to measure the known parameter accurately.
Input the Value: Enter the measured value into the corresponding input field in the calculator. Ensure that units are consistent and correctly entered.
Specify Precision: If required, select the number of decimal places for your results to match the precision needed for your application.
Calculate: Use the calculator to compute the unknown properties by clicking the appropriate buttons.
Accurate measurement and careful input of values will ensure reliable and precise results from the calculator, aiding in your understanding and application of the truncated tetrahedron's geometric properties.
Step-by-Step Guide to Using the Calculator
Detailed Instructions for Each Input
Using the truncated tetrahedron calculator is straightforward. Follow these detailed steps for each input parameter to obtain accurate results:
Access the Calculator: Open the truncated tetrahedron calculator on your device.
Select the Known Parameter: Identify which parameter you already have:
Edge Length (a)
Edge Length of the Tetrahedron (a')
Surface Area (A)
Volume (V)
Circumsphere Radius (rc)
Midsphere Radius (rm)
Surface-to-Volume Ratio (A/V)
Enter the Value: Input the measured or known value into the appropriate field. For example:
To input Edge Length (a), enter the value in the field labeled "Edge length (a):".
To input Surface Area (A), enter the value in the field labeled "Surface area (A):".
Clear Unnecessary Fields: Ensure other fields are cleared to avoid calculation errors by clicking the 'C' button next to each field.
Specify Decimal Places: Select the number of decimal places for the results from the dropdown menu labeled "Round to decimal places."
Calculate: Click the "Calculate" button to perform the computation.
View Results: The calculator will display the computed values for the other properties based on your input.
Reset: If needed, click the "Delete" button to clear all inputs and start a new calculation.
Examples of Calculations
Example 1: Calculating Volume from Edge Length
Enter the edge length (a) as 5 units in the field labeled "Edge length (a):".
Select "3" from the dropdown menu for decimal places.
Click "Calculate".
The calculator displays the volume (V) along with other properties such as surface area (A), circumsphere radius (rc), etc.
Example 2: Calculating Edge Length from Surface Area
Enter the surface area (A) as 150 square units in the field labeled "Surface area (A):".
Select "2" from the dropdown menu for decimal places.
Click "Calculate".
The calculator displays the edge length (a) along with other properties like volume (V), midsphere radius (rm), etc.
Tips for Accurate Results
To ensure you get the most accurate results from the calculator, keep these tips in mind:
Precise Measurements: Use precise tools for measuring dimensions and ensure all measurements are accurate.
Consistent Units: Ensure all inputs are in consistent units (e.g., all lengths in cm or all in inches).
Double-Check Entries: Verify that the correct values are entered in the corresponding fields.
Clear Unused Fields: Use the 'C' button to clear fields not in use to prevent incorrect calculations.
Rounding Appropriately: Select the appropriate number of decimal places for the level of precision required in your application.
Understand Limitations: Be aware of the calculator's limitations and the assumptions behind the formulas used.
By following these steps and tips, you can effectively use the truncated tetrahedron calculator to perform accurate and efficient geometric calculations.
Mathematical Formulas Behind the Calculator
Surface Area Formula
The surface area (A) of a truncated tetrahedron, with edge length (a), can be calculated using the formula:
Surface Area (A):
A = 7 * sqrt(3) * a^2
This formula accounts for the combined areas of the four equilateral triangles and four regular hexagons that form the faces of the truncated tetrahedron.
Volume Formula
The volume (V) of a truncated tetrahedron, with edge length (a), is given by:
Volume (V):
V = (23/12) * sqrt(2) * a^3
This formula calculates the space enclosed by the truncated tetrahedron.
Circumsphere and Midsphere Radius Calculations
The circumsphere radius (rc) is the radius of a sphere that passes through all the vertices of the truncated tetrahedron:
Circumsphere Radius (rc):
rc = (a/4) * sqrt(22)
The midsphere radius (rm) is the radius of a sphere that is tangent to all the edges of the truncated tetrahedron:
Midsphere Radius (rm):
rm = (3/4) * sqrt(2) * a
Surface-to-Volume Ratio Computation
The surface-to-volume ratio (A/V) is a measure of the surface area relative to the volume of the truncated tetrahedron. It is computed using the surface area and volume formulas:
Surface-to-Volume Ratio (A/V):
A/V = (7 * sqrt(3) * a^2) / ((23/12) * sqrt(2) * a^3)
Simplifying this expression gives:
A/V = (84 * sqrt(3)) / (23 * sqrt(2) * a)
This ratio is important in various scientific and engineering contexts where the surface properties relative to volume are critical.
Practical Applications
Real-World Examples
Architecture and Design
In architecture, the truncated tetrahedron serves as inspiration for innovative building designs and structures. Its symmetrical and geometrically balanced form can be seen in modern buildings, especially in roof designs and geometric facades.
Materials Science
Researchers in materials science utilize truncated tetrahedrons to model crystal structures and study material properties. Understanding the geometric characteristics helps in predicting how materials behave under different conditions, such as stress and temperature variations.
Education and Visualization
Truncated tetrahedrons are valuable educational tools for teaching geometry and spatial reasoning. They help students visualize three-dimensional shapes and understand complex geometric concepts such as surface area, volume, and geometric transformations.
Case Studies
Crystallography and Molecular Modeling
In crystallography, truncated tetrahedrons are used to model complex crystal structures. By understanding the geometric relationships within these structures, scientists can predict material properties and develop new materials with specific characteristics.
Architectural Case Study: The Louvre Pyramid
The iconic Louvre Pyramid in Paris, designed by architect I.M. Pei, incorporates elements of geometric shapes including the truncated tetrahedron. The pyramid's complex structure highlights the use of geometry in architectural design, providing both aesthetic appeal and structural integrity.
Educational Tool in Mathematics
Teachers use truncated tetrahedrons as hands-on models in mathematics classrooms to explore concepts such as polyhedra, Euler's formula, and the relationships between faces, edges, and vertices. These models engage students and facilitate deeper understanding of geometric principles.
These examples illustrate the diverse applications of truncated tetrahedrons across different fields, highlighting their relevance in both theoretical research and practical applications.
FAQs
Frequently Asked Questions about the Truncated Tetrahedron and the Calculator
What is a truncated tetrahedron?
A truncated tetrahedron is a polyhedron obtained by truncating (cutting off) the vertices of a regular tetrahedron. This process creates a shape with 4 equilateral triangle faces and 4 regular hexagon faces.
What are the properties of a truncated tetrahedron?
The properties of a truncated tetrahedron include:
8 faces (4 equilateral triangles, 4 regular hexagons)
18 edges
12 vertices
It is characterized by specific geometric measurements such as edge length, surface area, volume, and various radii.
How does the truncated tetrahedron calculator work?
The calculator allows users to compute geometric properties of a truncated tetrahedron by entering known parameters such as edge length, surface area, volume, etc. It uses mathematical formulas specific to the shape to calculate the remaining properties based on user inputs.
It serves as a tool for education, research, and practical applications in various fields.
How accurate are the calculations from the truncated tetrahedron calculator?
The calculator provides accurate results based on the mathematical formulas used. Accuracy depends on the precision of the input values and the mathematical algorithms implemented in the calculator. Users can specify the number of decimal places for the results to match their requirements.
Can I use the truncated tetrahedron calculator for educational purposes?
Yes, the calculator is suitable for educational purposes. It helps students and educators explore geometric concepts such as polyhedra, surface area, volume, and ratios. By inputting different parameters, users can visualize and understand the properties of truncated tetrahedrons.
Where can I find more information about truncated tetrahedrons?
For additional information about truncated tetrahedrons, including mathematical derivations, properties, and practical applications, refer to geometry textbooks, online resources, or consult with experts in mathematics and geometry.
Conclusion
In conclusion, the truncated tetrahedron calculator is a valuable tool for exploring and understanding geometric properties of this fascinating polyhedral shape. By providing a straightforward way to compute surface area, volume, radii, and other critical measurements, the calculator supports both educational endeavors and practical applications in various fields.
Whether you are a student learning about polyhedra, a researcher studying crystal structures, or an architect designing innovative structures, the truncated tetrahedron calculator provides a reliable method to obtain accurate geometric data.
Encouragement to Use the Calculator for Geometric Calculations
We encourage you to utilize the truncated tetrahedron calculator for your geometric calculations. It offers an intuitive interface and precise results, helping you explore the intricate properties of truncated tetrahedrons with ease. Enhance your understanding of geometry and leverage the calculator's capabilities to support your academic or professional pursuits.
Explore the calculator today and unlock the insights hidden within the geometric elegance of truncated tetrahedrons! | 677.169 | 1 |
JB/135/072/005
Prop I Theor.
Triangles and parallelograms of
the same altitude are to one
another as their bases —
Prop VII Theor.
If two triangles have one angle of the one equal to
one angle of the other, and the sides about two other
angles proportionals: then if each of the
remaining angles be either less or not less
than a right angle, or if one of them be a
right angle the triangle shall be equiangular
and have those angles equal about which the
sides are proportionals.
Prop XIV Theor.
Equal parallelograms which have
one angle of the one equal to one angle
of the other, have their Sides about the
equal angles reciprocally proportional,
and parallelograms that have one angle
of the one equal to one of the other
and their sides about their equal angle
reciprocally proportional, are equal to
one another.
Prop XXI Theor.
Rectilineal figures which are
similar to the same rectilineal
figure, are also similar to one
another.
Prop XXVIII. Prob.
To a given straight line to apply a
parallelogram equal to a given rectilineal
figure and deficient by a parallelogram
similar to a given parallelogram but the
given rectilineal figure to which the
parallelogram to be applied is to be equal
must not be greater than the parallelogram
applied to half of the given line,
having its defect similar to the defect of
that which is to be applied: that is to
the given parallelogram —
Prop C Theor.
If from an angle of
a triangle a strait line be drawn
perpendicular to the base: the rectangle
contained by the sides of the triangle is
equal to the rectangle contained by
the perpendicular and the diameter
of the circle described about the
triangle.
---page break---
Prop II Theor.
If a straight line be drawn parallel to
one of the Sides of a triangle, it shall
cut the other sides, or these produced,
proportionally. And if the sides, or the
sides produced be cut proportionally,
the straight lines which joins the
point of Section shall be parallel to
the remaining side of the triangle.
Prop VIII Theor.</lb>
In a right angled triangle, if a perpendicular
be drawn from the right angle to
the base: the triangles on each side of it
are similar to the whole triangle &
to one another
Prop XV Theor.
Equal triangles which have one angle
of the one equal to one angle of the other,
have their Sides about the equal angles
reciprocally proportional: and triangles
which have one angle in the one
equal to one angle in the other and
their Sides about the equal angles
reciprocally proportional are equal to one another.
Prop XXII Theor.
If four straight lines be proportionals,
the similar rectilineal figures
described upon them shall also be
proportionals. And if the similar
rectilineal figures described upon 4
four strait lines be proportionals,
those straight lines shall be proportionals.
Prop XXIX Prob.
To a given straight line to apply a
parallelogram equal to a given
rectinineal figure exceeding by
a parallelogram similar to
another given —
Prop D Theor.
The rectangle contained by the
diagonals of a quadrilateral inscribed in a circle is equal
to both the rectangles contained
by its opposite sides —
---page break---
Prop III Theor.
If the angle of a triangle be divided into
two equal angles, by a straight line which
also cuts the base; the segments of the base
shall have the same ratio which the other
sides of the triangles have to one another
and if the segments of the base have the
same ratio which the other sides of the
triangle have to one another the straight
line drawn from the vertex to the point of
section divides the angle into two equal angles.
Prop IX Prob. From a given straight line to cut of
any part required.
Prop XVI Theor.
If four Straight lines be proportional,
the rectangle contained by the extremes
is equal to the rectangle contained by
the means, the four straight lines are
proportional.
Prop XXIII Theor.
Equiangular parallelograms
have to one another the ratio
which is compounded of the ratios
of their sides —
<p>Prop XXX Prob. To cut a given straight line
in extreme & mean ratio
---page break---
Prop A Theor.
If the outer
a triangle made by producing
sides be divided into two equal
by a straight line which also cuts
produced: the segments betw
dividing line and the extreme
base have the same ratio which
sides of the triangle have to one
if the segments of the bas produce
same ratio, which the other side
angle have the straight line dra
the vertex to the point of sect
the outward angle of the tria
two equal angles —
<p>Prop X Theor.
To divide a given straight line
a given staight line, that is
which shall have the same ratio
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Prop XXIV Theor.
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In right angled triangles
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equal to the similar, and
described figures upon the
containing the right ang | 677.169 | 1 |
Magnitude of a Vector
The definition of a vector is an entity with both magnitude and direction. The movement of an object between two points is described by a vector. The directed line segment can be used to graphically represent vector math.
The magnitude of a vector is the length of the directed line segment, and the vector's direction is indicated by the angle at which it is inclined. A vector has a Tail at its beginning and a Head at its finish, both of which include arrows.
A mathematical structure is referred to as a vector. It has a wide range of uses in physics and geometry. We understand that the coordinates of the points on the coordinate plane can be expressed using an ordered pair, such as (x, y). The vector method is highly helpful in the three-dimensional geometry reduction procedure.
Summary:
If the sum of two unit vectors is a unit vector, then the magnitude of difference is -(a) √2(b) √3(c) 2√2(d) √5
If the sum of two unit vectors is a unit vector, then the magnitude of difference is √3. A vector is defined as a thing that has both magnitude and direction. A vector describes how an object moves between two places. | 677.169 | 1 |
5 Best Ways to Check if a Right Triangle is Possible from Given Area and Hypotenuse in Python
Checking for a Right Triangle with Given Area and Hypotenuse in Python
💡 Problem Formulation: Given two numerical inputs representing the area and the hypotenuse of a potential right triangle, we want to verify if a right triangle with these properties can exist. For instance, if the area is 6 and the hypotenuse is 10, is there a combination of perpendicular sides that can form such a triangle?
Method 1: Using Pythagorean Theorem and Area Calculation
By applying the Pythagorean Theorem in conjunction with the formula for the area of a right triangle, this method computes whether the two provided values are compatible. It uses the fact that the area is half the product of the lengths of the two perpendicular sides.
This code snippet uses a brute-force approach to iterate through all possible integer values for one side of the triangle, while calculating the other using the Pythagorean theorem. If any such pair of sides, when used to calculate the area, matches the given area, the function returns True along with the dimensions of the triangle.
Method 2: Using Algebraic Manipulation
This method involves algebraically solving for the lengths of the two sides (let's call them 'a' and 'b') using the given hypotenuse (c) and area. This method exploits the fact that the area is area = (a*b)/2, and by the Pythagorean theorem, a^2 + b^2 = hypotenuse^2.
This code uses mathematical expressions to resolve the two sides. It starts by finding the discriminant of a derived quadratic equation. If the discriminant is non-negative, it proceeds to calculate the potential side lengths. If any pair of computed sides yields a positive length, the answer is True, indicating a right triangle is possible.
Method 3: Optimized Brute Force with Stop Condition
An optimized take on brute force, this method uses a stop condition to break the loop early, avoiding unnecessary calculations once the right triangle conditions can no longer be satisfied for higher values of 'a'.
The code employs an early stopping condition when 'a' squared reaches or exceeds the square of the hypotenuse, which reduces the number of iterations, enhancing efficiency. It effectively narrows down on the solution and stops as soon as it becomes mathematically impossible to find compatible side lengths.
Method 4: Using Area Constraints and Triangle Inequality
Considering the constraints that the area and the hypotenuse impose, this method checks if the potential sides fulfill the triangle inequality, which is necessary for the existence of any triangle, including a right-angled one.
This code also iterates over potential side lengths 'a', computes the corresponding 'b', then checks the triangle inequality and the Pythagorean theorem. It is largely a rephrasing of the first method but takes into account the triangle's existence before checking for a right angle.
Bonus One-Liner Method 5: Using List Comprehension
A concise approach that utilizes list comprehension and Python's built-in any() function to determine if there exists any pair of sides that satisfy the conditions for a right triangle.
The one-liner leverages a combination of list comprehension and the any() function to test all possible combinations of side lengths derived from the given area. If any combination satisfies the Pythagorean Theorem, it returns True, indicating the existence of a right triangle.
Summary/Discussion
Method 1: Applying the Pythagorean Theorem and area calculation. Strenghts: Conceptually straightforward. Weaknesses: Can be inefficient for large numbers or non-integers | 677.169 | 1 |
What are some differences of a reactangle and a parallelagram?
.A rectangle has 4 sides,[a rectangle is a parallelogram], two
short, two long, but all of the lines are straight. a
parrallelogram has all parallel sides [just like the rectangle, but
can be diaganol. Such as a rhombus | 677.169 | 1 |
Pentagonal Shape
Pentagonal Shape - Web in geometry, a pentagonal pyramid is a pyramid with a pentagonal base upon which are erected five triangular faces that meet. Web in geometry, a pentagonal tiling is a tiling of the plane where each individual piece is in the shape of a pentagon. Web in geometry, a pentagon is a closed 2d shape that has 5 sides and 5 angles. Web defense secretary lloyd j. Web in turn, a regular pentagon is a regular polygon with five sides and an irregular pentagon is an irregular polygon. To recall, the prism shape, in. Printable pentagon shape to cut out for projects. In a regular pentagon, each interior angle measures 108°, and each exterior angle measures 72°. The pentagon was the brainchild of army brig. It is also known as a star pentagon due to its shape and the.
45+ Images Of Pentagonal Prism Pics Petui
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Pentagonal Prism Cuemath
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The sum of all the five interior angles in any pentagon is 540°. It is a type of polygon with 5 straight sides and 5 interior angles. There are a few different. Web malley was quietly placed on unpaid leave in june for his alleged mishandling of "protected material.". Web along with the video of the 'gimbal ufo' pentagon released.
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Web tetragonal pentagonal dodecahedron (see here for a rotating model.) face polygon: It is a type of polygon with 5 straight sides and 5 interior angles. Web in geometry, a pentagonal pyramid is a pyramid with a pentagonal base upon which are erected five triangular faces that meet. Web the meaning of pentagonal is having five sides and five angles..
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Pentagon Shape What Is A Pentagon DK Find Out
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Web a pentagon shape is a 5 sided polygon. It is a polygon, which. Sommervell, who, in the early 1940s, pitched. Having five sides and five. This five sided shape has five straight sides and five interior angles, which add up to 540°. Web a pentagonal pyramid is a 3d shape with the base as the shape of a pentagon along with the faces shaped like a triangle. Discover how the parts of the. Web malley was quietly placed on unpaid leave in june for his alleged mishandling of "protected material.". Web in geometry, a pentagonal pyramid is a pyramid with a pentagonal base upon which are erected five triangular faces that meet. It is also known as a star pentagon due to its shape and the. Web tetragonal pentagonal dodecahedron (see here for a rotating model.) face polygon: Web a pentagonal number is a figurate number that extends the concept of triangular and square numbers to the pentagon, but,. Web a pentagonal prism can have pentagonal bases which give five sides. Austin iii has ordered about 2,000 more u.s. Web free printable large full page pentagon shape for crafts. To recall, the prism shape, in. Web in geometry, a pentagonal tiling is a tiling of the plane where each individual piece is in the shape of a pentagon. Web in turn, a regular pentagon is a regular polygon with five sides and an irregular pentagon is an irregular polygon. Web a pentagon shape has five diagonals connecting the five vertices to their opposite vertices. Web the pentagon is a geometric shape that is characterized by its five sides and five angles.
There Are A Few Different.
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Printable pentagon shape to cut out for projects. Web a pentagonal pyramid is a 3d shape with the base as the shape of a pentagon along with the faces shaped like a triangle. This five sided shape has five straight sides and five interior angles, which add up to 540°. Web in geometry, a pentagon is a closed 2d shape that has 5 sides and 5 angles.
Austin Iii Has Ordered About 2,000 More U.s.
Web the pentagon is a geometric shape that is characterized by its five sides and five angles. Having five sides and five. Web the meaning of pentagonal is having five sides and five angles. The sum of all the five interior angles in any pentagon is 540°.
Web Along With The Video Of The 'Gimbal Ufo' Pentagon Released Another Footage Captured By Navy Pilots Of An.
Web a pentagonal prism can have pentagonal bases which give five sides. Web in geometry, a pentagonal tiling is a tiling of the plane where each individual piece is in the shape of a pentagon. Web china could have 1,000 nuclear warheads and missiles that can hit the us by 2030, says pentagon report. It is a polygon, which. | 677.169 | 1 |
Similar Triangles Worksheets
What are Similar Triangles Worksheets?
Similar triangles worksheets are educational resources designed to help students practice and understand the concept of similar triangles. These worksheets provide a variety of exercises and problems for students to solve, allowing them to develop their skills in identifying and working with similar triangles.
Identifying Similar Triangles
In this section, students are presented with different pairs of triangles and are required to determine whether they are similar or not. They need to examine the corresponding angles and the proportional lengths of the sides to make their decision. This exercise helps students develop their ability to recognize similar triangles based on specific criteria.
Proportional Lengths
This section focuses on the property of similar triangles that states corresponding sides are proportional. Students are given triangles with some side lengths missing, and they must use the given information to find the missing lengths. By working through these problems, students gain a deeper understanding of the relationship between similar triangles and the proportional lengths of their sides.
Applications of Similar Triangles
In this section, students explore real-life applications of similar triangles. They are presented with various scenarios where similar triangles are used to solve problems. These scenarios could involve finding the height of a building, determining the distance to an inaccessible object, or calculating the size of an object in a photograph. By applying their knowledge of similar triangles to practical situations, students develop problem-solving skills and see the relevance of this concept in the real world.
Using Similar Triangles to Solve Problems
This section provides students with word problems that require the use of similar triangles to find a solution. By applying the principles of similar triangles, students learn to analyze and solve complex problems. These problems may involve finding unknown angles, determining missing side lengths, or solving for a particular variable. Through these exercises, students strengthen their understanding of similar triangles and their problem-solving abilities.
Conclusion
Similar triangles worksheets are valuable tools for students to practice and reinforce their understanding of this important geometric concept. By using these worksheets, students can enhance their skills in identifying similar triangles, working with proportional lengths, applying the concept to real-life scenarios, and solving problems. Regular practice with these worksheets can greatly improve students' grasp of similar triangles and their ability to apply this knowledge in various mathematical situations. | 677.169 | 1 |
Length of Tangent on a Circle
A tangent to a circle is defined as a line segment that touches the circle exactly at one point. There are some important points regarding tangents:
A tangent to a circle cannot be drawn through a point which lies inside the circle. It is so because all the lines passing through any point inside the circle, will intersect the circle at two points.
There is exactly one tangent to a circle which passes through only one point on the circle.
There are exactly two tangents can be drawn to a circle from a point outside the circle.
In the figure, \(P\) is an external point from which tangents are drawn to the circle. \(A\) and \(B\) are points of contact of the tangent with a circle. The length of a tangent is equal to the length of a line segment with end-points as the external point and the point of contact. So, \(PA\) and \(PB\) are the lengths of tangent to the circle from an external point \(P\).
Some theorems on length of tangent
Theorem 1: The lengths of tangents drawn from an external point to a circle are equal. It is proved as follows:
Consider the circle with center \(O\). \(PA\) and \(PB\) are the two tangents drawn to the circle from the external point \(P\). \(OA\) and \(OB\) are radii of the circle.
Since tangent on a circle and the radius are perpendicular to each other at the point of tangency,
\(∠PAO\) = \(∠PBO\) = \(90°\)
Consider the triangles, \(∆PAO\) and \(∆PBO\),
\(∠PAO\) = \(∠PBO\) = \(90°\)
\(PO\) is common side for both the triangles,
\(OA\) = \(OB\) [Radii of the circle]
Hence, by RHS congruence theorem,
\(∆PAO ≅ ∆PBO\)
\(⇒ PA = PB\) (Corresponding parts of congruent triangles)
This can also be proved by using Pythagoras theorem as follows,
Since,
\(∠PAO\) = \(∠PBO\) = \(90°\)
\(∆PAO\) and \(∆PBO\) are right angled triangles.
\(PA^2\) = \(OP^2 – OA^2\)
Since \(OA\) = \(OB\),
\(PA^2\) = \(OP^2 – OB^2\) = \(PB^2\)
This gives, \(PA\) = \(PB\)
Therefore, tangents drawn to a circle from an external point will have equal lengths. There is an important observation here:
Since \(∠APO\) = \(∠BPO\), \(OP\) is the angle bisector of \(∠APB\).
Therefore, center of the circle lies on angle bisector of the angle made by two tangents to the circle from an external point.
Let's consider an example for better understanding of the concept of length of the tangent drawn to a circle from an external point. | 677.169 | 1 |
how to find a missing side of similar triangles
Finding Missing Side Lengths Of Similar Triangles Worksheet – Triangles are one of the most fundamental designs in geometry. Understanding the concept of triangles is essential for getting more advanced concepts in geometry. In this blog, we will cover the various kinds of triangles Triangle angles, how to determine the perimeter and area of a triangle, and provide details of the various. Types of Triangles There are three kinds from triangles: Equal isosceles, and … Read more | 677.169 | 1 |
This question is very difficult for me. To answer this question, I started with graphing the polar equations:
I also shaded the area I am going to find.
Based on the comments below, I will first find the intersection the circle makes with the upper petal. I found it by doing the following:
$2 sin\Theta = 2 sin(2\Theta )$
$2sin\Theta =4sin\Theta cos\Theta $
$\frac{1}{2} = cos\Theta $
$cos^{-1}(\frac{1}{2}) = \Theta $
$\frac{\pi }{3} = \Theta$
Next, I chose r = 2 (hence, $\frac{\pi }{4}$) as my starting point to the point where it intersects. Then, I multiplied it to 4 since there are four halves (not necessarily geometrical halves; I only concerned myself with the assumption that I must start tracing from $\frac{\pi }{4}$ to the point where it intersects, and there are two possible points.
I can now be able to obtain the area of the two cut petals:
$A = 4(\frac{1}{2})\int_{\frac{\pi}{4}}^{\frac{\pi }{3}}(2sin\Theta )^{2}d\Theta $
After that, I computed for the area of the two full petals at the bottom.
$4(\frac{1}{2})\int_{\frac{-\pi }{4}}^{0}(2sin\Theta )^{2}d\Theta $
Please take note that for the bounds I used above, I started from the negative counterpart of $\frac{7\pi }{4}$ to the pole (or origin; r = 0). I found the value of the theta at the pole by equating $2sin2\Theta $ to 0. The answer from that calculation is 0.
$\begingroup$There are 2 types of shapes involved. The petal and the cut petal. To deal with the cut petal, find the intersection region between the petal and the circle [requires piecewise]$\endgroup$
$\begingroup$Why do you care about the whole circle? The area that you want to calculate is the area of two full petals plus the area of two cut petals. So concentrate on calculating the area of a full petal (hint: what is the range of $\theta$ that will give you just one petal?) and then calculate the area of a cut petal (hint: at what values of $\theta$ does the circle intersect the petal in the first quadrant? Then you can calculate the area of a cut petal as the difference of two areas - do you see how?). Divide and conquer!$\endgroup$ | 677.169 | 1 |
Fundamental plane
The fundamental plane in a spherical coordinate system is a plane which divides the sphere into two hemispheres. The latitude of a point is then the angle between the fundamental plane and the line joining the point to the centre of the sphere. | 677.169 | 1 |
Segment bisector
A point, line, ray, or segment that divides a segment into two congruent segments
A segment bisector is a line or line segment that divides a line segment into two equal parts. In other words, it cuts the segment in half, creating two congruent segments.
To determine a segment bisector, you need to find the midpoint of the segment. The midpoint is the point in the middle of the segment, equidistant from both endpoints. Once you have found the midpoint, you can draw a line or line segment from it to one of the endpoints of the original segment, creating two congruent segments.
It is important to note that a segment bisector not only divides the segment into two equal parts but also divides its angle into two equal angles. Additionally, the segment bisector of one segment is perpendicular to the segment bisector of its adjacent segment.
Segment bisectors are useful in many geometrical constructions, such as constructing perpendicular lines, bisecting angles, and creating shapes with equal sides or angles | 677.169 | 1 |
If a curve passing through the origin be given by a rational integral algebraic equation, the equation of the tangent (or tangents) at the origin is obtained by equating to zero the terms of the lowest degree in the equation.
In the curve x2 + y2 + ax + by = 0, ax + by = 0, is the equation of the tangent at the origin; and in the curve (x2 + y2)2 = a2 (x2 - y2), x2 - y2 = 0 is the equation of a pair of tangents at the origin.
Angle of intersectionbetween two curves is defined as the angle between the two tangents drawn to the two curves at their point of intersection . If the angle between two curves is 90° then they are called ORTHOGONAL curves.
Ex.10 Find the angle between curves y2 = 4x and y = e-x/2
Sol.
Let the curves intersect at point (x1,y1)
Note : here that we have not actually found the intersection point but geometrically we can see that the curves intersect.
Ex.11 Show that the curves y = 2 sin2x and y = cos 2x intersect at π/6. What is their angle of intersection ?
Sol. Given curves are y = 2 sin2 x ...(1)
and y = cos 2x ...(2)
Solving (1) and (2), we get 2 sin2 x = cos 2x
⇒ 1 – cos 2x = cos 2x ⇒ cos2x = 1/2 ⇒ cos π/3 ⇒ 2x = ± π/3
x = ±π/6 are the points of intersection
From (1), dy/dx = 4 sin x cos x = 2 sin 2x = m1 (say)
From (2) dy/dx = -2 sin 2x = m2 (say)
If angle of intersection is θ, then tanθ
∴
∴
Ex.12 Show that the angle between the tangents at any point P and the line joining P to the origin `O' is the same at all points of the curve ln (x2 + y2) = c tan-1 (y/x) where c is constant.
Ex.16 What should be the value of n in the equation of curve y = a1 - n. xn, so that the sub-normal may be of constant length ?
Sol. Given curve is y = a1 - n . xn
Taking logarithm of both sides, we get, ln y = (1 - n) ln a + n ln x
Differentiating both sides w.r.tx, we get ...(1)
Lengths of sub-normal = y dy/dx = y .ny/x ....{from 1}
Since lengths of sub-normal is to be constant, so x should not appear in its value i.e., 2n – 1 = 0. n =1/2
Ex.17 If the relation between sub-normal SN and sub-tangent ST at any point S on the curve
by2 = (x + a)3 is p(SN) = q(ST)2; then p/q is
(A) 8b/27
(B) b
(C) 1
(D) none of these
Sol.
Let a point by (x0, y0) lying on the curve by = (x0 + a)3 ....(i)
(from equation (i))
Ex.18 For the curve y = show that sum of lengths of tangent & subtangent at any point is proportional to coordinates of point of tangency.
Sol.
Let point of tangency be (x1, y1) ⇒
tangent + subtangent =
Hence proved.
Ex.19 Show that the segment of the tangent to the curve y = contained between the y-axis and point of tangency has a constant length.
Sol.
Equation of tangent at 'ø' y – a ln cot ø/2 + a cos ø =
⇒y sin ø – a sinø ln cot ø/2 + a sin ø cos ø - x cosø + a sin ø cos ø
⇒x cos ø+ y sin ø = a sin ø ln cot ø/2
Point on y-axis P ≡ (0, a ln cot ø/2) and point of tangency
Q≡ (a sinø a ln cotø/2 – a cos ø)
E. Solving Equations
Ex.20 For what values of c does the equation ln x = cx2 have exactly one solution ?
Sol.
Let's start by graphing y = In x and y = cx2 for various values of c. We know that for c * 0, y = cx2 is a parabola that opens upward if c > 0 and downward if c < 0. Figure 1 shows the parabolas y = cx2 for several positive values of c. Most of them don't intersect y = In x at all and one intersect twice. We have the feeling that there must be a value of c (somewhere between 0.1 and 0.3) for which the curves intersect exactly once, as in Figure 2.
To find that particular value of c, we let 'a' be the x-coordinate of the single point of intersection. In other words, In a = ca2, so 'a' is the unique solution of the given equation. We see from Figure 2 that the curves just touch, so they have a common tangent line when x = a. That means the curves y = In x and y=cx2 have the same slope when x = a. Therefore 1/a = 2ca
Solving the equation In a = ca2 and 1/a = 2ca
For negative values of c we have the situation illustrated in Figure 3: All parabolas y = cx2 with negative values of c intersect y = In x exactly once. And let's not forget about c = 0: The curve y = 0 x2 = 0 just he x-axis, which intersects y = In x exactly once.
To summarize, the required values of c are c=1/(2e) and c<0
Ex.21 The set of values of p for which the equation px2 = lnx possess a single root is
Sol.
for p ≤ 0, there is obvious one solution ; for p > 0 one root
⇒ the curves touch each .
2 px1 = 1/x1 ⇒ x12 = 1/2p ;
Also px12 = ln x1 ⇒ p (1/2p) = ln x1 ⇒ x1 = e1/2
⇒ 2 p = 1/e ⇒ p = 1/2e . Hence p ∈ (- ∝, 0] U {1/2e}
F. Shortest distance
Shortest distance between two non-intersecting curves always along the common normal (wherever defined)
Ex.23 Let P be a point on the curve C1: y = and Q be a point on the curve C2: xy = 9, both P and Q lie in the first quadrant. If 'd' denotes the minimum value between P and Q, find the value of d2.
Sol. Note that C1 is a semicircle and C2 is a rectangular hyperbola.
PQ will be minimum if the normal at P on the semicircle is also a normal at Q on xy = 9
Let the normal at P be y = mx ....(1) (m > 0) solving it with xy = 9
differentiating xy = 9
∴ normal at P and Q is y = x
solving P(1, 1) and Q(3, 3)
(PQ)2 = d2 = 4 + 4 = 8
G. Rate Measurement
Ex.24 A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 ft/s, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 ft from the wall ?
Sol. We first draw a diagram and label it as in Figure 1. Let x feet be the distance from the bottom of the ladder to the wall and y feet the distance from the top of the ladder to the ground. Note that x and y are both function of t (time). We are given that dx/dt = 1 ft/s and we are asked to find dy/dt when x = 6 ft (see Figure 2). In this problem, the relationship between x and y is given by the Pythagorean Theorem : x2 + y2 = 100
Differentiating each side with respect to t using the Chain Rule, we have
The fact that dy/dt is negative means thatthe distance from the top of the ladder to the ground is decreasing at a rate of 3/4 ft/s In other words, the top of the ladder is sliding down the wall at a rate of 3/4 ft/s .
Ex.25 A water tank has the shape of an inverted circular cone with base radius 2 m and height 4 m. If water is being pumped into the tank at a rate of 2 m3/min, find the rate at which the water level is rising when the water is 3 m deep.
Sol.
We first sketch the cone and label it as in Figure. Let V, r, and h be the volume of the water, the radius of the surface, and the height at time t, where t is measured in minutes.
We are given that dV/dt = 2m3/min and we are asked to find dh/dt when h is 3 m. The quantities V and h are related by the equation V = 1/3pr2hBut it is the very useful to express V as a function of h alone.
In order to eliminate r, we use the similar triangles in Figure to write and the expression for V becomes
Substituting h = 3 m and dV/dt = 2m3/min, we have
The water level is rising at a rate of 8/(9p) ≈ 0.28 m/min.
Ex.26 A man walks along a straight path at the speed of 4 ft/s. A searchlight is located on the ground 20 ft from the path and is kept focused on the man. At what rate is the searchlight rotating when the man is 15 ft from the point on the path closest to the searchlight ?
Sol. We draw Figure and let x be the distance from the man to the point on the path closest to the searchlight. We let θ be the distance from the man to the point on the path closest to the searchlight and the perpendicular to the path.
We are given that dx/dt = 4 ft/s and are asked to find dq/dt when x = 15. The equation that relates
Differentiating each side with respect to t, we get dx/dt = 20sec2 θ dθ/dt
when x = 15, the length of the beam is 25, so cosθ = 4/5
The searchlight is rotating at a rate of 0.128 rad/s.
Ex.27 Two men A and B start with velocities v at the same time from the junction of two roads inclined at 45º to each other. If they travel by different roads, find the rate at which they are being separated.
Sol.
Let L and M be the positions of men A and B at any time t,
Let OL = x and LM = y. Then OM = x
given, dx/dt = v; to find dy/dt from ΔLOM,
∴ they are being separated from each other at the rate
Ex.28 A variable triangle ABC in the xy plane has its orthocentre at vertex 'B', a fixed vertex 'A' at the origin & the third vertex 'C' restricted to lie on the parabola y = The point B starts at the point (0, 1) at time t = 0 & moves upward along the y axis at a constant velocity of 2 cm/sec. How fast is the area of the triangle increasing when t = 7/2 sec ?
FAQs on Tangents at the Origin - Mathematics (Maths) Class 12 - JEE
Ans. Tangents at the origin refer to the lines that touch a curve at only one point, which is the origin (0,0). These tangents have a slope equal to the derivative of the curve at that point.
2. How are tangents at the origin calculated?
Ans. To calculate the tangents at the origin, we need to find the derivative of the curve at the origin. This can be done by finding the slope of the curve at the origin using calculus techniques such as differentiation.
3. Why are tangents at the origin important in calculus?
Ans. Tangents at the origin play a crucial role in calculus as they help us determine the behavior of a curve near the origin. By analyzing the slope and direction of the tangents, we can understand the rate of change at the origin and make predictions about the curve's behavior in the vicinity of the origin.
4. Can tangents at the origin be used to approximate the curve?
Ans. Yes, tangents at the origin can be used to approximate the curve near the origin. By using the slope and the point of tangency, we can create a linear function that closely approximates the curve in the vicinity of the origin.
5. What information can be inferred from tangents at the origin?
Ans. Tangents at the origin provide us with valuable information about the curve's behavior at that point. They give us insights into the curve's slope, concavity, and the rate of change at the origin. Additionally, they help us understand the shape and direction of the curve in the vicinity of the origin.
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Secant
In mathematics, a secant is a trigonometric function that is the reciprocal of the cosine function
In mathematics, a secant is a trigonometric function that is the reciprocal of the cosine function. The word "secant" originates from the Latin word "secare," which means "to cut." The function gets its name because it represents the length of the line segment that cuts the unit circle at a given angle and extends to the x-axis.
The secant function is denoted as sec(x) or secant of x, where x represents the angle in radians or degrees.
The secant function can be defined using the cosine function as follows:
sec(x) = 1 / cos(x)
To understand how secant is related to the properties of a triangle, particularly in right triangles, consider a right triangle with an angle x. The secant of angle x is defined as the ratio of the length of the hypotenuse to the length of the adjacent side:
sec(x) = hypotenuse / adjacent
This concept can be visualized in the unit circle. As the angle x varies, the secant of that angle represents the x-coordinate of the point where the line extending from the center of the unit circle intersects the line segment on the x-axis.
It is important to note that the secant function is undefined for certain values of the angle. For example, when the cosine function is equal to zero, the secant function becomes undefined, as division by zero is not permissible.
The secant function is related to several other trigonometric functions, such as the sine, cosine, tangent, cosecant, and cotangent functions, through various identities and formulas. These relationships allow for the evaluation and simplification of trigonometric expressions and equations.
In summary, the secant function is a fundamental trigonometric function that represents the reciprocal of the cosine function. It is used to calculate the length of a line segment that cuts the unit circle at a given angle and extends to the x-axis. The function has applications in various fields, including physics, engineering, and computer science, where trigonometry is often employed to model and solve real-world | 677.169 | 1 |
I have the following homework question:
find the length of side h. Assume the largest triangle is isosceles.
The way I'm thinking about it is that the 2 same angles could be any of the following options:
The answer is assuming that the 2 angles in Option A are the same, h=5.66.
I don't understand why Option B and Option C cannot be viable options? Why can't the 2 angles in either Option B or Option C be the same?
1 Answer
1
The other cases would imply $\frac{a}{2} + a + 90º = 180º$, or that $a = 60º$. This would mean that the triangle is equilateral. However, one of the sides is $2 + 2 = 4$, and one of the sides is $6$, so the triangle is not equilateral – a contradiction. | 677.169 | 1 |
Name The Line And Plane Shown In The Diagram
In geometry, lines and planes are fundamental concepts that define the structure and relationships within shapes. Identifying these elements correctly is crucial for solving complex geometric problems. Let's delve into a specific diagram and dissect the lines and planes it presents.
Pain Points:
Navigating geometric diagrams can be challenging, especially when faced with multiple lines and planes intersecting at various points. Without a clear understanding of their names and properties, it becomes difficult to analyze the relationships between them. This confusion hinders problem-solving and limits geometric comprehension.
Names of Lines and Planes in the Diagram:
Referring to the provided diagram, we can identify the following lines and planes:
Line AB: A straight line passing through points A and B.
Line CD: A straight line passing through points C and D.
Plane P: A flat surface containing points A, B, and C.
Plane Q: A flat surface containing points D, C, and E.
Summarizing the Main Points:
Identify lines by naming the points they pass through (e.g., Line AB).
Describe planes by naming the points they contain (e.g., Plane P).
Understanding the names of lines and planes is crucial for analyzing geometric relationships.
The plane is perpendicular to line segment AB at the point C, called the point of intersection.
Point of Intersection C
Point C is the point where line segment AB and plane PQRS intersect.
The line segment AB pierces the plane at this point, creating two half-planes.
The line segment AC lies in the plane, while the line segment BC extends outside the plane.
Orthogonal Projection D
Point D is the orthogonal projection of point A onto plane PQRS.
Line segment AD is perpendicular to the plane, connecting point A to the plane's surface.
The distance between points A and D represents the height of point A above the plane.
Angle of Inclination θ
The angle θ represents the angle between line segment AB and the plane PQRS.
It is measured from the normal to the plane (a line perpendicular to the plane) to the line segment.
The angle θ determines the inclination of the line with respect to the plane.
True Distance AC
The true distance between points A and C is the actual length of line segment AC.
It is the hypotenuse of right triangle ADC, where AD is the height and DC is the horizontal distance.
The true distance AC is longer than the projected distance AD due to the angle of inclination.
Horizontal Distance DC
The horizontal distance between points D and C is the distance along the plane's surface.
It is the length of line segment DC, which is perpendicular to line segment AD.
The horizontal distance DC is shorter than the true distance AC due to the angle of inclination.
Perpendicular Line CD
Line segment CD is perpendicular to line segment AB at point C.
It lies entirely in the plane PQRS and intersects line segment AB at a right angle.
The perpendicular line CD divides the line segment AB into two segments: AC and BC.
Plane Equation
The plane PQRS can be represented by an equation of the form ax + by + cz + d = 0.
This equation defines the plane's orientation and position in space.
The coefficients a, b, c, and d are constants that determine the plane's properties.
Parametric Equations
Parametric equations of line segment AB are given by:
x = x_A + t(x_B - x_A)
y = y_A + t(y_B - y_A)
z = z_A + t(z_B - z_A)
Where (x_A, y_A, z_A) and (x_B, y_B, z_B) are the coordinates of points A and B, respectively, and t is a parameter.
Conclusion
The diagram of the line and plane intersection provides valuable information about the spatial relationships and orientations between these two geometric entities. Understanding the concepts discussed in this article is essential for various applications in mathematics, engineering, and spatial analysis.
FAQs
What is the difference between a line and a plane?
A line is a one-dimensional path that extends infinitely in both directions, while a plane is a two-dimensional surface that extends infinitely in all directions.
How do you find the intersection point of a line and a plane?
To find the intersection point, you need to solve the system of equations formed by the equation of the line and the equation of the plane.
What is the orthogonal projection of a point onto a plane?
The orthogonal projection is the point on the plane directly below the given point, with the line connecting them perpendicular to the plane.
What is the significance of the angle of inclination?
The angle of inclination determines the angle between the line and the plane, affecting the length and orientation of the line within the plane.
How can you determine if a line is parallel or perpendicular to a plane?
If the line is parallel to the plane, it will have a different normal vector than the plane. If the line is perpendicular to the plane, its direction vector will be perpendicular to the plane's normal vector. | 677.169 | 1 |
Choose an answer
An airplane flying at height of 300 meters above the ground passes vertically above another plane at an instant when the angle of elevation of the two planes from the same point on the ground are 60° and 45° respectively. Then the height of the lower plane from the ground is (in meters).
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A tower subtends an angleαat a point on the same level as the root of the tower and at a second point, b meters above the first, the angle of depression of the foot of the tower isβ. The height of the tower is
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The upper 3/4 the portion of a vertical pole subtends an angle tan-13/5 at a point in the horizontal plane through its foot and at a distance 40 m from the foot. A possible height of the vertical pole is
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Choose an answer
A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank of the river is 60° and when he retires 40 meters away from the tree the angle of elevation becomes 30°. The breadth of the river is | 677.169 | 1 |
Review of Triangles
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Everyone. Welcome back. So we're going to spend a lot of time in this course talking about angles and triangles. And I want to give you a really good solid foundation for this because we'll be talking about them a lot later on. So in this video, I'm just gonna walk you through a basic sort of refresher on the basics of triangles. There's a couple of important conceptual and mathematical things you need to know some of them will do some examples together. Let's get started here. A triangle is really just a geometric shape with three sides. That's a really sort of basic definition over here. All these things here have three sides and they all sort of close together to form a shape and that's a triangle. Now, uh there's actually three types of triangles that we can classify based on the lengths of their sides. This is, this is just some vocab that you'll need to know. All right, the first one is called an equilateral triangle. And this is where if you'll see all of the lengths of the triangle are the same, all of them have a length of three equilateral, just means that three sides have equal length. That's actually what that word means. All right. The way that we indicate this in diagrams is you'll see these little tick marks next to the numbers that just means that those two or three measurements are all the same. That's an equilateral and isoli triangles. The next type. And this is where actually two of the side lengths have equal length. Notice how the bottom side is three and these two top sides over here are five. That's an isosceles triangle. The last one is called a scaling triangle. This is where actually none of the sides have equal length. Notice how this is a three, this is a five and this is a six. So you'll see no tick marks anywhere. Another way of saying this is that all the sides are different in the scaling tribe. All right, that's really all you need to know about their sides. Now, whenever sides meet in a triangle, they actually form angles. So the way we indicate this is by using a little curved arc symbol over here, and we express that angle in terms of degrees. So whenever you have two sides of a triangle meets, they form angles and there's three other types of triangles that we can classify based on those angles. All right. So again, this is just more vocabulary over here. An acute triangle over here is one in which all of the angles are less than 90 degrees. All of these words that you'll see acute obtuse and right have to do with what those angles are relative to 90. So look at this angle, the these uh these angles over here, all of these things are less than 90 degrees. So this is an acute triangle, right? So this next one over here, you'll see that there's 22 angles that measure 35 degrees, but there's one of them that measures 100 and 10. And this is an example of an obtuse triangle because one angle is greater than 90 degrees. So that's an obtuse triangle. The last one is called a right triangle. We're gonna be spending a lot of time talking about these and these are special triangles where one of the angles is exactly equal to 90 degrees. All right. Now, regardless of any type of triangle, whether we're looking at the sides or the angles. One really important thing you need to know is that in any type of triangle, all angles will always add up to 180 degrees. That is a fundamental property of triangles. So you would look at all of the triangles over here. All of these three numbers will add up to 100 and 80 same thing for this and same thing for this. All right. So that's uh the really important because if you know that all of the angles add up to 100 and 80 if you're ever missing one of them, then you can always find the other one. All right. So that's actually really important. Let's go ahead and take a look at our first example over here, we're gonna, we're gonna um for each of these triangles, we're gonna figure out the missing angle or the missing side. All right. So in this case over here, for this first example, we have a missing side represented by a variable over here. This is X. How do we find that? Well, we haven't discussed any mathematical ways of calculating this. But one of the things you could notice here is that these tick marks uh mean that the measurements have to be the same. So in other words, if the left side is four, then that means that this also has to equal four. All right, that's just something that you might need to know. Let's take a look at the next one example. B this is one where we have two of the angles, this is 40 degrees and 40 degrees, but we're actually missing one of the other ones. How can I find that? Well, again, remember all of the angles have to add up to 100 and 80 degrees in any triangle. So if you're ever missing one of them, you can always find for the other, uh I'm just gonna set up a simple equation over here. This is gonna be 40 plus 40 plus theta. So in other words, if I add up all of the angles I have to get 100 and 80. If I just subtract 40 from both of the sides over here, this is basically the same thing as subtracting 80. What we're gonna find here is this angle is equal to 100 degrees. All right. So this angle over here is 100 degrees. And therefore, this would actually be an obtuse triangle, but that's not what the question asked us. All right. But that's the answer, theta equals 100. So that's it. That's just a basic introduction. Let's go ahead and get some, get some practice.
2
Problem
Problem
Classify the triangle below according to its sides and angles.
I. Equilateral
II. Isosceles
III. Scalene
IV. Acute
V. Obtuse
VI. Right
A
I and IV
B
I and V
C
II and V
D
II and IV
E
III and VI
F
Only one of I, II, III, IV, V, and V
3
Problem
Problem
Find the missing angle θ\thetaθ for this right triangle.
A
30°
B
60°
C
90°
D
120°
4
concept
Solving Right Triangles with the Pythagorean Theorem
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Everyone. So in a previous video, we discussed the basics of triangles and I mentioned that we would very commonly be working with right triangles. But one of the most common situations that you'll see is where you have two sides of a right triangle that are known, but you have an unknown side. So for example, we have three and four that are known here, but this side X here is missing. Well, don't worry because in these kind of situations we can always solve for this missing side by using something called the Pythagorean theorem. It's probably something that you've heard before in a math class, but we're gonna be using it a lot in this course and you'll need to know it. So I'm gonna go ahead and explain it to you. What I'm gonna show you. It is that it's really just an equation relating the three sides of a right triangle. Let's go ahead and get started. We'll do some examples together, all right. The first thing you need to know about the Pythagorean theorem is you can only use it when you have a right triangle. All right. So you can only use it when you can assume or when you know that one of the angles over here is 92. If you don't know that, then this equation won't work. So what is the equation? Well, it's really just A squared plus B squared equals C squared again. You've probably heard that before. But what it just really means is if I take these two numbers over here A and B, they'll just be numbers and I square them and add them together. That's the same exact value as this side over here, squared as well. Let's take a look at our first example. So we can actually just get some practice with us and do it together. All right. So we have three and four that are known over here and you have X that's unknown. This is the right triangle. So I'll be able to use the Pythagorean theorem to solve that missing side. I just have my equation over here. A squared plus B squared equals C squared. All right. So how does this work? Well, what's really, really important about the Pythagorean theorem as well is that you always have to take mind, uh take notes that you're A and B need to be the shorter legs of the triangle. Always set your A and B as the shorter legs that form the corner, that form that 90 degree angle. And then you wanna set C as the hypotenuse the hypotenuse of a triangle is always the longest side, which usually is gonna be the diagonal. Not always, but it's almost, almost gonna be the diagonal one. All right. So in this right triangle over here, what we can see is that these two form the sort of corner like this, that's A and B and C is gonna be the diagonal, the longer one. So that's what we set as C and when it comes to A and B, it actually doesn't matter which one you pick as A or B, I'm gonna go ahead and just pick this one as my A and this one as my B. So what this equation says is that A squared plus B squared is C squared. So in other words, if I take four and I square it and I add it to three and I square it and I figure that out, that's gonna give me this missing side squared. That's gonna be X squared. All right. So four squared plus three squared. This actually just ends up being 16 plus nine, that's gonna equal uh X squared. And if you actually just go ahead and work that out, that's gonna be 25. So are we done here is the answer just 25? Well, no, a lot of students will mess this up. You have one last step to do here, which is, you have to take the square roots of both sides because you want X not X squared. If you do that, what you're gonna get is that X is equal to five and that is the answer. So the hypotenuse of this triangle is equal to five. All right. One really quick way to sort of check your work is that number of hypo news has to be the longest side. So notice how five is longer than three and four. If I got something that was lower than three or four, then I know I, I would have messed something up. All right. But that's really all it is. All right. That's the Pythagorean theorem. Go ahead and pause the video and see if you can find the answer to this problem over here. Example B All right. So let's go ahead and work it out. So we've got this triangle over here. Notice how it's actually a little bit different because we have a right triangle just like we did over here. But this time, we actually know what the hypotenuse is that diagonal, longest side is something we already know. And in fact, one of the shorter legs is actually one of our unknown values. But we can, we can still use the Pythagorean theorem because remember it's just we know two sides out of three. I'm gonna start off with my equation over here. A square plus B squared to C squared. I set my C to be the hypotenuse. So the word C is 10. And then again, it doesn't matter which one is my uh missing variable right or sorry. Which one is my A and B, either my A and B will just make up that 90 degree angle. So I'm gonna just go ahead and set this one to be A and this one to be B, we'd get the exact same answer if you did it the other way. All right. So this is A B equals six and A equals Y. So what the Pythagorean theorem says is that A squared plus B squared, some of the words Y squared plus my B which is six squared, that's gonna equal C squared and C squared is just 10 squared. That's gonna be 10 squared over here. Now, you just plug in the values that you already know and just sort of calculate right. So this is gonna be Y squared. Uh This is gonna be plus 36 that's 36 over here. That's gonna equal 10 squared, which is 100 you can subtract 36 from both sides like this, subtract 36. What you're gonna get over here is that Y squared is equal to 100 minus 36 which equals 64 that's equals 64. Now, the last thing you're gonna have to do, let me actually just write that in blue. Last thing you have to do is you just have to take the square root of both sides. You have Y is equal to the square root of 64 and that's equal to eight All right. So that is equal to eight over here. So that means we go back into our diagram and this is gonna equal eight. Notice how again the hypotenuse the 10 is still longer than both of the other, shorter sides over here. And that perfectly makes sense. All right. So that's it for this one, folks. Thanks for watching and I'll see you in the next one. | 677.169 | 1 |
Equilateral \triangle A B C is inscribed in a circle of radius 2. Extend \overline{A B} through B to point D so that A D=13, and extend \overline{A C} through C to point E so that A E=11.
Through D, draw a line \ell_{1} parallel to \overline{A E}, and through E, draw a
line \ell_{2} parallel to \overline{A D}. Let F be the intersection of \ell_{1} and \ell_{2}. Let G be
the point on the circle that is collinear with A and F and distinct from A. Given that the area of \triangle C B G can be expressed in the form p \sqrt{q} / r, where p, q, and r are positive integers, p and r are relatively prime, and q is not divisible by the square of any prime, find p+q+r. | 677.169 | 1 |
(b) The magnitude of |AB→|,|AB→|=(−6)2+(8)2=10∴The unit vector in the direction of AB→,AB→|AB→|=110(−6i˜+8j˜)=−35i˜+45j˜
Question 2: Given that A (–3, 2), B (4, 6) and C (m, n), find the value of m and of n such that 2AB→+BC→=(12−3)
Solution: A=(−32),B=(46) and C=(mn)AB→=AO→+OB→AB→=−(−32)+(46)=(74)BC→=BO→+OC→BC→=−(46)+(mn)=(−4+m−6+n)Given 2AB→+BC→=(12−3)2(74)+(−4+m−6+n)=(12−3)(14−4+m8−6+n)=(12−3)10+m=12m=22+n=−3n=−5
Question 3:
Diagram below shows a rectangle OABC and the point D lies on the straight line OB. | 677.169 | 1 |
Help me Find my Relationship!
In this lesson, students will investigate the relationship between angles when parallel lines are cut by a transversal. Students will identify angles, and find angle measures, and they will use the free application GeoGebra (see download link under Suggested Technology) to provide students with a visual representation of angle relationships. | 677.169 | 1 |
Using this approach the average azimuth is ~72.22 when I would expect it to be around ~50 degrees.
I realized that it doesn't take into consideration the length (magnitude) of each line segment, and the 2 dominant segments (both ~45 degree azimuth) only accounts for 2/10 of the average (2 point pairs of the total 10 point pairs).
So it seems liked I need to use the length of each segment as a weight to the average.
That's when I thought maybe vector math (dot product) is whats needed because I have a magnitude (length) and angle (azimuth); However, because the segments (vectors) are already lined up head to tail, the resultant vector's angle would just be the angle between the start point and end point (which is not what I am after)
I am looking for approaches to mathematically solve this problem, or references to what it's called that I am after.
Why isn't the average direction for the track equal to the direction from the first to the last point? The wiggle in the middle should cancel itself out. Imagine a semicircle - what's the average direction of that? Its the diameter across the semicircle. For any arc, the average direction is the direction of the chord from start to finish.
Calculating bearing across a large area in units of degrees requires the use of geodetic functions, to properly handle the curvature of the spheroid, since the back angle isn't .equivalent to the angle that got there. The name of this task is the second Geodetic Problem (aka Inverse aka Reverse). There are a multitude of libraries to solve this.
@Spacedman I agree that's true for an arc, but I am not sure about a piece-wise line segment. Consider this line segment i.vgy.me/HpNkVS.png Is the overall azimuth 90 degrees? It doesn't seem like it because I agree that the the y-components cancel each other out the x-components do not; You could be right though. When I think of an object starting at one point and maintain a constant speed, what bearing will that object spend most of its time? To me it would be the long part pointing ~135
@Vince I am not concerned with large area in units, I used such large corrdinates in the example because it was easy to make the Well Known Text easily with arthur-e.github.io/Wicket/sandbox-gmaps3.html - An approximation is suitable for my needs, just looking for methodologies and less reference to libraries (because I want to implement a SQL solution)
@CuriousDeveloper yes the mean direction for that path is heading East. Sure it spent a lot of time going in SE direction but then it turned round and went due N to get back on track. I'm pretty sure that will integrate out to an overall Easterly average. | 677.169 | 1 |
How to Calculate the Angle Between Two Vectors in MATLAB
In MATLAB, the angle between two vectors can be calculated using the `atan2()` function. This function takes two vectors as input, and returns the angle between them in radians. The angle is measured from the first vector to the second vector, with a positive angle indicating that the second vector is counterclockwise from the first vector, and a negative angle indicating that the second vector is clockwise from the first vector.
The `atan2()` function has the following syntax:
atan2(y, x)
where `y` and `x` are the two vectors whose angle you want to calculate.
For example, the following code calculates the angle between the vectors `v1 = [1, 2]` and `v2 = [3, 4]`:
angle = atan2(v2(2), v2(1)) – atan2(v1(2), v1(1))
This code returns the value `1.1071487177940904`, which is the angle between the two vectors in radians.
The angle between two vectors can be used for a variety of tasks, such as determining the direction of a vector, or calculating the similarity between two vectors.
Angle
Formula
Example
Angle between two vectors
arccos(dot(v1, v2) / (norm(v1) * norm(v2)))
angle = arccos(dot(v1, v2) / (norm(v1) * norm(v2)))
Angle between a vector and the x-axis
atan2(y, x)
angle = atan2(y, x)
Angle between a vector and the y-axis
-atan2(x, y)
angle = -atan2(x, y)
In this tutorial, we will learn how to calculate the angle between two vectors in MATLAB. We will start by discussing the mathematical formula for calculating the angle between vectors, and then we will provide an example of how to calculate the angle between two vectors in MATLAB.
Formula for the Angle Between Vectors
The formula for calculating the angle between two vectors $\vec{a}$ and $\vec{b}$ is given by:
In this tutorial, we learned how to calculate the angle between two vectors in MATLAB. We started by discussing the mathematical formula for calculating the angle between vectors, and then we provided an example of how to calculate the angle between two vectors in MATLAB.
2. Formula for the Angle between Vectors in MATLAB
The angle between two vectors in MATLAB can be calculated using the following formula:
angle(v1, v2)
where `v1` and `v2` are the two vectors. The output of the `angle()` function is a scalar value, which is the angle between the two vectors in radians.
To illustrate how to use the `angle()` function, let's consider the following two vectors:
v1 = [1, 2, 3]
v2 = [4, 5, 6]
The angle between these two vectors can be calculated using the following code:
angle(v1, v2)
The output of this code is `1.1071487177940904`. This means that the angle between the two vectors is approximately 1.1 radians, or 63.4 degrees.
3. Applications of the Angle between Vectors in MATLAB
The angle between vectors can be used for a variety of applications in MATLAB, including:
Computing the dot product of two vectors. The dot product of two vectors is a scalar value that is equal to the sum of the products of the corresponding elements of the two vectors. The angle between two vectors can be used to calculate the dot product of the vectors, as follows:
dot(v1, v2) = |v1| * |v2| * cos()
where " is the angle between the two vectors.
Determining the direction of a vector. The angle between a vector and the positive x-axis can be used to determine the direction of the vector. For example, if the angle between a vector and the positive x-axis is 0 degrees, then the vector is pointing in the positive x-direction. If the angle between a vector and the positive x-axis is 90 degrees, then the vector is pointing in the positive y-direction.
Finding the shortest distance between two vectors. The shortest distance between two vectors is the length of the line segment that connects the two vectors. The angle between the two vectors can be used to calculate the shortest distance between the vectors, as follows:
d = |v1 – v2| = |v1| * sin()
where `d` is the shortest distance between the two vectors, `v1` and `v2` are the two vectors, and " is the angle between the two vectors.
Determining the orientation of an object. The angle between two vectors can be used to determine the orientation of an object. For example, if the angle between two vectors that define the axes of an object is 90 degrees, then the object is oriented in a rectangular fashion. If the angle between two vectors that define the axes of an object is 180 degrees, then the object is oriented in a parallel fashion.
4. References
A: To find the angle between two vectors, you can use the atan2() function. This function takes two vectors as input, and returns the angle between them in radians. The following code shows how to use atan2() to find the angle between two vectors:
v1 = [1, 2, 3];
v2 = [4, 5, 6];
angle = atan2(v2(2) – v1(2), v2(1) – v1(1));
The output of this code will be the angle between the two vectors in radians. To convert the angle to degrees, you can multiply it by 180/pi.
Q: What is the difference between the dot product and the angle between two vectors?
A: The dot product of two vectors is a scalar, while the angle between two vectors is a radians. The dot product of two vectors is equal to the product of the magnitudes of the vectors and the cosine of the angle between them. The following formula shows the relationship between the dot product and the angle between two vectors:
dot_product = ||v1|| * ||v2|| * cos(theta)
where:
v1 and v2 are the two vectors
||v1|| and ||v2|| are the magnitudes of the vectors
theta is the angle between the vectors
Q: How can I find the angle between two vectors in degrees?
A: To find the angle between two vectors in degrees, you can multiply the angle between them in radians by 180/pi. The following code shows how to do this:
The output of this code will be the angle between the two vectors in degrees.
Q: What are the units of the angle between two vectors?
A: The angle between two vectors is measured in radians. Radian is a dimensionless unit, so the angle between two vectors does not have any units.
In this blog post, we discussed the concept of the angle between vectors and how to calculate it in MATLAB. We also provided several examples of how to use the `angle` function to find the angle between vectors in different scenarios Abseil with Just a Rope Abseiling is a thrilling activity that can be enjoyed by people of all ages and abilities. It's a great way to get outdoors and experience the thrill of heights. But what if you don't have access to a climbing wall or an outdoor course? Don't worry, you can…
Power BI Freeze Column: A Quick Guide Power BI is a powerful business intelligence tool that can help you visualize and analyze your data. One of the most useful features of Power BI is the ability to freeze columns. Freezing a column keeps it visible when you scroll through your data, making it easier to…
Piloncillo: The Sweet Gold of Mexico Piloncillo is a type of unrefined whole cane sugar that is produced in Mexico. It is made by boiling sugarcane juice until it reaches a thick, syrupy consistency. The syrup is then poured into molds and allowed to cool and harden. Piloncillo is a dark brown color and has…
Solving Limits with Square Roots Limits are an important concept in calculus, and they can be used to solve a variety of problems. In this article, we will discuss how to solve limits with square roots. We will start by reviewing the basics of limits, and then we will show you how to apply these…
Golang int64 to int: A Guide In Golang, the `int64` and `int` data types are both integers. However, they have different sizes and ranges. The `int64` type is a 64-bit integer, while the `int` type is a 32-bit integer. This means that the `int64` type can store larger values than the `int` type. Converting an…
Have you ever started a process in the background with nohup and then regretted it later? Maybe you realized that you needed to pass different arguments to the program, or maybe you just wanted to stop the process altogether. Whatever the reason, terminating a nohup process can be tricky. In this article, I will show… | 677.169 | 1 |
Activities to Teach Students Proofs Involving Parallel Lines
Parallel lines and theorems have been a fundamental concept in mathematics for several centuries. As a student, learning proofs involving parallel lines can be challenging, but it is essential to understand this concept for further mathematics study. To help students gain a better understanding of proofs involving parallel lines, several activities can be incorporated into the classroom.
1. The Angle Game
The angle game is a fun way to help students understand the relationship between parallel lines and the angle measures. To begin, draw two parallel lines on the board or worksheet and label them as l and m. Next, shade in an angle at the intersection of the two lines and label it as A. Then, ask students to identify all the other angles in the figure that are congruent (same size) to angle A. By doing so, they will be able to identify the corresponding angles, which are equal in size due to the parallel lines.
2. The Parallel Lines Cut By a Transversal Puzzle
Another bedrock activity to teach students proofs involving parallel lines is the parallel lines cut by a transversal puzzle. To start, draw two parallel lines and a transversal that intersects the two lines. Cut out several angle diagrams and figure out which angles correspond with each other based on the rules, such as alternate interior angles are identical.
3. Parallel Line Collaborative Activity
This activity involves placing students in groups of 3-4 and assigning them different roles, such as a recorder, a speaker, a problem solver, and a checker. To start, provide each group with a set of parallel lines and a figure, which they will have to solve. The recorder notes down the steps taken to find the solution, while the problem solver uses their knowledge of theorems and angles to solve the puzzle. The speaker will then present the solution to the rest of the class. Lastly, the checker will review the solution critically to check for mistakes.
4. Parallel Parking Proofs
Parallel parking is a practical application involving parallel lines, and it makes a fun activity to teach students proofs involving parallel lines. Draw two parallel lines on the classroom board and shade in a car situated between the two lines. Then, ask the students to draw two additional lines representing the wheels of the car, parallel to the two original lines. By doing so, students will understand how parallel lines work in practical use and gain a better understanding of proof concepts.
5. Obtuse and Acute Angles in Parallel Lines
This activity involves covering parallel lines with a piece of paper, leaving two opens spaces on opposite sides. Then, mark a point on the paper, and draw an acute angle and an obtuse angle, starting from that point and each of the open spaces to the other side of the paper. After removing the paper, students would be able to see that the acute angles and obtuse angles on one side would always sum up to a straight line. This activity helps students to understand subtleties associated with parallel lines and proofs.
In conclusion, incorporating various activities in teaching proofs involving parallel lines will make the learning process more engaging and easier for students. The activities will help students learn the concept in a more practical way and be more confident when everyone is involved. Now that the benefits of incorporating these activities have been highlighted, teachers have a plethora of options to choose from and help their students better understand the importance of parallel lines and theorems | 677.169 | 1 |
Which is very important for all the students of class six to understand matriculation and we have explained the geometry of chapter 4 of NCERT class six mathematics in a very good way for the students. Solved in this way, I have tried my best to explain the geometry to the students in visual form, we have started all the NCERT questions in the solution of NCERT through diagram
Question 1. Use the figure to name: (a) Five points (b) A line (c) Four rays (d) Five line segments.
Solution : (a) O, B, C. D, E
Question 2.
Name the line given in all possible (twelve) ways, choosing only two letters at a time from the four given.
Solution :
Question 3. Use the figure to name: (a) The line containing point E. (b) The line passing through A. (c) The line on which O lies (d) Two pairs of intersecting lines.
Solution : (a) AE←→, etc. (b)AE←→, etc. (c) CO←→orOC←→ (d)CO,←→−AE←→;AE←→,EF←→.
Question 4. How many lines can pass through (a) one given point? (b) two given points?
Solution : (a) Countless lines can pass through one given point. (b) One and only one line can pass through two given points.
Question 5. Draw a rough figure and label suitably in each of the following cases : (a) Point P lies on AB¯ (b)XY←→ and PQ←→ intersect at M. (c) Line contains E and F but not D. (d)Op¯ and OQ¯ meet at O.
Solution : (a) (b) (c) (d) Question 6. Consider the following figure of line MN¯ Say whether following statements are true or false in context of the given figure. (a) Q, M, O, N, P are points on the line MN¯ (b) M, O, N are points on a line segment MN¯. (c) M and N are end points of line segment MN¯. (d) O and N are end points of line segment OP¯. (e) M is one of the end points of line segment QO¯. (f) M is point on ray OP−→−. (g) Ray OP−→− is different from ray QP−→−. (h) Ray OP−→− is same as ray OM−→−. (i) Ray OM−→− is not opposite to ray OP−→−. (j) O is not an initial point of OP−→−. (k) N is the initial point of NP−→− and NM−→−.
Solution : (a) True (b) True (c) True (d) False (e) False (f) False (g) True (h) False (i) False (j) False (k) True. | 677.169 | 1 |
A point is taken at random in each of the two adjacent sides of a square. Show that the average area of the triangle formed by joining them is one eighth of the area of the square. Average area of triangle formed 1/8 that of square
la1noxz
Answered question
2022-10-03
Average area of triangle formed 18 that of square. A point is taken at random in each of the two adjacent sides of a square. Show that the average area of the triangle formed by joining them is one eighth of the area of the square.
Answer & Explanation
procjenomuj
Beginner2022-10-04Added 8 answers
Step 1 In compound probability product of two different probabilities Area=12x⋅y;pArea=12pxpy Step 2 Here px=py=12 for the full range of side length (0,1) considered We can directly compare the area of gray triangle area to that of the square 12⋅(12)212=18
0
seguitzla
Beginner2022-10-05Added 4 answers
Step 1 Inside your square, draw the quadrilateral with vertices (p,0), (0,q), (p,1), (1,q). The triangles in the corners of the square have total area 1/2. Proof: Draw the dashed lines below. Each triangle is half the area of the rectangle containing it, and those rectangles add up to area 1. Step 2 Thus, the average area of these 4 triangles is 1/8. By the symmetry of the problem, therefore, the expected area of each of the triangles is 1/8. | 677.169 | 1 |
Thomson Cubic
The Thomson cubic
of a triangle
is the locus the centers of circumconics whose normals at the vertices are concurrent.
It is a self-isogonal cubic with pivot point
at the triangle centroid, so its parameter is
and its trilinear equation is given
by | 677.169 | 1 |
Given some points $X=\{x_i:||x_i||=1,i=1,\ldots,n\} $ located on the sphere, how to calculate the point $\tilde{x}$ on the sphere that is nearest to these given points. That is to say $$\tilde{x}=\arg\min_{\tilde{x}}\sum_{x_i\in X}d(\tilde{x},x_i), s.t. ||\tilde{x}||=1,$$ where $d(\tilde{x},x_i)=\arccos(\tilde{x}\cdot x_i)$ is the spherical distance between $\tilde{x}$ and $x_i$.
The question is very similar to How to calculate center point in geographic coordinates?. In that question, someone suggested to compute the mean $\bar{x}=\frac{1}{n}\sum x_i$ of the given points, and normalize the mean vector. As far as I know, the mean $\bar{x}$ is optimal under the squared L2 norm. But I am not sure that whether the normalized mean is optimal under the spherical distance measure.
1 Answer
1
This is a problem of finding a geometric median on a sphere. While the mean provides a minimizer for the sum of squared distances (in a Euclidean setting), the $L^1$ minimum is generally not expressible in such an explicit form. For three points, the Euclidean solution is known as a Fermat point, and for four coplanar points an explicit solution is known.
Added: Let me point out a note on the three point case by K. Ghalieh and M. Hajja, The Fermat point of a spherical triangle in The Mathematical Gazette of Nov. 1996 (pp. 561-564). Although it is "behind a pay wall", you can nonetheless take advantage of JSTOR's free-of-charge program (registration required) to read the article online (see site for details), as I have done.
The authors show that when the sides of a spherical triangle $ABC$ are sufficiently short, each less than $\pi/2$ on a sphere of unit radius (implying that they are not all on the same great circle and that no two vertices are antipodal), then a Fermat point (minimizing the sum of spherical distances to three vertices) exists, is unique, and shares some defining properties with the case of a triangle in the Euclidean plane:
If all three vertex angles are less than 120°, then the Fermat point $P$ is "inside" the spherical triangle (in the smaller region of the sphere bounded by the triangle) and the three sides subtend equal angles around $P$:
$$ \angle APB = \angle BPC = \angle CPA = 120° $$
If one of the vertex angles is 120° or more, that vertex is the Fermat point.
$\begingroup$Thank you very much for providing the useful references. I will read them carefully. However as you mentioned, these approaches are iterative ones. I wonder whether there are analytic forms for my problem.$\endgroup$
$\begingroup$Unfortunately I could not find sphere specific papers not behind pay walls, but I saw the other answer and hurried to get it out that the problem is studied and not trivial. If there are special circumstances that you think make the problem easier, please add them to your Question.$\endgroup$
$\begingroup$Regarding your third paragraph: the authors do not actually show all of these things, rather leave it to the reader. In fact they only show that the Fermat point is either a vertex or the point with equiangular property.$\endgroup$ | 677.169 | 1 |
A pyramid has a polygonal base and flat triangular faces, which join at a common point called the apex. A pyramid is formed by connecting the bases to an apex. Each edge of the base is connected to the apex, and forms the triangular face, called the lateral face. Pyramid classifications. We typically name a pyramid based on the shape of its polygonal base. The following are some examples.
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The tallest of the three pyramids stands 71 feet (21.6 meters) tall and has a width at the base of 115 feet (35 meters). 673 glass panes enclose the structure, held together by a steel frame,
If you're cutting ply (or any thick material) for your pyramid, when you cut the triangle segment's hip edges, set your circular saw bevel (tilt) angle to the Triangle Segment Edge Bevel Angle so the edges join neatly. Set Palliativ vård uppsala
Book. GeoGebra Team German . Sections of Rectangular
The geometric shape that has 1 square face and 4 triangular faces is a pyramid. A pyramid also has five vertices and eight edges. Habiliteringen varbergs sjukhus
Pyramid, Pyramid A pyramid is an artificially made geometric solid of the shape made famous by the royal tombs of ancient Egypt. It is a solid with a base of… Prism , Prism In Euclidean geometry, a prism is a three-dimensional figure, or solid, having five or more faces, each of which is a polygon.
Set There is debate as to the geometry used in the design of the Great Pyramid of Giza in Egypt. Built around 2560 BC, its once flat, smooth outer shell is gone and all that remains is the roughly-shaped inner core, so it is difficult to know with absolute certainty. The pyramid contains a huge amount of information about the structure of the universe, the solar system and the man, encoded in its geometric shape in the form of an octahedron, which half it represents. Aerial photos show that the pyramids at Giza are located on a line that clearly correspond to the Fibonacci spiral.
For a pyramid to look like a pyramid, each of the four triangular-shaped sides must
Check out our pyramid shape selection for the very best in unique or custom, 5- sided Pyramid shape/Geometry shape aroma candle mold/ plaster mold /Diy
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Aug 18, 2012 The interesting question is "why did they choose this specific shape geometry and configuration of three pyramids for the Great Pyramid? Geometric shapes: 3-dimensional Triangular Pyramid. Hexagonal Pyramid 3 -D Objects (2 of 2) e.g. Identifying prisms, pyramids, cylinders, cones, etc. | 677.169 | 1 |
When it comes to converting units of measurement, understanding the relationship between different metrics can be quite challenging. One common conversion that often perplexes indi...Academic level: User ID: 407841. Gombos Zoran. #21 in Global Rating. William. ID 5683. Unit One Geometry Basics Homework 5 Angle Relationships -.measures. 4. If the measure of an angle is 13°, find the measure of its supplement. 5. If the measure of an angle is 38°, find the measure of its complement. 6. 1 and 2 form a linear pair. If m 1 = (5x + 9)° and m 2 = (3x + 11)°, find the measure of each angle. 7. 1 and 2 are vertical angles. If m 1 x= (17 m+ 1)° and m ... WebAngles Naming | 677.169 | 1 |
Cot Inverse Calculator
In the realm of trigonometry, the Cotangent Inverse Calculator emerges as a beacon, inviting curious minds to explore angles from a different perspective. This article delves into the intricacies of this calculator, unraveling its importance, providing a practical guide on its usage, and addressing frequently asked questions to demystify the world of cotangent inverses.
Importance
1. Unlocking Inverse Cotangents
While traditional trigonometry focuses on sine, cosine, and tangent, the cotangent inverse, or arccot(x), opens the door to a new set of angles and mathematical possibilities.
2. Beyond Traditional Trigonometry
The cotangent inverse extends trigonometric capabilities by providing a unique approach to calculating angles, offering a fresh perspective for various mathematical applications.
3. Complex Problem Solving
In fields such as physics, engineering, and computer science, the cotangent inverse plays a crucial role in solving complex problems that involve angles and angular relationships.
How to Use the Cotangent Inverse Calculator
1. Input x Value
Begin by entering the value of x into the designated input field. This represents the cotangent for which you want to find the inverse.
5. Can the Cotangent Inverse be Used for Complex Numbers?
While the cotangent inverse is primarily designed for real numbers, it can be extended to handle complex numbers with appropriate considerations.
6. What Happens if I Enter a Non-Numeric Value for x?
The calculator prompts an alert, reminding users to input a valid numeric value for x. Non-numeric entries are not processed.
7. Does the Calculator Provide Radian or Degree Results?
By default, the calculator provides results in radians. Users can convert the result to degrees if needed, keeping in mind the relationship between radians and degrees.
8. How Accurate Are the Results?
The accuracy depends on the precision of the input and the calculator. Adjusting the decimal places in the result can enhance precision as required.
9. Can I Use the Cotangent Inverse in Education?
Absolutely, the cotangent inverse is a valuable tool for educators and students studying advanced trigonometry. It aids in illustrating the concepts of inverse trigonometric functions.
10. Does the Calculator Work on Mobile Devices?
Yes, the calculator is designed to be responsive and user-friendly, ensuring compatibility with various devices, including mobile phones and tablets.
Conclusion
As we navigate the angles of trigonometry, the Cotangent Inverse Calculator becomes a compass, guiding us into uncharted territories. Its importance lies not only in unlocking new mathematical perspectives but also in its practical applications across diverse disciplines. Whether solving complex problems in physics or aiding students in grasping advanced trigonometric concepts, the cotangent inverse proves its significance. As the calculator unveils arccot(x) with a click, it invites us to explore angles beyond the familiar, creating a dynamic bridge between traditional trigonometry and the unexplored realms of inverse cotangents. | 677.169 | 1 |
Activities to Teach Students to Find Trigonometric Ratios Using the Unit Circle
Trigonometry is an important branch of mathematics that deals with the relationships between the sides and angles of triangles. It is a subject that is widely used in many fields, including engineering, science, and even architecture. One of the fundamental concepts in trigonometry is the use of trigonometric ratios, which help in finding the values of unknown sides and angles of a triangle. One useful tool for finding these ratios is the unit circle. Here are some activities that can help teach students how to find trigonometric ratios using the unit circle.
1. Introduce the Unit Circle
The first step is to introduce students to the concept of the unit circle. The unit circle is a circle with a radius of one unit. It is centered at the origin of a coordinate plane and is divided into four quadrants. Each quadrant represents a 90-degree movement in a counter-clockwise direction.
2. Identify the Coordinates
Next, ask students to identify the coordinates of points on the unit circle. The coordinates of a point on the unit circle are represented by the trigonometric ratios of sine, cosine, and tangent. In the first quadrant, the sine ratio is equal to the y-coordinate, while the cosine ratio is equal to the x-coordinate. The tangent ratio is equal to the ratio of the y-coordinate to the x-coordinate.
3. Practice Finding the Trigonometric Ratios
After introducing the concept of the unit circle and its coordinates, students can practice finding the trigonometric ratios of points on the circle. Provide them with a worksheet that includes points on the unit circle and ask them to find the sine, cosine, and tangent ratios. This will not only reinforce the concept of the unit circle but also help students develop their problem-solving skills.
4. Use Board Games
Another way to teach students how to find trigonometric ratios using the unit circle is by playing board games. Create a board game that allows students to move along the unit circle by rolling a dice. They can then identify the coordinates of the points they land on and find the corresponding trigonometric ratios.
5. Use Technology
With the availability of technology, students can also learn how to find trigonometric ratios using the unit circle through interactive online games and simulations. There are many online resources available that provide interactive tools to help students practice their skills.
In conclusion, learning how to find trigonometric ratios using the unit circle is an important concept in trigonometry. By introducing the unit circle and providing students with hands-on activities and practice problems, they can develop a strong understanding of these ratios. The use of board games and online resources can make learning fun and engaging, while also helping students to develop | 677.169 | 1 |
Given a circle with center $(a,b)$ and radius $r$, oriented counter-clockwise, and two points that sit along the circle, $(x_1,y_1)$ and $(x_2,y_2)$, what the is the great circle distance (GCD) between them.
I have something like
$\theta_1=\arccos\left(\frac{x_1-a}{\sqrt{(x_1-a)^2+(y_1-b)^2}}\right)$ and $\theta_2=\arccos\left(\frac{x_2-a}{\sqrt{(x_2-a)^2+(y_2-b)^2}}\right)$
$\theta^* = \theta_1-\theta_2$
$GCD=r \theta^*$
This is assuming that $(x_1,y_1)$ comes before $(x_2,y_2)$ along the circle (in the given orientation). However, this formulation is incorrect. It does not take into account that the range of arccosine is $0 \leq \theta \leq \pi$, and I'm not even sure if I'm orienting my angle properly. As a matter of fact, please include if you have a way to test which point comes first in the orientation.
For example, can you handle the following case for me: Say you have $(2,3)$ and $(4,5)$ sitting on a circle with center $(3,4)$. What is the central angle between them?
1 Answer
1
Translate your circle to the origin. The coordinates of $(x_{i},y_{i})$ become $(x_{i}-a,y_{i}-b)$ for $i=1,2$.
You know that if $X_{1}$ and $X_{2}$ are two vectors of the euclidean space $\mathbb{R}^{2}$, the angle $\theta$ formed by $X_{1}$ and $X_{2}$ is obtained as follows (see the remark at the bottom of this answer):
And you have to solve the previous equation. As you look for the GCD, take the greatest value for $\theta$.
Here, you have $(2-3,3-4)=(-1,-1)$, whose norm is $\sqrt{2}$ and $(4-3,5-4)=(1,1)$, whose norm is $\sqrt{2}$. For this points, it is easy to know what is $\theta$ because they are symmetric with respect to the origin, but let's try the previous "method".
so that $\cos(\theta)=-1$, which means $\theta=\pi+2k\pi$ ($k\in\mathbb{Z}$) and here we only consider $\theta\in]0,2\pi[$, so that $\theta = \pi$. Note that I take $\theta$ in the open interval $]0,2\pi[$ because $\theta=2k\pi$ only if $X_{1}=X_{2}$ but the GCD is not well defined in this case, as far as I know.
Of course, there are often two possible values lying in $]0,2\pi[$ for $\theta$. Take the bigger one as you look for the GCD.
Suppose you want to test whether $X_{1}$ or $X_{2}$ "comes first" when running along the circle counter-clockwise. Let $X_{i}=(x_{i}-a,y_{i}-b)$ ($i=1,2$) Take a look at
$$\frac{X_{i}}{\Vert X_{i}\Vert}=\frac{X_{i}}{r} \tag{$i=1,2$}$$
It lies on a circle of radius $1$ and center $(0,0)$. It can be considered as the trigonometric circle. Then, the abscissa of $\frac{X_{i}}{r}$ is $\cos(\alpha)$ and its ordinate is $\sin(\alpha)$ or equivalently $X_{i}=(r\cos(\alpha),r\sin(\alpha))$. Solve for $\alpha\in[0,2\pi]$, which is the angle as considered in the following representation:
Calculate the angle for $X_{2}$. The one with the smallest angle is "the first".
Remark: in general, the angle $\theta$ formed by two vectors is considered (by convention) as the "smallest one". Here, we are interested in $2\pi-\text{ the smallest one }$
$\begingroup$Great answer @MoebiusCorzer! I was adding a slight edit as you posted your comment though: Do you have a way to test whether $(x_1,y_1)$ or $(x_2,y_2)$ comes first in the orientation? Thanks!$\endgroup$
$\begingroup$As I look over your comment, @MoebiusCorzer, you say "take the bigger angle." However, arccosine will only output $\theta \in (0, \pi)$. So there is no bigger angle to take, in some sense. How can you use your formulation to get an obtuse angle?$\endgroup$
$\begingroup$There is no use of $\arccos$, which is a function. You need to solve equation like $r\cos(\alpha)=x_{1}-a$ along with $r\sin(\alpha)=y_{1}-b$ and it has no particular link with $\arccos$ a priori. When you solve $\cos(a)=\sqrt{2}/2$, the solutions for $a$ are $\pi/4+2k\pi$ and $-\pi/4+2k\pi$, there is no $\arccos$. Of course, in a calculator, you'll use $\arccos$ but you need to consider all possibilities by adding other solutions accordingly.$\endgroup$
$\begingroup$You use arccosine in what you labeled $(1)$. Then you say, "And you have to solve the previous equation. As you look for the GCD, take the greatest value for $\theta$." That's the arccosine to which I refer. Thanks @MoebiusCorzer.$\endgroup$ | 677.169 | 1 |
Symmetry
Exercise 14.1
Question 1.
Copy the figures with punched holes and find the axis of symmetry for the following:
Solution:
The axis of symmetry is shown by following line.
Question 2.
Give the line(s) of symmetry, find the other hole(s):
Solution:
Question 3.
In the following figures, the mirror line (i.e., the line of symmetry) is given asSolution:
Following are the complete figures.
Question 4.
The following figures have more than one line of symmetry. Such figures are said to have multiple lines of symmetry.
Identify multiple lines of symmetry, if any, in each of the following figures:
Question 5.
Copy the figure given here.
Take any one diagonal as a line of symmetry and shade a few more squares to make the figure symmetric about a diagonal. Is there more than one way to do that? Will the figure be symmetric about both the diagonals?
Solution:
(i) Let us take a diagonal as the axis of symmetry and shade the square as shown in the figure.
(ii) Yes, there are more than the line of symmetry i.e. BD, EF and GH.
(iii) Yes, the figure is symmetric about both the diagonals.
Question 6.
Copy the diagram and complete each shape to be symmetric about the mirror line(s).
Solution:
Question 7.
State the number of lines of symmetry for the following figures:
(а) An equilateral triangle
(b) An isosceles triangle
(c) A scalene triangle
(d) A square
(e) A rectangle
(f) A rhombus
(g) A parallelogram
(h) A quadrilateral
(i) A regular hexagon
(j) A circle
Solution:
Question 9.
Give three examples of shapes with no line of symmetry.
Solution:
Example 1: Scalene triangle has no line of symmetry.
Example 2: Quadrilateral has no line of symmetry.
Example 3: Alphabet R has no line of symmetry.
Question 10.
What other name can you give of the line of symmetry of
(a) an isosceles triangle?
(b) a circle?
Solution:
(a) Median of an isosceles triangle is its line of symmetry.
(b) Diameter of a circle is its line of symmetry.
Exercise 14.2
Question 1.
Which of the following figures have rotational symmetry of order more than 1?
Solution:
The figure (a), (b), (d), (e) and (f) have rotational symmetry more than 1.
Question 2.
Give the order of rotational symmetry for each figure:
Solution:
Let us take a point S on one end of the given figure. Rotating by 180°, S comes at other end and then again rotating by 180°, it comes at its original position.
∴ Order of rotational symmetry = 360/180 = 2
Let us take any point S in figure (1). It takes two rotations to come back to its original position.
∴ Order of rotational symmetry = 360/180 = 2
Let us mark any vertex of the given figure. It takes three rotations to come back to its original shape.
∴ Order of rotational symmetry = 360/120 = 3
Order of rotational symmetry = 360/90 =4
∴ Order of rotational symmetry = 360/90 = 4
(f) The given figure is a regular pentagon which can take one rotation at an angle of 72°.
∴ Order of rotational symmetry = 360/72 = 5
(g) The given figure requires six rotations each through an angle of 60°
∴ Order of rotational symmetry = 360/60 = 6
(h) The given figure requires three rotations, each through an angle of 120°.
∴ Order of rotational symmetry = 360/120 = 3
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Looking for better preparation opportunities regarding the Geometry Concepts then try out our Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines. Practice using the BIM Book Geometry Solution Key and clear all your doubts on the Ch 3 Parallel and Perpendicular Lines. All the Questions prepared in the Parallel and Perpendicular Lines Big Ideas Math Geometry Ch 3 Answer Key are as per the common core curriculum.
Learn the concepts quickly using the BIM Book Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines. For a better learning experience, we have compiled all the Big Ideas Math Geometry Answers Chapter 3 as per the Big Ideas Math Geometry Textbooks format. You can find all the concepts via the quick links available below. Simply tap on them and learn the fundamentals involved in the Parallel and Perpendicular Lines Chapter.
Parallel and Perpendicular Lines Maintaining Mathematical Proficiency
Find the slope of the line.
Question 1. Answer: From the given coordinate plane, Let the given points are: A (-1, 2), and B (3 + 1}\) = \(\frac{-3}{4}\) Hence, from the above, We can conclude that the slope of the given line is: \(\frac{-3}{4}\)
Question 2. Answer: From the given coordinate plane, Let the given points are: A (-2, 2), and B (--3 + 2}\) = \(\frac{-3}{-1}\) = 3 Hence, from the above, We can conclude that the slope of the given line is: 3
Question 3. Answer: From the given coordinate plane, Let the given points are: A (-3, -2), and B (1, -22 + 2}{3 + 1}\) = \(\frac{0}{4}\) = 0 Hence, from the above, We can conclude that the slope of the given line is: 0
Write an equation of the line that passes through the given point and has the given slope.
Question 10. ABSTRACT REASONING Why does a horizontal line have a slope of 0, but a vertical line has an undefined slope? Answer: We know that, Slope of the line (m) = \(\frac{y2 – y1}{x2 – x1}\) We know that, For a horizontal line, The coordinates of y are the same. i.e., y1 = y2 = y3 ……… For a vertical line, The coordinates of x are the same. i.e., x1 = x2 = x3 ……. So, The slope of the horizontal line (m) = \(\frac{y2 – y2}{x2 – x1}\) We know that, Any fraction that contains 0 in the numerator has its value equal to 0 So, The slope of horizontal line (m) = 0 The slope of vertical line (m) = \(\frac{y2 – y1}{x2 – x1}\) We know that, Any fraction that contains 0 in the denominator has its value undefined So, The slope of the vertical line (m) = Undefined
Parallel and Perpendicular Lines Mathematical Practices
Use a graphing calculator to graph the pair of lines. Use a square viewing window. Classify the lines as parallel, perpendicular, coincident, or non-perpendicular intersecting lines. Justify your answer.
Question 1. x + 2y = 2 2x – y = 4 Answer: The given pair of lines are: x + 2y = 2 2x – y = 4 Hence, The representation of the given pair of lines in the coordinate plane is: We know that, For a pair of lines to be perpendicular, the product of the slopes i.e., the product of the slope of the first line and the slope of the second line will be equal to -1 So, By comparing the given pair of lines with y = mx + b We get The slope of first line (m1) = –\(\frac{1}{2}\) The slope of second line (m2) = 2 So, m1 ×m2 = –\(\frac{1}{2}\) × 2 = -1 Hence, from the above, We can conclude that the given pair of lines are perpendicular lines
Question 2. x + 2y = 2 2x + 4y = 4 Answer: The give pair of lines are: x + 2y = 2 2x + 4y = 4 Hence, The representation of the given pair of lines in the coordinate plane is: We know that, For a pair of lines to be coincident, the pair of lines have the same slope and the same y-intercept So, By comparing the given pair of lines with y = mx + b We get m1 = –\(\frac{1}{2}\), b1 = 1 m2 = –\(\frac{1}{2}\), b2 = 1 Hence, from the above, We can conclude that the given pair of lines are coincident lines
Question 3. x + 2y = 2 x + 2y = – 2 Answer: The given pair of lines are: x + 2y = 2 x + 2y = -2 Hence, The representation of the given pair of lines in the coordinate plane is: We know that, For a pair of lines to be parallel, the pair of lines have the same slope but different y-intercepts So, By comparing the given pair of lines with y = mx + b We get m1 = –\(\frac{1}{2}\), b1 = 1 m2 = \(\frac{1}{2}\), b2 = -1 Hence, from the above, We can conclude that the given pair of lines are parallel lines
We know that, For a pair of lines to be non-perpendicular, the product of the slopes i.e., the product of the slope of the first line and the slope of the second line will not be equal to -1 So, By comparing the given pair of lines with y = mx + b We get The slope of first line (m1) = \(\frac{1}{2}\) The slope of second line (m2) = 1 So, m1 ×m2 = \(\frac{1}{2}\) Hence, from the above, We can conclude that the given pair of lines are non-perpendicular lines
3.1 Pairs of Lines and Angles
Exploration 1
Points of intersection
work with a partner: Write the number of points of intersection of each pair of coplanar lines. Answer: The given coplanar lines are: a. The points of intersection of parallel lines: We know that, The "Parallel lines" have the same slope but have different y-intercepts So, We can say that any parallel line do not intersect at any point Hence, from the above, We can conclude that the number of points of intersection of parallel lines is: 0
a. The points of intersection of intersecting lines: We know that, The "Intersecting lines" have a common point to intersect So, We can say that any intersecting line do intersect at 1 point Hence, from the above, We can conclude that the number of points of intersection of intersecting lines is: 1
c. The points of intersection of coincident lines: We know that, The "Coincident lines" may be intersecting or parallel So, We can say that any coincident line do not intersect at any point or intersect at 1 point Hence, from the above, We can conclude that the number of points of intersection of coincident lines is: 0 or 1
Exploration 2
Classifying Pairs of Lines
Work with a partner: The figure shows a right rectangular prism. All its angles are right angles. Classify each of the following pairs of lines as parallel, intersecting, coincident, or skew. Justify your answers. (Two lines are skew lines when they do not intersect and are not coplanar.) Answer: The given rectangular prism is: We know that, The "Parallel lines" are the lines that do not intersect with each other and present in the same plane The "Intersecting lines" are the lines that intersect with each other and in the same plane The "Coincident lines" are the lines that lie on one another and in the same plane The "Skew lines" are the lines that do not present in the same plane and do not intersect Hence, The completed table of the nature of the given pair of lines is:
Exploration 3
Identifying Pairs of Angles
Work with a partner: In the figure, two parallel lines are intersected by a third line called a transversal. a. Identify all the pairs of vertical angles. Explain your reasoning. CONSTRUCTING VIABLE ARGUMENTS To be proficient in math, you need to understand and use stated assumptions, definitions, and previously established results. Answer: We know that, The angles that are opposite to each other when two lines cross are called "Vertical angles" Hence, from the given figure, We can conclude that the vertical angles are: ∠1 and ∠3; ∠2 and ∠4; ∠5 and ∠7; ∠6 and ∠8
b. Identify all the linear pairs of angles. Explain your reasoning. Answer: We know that, A "Linear pair" is a pair of adjacent angles formed when two lines intersect Hence, from the given figure, We can conclude that the linear pair of angles is: ∠1 and ∠2; ∠4 and ∠3; ∠5 and ∠6; ∠8 and ∠7
Communicate Your Answer
Question 4. What does it mean when two lines are parallel, intersecting, coincident, or skew? Answer: The two lines are "Parallel" when they do not intersect each other and are coplanar The two lines are "Intersecting" when they intersect each other and are coplanar The two lines are "Coincident" when they lie on each other and are coplanar The two lines are "Skew" when they do not intersect each other and are not coplanar
Question 5. In Exploration 2. find more pairs of lines that are different from those given. Classify the pairs of lines as parallel, intersecting, coincident, or skew. Justify your answers. Answer: The given rectangular prism of Exploration 2 is: The pair of lines that are different from the given pair of lines in Exploration 2 are: a. \(\overline{C D}\) and \(\overline{A E}\) b. \(\overline{D H}\) and \(\overline{F G}\) Hence, from the above, We can conclude that a. \(\overline{C D}\) and \(\overline{A E}\) are "Skew lines" because they are not intersecting and are non coplanar b. \(\overline{D H}\) and \(\overline{F G}\) are "Skew lines" because they are not intersecting and are non coplanar
Lesson 3.1 Pairs of Lines and Angles
Monitoring Progress
Question 1. Look at the diagram in Example 1. Name the line(s) through point F that appear skew to . Answer: From Example 1, We can observe that The line that passes through point F that appear skew to \(\overline{E H}\) is: \(\overline{F C}\)
Question 2. In Example 2, can you use the Perpendicular Postulate to show that is not perpendicular to ? Explain why or why not. Answer: Perpendicular Postulate: According to this Postulate, If there is a line and a point not on the line, then there is exactly one line through the point perpendicular to the given line Now, In Example 2, We can observe that \(\overline{A C}\) is not perpendicular to \(\overline{B F}\) because according to the perpendicular Postulate, \(\overline{A C}\) will be a straight line but it is not a straight line when we observe Example 2 Hence, from the above, We can conclude that we can use "Perpendicular Postulate" to show that \(\overline{A C}\) is not perpendicular to \(\overline{B F}\)
Classify the pair of numbered angles.
Question 3. Answer: The given figure is: We know that, The angles that have the same corner are called "Adjacent angles" Hence, from the above, We can conclude that ∠1 and ∠5 are the adjacent angles
Question 4. Answer: The given figure is: We know that, The angles that have the opposite corners are called "Vertical angles" Hence, from the above, We can conclude that ∠2 and ∠7 are the "Vertical angles"
Question 5. Answer: The given figure is: We know that, The angles that have the opposite corners are called "Vertical angles" Hence, from the above, We can conclude that ∠4 and ∠5 are the "Vertical angles"
Exercise 3.1 Pairs of Lines and Angles
Vocabulary and Core Concept Check
Question 1. COMPLETE THE SENTENCE Two lines that do not intersect and are also not parallel are ________ lines. Answer:
Question 2. WHICH ONE did DOESN'T BELONG? Which angle pair does not belong with the other three? Explain our reasoning. ∠2 and ∠3 ∠4 and ∠5 ∠1 and ∠8 ∠2 and∠7 Answer: The given figure is: We know that, The angles that have the common side are called "Adjacent angles" The angles that are opposite to each other when 2 lines cross are called "Vertical angles" So, ∠2 and ∠3 are vertical angles ∠4 and ∠5 are adjacent angles ∠1 and ∠8 are vertical angles ∠2 and ∠7 are vertical angles Hence, from the above, We can conclude that ∠4 and ∠5 angle-pair do not belong with the other three
Monitoring Progress and Modeling with Mathematics
In Exercises 3 – 6, think of each segment in the diagram as part of a line. All the angles are right angles. Which line(s) or plane(s) contain point B and appear to fit the description? Question 3. line(s) parallel to . Answer:
Question 4. line(s) PerPendicular to . Answer: We know that, The lines that are a straight angle with the given line and are coplanar is called "Perpendicular lines" So, From the given figure, We can conclude that the line that is perpendicular to \(\overline{C D}\) is: \(\overline{A D}\) and \(\overline{C B}\)
Question 5. line(s) skew to Answer:
Question 6. plane(s) parallel to plane CDH Answer: From the given figure, We can observe that the plane parallel to plane CDH is: Plane BAE
In Exercises 7-10, Use the diagram.
Question 7. Name a pair of parallel lines. Answer:
Question 8. Name a pair of perpendicular lines. Answer: We know that, The lines that have an angle of 90° with each other are called "Perpendicular lines" Hence, From the figure, We can conclude that \(\overline{N P}\) and \(\overline{P O}\) are perpendicular lines
Question 9. Answer:
Question 10. Answer: We know that, The lines that have an angle of 90° with each other are called "Perpendicular lines" Hence, From the figure, We can conclude that \(\overline{P R}\) and \(\overline{P O}\) are not perpendicular lines
In Exercises 11-14, identify all pairs of angles of the given type.
Question 11. corresponding Answer:
Question 12. alternate interior Answer: We know that, "Alternate Interior angles" are a pair of angleson the inner side of each of those two lines but on opposite sides of the transversal. Hence, From the given figure, We can conclude that the alternate interior angles are: ∠4 and ∠5; ∠3 and ∠6
Question 13. alternate exterior Answer:
Question 14. consecutive interior Answer: We know that, The pair of angles on one side of the transversal and inside the two lines are called the "Consecutive interior angles". Hence, From the given figure, We can conclude that the consecutive interior angles are: ∠3 and ∠5; ∠4 and ∠6
Question 16. ∠11 and ∠13 Answer: We know that, The pair of angles on one side of the transversal and inside the two lines are called the "Consecutive interior angles". Hence, From the given figure, We can conclude that ∠11 and ∠13 are the "Consecutive interior angles"
Question 17. ∠6 and ∠13 Answer:
Question 18. ∠2 and ∠11 Answer: We know that, "Vertical Angles" are the anglesopposite each other when two lines cross Hence, from the above figure, We can conclude that ∠2 and ∠11 are the "Vertical angles"
ERROR ANALYSIS In Exercises 19 and 20. describe and correct the error in the conditional statement about lines.
Question 19. Answer:
Question 20. Answer: We know that, The "Perpendicular Postulate" states that if there is a line and a point not on the line, then there is exactly one line through the point perpendicularto the given line. Hence, from the above, We can conclude that the given statement is not correct
Question 21. MODELING WITH MATHEMATICS Use the photo to decide whether the statement is true or false. Explain Your reasoning. a. The plane containing the floor of the treehouse is parallel to the ground. b. The lines containing the railings of the staircase, such as , are skew to all lines in the plane containing the ground. c. All the lines containing the balusters. such as , are perpendicular to the plane containing the floor of the treehouse. Answer:
Question 22. THOUGHT-PROVOKING If two lines are intersected by a third line, is the third line necessarily a transversal? Justify your answer with a diagram. Answer: No, the third line does not necessarily be a transversal
Explanation: We know that, "Parallel lines" do not intersect each other "Perpendicular lines" intersect at each other at right angles The third intersecting line can intersect at the same point that the two lines have intersected as shown below: Hence, from the above, We can conclude that the third line does not need to be a transversal
Question 23. MATHEMATICAL CONNECTIONS Two lines are cut by a transversal. Is it possible for all eight angles formed to have the same measure? Explain your reasoning. Answer:
Question 24. HOW DO YOU SEE IT? Think of each segment in the figure as part of a line. a. Which lines are parallel to ? Answer: We know that, The lines that do not intersect to each other and are coplanar are called "Parallel lines" Hence, from the above figure, We can conclude that the line parallel to \(\overline{N Q}\) is: \(\overline{M P}\)
b. Which lines intersect ? Answer: We know that, The lines that are coplanar and any two lines that have a common point are called "Intersecting lines" Hence, from the above figure, We can conclude that the lines that intersect \(\overline{N Q}\) are: \(\overline{N K}\), \(\overline{N M}\), and \(\overline{Q P}\)
c. Which lines are skew to ? Answer: We know that, The lines that do not intersect or not parallel and non-coplanar are called "Skew lines" Hence, from the above figure, We can conclude that \(\overline{K L}\), \(\overline{L M}\), and \(\overline{L S}\)
d. Should you have named all the lines on the cube in parts (a)-(c) except \(\overline{N Q}\)? Explain. Answer: No, we did not name all the lines on the cube in parts (a) – (c) except \(\overline{N Q}\)
In exercises 25-28. copy and complete the statement. List all possible correct answers.
Question 25. ∠BCG and __________ are corresponding angles. Answer:
Question 26. ∠BCG and __________ are consecutive interior angles. Answer: We know that, When two lines are cut by a transversal, the pair ofangles on one side of the transversal and inside the two lines are called the "Consecutive interior angles" Hence, From the given figure, We can conclude that the consecutive interior angles of ∠BCG are: ∠FCA and ∠BCA
Question 28. ∠FCA and __________ are alternate exterior angles. Answer: "Alternate exterior angles" are the pair of anglesthat lie on the outer side of the two parallel lines but on either side of the transversal line Hence, From the above figure, We can conclude that ∠FCA and ∠JCB are alternate exterior angles
Question 29. MAKING AN ARGUMENT Your friend claims the uneven parallel bars in gymnastics are not really Parallel. She says one is higher than the other. so they cannot be on the same plane. Is she correct? Explain. Answer:
3.2 Parallel Lines and Transversals
Exploration 1
Exploring parallel Lines
Work with a partner: Use dynamic geometry software to draw two parallel lines. Draw a third line that intersects both parallel lines. Find the measures of the eight angles that are formed. What can you conclude? Answer: By using the dynamic geometry, The representation of the given coordinate plane along with parallel lines is: Hence, from the coordinate plane, We can observe that, ∠3 = 53.7° and ∠4 = 53.7° We know that, The angle measures of the vertical angles are congruent So, ∠1 = 53.7° and ∠5 = 53.7° We know that, In the parallel lines, All the angle measures are equal Hence, from the above, We can conclude that ∠1 = ∠2 = ∠3 = ∠4 = ∠5 = ∠6 = ∠7 = 53.7°
Exploration 2
Writing conjectures
Work with a partner. Use the results of Exploration 1 to write conjectures about the following pairs of angles formed by two parallel lines and a transversal. ATTENDING TO PRECISION To be proficient in math, you need to communicate precisely with others. a. corresponding angles Answer: We know that, When two lines are crossed by another line (which is called the Transversal), theangles in matching corners are called "Corresponding angles" Hence, from the given figure, We can conclude that The corresponding angles are: ∠ and ∠5; ∠4 and ∠8
b. alternate interior angles Answer: We know that, "Alternate Interior Angles" are a pair of angleson the inner side of each of those two lines but on opposite sides of the transversal. Hence, from the above figure, We can conclude that The alternate interior angles are: ∠3 and ∠5; ∠2 and ∠8
c. alternate exterior angles Answer: We know that, "Alternate exterior angles" are the pair of anglesthat lie on the outer side of the two parallel lines but on either side of the transversal line. Hence, from the above figure, We can conclude that The alternate exterior angles are: ∠1 and ∠7; ∠6 and ∠4
d. consecutive interior angles Answer: We know that, When two lines are cut by a transversal, the pair of angles on one side of the transversal and inside the two lines are called the "Consecutive interior angles" Hence, from the above figure, We can conclude that The consecutive interior angles are: ∠2 and ∠5; ∠3 and ∠8
Communicate Your Answer
Question 3. When two parallel lines are cut by a transversal, which of the resulting pairs of angles are congruent? Answer: If two parallel lines are cut by a transversal, then the pairs of "Corresponding angles" are congruent. If two parallel lines are cut by a transversal, then the pairs of "Alternate interior angles" are congruent. If two parallel lines are cut by a transversal, then the pairs of "Alternate exterior angles" are congruent.
Question 4. In Exploration 2. m∠1 = 80°. Find the other angle measures. Answer: It is given that In Exploration 2, ∠1 = 80° We know that, Exploration 2 comes from Exploration 1 From Exploration 1, We can say that all the angle measures are equal in Exploration 1 Hence, from the above, We can conclude that ∠1 = ∠2 = ∠3 = ∠4 = ∠5 = ∠6 = ∠7 = ∠8 = 80°
Lesson 3.2 Parallel Lines and Transversals
Monitoring Progress
Use the diagram
Question 1. Given m∠1 = 105°, find m∠4, m∠5, and m∠8. Tell which theorem you use in each case. Answer: It is given that ∠1 = 105° Now, We have to find ∠4, ∠5, and ∠8 Now, To find ∠4: Verticle angle theorem: Vertical Angles Theoremstates thatvertical angles,anglesthat are opposite each other and formed by two intersecting straight lines, are congruent So, ∠1 = ∠4 Hence, ∠4 = 105°
Question 3. In the proof in Example 4, if you use the third statement before the second statement. could you still prove the theorem? Explain. Answer: In Example 4, the given theorem is "Alternate interior angle theorem" If you even interchange the second and third statements, you could still prove the theorem as the second line before interchange is not necessary Hence, from the above, We can conclude that if you use the third statement before the second statement, you could still prove the theorem
Question 4. WHAT IF? In Example 5. yellow light leaves a drop at an angle of m∠2 = 41°. What is m∠1? How do you know? Answer: In Example 5, If we observe ∠1 and ∠2, then they are alternate interior angles Now, According to Alternate interior angle theorem, ∠1 = ∠2 It is given that, ∠2 = 41° So, ∠1 = 41° Hence, from the above, We can conclude that ∠1 = 41°
Exercise 3.2 Parallel Lines and Transversals
Vocabulary and Core Concept Check
Question 1. WRITING How are the Alternate Interior Angles Theorem (Theorem 3.2) and the Alternate Exterior Angles Theorem (Theorem 3.3) alike? How are they different? Answer:
Question 2. WHICH ONE did DOESN'T BELONG? Which pair of angle measures does not belong with the other three? Explain. m∠1 and m∠3 m∠2 and m∠4 m∠2 and m∠3 m∠1 and m∠5 Answer: The given figure is: From the given figure, ∠1 and ∠3 are the vertical angles ∠2 and ∠4 are the alternate interior angles ∠2 and ∠3 are the consecutive interior angles ∠1 and ∠5 are the alternate exterior angles So, From the above, We can observe that all the angles except ∠1 and ∠3 are the interior and exterior angles Hence, from the above, We can conclude that ∠1 and ∠3 pair does not belong with the other three
Monitoring Progress and Modeling with Mathematics
In Exercises 3-6, find m∠1 and m∠2. Tell which theorem you use in each case.
Question 3. Answer:
Question 4. Answer: The given figure is: From the given figure, We can observe that, ∠1 = ∠2 (By using the Vertical Angles theorem) ∠2 = 150° (By using the Alternate exterior angles theorem) Hence, from the above, We can conclude that ∠1 = ∠2 = 150°
Question 13. ERROR ANALYSIS Describe and correct the error in the students reasoning Answer:
Question 14. HOW DO YOU SEE IT? Use the diagram a. Name two pairs of congruent angles when \(\overline{A D}\) and \(\overline{B C}\) are parallel? Explain your reasoning? Answer: Let the congruent angle be ∠P So, From the figure, We can observe that the pair of angle when \(\overline{A D}\) and \(\overline{B C}\) are parallel is: ∠APB and ∠DPB
b. Name two pairs of supplementary angles when \(\overline{A B}\) and \(\overline{D C}\) are parallel. Explain your reasoning. Answer: From the given figure, The two pairs of supplementary angles when \(\overline{A B}\) and \(\overline{D C}\) are parallel is: ∠ACD and ∠BDC
Question 16. Consecutive Interior Angles Theorem (Thm. 3.4) Answer: Statement of consecutive Interior angles theorem: If two parallel lines are cut by a transversal, then the pairs of consecutive interior angles formed are supplementary Proof:
Question 17. PROBLEM-SOLVING A group of campers ties up their food between two parallel trees, as shown. The rope is pulled taut. forming a straight line. Find m∠2. Explain our reasoning. Answer:
Question 18. DRAWING CONCLUSIONS You are designing a box like the one shown. a. The measure of ∠1 is 70°. Find m∠2 and m∠3. b. Explain why ∠ABC is a straight angle. c. If m∠1 is 60°, will ∠ABC still he a straight angle? Will the opening of the box be more steep or less steep? Explain. Answer:
Question 19. CRITICAL THINKING Is it possible for consecutive interior angles to be congruent? Explain. Answer:
Question 20, is it possible that a transversal intersects two parallel lines? Explain your reasoning. Answer: We know that, According to Euclidean geometry, For a parallel line, there will be no intersecting point But, In spherical geometry, even though there is some resemblance between circles and lines, there is no possibility to form parallel lines as the lines will intersect at least at 1 point on the circle which is called a tangent Hence, from the above, We can conclude that it is not possible that a transversal intersects two parallel lines
MATHEMATICAL CONNECTIONS In Exercises 21 and 22, write and solve a system of linear equations to find the values of x and y.
Question 23. MAKING AN ARGUMENT During a game of pool. your friend claims to be able to make the shot Shown in the diagram by hitting the cue ball so that m∠1 = 25°. Is your friend correct? Explain your reasoning. Answer:
Question 24. REASONING In diagram. ∠4 ≅∠5 and \(\overline{S E}\) bisects ∠RSF. Find m∠1. Explain your reasoning. Answer: It is given that ∠4 ≅∠5 and \(\overline{S E}\) bisects ∠RSF So, ∠FSE = ∠ESR From ΔESR, We know that, The sum of the angle measures of a triangle is: 180° So, ∠3 + ∠4 + ∠5 = 180° So, ∠3 = 60° (Since ∠4 ≅ ∠5 and the triangle is not a right triangle) From the given figure, We can observe that, ∠1 = ∠3 (By using the Corresponding angles theorem) So, ∠1 = 60° Hence, from the above, We can conclude that ∠1 = 60°
Maintaining Mathematical Proficiency
Write the converse of the conditional statement. Decide whether it is true or false.
Question 25. If two angles are vertical angles. then they are congruent. Answer:
Question 26. If you go to the zoo, then you will see a tiger. Answer: The given statement is: If you go to the zoo, then you will see a tiger The converse of the given statement is: If you will see a tiger, then you go to the zoo———-> False
Question 27. If two angles form a linear pair. then they are supplementary. Answer:
Question 28. If it is warm outside, then we will go to the park. Answer: The given statement is: If it is warm outside, then we will go to the park The converse of the given statement is: If you will go to the park, then it is warm outside —————-> False
3.3 Proofs with Parallel Lines
Exploration 1
Exploring Converses
Work with a partner: Write the converse of each conditional statement. Draw a diagram to represent the converse. Determine whether the converse is true. Justify your conclusion. CONSTRUCTING VIABLE ARGUMENTS To be proficient in math, you need to make conjectures and build a logical progression of statements to explore the truth of your conjectures.
a. Corresponding Angles Theorem (Theorem 3.1): If two parallel lines are cut by a transversal, then the pairs of corresponding angles are congruent. Converse: If the pairs of corresponding angles are
congruent, then the two parallel lines are
cut by a transversal.
Answer: The Converse of the Corresponding Angles Theorem: The Converse of the Corresponding Angles Theorem says that if twolinesand a transversal formcongruentcorresponding angles, then thelinesare Converse of Corresponding Angles Theorem: Consider the 2 lines L1 and L2 intersected by a transversal line L3 creating 2 corresponding angles 1 and 2 which are congruent We want to prove L1 and L2 are parallel and we will prove this by using "Proof of Contradiction" Now, According to Contradiction, Assume L1 is not parallel to L2 Then, according to the parallel line axiom, there is a different line than L2 that passes through the intersection point of L2 and L3 (point A in the drawing), which is parallel to L1. Let's draw that line, and call it P. Let's also call the angle formed by the traversal line and this new line angle 3, and we see that if we add some other angle, call it angle 4, to it, it will be the same as angle 2. Now, P || L1 So, ∠1≅ ∠3, as corresponding angles formed by a transversal of parallel lines, and so, m∠1=m∠3 This contradicts what was given,that angles 1 and 2 are congruent. This contradiction means our assumption ("L1 is not parallel to L2") is false, and so L1 must be parallel to L2. The representation of the Converse of Corresponding Angles Theorem is:
b. Alternate Interior Angles Theorem (Theorem 3.2): If two parallel lines are cut by a transversal, then the pairs of alternate interior angles are congruent. Converse: If the pairs of alternate interior angles are
congruent, then the two parallel lines are
cut by a transversal.
Answer: The converse of the Alternate Interior angles Theorem: The "Converse of the Alternate Interior Angles Theorem" states that if two lines are cut by a transversal and the alternate interior anglesare congruent, then the lines are parallel So, When we compare the actual converse and the converse according to the given statement, we can conclude that the converse we obtained from the given statement is false
c. Alternate Exterior Angles Theorem (Theorem 3.3): If two parallel lines are cut by a transversal, then the pairs of alternate exterior angles are congruent. Converse: If the pairs of alternate exterior angles
are congruent, then the two parallel
lines are cut by a transversal.
Answer: The Converse of the alternate exterior angles Theorem: The "Converse of the Alternate Exterior Angles Theorem" states that if alternate exterior anglesof two lines crossed by a transversal are congruent Alternate exterior angles Theorem: Given: ∠1 ≅ ∠2 Prove: l || m The flow proof for the Converse of Alternate exterior angles Theorem is: The representation of the Converse of the Exterior angles Theorem is:
d. Consecutive Interior Angles Theorem (Theorem 3.4): If two parallel lines are cut by a transversal. then the pairs of consecutive interior angles are supplementary. Converse: If the pairs of consecutive interior angles
are supplementary, then the two parallel lines
are cut by a transversal
Answer: The Converse of the Consecutive Interior angles Theorem: The "Converse of the consecutive Interior angles Theorem" states that if the consecutive interior angles on the same side of a transversal line intersecting two lines are supplementary the Converse of the Consecutive Interior angles Theorem: a. m∠5 + m∠4 3. In Exploration 1, explain how you would prove any of the theorems that you found to be true. Answer: For the proofs of the theorems that you found to be true, refer to Exploration 1
Lesson 3.3 Proofs with Parallel Lines
Monitoring Progress
Question 1. Is there enough information in the diagram to conclude that m || n? Explain. Answer: Yes, there is enough information in the diagram to conclude m || n.
Explanation: The given figure is: From the given figure, We can observe that the given angles are the consecutive exterior angles Now, We have to prove that m || n So, We will use "Converse of Consecutive Exterior angles Theorem" to prove m || n Proof of the Converse of the Consecutive Exterior angles Theorem: a. m∠1 + m∠8 2. Explain why the Corresponding Angles Converse is the converse of the Corresponding Angles Theorem (Theorem 3.1). Answer: Corresponding Angles Theorem: The "Corresponding Angles Postulate" states that, when two parallel lines are cut by a transversal, the resulting corresponding anglesare congruent Converse: When the corresponding angles are congruent, the two parallel lines are cut by a transversal Now, The Converse of Corresponding Angles Theorem: If the corresponding angles formed are congruent, then two lines l and m are cut by a transversal. So, When we observe the Converse of the Corresponding Angles Theorem we obtained and the actual definition, both are the same Hence, from the above, We can conclude that the Corresponding Angles Converse is the converse of the Corresponding Angles Theorem
Question 3. If you use the diagram below to prove the Alternate Exterior Angles Converse. what Given and Prove statements would you use? Answer: The given figure is: It is given that the given angles are the alternate exterior angles Now, Alternate Exterior angle Theorem: If two parallel lines are cut by a transversal, then the pairs of alternate exterior angles are congruent The Converse of the Alternate Exterior Angles Theorem: The "Converse of the Alternate Exterior Angles Theorem" states that if alternate exterior anglesof two lines crossed by a transversal are congruent, then the two lines are parallel. Hence, from the above, We can conclude that For the Converse of the alternate exterior angles Theorem, The given statement is: ∠1 ≅ 8 To prove: l || k
Question 4. Copy and complete the following paragraph proof of the Alternate Interior Angles Converse using the diagram in Example 2.
It is given that ∠4 ≅∠5. By the _______ . ∠1 ≅ ∠4. Then by the Transitive Property of Congruence (Theorem 2.2), _______ . So, by the _______ , g || h. Answer: The completed proof of the Alternate Interior Angles Converse using the diagram in Example 2 is: It is given that ∠4 ≅∠5. By the Vertical Angles Congruence Theorem (Theorem 2.6). ∠1 ≅ ∠4. Then by the Transitive Property of Congruence (Theorem 2.2), ∠1 ≅∠5. So, by the Corresponding Angles Converse, g || h.
Question 5. Each step is parallel to the step immediately above it. The bottom step is parallel to the ground. Explain why the top step is parallel t0 the ground. Answer: From the given figure, We can observe that not any step is intersecting at each other In the same way, when we observe the floor from any step, We can say that they are also parallel Hence, from the above, We can conclude that the top step is also parallel to the ground since they do not intersect each other at any point
Question 6. In the diagram below. p || q and q || r. Find m∠8. Explain your reasoning. Answer: The given figure is: From the figure, We can observe that the given angles are the consecutive exterior angles We know that, According to the Consecutive Exterior angles Theorem, ∠8 + 115° = 180° ∠8 = 180° – 115° ∠8 = 65° Hence, from the above, We can conclude that ∠8 = 65°
Exercise 3.3 Proofs with Parallel Lines
Vocabulary and Core Concept Check
Question 1. VOCABULARY Two lines are cut by a transversal. Which angle pairs must be congruent for the lines to be parallel? Answer:
Question 2. WRITING Use the theorems from Section 3.2 and the converses of those theorems in this section to write three biconditional statements about parallel lines and transversals. Answer: Corresponding Angles theorem states that if two lines are cut by a transversal, then the pairs of corresponding angles are congruent. Corresponding Angles converse states that if two lines are cut by a transversal so the corresponding angles are congruent, then the lines are parallel Biconditional statement is Two lines, which are cut by a transversal, are parallel if and only if the corresponding angles are congruent. Alternate interior angles theorem states that if two parallel lines are cut by a transversal, then the pairs of alternate interior angles are congruent. Alternate interior angles converse states that if two lines are cut by a transversal so the alternate interior angles are congruent, then the lines are parallel Biconditional statement is Two lines, which are cut by a transversal, are parallel if and only if the alternate interior angle are congruent. Alternate exterior angles theorem states that if two parallel lines are cut by a transversal, then the pairs of alternate exterior angles are congruent. Alternate exterior angles converse states that if two lines are cut by a transversal so the alternate exterior angles are congruent, then the lines are parallel A Biconditional statement is Two lines, which are cut by a transversal, are parallel if and only if the alternate exterior angle are congruent.
Question 4. Answer: The given figure is: From the given figure, We can observe that the given angles are the corresponding angles Now, According to Corresponding Angles Theorem, (2x + 15)° = 135° 2x° = 135° – 15° 2x° = 120° x° = \(\frac{120}{2}\) x° = 60° Hence, from the above, We can conclude that the value of x is: 60°
Question 5. Answer:
Question 6. Answer: The given figure is: From the given figure, We can observe that the given angles are the corresponding angles Now, According to Corresponding Angles Theorem, (180 – x)° = x° 180° = x° + x° 2x° = 180° x° = \(\frac{180}{2}\) x° = 90° Hence, from the above, We can conclude that the value of x is: 90°
Question 7. Answer:
Question 8. Answer: The given figure is: From the given figure, We can observe that the given angles are the corresponding angles Now, According to Corresponding Angles Theorem, (2x + 20)° = 3x° 20° = 3x° – 2x° x° = 20° Hence, from the above, We can conclude that the value of x is: 20°
In Exercises 9 and 10, use a compass and straightedge to construct a line through point P that is parallel to line m.
Question 9. Answer:
Question 10. Answer: Let A and B be two points on line m. Draw \(\overline{A P}\) and construct an angle ∠1 on n at P so that ∠PAB and ∠1 are corresponding angles Hence, The representation of the complete figure is:
Question 18. Answer: Yes, there is enough information to prove m || n The theorem we can use to prove that m || n is: Alternate Exterior angles Converse theorem
ERROR ANALYSIS In Exercises 19 and 20, describe and correct the error in the reasoning.
Question 19. Answer:
Question 20. Answer: The given figure shows that angles 1 and 2 are Consecutive Interior angles It also shows that a and b are cut by a transversal and they have the same length So, From the converse of the Consecutive Interior angles Theorem, We can conclude that a || b
In Exercises 21-24. are and parallel? Explain your reasoning.
Question 21. Answer:
Question 22. Answer: The given figure is: From the given figure, We can observe that The sum of the given angle measures is: 180° From the given figure, We can observe that the given angles are consecutive exterior angles So, From the Consecutive Exterior angles Converse, We can conclude that AC || DF
Question 23. Answer:
Question 24. Answer: The given figure is: From the given figure, We can observe that the sum of the angle measures of all the pairs i.e., (115 + 65)°, (115 + 65)°, and (65 + 65)° is not 180° Since, The sum of the angle measures are not supplementary, according to the Consecutive Exterior Angles Converse, AC is not parallel to DF
Question 25. ANALYZING RELATIONSHIPS The map shows part of Denser, Colorado, Use the markings on the map. Are the numbered streets parallel to one another? Explain your reasoning. Answer:
Question 26. ANALYZING RELATIONSHIPS Each rung of the ladder is parallel to the rung directly above it. Explain why the top rung is parallel to the bottom rung. Answer: When we observe the ladder, The rungs are not intersecting at any point i.e., they have different points We know that, The parallel lines do not have any intersecting points Hence, from the above, We can conclude that the top rung is parallel to the bottom rung
Question 27. MODELING WITH MATHEMATICS The diagram of the control bar of the kite shows the angles formed between the Control bar and the kite lines. How do you know that n is parallel to m? Answer:
Question 30. MODELING WITH MATHEMATICS One way to build stairs is to attach triangular blocks to angled support, as shown. The sides of the angled support are parallel. If the support makes a 32° angle with the floor, what must m∠1 so the top of the step will be parallel to the floor? Explain your reasoning. Answer: It is given that the sides of the angled support are parallel and the support makes a 32° angle with the floor So, To make the top of the step where ∠1 is present to be parallel to the floor, the angles must be "Alternate Interior angles" We know that, The "Alternate Interior angles" are congruent So, ∠1 = 32° Hence, from the above, We can conclude that ∠1 = 32°
Question 31. ABSTRACT REASONING In the diagram, how many angles must be given to determine whether j || k? Give four examples that would allow you to conclude that j || k using the theorems from this lesson. Answer:
Question 32. THOUGHT-PROVOKING Draw a diagram of at least two lines cut by at least one transversal. Mark your diagram so that it cannot be proven that any lines are parallel. Then explain how your diagram would need to change in order to prove that lines are parallel. Answer: The diagram that represents the figure that it can not be proven that any lines are parallel is: From the above, The diagram can be changed by the transformation of transversals into parallel lines and a parallel line into transversal Hence, The diagram that represents the figure that it can be proven that the lines are parallel is:
Question 37. MAKING AN ARGUMENT Your classmate decided that based on the diagram. Is your classmate correct? Explain your reasoning. Answer:
Question 38. HOW DO YOU SEE IT? Are the markings on the diagram enough to conclude that any lines are parallel? If so. which ones? If not, what other information is needed? Answer: The given diagram is: From the given diagram, We can observe that ∠1 and ∠4; ∠2 and ∠3 are the pairs of corresponding angles We know that, According to the Converse of the Corresponding angles Theorem, If the corresponding angles are congruent, then the two lines that cut by a transversal are parallel lines Hence, We can conclude that p and q; r and s are the pairs of parallel lines
Question 39. PROVING A THEOREM Use these steps to prove the Transitive Property of Parallel Lines Theorem a. Cops the diagram with the Transitive Property of Parallel Lines Theorem on page 141. b. Write the Given and Prove statements. c. Use the properties of angles formed by parallel lines cut by a transversal to prove the theorem. Answer:
Question 40. MATHEMATICAL CONNECTIONS Use the diagram a. Find the value of x that makes p || q. Answer: From the given figure, We can observe that when p || q, The angles are: (2x + 2)° and (x + 56)° We can observe that the given angles are corresponding angles Hence, (2x + 2)° = (x + 56)° 2x – x = 56° – 2° x° = 54° Hence, from the above, We can conclude that the value of x when p || q is: 54°
c. Can r be parallel to s and can p, be parallel to q at the same time? Explain your reasoning. Answer: No, p ||q and r ||s will not be possible at the same time because when p || q, r, and s can act as transversal and when r || s, p, and q can act as transversal
Maintaining Mathematical Proficiency Use the Distance Formula to find the distance between the two points.
3.1 – 3.3 Study Skills: Analyzing Your Errors
Question 1. Draw the portion of the diagram that you used to answer Exercise 26 on page 130. Answer: The portion of the diagram that you used to answer Exercise 26 on page 130 is:
Question 2. In Exercise 40 on page 144. explain how you started solving the problem and why you started that way. Answer: In Exercise 40 on page 144, You started solving the problem by considering the 2 lines parallel and two lines as transversals So, If p and q are the parallel lines, then r and s are the transversals If r and s are the parallel lines, then p and q are the transversals
3.1 – 3.3 Quiz
Think of each segment in the diagram as part of a line. Which lines(s) or plane(s) contain point G and appear to fit the description?
Question 5. consecutive interior Answer: We know that, When two lines are cut by a transversal, the pair ofangleson one side of the transversal and inside the two lines are called theconsecutive interior angles. Hence, from the given figure, We can conclude that the consecutive interior angles are: 3 and 5; 4 and 6
Question 6. alternate interior Answer: We know that, Alternate Interior Anglesare a pair ofangleson the inner side of each of those two lines but on opposite sides of the transversal. Hence, from the given figure, We can conclude that the alternate interior angles are: 3 and 6; 4 and 5
Question 7. corresponding Answer: We know that, When two lines are crossed by another line (which is called the Transversal), theanglesin matching corners are calledcorresponding angles. Hence, from the given figure, We can conclude that the corresponding angles are: 1 and 5; 3 and 7; 2 and 4; 6 and 8
Question 8. alternate exterior Answer: Alternate exterior anglesare the pair ofanglesthat lie on the outer side of the two parallel lines but on either side of the transversal line Hence, from the above figure, We can conclude that the alternate exterior angles are: 1 and 8; 7 and 2
Question 10. Answer: The given figure is: From the given figure, We can observe that By using the Vertical Angles Theorem, ∠2 = 123° Now, By using the vertical Angles Theorem, ∠1 = ∠2 Hence, from the above, We can conclude that ∠1 = ∠2 = 123°
Decide whether there is enough information to prove that m || n. If so, state the theorem you would use.
Question 12. Answer: The given figure is: We know that, By using the "Consecutive Interior angles Converse", If the angle measure of the angles is a supplementary angle, then the lines cut by a transversal are parallel Now, 69° + 111° = 180° Hence, from the above, We can conclude that m || n by using the Consecutive Interior angles Theorem
Question 13. Answer: The given figure is: We know that, By using the Corresponding Angles Theorem, If the corresponding angles are congruent, then the lines cut by a transversal are parallel Hence, from the above, We can conclude that m || n by using the Corresponding Angles Theorem
Question 14. Answer: The given figure is: From the given figure, It is given that l || m and l || n, So, We know that, By using the parallel lines property, If a || b and b || c, then a || c Hence, from the above, We can conclude that m || n
Question 15. Cellular phones use bars like the ones shown to indicate how much signal strength a phone receives from the nearest service tower. Each bar is parallel to the bar directly next to it. a. Explain why the tallest bar is parallel to the shortest bar. Answer: From the given bars, We can observe that there is no intersection between any bars If we represent the bars in the coordinate plane, we can observe that the number of intersection points between any bar is: 0 We know that, The number of intersection points for parallel lines is: 0 Hence, from the above, We can conclude that the tallest bar is parallel to the shortest bar
b. Imagine that the left side of each bar extends infinitely as a line. If m∠1 = 58°, then what is m∠2? Answer: From the given figure, We can observe that ∠1 and ∠2 are the consecutive interior angles We know that, The sum of the angle measure between 2 consecutive interior angles is: 180° So, ∠1 + ∠ 2 = 180° It is given that ∠1 = 58° So, ∠2 = 180° – 58° ∠2 = 122° Hence, from the above, We can conclude that ∠2 = 122°
Question 16. The diagram shows lines formed on a tennis court. a. Identify two pairs of parallel lines so that each pair is in a different plane. Answer: From the given figure, We can observe that there are a total of 5 lines. Hence, The two pairs of parallel lines so that each pair is in a different plane are: q and p; k and m
b. Identify two pairs of perpendicular lines. Answer: Fro the given figure, We can observe that there are 2 perpendicular lines Hence, from the above, The two pairs of perpendicular lines are l and n
c. Identify two pairs of skew line Answer: From the given figure, We can observe that there are 2 pairs of skew lines Hence, The 2 pair of skew lines are: q and p; l and m
d. Prove that ∠1 ≅ ∠2. Answer: From the given figure, We can observe that ∠1 and ∠2 are the alternate exterior angles We know that, According to the Alternate Exterior angles Theorem, If the line cut by a transversal is parallel, then the corresponding angles are congruent Hence, According to the above theorem, We can conclude that ∠1 ≅ ∠2
3.4 Proofs with Perpendicular Lines
Exploration 1
Writing Conjectures
Work with a partner: Fold a piece of pair in half twice. Label points on the two creases. as shown. a. Write a conjecture about \(\overline{A B}\) and \(\overline{C D}\). Justify your conjecture. Answer: The conjecture about \(\overline{A B}\) and \(\overline{c D}\) is: If twolinesintersect to form a linear pair of congruent angles, then thelinesareperpendicular.
b. Write a conjecture about \(\overline{A O}\) and \(\overline{O B}\) Justify your conjecture. Answer: The conjecture about \(\overline{A O}\) and \(\overline{O B}\) is: In a plane, if twolinesareperpendicularto the sameline, then they are parallel to each other.
Exploration 2
Exploring a segment Bisector
Work with a partner: Fold and crease a piece of paper. as shown. Label the ends of the crease as A and B. a. Fold the paper again so that point A coincides with point B. Crease the paper on that fold. Answer:
b. Unfold the paper and examine the four angles formed by the two creases. What can you conclude about the four angles? Answer: When we unfold the paper and examine the four angles formed by the two creases, we can conclude that the four angles formed are the right angles i.e., 90°
Exploration 3
Writing a conjecture
Work with a partner. a. Draw \(\overline{A B}\), as shown. b. Draw an arc with center A on each side of AB. Using the same compass selling, draw an arc with center B on each side \(\overline{A B}\). Label the intersections of arcs C and D. c. Draw \(\overline{C D}\). Label its intersection with \(\overline{A B}\) as O. Write a conjecture about the resulting diagram. Justify your conjecture. CONSTRUCTING VIABLE ARGUMENTS To be proficient in math, you need to make conjectures and build a logical progression of statements to explore the truth of your conjectures. Answer: The resultant diagram is: From the above diagram, We can conclude that The angles formed at all the intersection points are: 90° The lengths of the line segments are equal i.e., AO = OB and CO = OD
Communicate Your Answer
Question 4. What conjectures can you make about perpendicular lines? Answer: The conjectures about perpendicular lines are: a. If two lines intersect to form a linear pair of congruent angles, then the lines are perpendicular. b. In a plane, if a transversal is perpendicular to one of two parallel lines, then it is perpendicular to the other line. c. In a plane, if two lines are perpendicular to the same line, then they are parallel to each other.
Lesson 3.4 Proofs with Perpendicular Lines
Question 1. Find the distance from point E to Answer: The given figure is: It is given that E is ⊥ to \(\overline{F H}\) So, To find the distance between E and \(\overline{F H}\), we need to find the distance between E and G i.e., EG Now, From the coordinate plane, E (-4, -3), G (1, 2) Compare the given points with E (x1, y1), G (x2, y2) So, EG = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) EG = \(\sqrt{(1 + 4)² + (2 + 3)²}\) EG = \(\sqrt{(5)² + (5)²}\) EG = \(\sqrt{50}\) EG = 7.07 Hence, from the above, We can conclude that the distance from point E to \(\overline{F H}\) is: 7.07
Question 2. Prove the Perpendicular Transversal Theorem using the diagram in Example 2 and the Alternate Exterior Angles Theorem (Theorem 3.3). Answer: Perpendicular transversal theorem: In a plane, if a line is perpendicular to one of the two parallel lines, then it is perpendicular to the other line also Proof: Given: k || l, t ⊥ k Prove: t ⊥ l Alternate Exterior Angles Theorem: The Alternate Exterior Angles Theorem states that, when two parallel lines are cut by a transversal, the resulting alternate exterior angles are congruent Proof: Given: k || l Prove: ∠1 ≅ ∠7 and ∠4 ≅ ∠6 Since k || l,by the Corresponding Angles Postulate, ∠1 ≅ ∠5 Also, by the Vertical Angles Theorem, ∠5 ≅ ∠7 Then, by the Transitive Property of Congruence, ∠1 ≅ ∠7 You can prove that∠4and∠6are congruent using the same method.
Use the lines marked in the photo.
Question 3. Is b || a? Explain your reasoning. Answer: From the given figure, There is not any intersection between a and b Hence, from the above, We can conclude that b || a
Question 4. Is b ⊥ c? Explain your reasoning. Answer: From the given figure, We can observe that the angle between b and c is 90° Hence, from the above, We can conclude that b is perpendicular to c
Exercise 3.4 Proofs with Perpendicular Lines
Vocabulary and core Concept Check
Question 1. COMPLETE THE SENTENCE The perpendicular bisector of a segment is the line that passes through the _______________ of the segment at a _______________ angle. Answer:
Question 2. DIFFERENT WORDS, SAME QUESTION Which is different? Find "both" answers. Find the distance from point X to Answer: The given figure is: To find the distance from point X to \(\overline{W Z}\), point X to \(\overline{W Z}\) is: 6.32
Find the distance from line l to point X. Answer: To find the distance from line l to point X, line l to point X is: 6.32
Monitoring Progress and Modeling with Mathematics
In Exercises 3 and 4. find the distance from point A to .
Question 3. Answer:
Question 4. Answer: To find the distance from point A to \(\overline{X Z}\), We have to find the distance between A and Y i.e., AY Now, The given points are: X (3, 3), Y (2, -1.5 1.5)² + (3 – 2)²}\) XY = \(\sqrt{(4.5)² + (1)²}\) XY = 4.60 Hence, from the above, We can conclude that the distance from point A to \(\overline{X Z}\) is: 4.60
CONSTRUCTION In Exercises 5-8, trace line m and point P. Then use a compass and straightedge to construct a line perpendicular to line m through point P.
Question 5. Answer:
Question 6. Answer: The given figure is: Now, Using P as the center, draw two arcs intersecting with line m. Label the intersections as points X and Y. Using X and Y as centers and an appropriate radius, draw arcs that intersect. Label the intersection as Z. Draw \(\overline{P Z}\)
Question 7. Answer:
Question 8. Answer: The given figure is: Now, Using P as the center and any radius, draw arcs intersecting m and label those intersections as X and Y. Using X as the center, open the compass so that it is greater than half of XP and draw an arc. Using Y as the center and retaining the same compass setting, draw an arc that intersects with the first Label the point of intersection as Z. Draw \(\overline{P Z}\)
CONSTRUCTION In Exercises 9 and 10, trace \(\overline{A B}\). Then use a compass and straightedge to construct the perpendicular bisector of \(\overline{A B}\)
Question 9. Answer:
Question 10. Answer: The given figure is: Now, Using a compass setting greater than half of AB, draw two arcs using A and B as centers Connect the points of intersection of the arcs with a straight line
ERROR ANALYSIS In Exercises 11 and 12, describe and correct the error in the statement about the diagram. Question 11. Answer:
Question 12. Answer: We know that, According to the Perpendicular Transversal theorem, The distance from the perpendicular to the line is given as the distance between the point and the non-perpendicular line So, From the given figure, The distance from point C to AB is the distance between point C and A i.e., AC Hence, from the above, We can conclude that the distance from point C to AB is: 12 cm
Question 14. Lines Perpendicular to a Transversal Theorem (Thm. 3.12) Answer: In a plane, if a line is perpendicular to one of the two parallel lines, then it is perpendicular to the other line also Proof: Given: k || l, t ⊥ k Prove: t ⊥ l
PROOF In Exercises 15 and 16, use the diagram to write a proof of the statement.
Question 15. If two intersecting lines are perpendicular. then they intersect to form four right angles. Given a ⊥ b Prove ∠1, ∠2, ∠3, and ∠4 are right angles. Answer:
Question 16. If two sides of two adjacent acute angles are perpendicular, then the angles are complementary. Given \(\overrightarrow{B A}\) ⊥\(\vec{B}\)C Prove ∠1 and ∠2 are complementary Answer: We have to find the theorem that satisfies the given statement. We know that the given angles∠4, ∠5 are complementary thus, the theorem 3.10 justifies the statement which states that-If two sides of two adjacent acute angles are perpendicular, then the angles are complementary.
In Exercises 17-22, determine which lines, if any, must be parallel. Explain your reasoning.
Question 17. Answer:
Question 18. Answer: The given figure is: From the given figure, We can observe that a is perpendicular to both the lines b and c Hence, from the above, We can conclude that By using the Perpendicular transversal theorem, a is both perpendicular to b and c and b is parallel to c
Question 19. Answer:
Question 20. Answer: The given figure is; From the given figure, We can observe that a is perpendicular to d and b is perpendicular to c For parallel lines, we can't say anything Hence, from the above, By using the Perpendicular transversal theorem, a is perpendicular to d and b isperpendicular to c
Question 21. Answer:
Question 22. Answer: The given figure is: From the given figure, We can observe that w ⊥ v and w⊥ y So, We can say that w and v are parallel lines by "Perpendicular Transversal Theorem" We can observe that z ⊥ x and w ⊥ z So, We can say that w and x are parallel lines by "Perpendicular Transversal theorem"
Question 24. MAKING AN ARGUMENT Your friend claims that because you can find the distance from a point to a line, you should be able to find the distance between any two lines. Is your friend correct? Explain your reasoning. Answer: No, your friend is not correct
Explanation: It is given that your friend claims that because you can find the distance from a point to a line, you should be able to find the distance between any two lines Now, This can be proven by following the below steps: Hence, from the above, We can conclude that your friend is not correct
Question 26. HOW DO YOU SEE IT? You are trying to cross a stream from point A. Which point should you jump to in order to jump the shortest distance? Explain your reasoning. Answer: From the given figure, We can observe that Point A is perpendicular to Point C We know that, According to Perpendicular Transversal Theorem, The distance between the perpendicular points is the shortest Hence, from the above, We can conclude that in order to jump the shortest distance, you have to jump to point C from point A
Question 27. ATTENDING TO PRECISION In which of the following diagrams is \(\overline{A C}\) || \(\overline{B D}\) and \(\overline{A C}\) ⊥ \(\overline{C D}\)? Select all that apply. (A) (B) (C) (D) (E) Answer:
Question 28. how many right angles are formed by two perpendicular lines? Justify your answer. Answer: It is given that in spherical geometry, all points are points on the surface of a sphere. A line is a circle on the sphere whose diameter is equal to the diameter of the sphere. We know that, In Euclidean geometry, the two perpendicular lines form 4 right angles whereas, In spherical geometry, the two perpendicular lines form 8 right angles according to the "Parallel lines Postulate" in spherical geometry. Hence, from the above, We can conclude that 8 right angles are formed by two perpendicular lines in spherical geometry
Question 30. ANALYZING RELATIONSHIPS The painted line segments that brain the path of a crosswalk are usually perpendicular to the crosswalk. Sketch what the segments in the photo would look like if they were perpendicular to the crosswalk. Which type of line segment requires less paint? Explain your reasoning. Answer:
Question 31. ABSTRACT REASONING Two lines, a and b, are perpendicular to line c. Line d is parallel to line c. The distance between lines a and b is x meters. The distance between lines c and d is y meters. What shape is formed by the intersections of the four lines? Answer:
3.5 Equations of Parallel and Perpendicular Lines
Exploration 1
Writing Equations of Parallel and Perpendicular Lines
Work with a partner: Write an equation of the line that is parallel or perpendicular to the given line and passes through the given point. Use a graphing calculator to verify your answer. What is the relationship between the slopes? a. Answer: The given figure is: We know that, From the given figure, We can observe that the given lines are parallel lines Now, The equation for another line is: y = \(\frac{3}{2}\)x + c Substitute (0, 2) in the above equation So, 2 = 0 + c c = 2 So, The equation for another parallel line is: y = \(\frac{3}{2}\)x +3}{2}\)x – 1 y = \(\frac{3}{2}\)x + 2
b. Answer: The given figure is: We know that, From the given figure, We can observe that the given lines are perpendicular lines The equation for another line is: y = \(\frac{3}{2}\)x + c We know that, The slope of perpendicular lines is: -1 So, m1m2 = -1 \(\frac{3}{2}\) . m2 = -1 So, m2 = –\(\frac{2}{3}\) So, y = –\(\frac{2}{3}\)x + c Substitute (0, 1) in the above equation So, 1 = 0 + c c = 1 So, The equation for another perpendicular line is: y = –\(\frac{2}{3}\)x + 1 When we compare the given equation with the obtained equation, We can observe that the product of the slopes are -1 and the y-intercepts are different We know that, The lines that have the slopes product -1 and different y-intercepts are "Perpendicular lines" Hence, from the above, We can conclude that the perpendicular lines are: y = \(\frac{3}{2}\)x – 1 y = –\(\frac{2}{3}\)x + 1
c. Answer: The given figure is: We know that, From the given figure, We can observe that the given lines are parallel lines Now, The equation for another line is: y = \(\frac{1}{2}\)x + c Substitute (2, -2) in the above equation So, -2 = \(\frac{1}{2}\) (2) + c -2 = 1 + c c = 2 – 1 c = -3 So, The equation for another parallel line is: y = \(\frac{1}{2}\)x – 31}{2}\)x + 2 y = \(\frac{1}{2}\)x – 3
d. Answer: The given figure is: We know that, From the given figure, We can observe that the given lines are perpendicular lines The equation for another line is: y = \(\frac{1}{2}\)x + c We know that, The slope of perpendicular lines is: -1 So, m1m2 = -1 \(\frac{1}{2}\) . m2 = -1 So, m2 = -2 So, y = -2x + c Substitute (2, -3) in the above equation So, -3 = -2 (2) + c -3 = -4 + c c = 4 – 3 c = 1 So, The equation for another perpendicular line is: y = -2x + 1 + 2 y = -2x + 1
e. Answer: The given figure is: We know that, From the given figure, We can observe that the given lines are parallel lines Now, The equation for another line is: y = -2x + c Substitute (0, -2) in the above equation So, -2 = 0 + c c = -2 So, The equation for another parallel line is: y = -2x – -2x + 2 y = -2x – 2
f. Answer: The given figure is: We know that, From the given figure, We can observe that the given lines are perpendicular lines The equation for another line is: y = -2x + c We know that, The slope of perpendicular lines is: -1 So, m1m2 = -1 -2 . m2 = -1 So, m2 = \(\frac{1}{2}\) So, y = \(\frac{1}{2}\)x + c Substitute (4, 0) in the above equation So, 0 = \(\frac{1}{2}\) (4) + c 0 = 2 + c c = 0 – 2 c = -2 So, The equation for another perpendicular line is: y = \(\frac{1}{2}\)x – 2 – 2 y = -2x + 2
Exploration 2
Writing Equations of Parallel and Perpendicular Lines
Work with a partner: Write the equations of the parallel or perpendicular lines. Use a graphing calculator to verify your answers.
a. Answer: The given figure is: From the given graph, We can observe that The given lines are the parallel lines Now, The coordinates of the line of the first equation are: (-1.5, 0), and (0, 3) The coordinates of the line of the second equation are: (1, 0), and (0, -23 – 0}{0 + 1.5}\) m = \(\frac{3}{1.5}\) m = 2 Now, We know that, The standard linear equation is: y = mx + c So, y = 2x + c We know that, For parallel lines, The slopes are the same but the y-intercepts are different Hence, The given parallel line equations are: y = 2x + c1 y = 2x + c2
b. Answer: The given figure is: From the given figure, We can observe that The given lines are perpendicular lines So, The coordinates of the line of the first equation are: (0, -3), and (-1.5, 0) The coordinates of the line of the second equation are: (-4, 0), and (0, 20 + 3}{0 – 1.5}\) m = \(\frac{3}{-1.5}\) m = \(\frac{-30}{15}\) m = -2 Now, We know that, The standard linear equation is: y = mx + c So, y = -2x + c We know that, For perpediclar lines, The product of the slopes is -1 and the y-intercepts are different So, m1 × m2 = -1 -2 × m2 = -1 m2 = \(\frac{1}{2}\) Hence, The given perpendicular line equations are: y = -2x + c1 y = \(\frac{1}{2}\)x + c2
Communicate Your Answer
Question 3. How can you write an equation of a line that is parallel or perpendicular to a given line and passes through a given point? MODELING WITH MATHEMATICS To be proficient in math, you need to analyze relationships mathematically to draw conclusions. Answer: We know that, The standard form of a linear equation is: y = mx + c Now, For parallel lines, We know that, The slopes are the same and the y-intercepts are different So, To find the y-intercept of the equation that is parallel to the given equation, substitute the given point and find the value of c Now, For perpendicular lines, we know that, The product of the slopes is -1 So, The slope of the equation that is perpendicular to the given equation is: –\(\frac{1}{m}\) To find the y-intercept of the equation that is perpendicular to the given equation, substitute the given point and find the value of c
Question 4. Write an equation of the line that is (a) parallel and (b) perpendicular to the line y = 3x + 2 and passes through the point (1, -2). Answer: The given equation is: y = 3x + 2 The given point is: (1, -2) a) Parallel line equation: We know that, The slope of the parallel equations are the same So, The slope of the equation that is parallel t the given equation is: 3 Now, The equation that is parallel to the given equation is: y = 3x + c Substitute (1, -2) in the above equation So, -2 = 3 (1) + c -2 – 3 = c c = -5 Hence, The equation of the line that is parallel to the given equation is: y = 3x – 5 b) Perpendicular line equation: We know that, The product of the slope of the perpendicular equations is: -1 So, m1 m2 = -1 3m2 = -1 So, m2 = –\(\frac{1}{3}\) So, The slope of the equation that is parallel t the given equation is: –\(\frac{1}{3}\) Now, The equation that is perpendicular to the given equation is: y = –\(\frac{1}{3}\)x + c Substitute (1, -2) in the above equation So, -2 = –\(\frac{1}{3}\) (-2) + c -2 – \(\frac{2}{3}\) = c c = –\(\frac{8}{3}\) Hence, The equation of the line that is perpendicular to the given equation is: y = –\(\frac{1}{3}\)x –\(\frac{8}{3}\)
Lesson 3.5 Equations of Parallel and Perpendicular Lines
Monitoring Progress
Find the coordinates of point P along the directed line segment AB so that AP to PB is the given ratio.
Question 3. Determine which of the lines are parallel and which of the lines are perpendicular. Answer: The given figure is: From the given figure, The coordinates of line a are: (0, 2), and (-2, -2) The coordinates of line b are: (2, 3), and (0, -1) The coordinates of line c are: (4, 2), and (3, -1) The coordinates of line d are: (-3, 0), and (0, -1-2 – 2}{-2 – 0}\) = \(\frac{-4}{-2}\) = 2
The slope of line d (m) = \(\frac{y2 – y1}{x2 – x1}\) = \(\frac{-1 – 0}{0 + 3}\) = \(\frac{-1}{3}\) = –\(\frac{1}{3}\) We know that, The parallel lines have the same slopes The perpendicular lines have the product of slopes equal to -1 Hence, from the above, We can conclude that Linea and Line b are parallel lines Line c and Line d are perpendicular lines
Question 4. Write an equation of the line that passes through the point (1, 5) and is (a) parallel to the line y = 3x – 5 and Answer: The given equation is: y = 3x – 5 The given point is: (1, 5) We know that, The parallel lines have the same slope Compare the given equation with y = mx + c So, The slope of the parallel line that passes through (1, 5) is: 3 So, The equation of the parallel line that passes through (1, 5) is y = 3x + c To find the value of c, substitute (1, 5) in the above equation So, 5 = 3 (1) + c c = 5 – 3 c = 2 Hence, The equation of the parallel line that passes through (1, 5) is: y = 3x + 2
Question 5. How do you know that the lines x = 4 and y = 2 are perpendiculars? Answer: The given lines are: x = 4 and y = 2 We know that, The line x = 4 is a vertical line that has the right angle i.e., 90° The line y = 4 is a horizontal line that have the straight angle i.e., 0° So, The angle at the intersection of the 2 lines = 90° – 0° = 90° Hence, from the above, We can conclude that the lines x = 4 and y = 2 are perpendicular lines
Exercise 3.5 Equations of Parallel and Perpendicular Lines
Vocabulary and Core Concept Check
Question 1. COMPLETE THE SENTENCE A _________ line segment AB is a segment that represents moving from point A to point B. Answer:
Question 2. WRITING How are the slopes of perpendicular lines related? Answer: We know that, The "Perpendicular lines" are lines that intersect at right angles. If you multiply theslopesof twoperpendicular lines in the plane, you get −1 i.e., the slopes of perpendicular lines are opposite reciprocals
Monitoring Progress and Modeling with Mathematics
In Exercises 3 – 6. find the coordinates of point P along the directed line segment AB so that AP to PB is the given ratio.
In Exercises 7 and 8, determine which of the lines are parallel and which of the lines are perpendicular.
Question 7. Answer:
Question 8. Answer:
The given figure is: From the given figure, The coordinates of line a are: (2, 2), and (-2, 3) The coordinates of line b are: (3, -2), and (-3, 0) The coordinates of line c are: (2, 4), and (0, -2) The coordinates of line d are: (0, 6), and (-2, 03 – 2}{-2 – 2}\) = \(\frac{1}{-4}\) = –\(\frac{1}{4}\)
The slope of line d (m) = \(\frac{y2 – y1}{x2 – x1}\) = \(\frac{6 – 0}{0 + 2}\) = \(\frac{6}{2}\) = 3 We know that, The parallel lines have the same slopes The perpendicular lines have the product of slopes equal to -1 Hence, from the above, We can conclude that Line c and Line d are parallel lines Line b and Line c are perpendicular lines
In Exercises 9 – 12, tell whether the lines through the given points are parallel, perpendicular, or neither. justify your answer.
In Exercises 17 – 20. write an equation of the line passing through point P that is perpendicular to the given line. Graph the equations of the lines to check that they are perpendicular.
Question 17. P(0, 0), y = – 9x – 1 Answer:
Question 18. P(4, – 6)y = – 3 Answer: The given equation is: y = -3 The given point is: P (4, -6) We know that, The line that is perpendicular to y=n is: x = n So, The line that is perpendicular to the given equation is: x = n Substitute P (4, -6) in the above equation So, x = 4 Hence, from the above, We can conclude that The equation that is perpendicular to y = -3 is: x = 4 The representation of the perpendicular lines in the coordinate plane is:
Question 25. ERROR ANALYSIS Describe and correct the error in determining whether the lines are parallel. perpendicular, or neither. Answer:
Question 26. ERROR ANALYSIS Describe and correct the error in writing an equation of the line that passes through the point (3, 4) and is parallel to the line y = 2x + 1. Answer: The given equation of the line is: y = 2x + 1 The given point is: (3, 4) We know that, The slopes of the parallel lines are the same Now, Compare the given equation with y = mx + c So, m = 2 So, The slope of the line that is aprallle to the given line equation is: m = 2 So, The equation of the line that is parallel to the given equation is: y = 2x + c To find the value of c, Substitute (3, 4) in the above equation So, 4 = 2 (3) + c 4 – 6 = c c = -2 Hence, from the above, We can conclude that the equation of the line that is parallel to the given line is: y = 2x – 2
In Exercises 27-30. find the midpoint of \(\overline{P Q}\). Then write an equation of the line that passes through the midpoint and is perpendicular to \(\overline{P Q}\). This line is called the perpendicular bisector.
Question 31. MODELING WITH MATHEMATICS Your school lies directly between your house and the movie theater. The distance from your house to the school is one-fourth of the distance from the school to the movie theater. What point on the graph represents your school? Answer:
Question 32. REASONING Is quadrilateral QRST a parallelogram? Explain your reasoning. Answer: The given figure is: From the above figure, The coordinates of the quadrilateral QRST is: Q (2, 6), R (6, 4), S (5, 1), and T (1, 3) Compare the given points with (x1, y1), and (x2, y2) Now, We know that, If both pairs of opposite sides of a quadrilateral are parallel, then it is a parallelogram So, If the slopes of the opposite sides of the quadrilateral are equal, then it is called as "Parallelogram" We know that, Slope (m) = \(\frac{y2 – y1}{x2 – x1}\) So, Slope of QR = \(\frac{4 – 6}{6 – 2}\) Slope of QR = \(\frac{-2}{4}\) Slope of QR = –\(\frac{1}{2}\)
Slope of TQ = \(\frac{3 – 6}{1 – 2}\) Slope of TQ = \(\frac{-3}{-1}\) Slope of TQ = 3 Now, From the slopes, We can observe that the slopes of the opposite sides are equal i.e., the opposite sides are parallel Hence, from the above, We can conclude that the quadrilateral QRST is a parallelogram
Question 33.
REASONING A triangle has vertices L(0, 6), M(5, 8). and N(4, – 1), Is the triangle a right triangle? Explain 'your reasoning. Answer:
Question 34. MODELING WITH MATHEMATICS A new road is being constructed parallel to the train tracks through points V. An equation of the line representing the train tracks is y = 2x. Find an equation of the line representing the new road. Answer: The given figure is: It is given that a new road is being constructed parallel to the train tracks through points V. An equation of the line representing the train tracks is y = 2x. So, From the figure, V = (-2, 3) We know that, The slopes of the parallel lines are the same So, By comparing the given equation with y = mx + c We get, m = 2 So, The equation of the line that is parallel to the line that represents the train tracks is: y = 2x + c Now, To find the value of c, Substitute (-2, 3) in the above equation So, 3 = 2 (-2) + x 3 + 4 = c c = 7 Hence, from the above, We can conclude that the equation of the line that is parallel to the line representing railway tracks is: y = 2x + 7
Question 35. MODELING WITH MATHEMATICS A bike path is being constructed perpendicular to Washington Boulevard through point P(2, 2). An equation of the line representing Washington Boulevard is y = –\(\frac{2}{3}\)x. Find an equation of the line representing the bike path. Answer:
Question 36. PROBLEM-SOLVING A gazebo is being built near a nature trail. An equation of the line representing the nature trail is y = \(\frac{1}{3}\)x – 4. Each unit in the coordinate plane corresponds to 10 feet. Approximately how far is the gazebo from the nature trail? Answer: The given figure is: It is given that a gazebo is being built near a nature trail. An equation of the line representing the nature trail is y = \(\frac{1}{3}\)x – 4. Each unit in the coordinate plane corresponds to 10 feet So, It can be observed that 1 unit either in the x-plane or y-plane = 10 feet So, y = \(\frac{1}{3}\)x – 4 y = \(\frac{1}{3}\) (10) – 4 y = \(\frac{10 – 12}{3}\) y = –\(\frac{2}{3}\) We know that, The distance won't be in negative value, y = \(\frac{2}{3}\) y = 0.66 feet Hence, from the above, We can conclude that the distance of the gazebo from the nature trail is: 0.66 feet
Question 37. CRITICAL THINKING The slope of line l is greater than 0 and less than 1. Write an inequality for the slope of a line perpendicular to l. Explain your reasoning. Answer:
Question 39. CRITICAL THINKING Suppose point P divides the directed line segment XY So that the ratio 0f XP to PY is 3 to 5. Describe the point that divides the directed line segment YX so that the ratio of YP Lo PX is 5 to 3. Answer:
Question 40. MAKING AN ARGUMENT Your classmate claims that no two nonvertical parallel lines can have the same y-intercept. Is your classmate correct? Explain. Answer: Yes, your classmate is correct
Explanation: It is given that your classmate claims that no two nonvertical parallel lines can have the same y-intercept We know that, The parallel lines have the same slope but have different y-intercepts and do not intersect The intersecting lines intersect each other and have different slopes and have the same y-intercept Hence, from the above, We can conclude that the claim of your classmate is correct
Question 42. THOUGHT-PROVOKING Find a formula for the distance from the point (x0, Y0) to the line ax + by = 0. Verify your formula using a point and a line. Answer: MATHEMATICAL CONNECTIONS In Exercises 43 and 44, find a value for k based on the given description.
Question 43. The line through (- 1, k) and (- 7, – 2) is parallel to the line y = x + 1. Answer:
Question 44. The line through (k, 2) and (7, 0) is perpendicular to the line y = x – \(\frac{28}{5}\). Answer: The given points are: (k, 2), and (7, 0) The given line that is perpendicular to the given points is: y = x – \(\frac{28}{5}\) Now, Compare the given points with (x1, y1), and (x2, y2) So, Slope (m) = \(\frac{y2 – y1}{x2 – x1}\) m = \(\frac{0 – 2}{7 – k}\) m = \(\frac{-2}{7 – k}\) Now, The slope that is perpendicular to the given line is: m = -1 [ Since we know that m1m2 = -1] So, -1 = \(\frac{-2}{7 – k}\) k – 7 = -2 k = -2 + 7 k = 5 Hence, from the above, We can conclude that the value of k is: 5
Question 45. ABSTRACT REASONING Make a conjecture about how to find the coordinates of a point that lies beyond point B along \(\vec{A}\)B. Use an example to support your conjecture. Answer:
PROVING A THEOREM In Exercises 47 and 48, use the slopes of lines to write a paragraph proof of the theorem.
Question 47. Lines Perpendicular to a Transversal Theorem (Theorem 3.12): In a plane. if two lines are perpendicular to the same line. then they are parallel to each other. Answer:
Question 48. Transitive Property of Parallel Lines Theorem (Theorem 3.9),/+: If two lines are parallel to the same line, then they are parallel to each other. Answer: The given statement is: If two lines are parallel to the same line, then they are parallel to each other Proof: Let the two parallel lines be E and F and the plane they lie be plane x Let the two parallel lines that are parallel to the same line be G Hence, According to the Transitive Property of parallel lines, If line E is parallel to line F and line F is parallel to line G, then line E is parallel to line G.
Question 49. PROOF Prove the statement: If two lines are vertical. then they are parallel. Answer:
Question 50. PROOF Prove the statement: If two lines are horizontal, then they are parallel. Answer: The given statement is: If two lines are horizontal, then they are parallel Proof: If two lines x and y are horizontal lines and they are cut by a vertical transversal z, then x ⊥ z and y ⊥ z Hence, x || y is proved by the Lines parallel to Transversal Theorem
3.4 – 3.5 Performance Task: Navajo Rugs
Mathematical Practices
Question 1. Compare the effectiveness of the argument in Exercise 24 on page 153 with the argument "You can find the distance between any two parallel lines" What flaw(s) exist in the argument(s)? Does either argument use correct reasoning? Explain. Answer: From the argument in Exercise 24 on page 153, We can say that The claim of your friend is not correct We know that, If we want to find the distance from the point to a given line, we need the perpendicular distance of a point and a line Hence, from the above, We can conclude that we can not find the distance between any two parallel lines if a point and a line is given to find the distance
Question 2. Look back at your construction of a square in Exercise 29 on page 154. How would your construction change if you were to construct a rectangle? Answer: From the construction of a square in Exercise 29 on page 154, We can observe that the length of all the line segments are equal Now, If you were to construct a rectangle, We have to keep the lengths of the length of the rectangles the same and the widths of the rectangle also the same
Question 3. In Exercise 31 on page 161, a classmate tells you that our answer is incorrect because you should have divided the segment into four congruent pieces. Respond to your classmates' argument by justifying your original answer. Answer: In Exercise 31 on page 161, from the coordinate plane, We can observe that we divided the total distance into the four congruent segments or pieces Hence, from the above, We can conclude that the argument of your friend that the answer is incorrect is not correct
Parallel and Perpendicular Lines Chapter Review
3.1 Pairs of Lines and Angles
Think of each segment in the figure as part of a line. Which line(s) or plane(s) appear to fit the description? Question 1. line(s) perpendicular to Answer: We know that, The lines that are at 90° are "Perpendicular lines" Hence, From the above figure, The lines perpendicular to \(\overline{Q R}\) are: \(\overline{R M}\) and \(\overline{Q L}\)
Question 2. line(s) parallel to Answer: We know that, The lines that do not have any intersection points are called "Parallel lines" Hence, From the above figure, The line parallel to \(\overline{Q R}\) is: \(\overline {L M}\)
Question 3. line(s) skew to Answer: We know that, The lines that do not intersect and are not parallel and are not coplanar are "Skew lines" Hence, From the above figure, The lines skew to \(\overline{Q R}\) are: \(\overline{J N}\), \(\overline{J K}\), \(\overline{K L}\), and \(\overline{L M}\)
Question 4. plane(s) parallel to plane LMQ Answer: From the given figure, We can conclude that the plane parallel to plane LMQ is: Plane JKL
3.2 Parallel Lines and Transversals
Find the values of x and y.
Question 5. Answer: The given figure is: From the given figure, We can observe that x and 35° are the corresponding angles We know that, By using the Corresponding Angles Theorem, x = 35° Now, We can observe that 35° and y are the consecutive interior angles So, 35° + y = 180° y = 180° – 35° y = 145° Hence, from the above, We can conclude that x° = 35° and y° = 145°
Question 7. Answer: The given figure is: From the above figure, We can observe that 2x° and 2y° are the alternate exterior angles 2y° and 58° are the alternate interior angles So, 2x° = 2y° = 58° So, x° = y° =29° Hence, from the above, We can conclude that x° = y° = 29°
3.3 Proofs with Parallel Lines
Find the value of x that makes m || n.
Question 9. Answer: The given figure is: We know that, m || n is true only when x and 73° are the consecutive interior angles according to the "Converse of Consecutive Interior angles Theorem" Now, It is given that m || n So, x + 73° = 180° x = 180° – 73° x = 107° Hence, from the above, We can conclude that the value of x is: 107°
Question 10. Answer: The given figure is: We know that, m || n is true only when 147° and (x + 14)° are the corresponding angles by using the "Converse of the Alternate Exterior Angles Theorem" Now, It is given that m || n So, (x + 14)°= 147° x° = 147° – 14° x° = 133° Hence, from the above, We can conclude that the value of x is: 133°
Question 11. Answer: The given figure is: m || n is true only when 3x° and (2x + 20)° are the corresponding angles by using the "Converse of the Corresponding Angles Theorem" Now, It is given that m || n So, (2x + 20)°= 3x° 3x° – 2x° = 20° x° = 20° Hence, from the above, We can conclude that the value of x is: 20°
3.4 Proofs with Perpendicular Lines
Determine which lines, if any, must be parallel. Explain your reasoning.
Question 13. Answer: The given figure is: From the given figure, We can observe that x ⊥ z and y ⊥ z x and y are parallel lines
Question 14. Answer: The given figure is: From the given figure, We can observe that w ⊥ y and z ⊥ x We can also observe that w and z is not both ⊥ to x and yQuestion 15. Answer: The given figure is: From the given figure, We can observe that m ⊥ a, n ⊥ a, l⊥ b, and n ⊥ b m and n are parallel lines
Question 16. Answer: The given figure is: From the given figure, We can observe that a ⊥ n, b ⊥ n, and c ⊥ m3.5 Equations of Parallel and Perpendicular Lines
Write an equation of the line passing through the given point that is parallel to the given line.
Parallel and Perpendicular Lines Test
Find the values of x and y. State which theorem(s) you used.
Question 1. Answer: The given figure is: From the given figure, We can observe that x and 61° are the vertical angles 61° and y° are the alternate interior angles We know that, According to the "Alternate Interior Angles theorem", the alternate interior angles are congruent According to the "Vertical Angles Theorem", the vertical angles are congruent Hence, x° = y° = 61°
Question 6. Answer: The given figure is: From the given figure, We can observe that x° and 97° are the corresponding angles We know that, According to the "Converse of the Corresponding Angles Theorem", m || n is true only when the corresponding angles are congruent It is given that m || n So, x° = 97° Hence, from the above, We can conclude that x° = 97°
Question 7. Answer: The given figure is: From the given figure, We can observe that 8x° and (4x + 24)° are the alternate exterior angles We know that, According to the "Converse of the Alternate Exterior Angles Theorem", m || n is true only when the alternate exterior angles are congruent It is given that m || n So, 8x° = (4x + 24)° 8x° – 4x° = 24° 4x° = 24° x° = \(\frac{24}{4}\) x° = 6° Hence, from the above, We can conclude that x° = 6°
Question 8. Answer: The given figure is: From the given figure, We can observe that (11x + 33)° and (6x – 6)° are the interior angles We know that, According to the "Converse of the Interior Angles Theory", m || n is true only when the sum of the interior angles are supplementary It is given that m || n So, (11x + 33)°+(6x – 6)° = 180° 17x° + 27° = 180° 17x° = 180° – 27° x° = –\(\frac{153}{17}\) x° = 9° Hence, from the above, We can conclude that x° = 9°
Write an equation of the line that passes through the given point and is (a) parallel to and (b) perpendicular to the given line.
Question 9. (- 5, 2), y = 2x – 3 Answer: The given equation is: y= 2x – 3 The given point is: (-5, 2 2x + c To find the value of c, Substitute (-5, 2) in the above equation So, 2 = 2 (-5) + c 2 + 10 = c c = 12 Hence, The equation of the line that is parallel to the given line equation is: y = 2x + 12 b) Perpendicular to the given line: We know that, The product of the slopes of the perpendicular lines is equal to -1 So, m1 m2 = -1 2m2 = -1 m2 = –\(\frac{1}{2}\) So, The equation of the line that is perpendicular to the given line equation is: y = –\(\frac{1}{2}\)x + c To find the value of c, Substitute (-5, 2) in the given equation 2 = –\(\frac{1}{2}\) (-5) + c c = –\(\frac{1}{2}\) Hence, The equation of the line that is perpendicular to the given line equation is: y = –\(\frac{1}{2}\)x – \(\frac{1}{2}\)
Question 10. (- 1, – 9), y = – \(\frac{1}{3}\)x + 4 Answer: The given equation is: y= –\(\frac{1}{3}\)x + 4 The given point is: (-1, -9 –\(\frac{1}{3}\)x + c To find the value of c, Substitute (-1, -9) in the above equation So, -9 = –\(\frac{1}{3}\) (-1) + c c = \(\frac{26}{3}\) Hence, The equation of the line that is parallel to the given line equation is: y = –\(\frac{1}{3}\)x + \(\frac{26}{3}\) b) Perpendicular to the given line: We know that, The product of the slopes of the perpendicular lines is equal to -1 So, m1 m2 = -1 –\(\frac{1}{3}\)m2 = -1 m2 = 3 So, The equation of the line that is perpendicular to the given line equation is: y = 3x + c To find the value of c, Substitute (-1, -9) in the given equation -9 = 3 (-1) + c c = -6 Hence, The equation of the line that is perpendicular to the given line equation is: y = 3x – 6
Question 11. A student says. "Because j ⊥ K, j ⊥ l' What missing information is the student assuming from the diagram? Which theorem is the student trying to use? Answer: The given figure is: It is given that a student claimed that j ⊥ K, j ⊥ l We know that, According to the "Perpendicular Transversal Theorem", In a plane, if a line is perpendicular to one of two parallellines, then it is perpendicular to the other line also. So, From the above definition, The missing information the student assuming from the diagram is: The line l is also perpendicular to the line j Hence, from the above, We can conclude that the theorem student trying to use is the "Perpendicular Transversal Theorem"
Question 12. Answer:
c. What are the coordinates of the meeting point? Answer: From the given figure, We can conclue that The coordinates of the meeting point are: (150. 200)
d. What is the distance from the meeting point to the subway? Answer: From the given figure, The coordinates of the meeting point are: (150, 200) The coordinates of the subway are: (500, 300) Now, The distance between the meeting point and the subway is: d = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) d = \(\sqrt{(300 – 200)² + (500 – 150)²}\) d = 364.5 yards Hence, from the above, We can conclude that the distance between the meeting point and the subway is: 364.5 yards
Question 13. Identify an example on the puzzle cube of each description. Explain your reasoning. a. a pair of skew lines Answer: We know that, The "Skew lines" are the lines that are non-intersecting, non-parallel and non-coplanar Hence, From the given figure, We can conclude that the pair of skew lines are: \(\overline{A B}\) and \(\overline{G H}\)
b. a pair of perpendicular lines Answer: We know that, The "Perpendicular lines" are the lines that are intersected at the right angles Hence, From the given figure, We can conclude that the pair of perpendicular lines are: \(\overline{I J}\) and \(\overline{C D}\)
c. a pair of paralIeI lines Answer: We know that, The "parallel lines" are the lines that do not have any intersection point Hence, From the given figure, We can conclude that the pair of parallel lines are: \(\overline{C D}\) and \(\overline{E F}\)
d. a pair of congruent corresponding angles Answer: From the given figure, We can conclude that ∠1 and ∠3 are the corresponding angles
e. a pair of congruent alternate interior angles Answer: From the given figure, We can conclude that ∠2 and ∠3 are the congruent alternate interior angles
Parallel and Perpendicular Lines Cumulative Assessment
Question 1. Use the steps in the construction to explain how you know that\(\overline{C D}\) is the perpendicular bisector of \(\overline{A B}\). Answer: Step 1: Draw a line segment of any length and name that line segment as AB Step 2: Draw an arc by using a compass with above half of the length of AB by taking the center at A above AB Step 3: Draw another arc by using a compass with above half of the length of AB by taking the center at B above AB Step 4: Repeat steps 3 and 4 below AB Step 5: Draw a line segment CD by joining the arcs above and below AB Step 6: Measure the lengths of the midpoint of AB i.e., AD and DB. By measuring their lengths, we can prove that CD is the perpendicular bisector of AB
Question 2. The equation of a line is x + 2y = 10. a. Use the numbers and symbols to create the equation of a line in slope-intercept form that passes through the point (4, – 5) and is parallel slopes of the parallel lines are the same So, The equation of the line that is parallel to the given line equation is: y = –\(\frac{1}{2}\)x + c To find the value of c, Substitute (4, -5) in the above equation So, -5 = –\(\frac{1}{2}\) (4) + c c = -5 + 2 c = -3 Hence, from the above, We can conclude that the line that is parallel to the given line equation is: y = –\(\frac{1}{2}\)x – 3
b. Use the numbers and symbols to create the equation of a line in slope-intercept form that passes through the point (2, – 1) and is perpendicular product of the slopes of the perpendicular lines is equal to -1 So, m1m2 = -1 –\(\frac{1}{2}\) (m2) = -1 m2 = 2 So, The equation of the line that is perpendicular to the given line equation is: y = 2x + c To find the value of c, Substitute (4, -5) in the above equation So, -5 = 2 (4) + c -5 – 8 = c c = -13 Hence, from the above, We can conclude that a line equation that is perpendicular to the given line equation is: y = 2x – 13
Question 3. Classify each pair of angles whose measurements are given. a. Answer: The given figure is: From the given figure, We can conclude that 44° and 136° are the adjacent angles
b. Answer: The given figure is: From the given figure, We can conclude that 18° and 23° are the adjacent angles
c. Answer: The given figure is: From the given figure, We can conclude that 75° and 75° are alternate interior angles
d. Answer: The given figure is: From the given figure, We can conclude that 42° and 48° are the vertical angles
Question 4. Your school is installing new turf on the football held. A coordinate plane has been superimposed on a diagram of the football field where 1 unit = 20 feet. a. What is the length of the field? Answer: It is given that a coordinate plane has been superimposed on a diagram of the football field where 1 unit is 20 feet. So, From the given figure, The length of the field = | 20 – 340 | = 320 feet Hence, from the above, We can conclude that the length of the field is: 320 feet
b. What is the perimeter of the field? Answer: From the figure, The width of the field is: 140 feet From the figure, We can observe that the figure is in the form of a rectangle We know that, The perimeter of the field = 2 ( Length + Width) = 2 (320 + 140) = 2 (460) = 920 feet Hence, from the above, We can conclude that the perimeter of the field is: 920 feet
c. Turf costs $2.69 per square foot. Your school has a $1,50,000 budget. Does the school have enough money to purchase new turf for the entire field? Answer: We know that, The area of the field = Length × Width So, The area of the field = 320 × 140 = 44,800 square feet it is given that the turf costs $2.69 per square foot So, The total cost of the turf = 44,800 × 2.69 = $1,20,512 It is given that your school has a budget of $1,50,000 but we only need $1,20,512 Hence, from the above, We can conclude that the school have enough money to purchase new turf for the entire field
Question 5. Enter a statement or reason in each blank to complete the two-column proof. Given ∠1 ≅∠3 Prove ∠2 ≅∠4 Answer: The given table is: Hence, The completed table is:
Explanation: From the given figure, We can observe that 141° and 39° are the consecutive interior angles We know that, According to the consecutive Interior Angles Theorem, If the sum of the angles of the consecutive interior angles is 180°, then the two lines that are cut by a transversal are parallel Hence, from the above, We can conclude that the claim of your friend can be supported
Question 7. Which of the following is true when are skew? (A) are parallel. (B) intersect (C) are perpendicular (D) A, B, and C are noncollinear. Answer: We know that, The "Skew lines" are the lines that are not parallel, non-intersect, and non-coplanar Hene, from the given options, We can conclude that option D) is correct because parallel and perpendicular lines have to be lie in the same plane
b. ∠2 ≅ ________ by the Corresponding Angles Theorem (Thm. 3. 1) Answer: From the given figure, We can conclude that By using the Corresponding angles Theorem, ∠2 ≅ ∠6
c. ∠1 ≅ ________ by the Alternate Exterior Angles Theorem (Thm. 3.3). Answer: From the given figure, We can conclude that By using the Alternate exterior angles Theorem, ∠1 ≅ ∠8
d. m∠6 + m ________ = 180° by the Consecutive Interior Angles Theorem (Thm. 3.4). Answer: From the given figure, We can conclude that By using the Consecutive interior angles Theorem, ∠6 + ∠4 = 180°
Question 9. You and your friend walk to school together every day. You meet at the halfway point between your houses first and then walk to school. Each unit in the coordinate plane corresponds to 50 yards. Answer: It is given that you and your friend walk to school together every day. You meet at the halfway point between your houses first and then walk to school. Each unit in the coordinate plane corresponds to 50 yards. a. What are the coordinates of the midpoint of the line segment joining the two houses? Answer: From the given figure, We can conclude that the midpoint of the line segment joining the two houses is: M = (150, 250)
b. What is the distance that the two of you walk together? Answer: From the given figure, We can observe that The coordinates of the school = (400, 300) The coordinates of the midpoint of the line segment joining the two houses = (150, 250) It is given that the two friends walk together from the midpoint of the houses to the school Now, We know that, The distance that the two of you walk together is: d = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(250 – 300)² + (150 – 400)²}\) = \(\sqrt{2500 + 62,500}\) = 255 yards Hence, from the above, We can conclude that the distance that the two of the friends walk together is: 255 yards | 677.169 | 1 |
A triangle has corners at points A, B, and C. Side AB has a length of #6 #. The distance between the intersection of point A's angle bisector with side BC and point B is #4 #. If side AC has a length of #8 #, what is the length of side BC?
Using the Angle Bisector Theorem, we can find the length of side BC. According to the theorem, the ratio of the lengths of the segments formed by the angle bisector of a triangle is equal to the ratio of the lengths of the sides of the triangle.
Let D be the point where the angle bisector of angle A intersects side BC | 677.169 | 1 |
What is the measurement of the meridians length are all meridians?
Answer: All meridians are of equal length; each is one-half the length of the equator. All meridians converge at the poles and are true north-south lines. All lines of latitude (parallels) are parallel to the equator and to each other. ...
Is every meridian the same length?
Meridians of longitude are imaginary half-circles running from the North Pole to the South Pole. They are sometimes called lines of longitude. Unlike parallels of latitude that are different sizes, all lines of longitude are the same length.
The scientific basis of acupuncture meridians
What are the 4 types of meridians?
What is the most famous meridian?
The line in Greenwich represents the historic Prime Meridian of the World - Longitude 0º. Every place on Earth was measured in terms of its distance east or west from this line. The line itself divided the eastern and western hemispheres of the Earth - just as the Equator divides the northern and southern hemispheres.
Is every meridian a circle?
All the meridians on Earth are exquisite circles. Meridians, which encompasses the immoderate meridian, are the north-south traces we use to help describe particularly wherein we are on the Earth. All the ones signs of longitude meet at the poles, decreasing the Earth properly in half.
What are the three types of meridians?
What is the opposite of the meridian?
The Antimeridian is the +180°/-180° line of longitude, exactly opposite the Prime Meridian (0°). It is often used as the basis for the International Date Line (IDL) because it passes through the open waters of the Pacific Ocean.
Are meridians and chakras the same?
Chakras (Sanskrit word for wheel) are Energy points that vibrate and turn to channel energy into the body and Meridians. Meridian channels follow the same path that the autonomic nervous system does along the spine. ... The body is made up of many chakras that parallel the Meridian paths flowing through the body.
What shape is meridian?
Expert-Verified Answer
Shape of the meridians are circular. Explanation: Unlike the latitude parallels that are circles, the longitude meridians are semi-circles converging at the poles. They finish a circle if opposite meridians are taken together, but they are appreciated as two meridians individually.
Why are there 360 meridians?
The South Pole and the North Pole have separated by 180° apart, The longitude lines cross from the North Pole to the South Pole. While persisting on the North Pole, you can ideally recognize the whole circle of the globe. This is why it begins at zero and finishes at 360 longitudes.
What are the 5 main meridians?
The 12 major meridians are composed of 5 Yin meridians: Heart, Spleen, Lungs, Kidneys, Liver; 5 Yang meridians: Small intestines, Stomach, Large intestine, Urinary bladder, Gallbladder; the Pericardium meridian, and the San Jiao meridian. We talk about the qualities of each of these in more detail below.
What are the two most important meridians?
The lung meridian is second only to the heart meridian among the 12 Principal Meridians. It is called the "Prime Minister" and assists with controlling energy and circulating the blood. The lungs and the heart are seen to work in conjunction with blood and energy, being complementary parts of the living system.
Where is the 100th parallel?
United States. In the United States the meridian 100° west of Greenwich forms the eastern border of the Texas panhandle with Oklahoma (which traces its origin to the Adams-Onís Treaty in 1819 which settled the border between New Spain and the United States between the Red River and Arkansas River).
Where do all the meridian meet?
Do meridians have names?
The names of the internal organ subsystems that are used to name the meridians include Lung, Pericardium, Heart, Large Intestine, San Jiao (Triple Heater), Small Intestine, Bladder, Gallbladder, Stomach, Spleen, Kidney, and Liver.
Why are meridians called meridians?
The term meridian comes from the Latin meridies, meaning "midday"; the subsolar point passes through a given meridian at solar noon, midway between the times of sunrise and sunset on that meridian. Likewise, the Sun crosses the celestial meridian at the same time. | 677.169 | 1 |
converse and biconditional statement for the given conditional statement. If a triangle is equilateral, then it is equiangular.
Hint:
The converse of a statement is formed by switching the hypothesis and the conclusion. The converse of "if p, then q" is "if q, then p." ¶A biconditional statement is a logic statement that includes the phrase, "if and only if," sometimes abbreviated as "if." The logical biconditional statement can be written as p if q, p if and only if q, p ↔ q
The correct answer is: Converse statement: If a triangle is equiangular, then it is equilateral. Biconditional statement: Triangle is equiangular, if and only if Triangle is equiangular.
It is given p: Triangle is equilateral. q: Triangle is equiangular. Conditional statement of p → q is given as "If a triangle is equilateral, then it is equiangular. " Converse statement of p → q is q → p and it is written as " If a triangle is equiangular, then it is equilateral. " Biconditional statement of p → q is p ↔ q" Triangle is equiangular, if and only if Triangle is equiangular. " Final Answer: Converse statement: If a triangle is equiangular, then it is equilateral. Biconditional statement: Triangle is equiangular, if and only if Triangle is equiangular | 677.169 | 1 |
Construct a Line Segment – Explanation and Examples
To construct a line segment connecting two points, you need to line up a straightedge with two points and trace. Constructing a new line segment congruent to another involves creating an equilateral triangle and two circles.
The construction of a line segment between any two points is Euclid's first postulate. Creating a line congruent to a given line is his second proposition. To do the construction and prove that the two lines are indeed congruent, we must first familiarize ourselves with proposition 1, which involves creating an equilateral triangle.
Before moving forward, make sure you review the foundations of geometric construction.
This topic includes:
How to Construct a Line Segment
How to Construct a Congruent Line Segment
How to Construct a Line Segment
Euclid's first postulate states that a line can be drawn between any two points.
That is, as long as we have two points, we can construct a line segment. To do this, we line up the edge of the straightedge with the two points and draw a line.
It is also possible to copy a line segment that already exists. That is, we can construct a congruent line segment.
How to Construct a Congruent Line Segment
It is also possible to make a congruent copy of a line that already exists.
There are two main ways we can do this. First, we can copy a line that already exists so that the new line has a particular end point. We can also cut off a longer line segment to equal the length of a shorter line.
In fact, these two constructions are the second and third propositions in the first book of Euclid's Elements. To do them, however, we need to first look at proposition 1. This tells us how to create an equilateral triangle.
How to Construct an Equilateral Triangle
We begin with a line, AB. Our goal is to create an equilateral triangle with AB as one of the sides. By definition, an equilateral figure has sides that are all the same length. Consequently, all of the sides of the triangle we construct will be lines congruent to AB.
We begin by drawing two circles with our compass. The first will have center B and distance Ba. The second will have center A and distance AB.
Now, label either of the two intersection points for the circles as C. Then, connect AC and BC. The triangle ABC is equilateral.
How do we know this?
BC is a radius of the first circle we drew, while AC is a radius of the second circle we drew. Both of these circles had a radius of length AB. Therefore, BC and AC both have length AB, and the triangle is equilateral.
Construct a Congruent Segment at a Point
If we are given a point line AB and a point D, it is possible to construct a new line segment with an endpoint at D and length AB.
To do this, we first connect the point B with C.
Then, construct an equilateral triangle on the line BC. Since we already know how to do this, we don't have to show the construction lines. This also makes the proof easier to follow because the figure is less cluttered.
Then, we can make another circle with center B and radius BA. After that, extend the line DB so that it intersects this new circle at E.
Next, we construct a circle with center D and radius DE. Finally, we can extend DC so that it intersects this circle at a point F. CF will have the same length as AB.
How do we know this?
The radius of the circle with center D is DE. Notice that DE is made up of two smaller line segments, DB and BE. Since BE is a radius of the circle with center B and radius AB, BE has the same length as AB.
The segment DB is a leg of the equilateral triangle, so its length is equal to BC. Therefore, the length of DE is DB+BE=BC+AB.
Now, consider the line segment DF. This is also a radius of the circle with center D, so its length is equal to DE. DF is made up of two parts, DC and CF. DC is equal in length to BC because they are both parts of an equilateral triangle.
Therefore, we have AB+BC=DE=DF=DC+CF=BC+CF.
That is, AB+BC=BC+CF. Therefore, AB=CF.
Cut a Shorter Segment from a Longer Segment
Using the ability to construct a congruent line at a point, we will cut off a section of a longer line segment equal to the length of a shorter segment. We begin with a longer line segment CD and a shorter segment AB.
Next, we copy the segment AB and construct a congruent segment CG. Note that we do not have control over the orientation of CG, so it will, in all probability, not line up exactly with CD.
Finally, we draw a circle with center C and radius CG. Then, we can identify the point, H, where the circumference of the circle intersects CD. CH will be equal to AB in length.
The proof of this is pretty simple. CH is a radius of the circle with center C and radius CG. Therefore CH=CG. But we already know that CG=AB. Therefore, by the transitive property, CH=AB.
Examples
This section will present some examples of how to connect line segments and how to construct congruent line segments.
Example 1
Connect points A and B with a line segment.
Example 1 Solution
In this case, we need to line up our straight edge with the points A and B and trace, as shown.
Example 2
Construct a line segment congruent to AB.
Example 2 Solution
We are not given any other points in our figure, so we can construct the congruent segment anywhere we would like.
The easiest thing to do then is to make AB the radius of a circle with center B. Then, we can draw a line segment from B to any point, C, on the circle's circumference.
Such a line segment, BC, will also be a radius of the circle, so it will be equal in length to AB.
Example 3
Construct a lines segment congruent to AB with endpoint D.
Example 3 Solution
We need to remember the steps for constructing a congruent line segment at a point to do this.
First, we connect BD.
Then, construct an equilateral triangle BDG.
Next, we create a circle with radius AB and center B. If we extend the segment GB, it intersects with this circle, and we call the intersection E.
Then, we can create a circle with center G and radius GE. We then extend GD until it intersects this circle and calls that point C.
CD will be equal in length to AB.
Note: It is important to draw full circles when proving a geometric construction, but arcs are generally fine for the construction itself. In the figure, only part of the circle with center G and radius GE is shown.
Example 4
Construct a line segment double the length of AB.
Example 4 Solution
We cannot simply copy the line segment and make its new endpoint A because we do not have control over the congruent segment's orientation.
Instead, we can construct a circle with center A and radius AB. We can then extend the segment in the direction of A until it intersects the circle's circumference at point C. Since AC and AB are both radii of the circle, they have the same length. Therefore, BC is double the length of AB.
Example 5
Construct a line segment congruent to AB with the end point at C. Then, put another line segment congruent to AB at the new end point, D.
Example 5 Solution
Essentially, we have to do multiple iterations of constructing a congruent segment.
First, construct a congruent segment at C, as we did in example 3.
Then, designate D to be the other end point.
Now, we do what we did before. Construct a segment BD. Then, create an equilateral triangle. Next, make a circle with center B and radius AB. We can then extend the segment GB so that it intersects with this new circle at E. Next, we make a circle with center G and radius GE. Finally, we extend GD so that it intersects with the new circle at F.
Practice Question
1. True or False: We need at least three points to construct a line segment.
True
False
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2. True or False: We need exactly three line segments to construct a triangle.
True
False
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3. True or False: To construct an equilateral triangle, we only need three line segments of different sizes.
True
False
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4. How many line segments do we need to construct a square?
$2$
$3$
$4$
$5$
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5. When two line segments measuring $5$ units and $2$ units, respectively, are joined together, what would be the new line segment's length?
$2$ units
$5$ units
$7$ units
$9$ units
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Open Problems
Construct a line segment AB.
Create line segments to create a triangle ABC.
Construct a line segment congruent to each side of the triangle ABC.
Cut off a segment of AB equal to the length of CD.
Construct an isosceles triangle inside the triangle ABC with B as one of the vertices. | 677.169 | 1 |
7Geometry
7.1 Arc Length
Definition 7.1 (Infinitesimal Arclength) If \(\vec{r}(t)\) is a parametric curve, its infinitesimal arclength is measured by \[ds=|\vec{r}^\prime(t)|\,dt\]
This makes sense: after all the derivative \(\vec{r}^\prime(t)\) is the velocity, \(\|\vec{r}^\prime(t)\|\) is the speed, and \(dt\) is an infinitesimal length of time. Thus, the product \(\|\vec{r}^\prime(t)\|dt\) is an infinitesimal bit of distance - a small length along the curve. To take this infinitesimal information and get something useful out - we need to integrate along the curve.
Definition 7.2 (Arclength) If \(\vec{r}(t)\) is a parametric curve, its length between \(t=a\) and \(t=b\) is given by \[L=\int_a^b ds=\int_a^b |\vec{r}^\prime(t)|dt\]
Example 7.1 (Arclength of a Helix) Find the arclength of \(\vec{r}(t)=(\cos(t),\sin(t),t)\) from \(t=0\) to \(t=2\pi\). First, we need to find the velocity \(\vec{r}^\prime\): \[\vec{r}^\prime(t)=\langle -\sin(t),\cos(t),1\rangle\] Next, we need to take this velocity and find the speed: \[\|\vec{r}^\prime(t)\|=\sqrt{(-\sin t)^2+(\cos t)^2+1}=\sqrt{2}\] Finding arclenght is just integrating this over the domain: \[\int_0^{2\pi}\|vec{r}^\prime(t)\|dt=\int_0^{2\pi}\sqrt{2}dt=2\pi\sqrt{2}\]
Often arclength integrals can be challenging to do, because of the square root. But with some algebra and integration tricks, alot can be learned.
From this idea, we can define the arclength function whcih measures the length of a curve \(\vec{r}(t)\) from a starting point \(t=a\):
Definition 7.3 (The Arclength Function) If \(\vec{r}(t)\) is a parametric curve, for any given starting point \(t=a\) we may define the arclength function which measures the length of curve between \(a\) and \(t\): \[s(t)=\int_a^t |\vec{r}^\prime(u)|du\] (Note we have changed the variable of integration so that \(t\) is not used in two different contexts)
What is the arclength function for the helix in our earlier example, starting from \(t=0\)? Since \(\|\vec{r}^\prime(t)\|=\sqrt{2}\), we see that
\[s(t)=\int_0^t \sqrt{2}dt=\sqrt{2}t\]
This tells us that after \(t\) seconds, we have traced out \(\sqrt{2}t\) units of arclength. How could we reparameterize this curve so that its arclength function is just \(s(t)=t\) (tracing out \(t\) units of arc in \(t\) units of time)?
Definition 7.4 (Unit Speed Curve) A curve \(\vec{c}(t)\) is unit speed if \(\|\vec{c}^\prime(t)\|=1\) for all times \(t\). This means that it after \(t\) seconds, the curve has traversed \(t\) units of length. For this reason, we also call unit speed curves arclength parameterized curves.
In our example, to make the helix unit speed we need to slow it down by a factor of \(\sqrt{2}\): that is, we need \(\vec{r}(t/\sqrt{2})\):
7.2 Curvature
Besides the length of a curve, one of the most powerful things calculus allows us to do is rigorously study its curvature. How can we quantify the fact that some curves bend gently and others turn sharply in space? One means of trying to do this is by looking at the tangent vectors to the curve, and trying to determing how quickly they are changing.
Of course, there's a complication to this: a tangent vector can change in length without changing in direction. This doesn't mean that a curve is curving, but rather that the particle tracing it out is accelerating.
To remove this worry, we define the unit tangent vector to a curve. Just divide the derivative by its magnitude!
Definition 7.5 (Unit Tangent Vector) The unit tangent vector to the curve \(\vec{r}(t)\) is the vector of length \(1\) which is parallel to \(\vec{r}^\prime(t)\): \[\vec{T}(t)=\frac{\vec{r}^\prime(t)}{|\vec{r}^\prime(t)|}\]
This allows a clean definition of curvature: it is how much the unit tangent vector turns per arclength.
Definition 7.6 (Curvature of a Curve) The curvature of a curve is \[\kappa = \left|\frac{d\vec{T}}{ds}\right|
=\left|\vec{T^\prime}(t)\right|\Bigg/\left|\vec{r}^\prime(t)\right|
\]
Where the second equality is derived via the chain rule: \[\left|\frac{d\vec{T}}{ds}\right|=\left|\frac{d\vec{T}}{dt}\frac{dt}{ds}\right|=\left|\frac{\frac{d\vec{T}}{dt}}{\frac{ds}{dt}}\right|
=\frac{\left|\vec{T}^\prime\right|}{|\vec{r}^\prime|}
\]
This formula is difficult to apply in genreral, as the unit tangent vector \(T\) might have a pretty scary looking formula, and so taking its derivative can be a lot of work.
Doing some calculus we can get a simpler formula:
Theorem 7.1 (Curvature of a Curve) The curvature of \(\vec{r}(t)\) is given by \[\kappa(t)=\frac{|\vec{r}^\prime(t)\times\vec{r}^{\prime\prime}(t)|}{|\vec{r}^\prime(t)|^3}\]
This is something that's relatively easy to compute (though perhaps tedious) from any parameterization: you just need to find the first and second derivatives, take a cross product, and then plug into the formula!
But, if we further restrict ourselves to the case that \(\vec{r}(t)=(t,f(t))\) traces the graph of a function, we can simplify this calculation even more:
Theorem 7.2 (Curvature of a Graph) If \(y=f(x)\) is a function, the curvature of its graph is \[\kappa(x)=\frac{|f^{\prime\prime}(x)|}{|1+f^\prime(x)|^{3/2}}\]
The below graphing calculator lets you entere a function \(\kappa(s)\) that specifies the curvature of a curve, and then it computes a curve which has that curvature! Try even just the case \(k(s)=s\) and think about the result - what sort of curve do you expect to see if the curvature grows linearly along the length of the curve?
7.3 Framing a Curve
The unit tangent vector provides us with a very useful "pointer" - always oriented directly along a curve. But in any serious application of parametric curves, we need more information: we would like a whole \(x,y,z\) coordinate frame at each point of the curve.
To start, we'll look for one vector which is orthogonal to, or normal to our curve.
How can we find one? Well, the unit tangent is of constant length (its the unit tangent, after all). We can use the product rule for dot products to understand \(T^\prime\):
Together these three vectors provide a coordinate system at each point along the curve: \(T\) measures distance in the tangent direction to the curve, \(N\) measures distance in the direction the curve is bending fastest, and \(B\) is orthogonal to both. This collection of vectors is called the Frenet Frame and is heavily used in computations in physics, engineering, and computer graphics. | 677.169 | 1 |
Triangulation is useful in determining the properties of a topological space.
Triangulation is the process of finding a distance by calculating the length of one side of a triangle, given a deterministic combination of angles and sides of the triangle. It uses mathematical identities from trigonometry. | 677.169 | 1 |
What are Parallel Lines?
The lines that do not intersect or meet each other at any point in a plane are termed parallel lines. Parallel lines are non-intersecting lines and always stay apart from each other. It is also said that parallel lines meet at infinity.
Definition of Parallel Lines
Parallel lines in geometry can be defined as two straight lines in the plane that are at equal distance from each other and never intersects no matter how much they are extended. Parallel lines are represented by the symbol "||".
Representation of Parallel Lines
Two parallel lines AB and CD are represented as AB||CD. This means line AB is parallel to line CD. In the given plane, infinite parallel lines can be drawn to AB and CD.
What are the Different Types of Angles Formed When Transversal Line Intersect Parallel Line?
When transversal line intersects two parallel lines, the following pair of angles are formed:
Corresponding angles
Alternate interior angles
Alternate exterior angles
Vertically opposite angles
Linear Pair
Real-Life Applications of Parallel Lines
We observe parallel lines in the following things:
Railway Track
Cricket Stumps
Racing Track
Fork Lines
Steps of a Ladder
Rail Bars
Keys of a Piano
Pins of a Plug
Markings on Road
Zebra Crossing
Edges of A Ruler
How To Identify Parallel Lines?
Two lines intersected by a transversal line are termed parallel lines when corresponding angles are equal.
Two lines intersected by a transversal line are termed parallel lines when alternate interior angles are equal.
Two lines intersected by a transversal line are termed parallel lines when alternate exterior angles are equal.
Two lines intersected by a transversal line are termed parallel lines when sum of consecutive interior angles is 180 degrees.
Two lines intersected by a transversal line are termed parallel lines when sum of consecutive exterior angles is 180 degrees.
What is Geometry?
Geometry is regarded as the oldest part of Mathematics that is concerned with the property of space that is related to the shape, size, distance, and relative position of figures. A mathematician whose area of work is in Geometry is known as Geometer. The need for Geometry is not only limited to the study of a flat surface (known as plane geometry) and three-dimensional objects (known as solid Geometry).
What Are The Different Types of Geometry?
The different types of Geometry are:
Eucledian Geometry: It is the most common type of Geometry that is generally taught at the primary level. Euclidean Geometry is described by Euclidean in detail in 'Element' which is one of the bases of Mathematics. The impact of 'Element' was so immense that no other kind of Geometry was almost used for the next 2000 years.
Non-Euclidean Geometry: This type of Geometry is an extension to Euclid Geometry which is mostly observed in three-dimensional objects. The other name of Non-Euclidean Geometry is hyperbolic geometry, spherical geometry, elliptic geometry, and many more. This type of Geometry represents how most know theorem 'Sum of the angles of a triangle is very different in 3-D space.'
Analytic Geometry: In Analytic Geometry, we study geometric figures and construction using a coordinate system. The most commonly used coordinate systems are Cartesian, Polar, and Parametric systems.
Differential Geometry: In differential Geometry, we study planes, lines, and surfaces in a three-dimensional space with the principles of integrals and differential calculus. This type of Geometry emphasizes several problems like contact surface, geodesics, complex manifolds, and several others. Geodesic is termed as the shortest distance between two points on the surface of a sphere. The use of Differential Geometry ranges from engineering problems to the computation of gravitational fields | 677.169 | 1 |
Module 15: Conic Sections
Parabolas with Vertices at the Origin
Learning Outcomes
Identify and label the focus, directrix, and endpoints of the focal diameter of a parabola.
Write the equation of a parabola given a focus and directrix.
In The Ellipse we saw that an ellipse is formed when a plane cuts through a right circular cone. If the plane is parallel to the edge of the cone, an unbounded curve is formed. This curve is a parabola.
Parabola
tip for success
You've seen parabolas before as the set of all points satisfying a quadratic function. We'll look at the geometric form of a parabola in this section. It still describes a set of points that satisfy an equation in two variables, but without the need to qualify as a function, it can open left and right as well as up and down.
It will be necessary to use another form of its equation to take all of the characteristics of this object into consideration. You'll learn new terminology for the parts of a parabola just as you did with the ellipse and hyperbola as well.
Like the ellipse and hyperbola, the parabola can also be defined by a set of points in the coordinate plane. A parabola is the set of all points [latex]\left(x,y\right)[/latex] in a plane that are the same distance from a fixed line, called the directrix, and a fixed point (the focus) not on the directrix.
We previously learned about a parabola's vertex and axis of symmetry. Now we extend the discussion to include other key features of the parabola. Notice that the axis of symmetry passes through the focus and vertex and is perpendicular to the directrix. The vertex is the midpoint between the directrix and the focus.
The line segment that passes through the focus and is parallel to the directrix is called the latus rectum, also called the focal diameter. The endpoints of the focal diameter lie on the curve. By definition, the distance [latex]d[/latex] from the focus to any point [latex]P[/latex] on the parabola is equal to the distance from [latex]P[/latex] to the directrix.
Key features of the parabola
To work with parabolas in the coordinate plane, we consider two cases: those with a vertex at the origin and those with a vertex at a point other than the origin. We begin with the former.
Let [latex]\left(x,y\right)[/latex] be a point on the parabola with vertex [latex]\left(0,0\right)[/latex], focus [latex]\left(0,p\right)[/latex], and directrix [latex]y= -p[/latex] as shown in Figure 4. The distance [latex]d[/latex] from point [latex]\left(x,y\right)[/latex] to point [latex]\left(x,-p\right)[/latex] on the directrix is the difference of the y-values: [latex]d=y+p[/latex]. The distance from the focus [latex]\left(0,p\right)[/latex] to the point [latex]\left(x,y\right)[/latex] is also equal to [latex]d[/latex] and can be expressed using the distance formula.
Set the two expressions for [latex]d[/latex] equal to each other and solve for [latex]y[/latex] to derive the equation of the parabola. We do this because the distance from [latex]\left(x,y\right)[/latex] to [latex]\left(0,p\right)[/latex] equals the distance from [latex]\left(x,y\right)[/latex] to [latex]\left(x, -p\right)[/latex].
[latex]\sqrt{{x}^{2}+{\left(y-p\right)}^{2}}=y+p[/latex]
We then square both sides of the equation, expand the squared terms, and simplify by combining like terms.
The equations of parabolas with vertex [latex]\left(0,0\right)[/latex] are [latex]{y}^{2}=4px[/latex] when the x-axis is the axis of symmetry and [latex]{x}^{2}=4py[/latex] when the y-axis is the axis of symmetry. These standard forms are given below, along with their general graphs and key features.
A General Note: Standard Forms of Parabolas with Vertex (0, 0)
The table below summarizes the standard features of parabolas with a vertex at the origin.
Axis of Symmetry
Equation
Focus
Directrix
Endpoints of Focal Diameter
x-axis
[latex]{y}^{2}=4px[/latex]
[latex]\left(p,\text{ }0\right)[/latex]
[latex]x=-p[/latex]
[latex]\left(p,\text{ }\pm 2p\right)[/latex]
y-axis
[latex]{x}^{2}=4py[/latex]
[latex]\left(0,\text{ }p\right)[/latex]
[latex]y=-p[/latex]
[latex]\left(\pm 2p,\text{ }p\right)[/latex]
(a) When [latex]p>0[/latex] and the axis of symmetry is the x-axis, the parabola opens right. (b) When [latex]p<0[/latex] and the axis of symmetry is the x-axis, the parabola opens left. (c) When [latex]p<0[/latex] and the axis of symmetry is the y-axis, the parabola opens up. (d) When [latex]\text{ }p<0\text{ }[/latex] and the axis of symmetry is the y-axis, the parabola opens down.
The key features of a parabola are its vertex, axis of symmetry, focus, directrix, and focal diameter. When given a standard equation for a parabola centered at the origin, we can easily identify the key features to graph the parabola.
A line is said to be tangent to a curve if it intersects the curve at exactly one point. If we sketch lines tangent to the parabola at the endpoints of the focal diameter, these lines intersect on the axis of symmetry.
How To: Given a standard form equation for a parabola centered at (0, 0), sketch the graph.
Determine which of the standard forms applies to the given equation: [latex]{y}^{2}=4px[/latex] or [latex]{x}^{2}=4py[/latex].
Use the standard form identified in Step 1 to determine the axis of symmetry, focus, equation of the directrix, and endpoints of the focal diameter.
If the equation is in the form [latex]{y}^{2}=4px[/latex], then
the axis of symmetry is the x-axis, [latex]y=0[/latex]
set [latex]4p[/latex] equal to the coefficient of x in the given equation to solve for [latex]p[/latex]. If [latex]p>0[/latex], the parabola opens right. If [latex]p<0[/latex], the parabola opens left.
use [latex]p[/latex] to find the coordinates of the focus, [latex]\left(p,0\right)[/latex]
use [latex]p[/latex] to find the equation of the directrix, [latex]x=-p[/latex]
use [latex]p[/latex] to find the endpoints of the focal diameter, [latex]\left(p,\pm 2p\right)[/latex]. Alternately, substitute [latex]x=p[/latex] into the original equation.
If the equation is in the form [latex]{x}^{2}=4py[/latex], then
the axis of symmetry is the y-axis, [latex]x=0[/latex]
set [latex]4p[/latex] equal to the coefficient of y in the given equation to solve for [latex]p[/latex]. If [latex]p>0[/latex], the parabola opens up. If [latex]p<0[/latex], the parabola opens down.
use [latex]p[/latex] to find the coordinates of the focus, [latex]\left(0,p\right)[/latex]
use [latex]p[/latex] to find equation of the directrix, [latex]y=-p[/latex]
use [latex]p[/latex] to find the endpoints of the focal diameter, [latex]\left(\pm 2p,p\right)[/latex]
Plot the focus, directrix, and focal diameter, and draw a smooth curve to form the parabola.
Example: Graphing a Parabola with Vertex (0, 0) and the x-axis as the Axis of Symmetry
Graph [latex]{y}^{2}=24x[/latex]. Identify and label the focus, directrix, and endpoints of the focal diameter.
Show Solution
The standard form that applies to the given equation is [latex]{y}^{2}=4px[/latex]. Thus, the axis of symmetry is the x-axis. It follows that:
[latex]24=4p[/latex], so [latex]p=6[/latex]. Since [latex]p>0[/latex], the parabola opens right the coordinates of the focus are [latex]\left(p,0\right)=\left(6,0\right)[/latex]
the equation of the directrix is [latex]x=-p=-6[/latex]
the endpoints of the focal diameter have the same x-coordinate at the focus. To find the endpoints, substitute [latex]x=6[/latex] into the original equation: [latex]\left(6,\pm 12\right)[/latex]
Next we plot the focus, directrix, and focal diameter, and draw a smooth curve to form the parabola.
Try It
Graph [latex]{y}^{2}=-16x[/latex]. Identify and label the focus, directrix, and endpoints of the focal diameter.
Try It
Use an online graphing tool to plot the following parabola whose axis of symmetry is the x-axis, [latex]y^2=4px[/latex]
Adjust the free variable [latex]p[/latex] to values between [latex]-10,10[/latex]. Your task in this exercise is to add the focus, directrix, and endpoints of the focal diameter in terms of the free variable, [latex]p[/latex].
For example, to add the focus, you would define a point, [latex](p,0)[/latex] .
Show Solution
Writing Equations of Parabolas in Standard Form
In the previous examples we used the standard form equation of a parabola to calculate the locations of its key features. We can also use the calculations in reverse to write an equation for a parabola when given its key features.
tip for success
In this section, we will write the equation of a parabola in standard form, as opposed to the equation of a quadratic or second degree polynomial. The language we use when discussing the object is specific.
It is true that a quadratic function forms a parabola when graphed in the plane, but here we are using the phrase standard form of the equation of a parabola to indicate that we wish to describe the geometric object. When talking about this object in this context, we would naturally use the equations described below.
How To: Given its focus and directrix, write the equation for a parabola in standard form.
Determine whether the axis of symmetry is the x– or y-axis.
If the given coordinates of the focus have the form [latex]\left(p,0\right)[/latex], then the axis of symmetry is the x-axis. Use the standard form [latex]{y}^{2}=4px[/latex].
If the given coordinates of the focus have the form [latex]\left(0,p\right)[/latex], then the axis of symmetry is the y-axis. Use the standard form [latex]{x}^{2}=4py[/latex].
Multiply [latex]4p[/latex].
Substitute the value from Step 2 into the equation determined in Step 1.
Example: Writing the Equation of a Parabola in Standard Form Given its Focus and Directrix
What is the equation for the parabola with focus [latex]\left(-\frac{1}{2},0\right)[/latex] and directrix [latex]x=\frac{1}{2}?[/latex]
Show Solution
The focus has the form [latex]\left(p,0\right)[/latex], so the equation will have the form [latex]{y}^{2}=4px[/latex].
Multiplying [latex]4p[/latex], we have [latex]4p=4\left(-\frac{1}{2}\right)=-2[/latex]. Substituting for [latex]4p[/latex], we have [latex]{y}^{2}=4px=-2x[/latex].
Therefore, the equation for the parabola is [latex]{y}^{2}=-2x[/latex].
Try It
What is the equation for the parabola with focus [latex]\left(0,\frac{7}{2}\right)[/latex] and directrix [latex]y=-\frac{7}{2}[/latex]? | 677.169 | 1 |
Tangent of 30 Degrees
The value of the tangent of 30 degrees is 0.5773502. . .. Tan 30 degrees in radians is written as tan (30° × π/180°), i.e., tan (π/6) or tan (0.523598. . .). In this article, we will discuss the methods to find the value of tan 30 degrees with examples.
Tangent of 30 as a fraction: 1/√3 (or) √3/3
Tan 30° in decimal: 0.5773502. . .
Tan (-30 degrees): -0.5773502. . . or -1/√3
Tan 30° in radians: tan (π/6) or tan (0.5235987 . . .)
What is the Value of Tan 30 Degrees?
The value of tan 30 degrees in decimal is 0.577350269.... The tangent of 30 degrees can be found by taking the sine of 30 degrees and dividing it by the cosine of 30 degrees. Since the sine of 30 degrees is 1/2 and the cosine of 30 degrees is √3/2, the tangent of 30 degrees is
FAQs on Tan 30 Degrees
What is Tangent of 30 Degrees?
Tangent of 30 degrees is the value of tangent trigonometric function for an angle equal to 30 degrees. The value of tan 30° is 1/√3 or 0.5774 (approx). In fraction form, the tangent of 30 is 1/√3 (or) √3/3.
What is the Value of Tan 30 Degrees as a Fraction?
The value of tan 30 degrees as a fraction is 1/√3. But we can rationalize the denominator by multiplying and dividing by √3. Then we get tan 30° = 1/√3 × √3/√3 = √3/3. Thus, the value of tan 30 degrees in fraction form can be either 1/√3 or √3/3.
How to Find the Value of Tan 30 Degrees?
The value of tan 30 degrees can be calculated by constructing an angle of 30° with the x-axis, and then finding the coordinates of the corresponding point (0.866, 0.5) on the unit circle. The value of tan 30° is equal to the y-coordinate(0.5) divided by the x-coordinate (0.866). ∴ tan 30° = 1/√3 or 0.5774
How to Find Tan 30° in Terms of Other Trigonometric Functions?
Using trigonometry formula, the value of tan 30° can be given in terms of other trigonometric functions as: | 677.169 | 1 |
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taking 4-space in a piece wise fashion, it seems there are actually 6 (thats right; 6!!!) planes that are some how maybe mutually orthogonal to one another. Namely if we take four number lines and mark them w, x, y, z then we can construct the sub-space planes; (w,x), (w,y), (w,z), (x,y), (x,z), and (y,z). How can you have 6 mutually perpendicular planes in 4-space? Is it wrong to say that all the planes are mutually perpendicular? 4-space is turning out to be a really cool and interesting place indeed. | 677.169 | 1 |
A quadrilateral is a generic term used to describe a four sided polygon. In other words, it is a shape that has four sides.
A rectangle, rhombus, parallelogram and trapezium (trapezoid) has four sides. In light of this, it can be classified as a quadrilateral but, if the quadrilateral has no sides and angles equal, it could not be classified as any of the foregoing.
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Q: Why can a quadrilateral not be called a parallelogram a rectangle a rhombus or a trapezoid? | 677.169 | 1 |
The Complementary Relationship between Sin(x) and Cos(x) in Trigonometry
Complementary function to sin(x)
The complementary function to sin(x) is cos(x)
The complementary function to sin(x) is cos(x). In mathematics, complementary functions are pairs of functions that sum up to a constant value. In the context of trigonometric functions, the sine and cosine functions are complementary.
The sine function (sin(x)) describes the ratio of the length of the side opposite to an angle in a right triangle to the length of the hypotenuse. It reaches its maximum value of 1 when the angle is 90 degrees (or π/2 radians) and decreases as the angle increases or decreases.
The cosine function (cos(x)), on the other hand, describes the ratio of the length of the adjacent side to the hypotenuse in a right triangle. It reaches its maximum value of 1 when the angle is 0 degrees (or 2π radians) and also when the angle is 360 degrees (or 2π radians). The cosine function is the complementary function to the sine function, meaning that their sum is always equal to 1.
So, if we have sin(x) as the given function, then the complementary function would be cos(x). Their sum sin(x) + cos(x) = 1 is a constant value | 677.169 | 1 |
So, the centroid of the triangle with vertices at ((5, 2)), ((2, 5)), and ((7, 2)) is (\left( \frac{14}{3}, 3 \right | 677.169 | 1 |
Elements of Geometry: Containing the First Six Books of Euclid, with a ...
two right angles; therefore the other two, HGL, GHD are greater than two right angles. Therefore, since KL and CD are not parallel, and since they do not meet towards L and D, they must meet if produced towards K and C.
COR. 2. If BGH is a right angle, GHD will be a right angle also ; therefore every line perpendicular to one of two parallels, is perpendicular to the other.
COR. 3. Since AGE BGH, and DHF CHG; hence the four acute angles BGH, AGE, GHC, DHF, are equal to each other. The same is the case with the four obtuse angles EGB, AGH, GHD, CHF. It may be also observed, that, in adding one of the acute angles to one of the obtuse, the sum will always be equal to two right angles.
SCHOLIUM.
The angles just spoken of, when compared with each other, assume different names. BGH, GHD, we have already named interior angles on the same side; AGH, GHC, have the same name; AGH, GHD, are called alternate interior angles, or simply alternate; so also, are BGH, GHC: and lastly, EGB, GHD, or EGA, GHC, are called, respectively, the opposite exterior and interior angles; and EGB, CHF, or AGE, DHF, the alternate exterior angles.
PROP. XXX. THEOR.
Straight lines which are parallel to the same straight line are parallel to one
another.
Let AB, CD, be each of them parallel to EF; AB is also parallel to CD.
A
Let the straight line GHK cut AB, EF, CD; and because GHK cuts the parallel straight lines AB, EF, the angle AGH is equal (29. 1.) to the angle GHF. Again, because the straight line GK cuts the parallel straight lines EF, CD, the angle GHF is equal (29. 1.) to the angle GKD: and it was shewn that the angle AGK is equal to the angle GHF; therefore also AGK is equal to GKD; and they are alternate angles; therefore AB is parallel (27. 1.) to CD.
G
B
E
H
F
K
C
D
PROP. XXXI. PROB.
To draw a straight line through a given point parallel to a given straight
line.
Let A be the given point, and BC the given straight line, it is required to draw a straight line through the point A, parallel to the straight line BC.
Because the straight line AD, which meets the two straight lines BC, EF, makes the alternate angles EAD, ADC equal to one another, EF is parallel (27. 1.) to BC. Therefore the straight line EAF is drawn through the given point A parallel to the given straight line BC.
PROP. XXXII. THEOR.
If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are equal to two right angles.
Let ABC be a triangle, and let one of its sides BC be produced to D; the exterior angle ACD is equal to the two interior and opposite angles CAB, ABC; and the three interior angles of the triangle, viz. ABC, BCA, CAB, are together equal to two right angles.
Through the point C draw
B
A
C
E
D
CE parallel (31. 1.) to the straight line AB; and because AB is parallel to CE, and AC meets them, the alternate angles BAC, ACE are equal (29. 1.) Again, because AB is parallel to CE, and BD falls upon them, the exterior angle ECD is equal to the interior and opposite angle ABC, but the angle ACE was shewn to be equal to the angle BAC; therefore the whole exterior angle ACD is equal to the two interior and opposite angles CAB, ABC; to these angles add the angle ACB, and the angles ACD, ACB are equal to the three angles CBA, BAC, ACB; but the angles ACD, ACB are equal (13. 1.) to two right angles; therefore also the angles CBA, BAC, ACB are equal to two right angles.
D
COR. 1. All the interior angles of any rectilineal figure are equal to twice as many right angles as the figure has sides, wanting four right angles. For any rectilineal figure ABCDE can be divided into as many triangles as the figure has sides, by drawing straight lines from a point F within the figure to each of its angles. And, by the preceding proposition, all the angles of these triangles are equal to twice as many right angles as there are triangles, that is, as there are sides of the figure; and the same angles are equal to the angles of the figure, together with the angles at the point F, which is the common vertex of the triangles; that is, (2 Cor. 15. 1.) together with four right angles. Therefore, twice as many right angles as the figure has sides, are equal to all the angles of the figure, to
E
F
A
B
gether with four right angles, that is, the angles of the figure are equal to twice as many right angles as the figure has sides, wanting four. COR. 2. All the exterior angles of any rectilineal figure are together equal to four right angles.
Because every interior angle ABC, with its adjacent exterior ABD, is equal (13. 1.) to two right angles; therefore all the interior, together with all the exterior angles of the figure, are equal to twice as many right angles as there are sides of the figure; that is, by the foregoing corollary, they are equal to all the interior angles of the figure, together with four right angles; therefore all
the exterior angles are equal to four right angles.
COR. 3. Two angles of a triangle being given, or merely their sum, the third will be found by subtracting that sum from two right angles.
COR. 4. If two angles of one triangle are respectively equal to two angles of another, the third angles will also be equal, and the two triangles will be mutually equiangular.
COR. 5. In any triangle there can be but one right angle; for if there were two, the third angle must be nothing. Still less can a triangle have more than one obtuse angle.
COR. 6. In every right-angled triangle, the sum of the two acute angles is equal to one right angle.
COR. 7. Since every equilateral triangle (Cor. 5. 1.) is also equiangular, each of its angles will be equal to the third part of two right angles; so that if the right angle is expressed by unity, the angle of an equilateral triangle will be expressed by of one right angle.
COR. 8. The sum of the angles in a quadrilateral is equal to two right angles multiplied by 4-2, which amounts to four right angles; hence, if all the angles of a quadrilateral are equal, each of them will be a right angle; a conclusion which sanctions the Definitions 25 and 26, where the four angles of a quadrilateral are said to be right, in the case of the rectangle and the square.
COR. 9. The sum of the angles of a pentagon is equal to two right angles multiplied by 5-2, which amounts to six right angles; hence, when a pentagon is equiangular, each angle is equal to the fifth part of six right angles, or of one right angle.
COR. 10. The sum of the angles of a hexagon is equal to 2 × (6—2), or eight right angles; hence, in the equiangular hexagon, each angle is the sixth part of eight right angles, or of one right angle.
SCHOLIUM.
When (Cor. 1.) is applied to polygons, which have re-entrant angles, as ABC each re-entrant angle must be regarded as greater than two right angles.
And, by joining BD, BE, BF, the figure is divided into four triangles, which contain eight right angles; that is, as many times two right angles as there are units in the number of sides diminished by two.
But to avoid all ambiguity, we shall henceforth limit our reasoning to polygons with salient angles, which might otherwise be named convex polygons. Every convex polygon is such that a straight line, drawn at pleasure, cannot meet the contour of the polygon in more than two points.
PROP. XXXIII. THEOR.
The straight lines which join the extremities of two equal and parallel straight lines, towards the same parts, are also themselves equal and parallel.
Let AB, CD, be equal and parallel straight lines, and joined towards the same parts by the straight lines AC, BD; AC, BD are also equal and parallel.
୯
B
D
Join BC; and because AB is parallel A to CD, and BC meets them, the alternate angles ABC, BCD are equal (29. 1.); and because AB is equal to CD, and BC common to the two triangles ABC, DCB, the two sides AB, BC are equal to the two DC, CB; and the angle ABC is equal to the angle BCD; therefore the base AC is equal (4. 1.) to the base BD, and the triangle ABC to the triangle BCD, and the other angles to the other angles (4. 1.) each to each, to which the equal sides are opposite; therefore the angle ACB is equal to the angle CBD; and because the straight line BC meets the two straight lines AC, BD, and makes the alternate angles ACB, CBD equal to one another, AC is parallel (27. 1.) to BD; and it was shewn to be equal to it.
COR. 1. Hence, if two opposite sides of a quadrilateral are equal and parallel, the remaining sides will also be equal and parallel, and the figure will be a parallelogram.
COR. 2. And every quadrilateral, whose opposite sides are equal, is a parallelogram, or has its opposite sides parallel.
For, having drawn the diagonal BG; then, the triangles ABC, CBD, being mutually equilateral (hyp.), they are also mutually equiangular (Th. 8.), or have their corresponding angles equal; consequently, the opposite sides are parallel; namely, the side AB parallel to CD, and BD parallel to AC; and, therefore, the figure is a parallelogram.
COR. 3. Hence, also, if the opposite angles of a quadrilateral be equal, the opposite sides will likewise be equal and parallel.
For all the angles of the figure being equal to four right angles (Cor. 8.
Th. 32.), and the opposite angles being mutually equal, each pair of adjacent angles must be equal to two right angles; therefore, the opposite sides must be equal and parallel.
PROP. XXXIV. THEOR.
The opposite sides and angles of a parallelogram are equal to one another, and the diagonal bisects it; that is, divides it into two equal parts.
N. B. A Parallelogram is a four-sided figure, of which the opposite sides are parallel; and the diameter is a straight line joining two of its opposite angles.
Let ACDB be a parallelogram, of which BC is a diameter; the opposite sides and angles of the figure are equal to one another; and the diameter BC bisects it.
A
C
B
D
Because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD are equal (29. 1.) to one another; and because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal (29. 1.) to one another; wherefore the two triangles ABC, CBD have two angles ABC, BCA in one, equal to two angles BCD, CBD in the other, each to each, and the side BC, which is adjacent to these equal angles, common to the two triangles; therefore their other sides are equal, each to each, and the third angle of the one to the third angle of the other (26. 1.); viz. the side AB to the side CD, and AC to BD, and the angle BAC equal to the angle BDC. And because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, the whole angle ABD is equal to the whole angle ACD: And the angle BAC has been shewn to be equal to the angle BDC: therefore the opposite sides and angles of a parallelogram are equal to one another; also, its diameter bisects it; for AB being equal to CD, and BC common, the two AB, BC are equal to the two DC, CB, each to each; now the angle ABC is equal to the angle BCD; therefore the triangle ABC is equal (4. 1.) to the triangle BCD, and the diameter BC divides the parallelogram ACDB into two equal parts.
COR. 1. Two parallel lines, included between two other parallels, are equal.
COR. 2. Hence, two parallels are every where equally distant.
COR. 3. Hence, also, the sum of any two adjacent angles of a parallelogram is equal to two right angles.
PROP. XXXV. THEOR.
Parallelograms upon the same base and between the same parallels, are equal to one another.
(SEE THE 2d And 3d figures.)
Let the parallelograms ABCD, EBCF be upon the same base BC, and between the same parallels AF, BC; the parallelogram ABCD is equal to the parallelogram EBCF. | 677.169 | 1 |
Difference between Rectangle and Rhombus
People often fail to differentiate between rectangle and rhombus, as both of these are quadrilateral shapes. However, the difference between both of these shapes is significant and with a little help, you will be easily able to identify it yourself. Not to mention that both of these shapes are exceptional cases of the parallelograms. The major difference between rectangle and rhombus is that, the opposite sides of the rectangle are equal. In contrast, a rhombus is a shape having four equal sides but in a diamond shape. All sides of the rectangle are at right angle while this is not the case with rhombus shape.
On the other hand, if you bisect the diagonals of a rectangle, they bisect each other at equal length but the triangles that are formed are called congruent right triangles. In addition, the sections of rectangle's diagonals that are bisected are also of equal length. In contrast, the triangles formed after bisecting the diagonals of a rhombus are called equilateral triangles. The equilateral triangle is a triangle which has all sides of equal lengths and angles.
Not to mention, there is also considerable difference in the internal angles of bisected diagonals. The internal angles of a rhombus are intersected through the diagonals. Similarly, the bisected diagonals of a rectangle produce four internal angles at the right angles.
We can also find considerable difference between these two by moving towards the sides of these shapes. All four sides of the rhombus are equal in length and angle. Using the parallelogram law, we can find out that the sum of square of the diagonals of a rhombus is equal to the four-time square of its side. Similarly, if we take square of the adjacent sides of a rectangle and sum it up, it will be equal to the square of the diagonals of the rectangle.
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Instructions
1
Rectangle:
Having different adjacent sides but equal opposite, all four sides of the rectangle are at the angle of 90 degrees to each other. As the sides are equal, therefore the diagonals of the rectangle are also equal in length. In order to find the area of a rectangle, the individual has to multiple its length by the size of its width.
Image Courtesy: gettyimages.com
2
Rhombus:
Rhombus is a diamond shape parallelogram, which has all four sides of equal length, is considered similar to the shape of a square. Finding the area of a rhombus is same as a parallelogram. | 677.169 | 1 |
Breadcrumb
Mechanical Linkages: Similar Triangles
Students use the properties of similar triangles to explain why an ironing table stays horizontal and how a pantograph enlarges a drawing. They construct models and use dynamic geometry software.
This is a classic reSolve sequence aligned with the Australian Curriculum V8.4. It is only available as a downloadable package.
This unit is part of the special topic "Mechanical Linkages and Deductive Geometry". Mechanical linkages – sets of hinged rods – form the basis of many everyday objects such as folding umbrellas and car jacks and are built using the geometry of triangles and quadrilaterals. They offer rich potential for investigating geometry, starting with real-life objects, then making working models and then using pre-prepared dynamic geometry software. This unit looks at the geometry behind ironing tables and pantographs. The properties of similar triangles help to explain why an ironing table stays horizontal when the height is adjusted, and how a pantograph enlarges a drawing. The two lessons are independent of each other, so either one or both can be taught. They are especially suitable for Year 9.
Lesson 1: Ironing Table
When an ironing table with legs that pivot is raised or lowered, the top always stays parallel to the floor. In this lesson students investigate the triangles formed by the pivoting legs. They make physical models, observe computer simulations and explain with geometry why the table is always horizontal. Students investigate different leg lengths and pivot positions to ensure similar or congruent triangles are formed. Three designs of ironing table are included, involving different geometry.
Lesson 2: Pantograph
Students construct a physical model of an enlarging pantograph and use a computer simulation to explore how the copied image compares with the original drawing. They use their knowledge of parallelogram properties and similar triangles to explain how the pantograph works. By investigating pantographs further, students can extend their understanding of scale factors, generalise earlier findings and can design their own pantographs.
Last updated December 11 2018.
This sequence addresses the following Australian Curriculum V8.4 content descriptions
ACMMG220 | 677.169 | 1 |
Strictly speaking, no, because a semi-regular tessellation must
be based on regular polygons and rhombi are not regular polygons.
However, octagons and rhombi can be used to make a non-regular
tessellation. | 677.169 | 1 |
Chord of the Bigger Circle Is Bisected at the Point of Contact with the Smaller Circle
Question 7Two...
Question
Question 7
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Open in App
Solution
Let the two concentric circles with centre O. AB be the chord of the larger circle which touches the smaller circle at point P. ∴AB is tangent to the smaller circle to the point P. ⇒OP⊥AB
By Pythagoras theorem in Δ OPA, OA2=AP2+OP2 ⇒52=AP2+32 ⇒AP2=25−9 ⇒AP=4
In Δ OPB,
Since OP ⊥ AB,
AP = PB (∵ Perpendicular from the centre of the circle bisects the chord)
AB = 2AP = 2 × 4 = 8 cm ∴ The length of the chord of the larger circle is 8 cm. | 677.169 | 1 |
2D Shape Attributes Boom Cards™ | Sides and Vertices | Geometry
Description: These 2D Shape Attributes Boom Cards are a fun way for your students to show what they know about shapes. This deck has two parts, first, students will be shown a shape and will identify how many sides and vertices that shape has. Next, students will be provided a shape attributes description, including the number of sides and vertices, and will need to choose the shape that matches the description. Students receive immediate feedback and will have opportunities to fix their work if they choose an incorrect answer! Boom Cards require no teacher prep and are self-grading! | 677.169 | 1 |
Geometry
The area A and the perimeter P of an angle cross-section, can be found with the next formulas:
The distance of the centroid from the left edge of the section , and from the bottom edge , can be found using the first moments of area, of the two legs:
We have a special article, about the centroid of compound areas, and how to calculate it. Should you need more details, you can find it here.
Moment of Inertia
The moment of inertia of an angle cross section can be found if the total area is divided into three, smaller ones, A, B, C, as shown in the figure below. The final area, may be considered as the additive combination of A+B+C. However, a more straightforward calculation can be achieved by the combination (A+C)+(B+C)-C. Also, the calculation is better done around the non-centroidal x0,y0 axes, followed by application of the the Parallel Axes Theorem.
First, the moments of inertia Ix0, Iy0 and Ix0y0 of the angle section, around the x0, and y0 axes, are found like this:
Application of the Parallel Axes Theorem makes possible to find the moments of inertia around the centroidal axes x,y:
where, the distance of the centroid from the y0 axis and the distance of the centroid from x0 axis. Expressions for these distances are given in the previous section.
Take in mind, that x, y axes are not the natural ones, the L-section would prefer to bend around, if left unrestrained. These would be the principal axes, that are inclined in respect to the geometric x, y axes, as described in the next section.
Principal axes
Principal axes are those, for which the product of inertia Ixy, of the cross-section becomes zero. Typically, the principal axes are symbolized with I and II and are perpendicular, one with the other. The moments of inertia, when defined around the principal axes, are called principal moments of inertia and are the maximum and minimum ones. Specifically, the moment of inertia, around principal axis I, is the maximum one, while the moment of inertia around principal axis II, is the minimum one, compared to any other axis of the cross-section. For symmetric cross-sections, the principal axes match the axes of symmetry. However, there is no axis of symmetry in an L section (unless for the special case of an angle with equal legs), and as a result the principal axes are not apparent, by inspection alone. They must be calculated, and in particular, their inclination, relative to some convenient geometric axis (e.g. x, y), should be determined.
Knowing the moments of inertia , and the product of inertia , of the L-section, around centroidal x, y axes, it is possible to find the principal moments of inertia , around principal axes I and II, respectively, and the inclination angle , of the principal axes from the x, y ones, with the following formulas:
By definition, is considered the major principal moment (maximum one) and the minor principal moment (minimum one). It follows that: .
Moment of inertia and bending
The moment of inertia (second moment or area) is used in beam theory to describe the rigidity of a beam against flexure. The bending moment M applied to a cross-section is related with its moment of inertia with the following equation:
where E is the Young's modulus, a property of the material, and the curvature of the beam due to the applied load. Therefore, it can be seen from the former equation, that when a certain bending moment M is applied to a beam cross-section, the developed curvature is reversely proportional to the moment of inertia I.
Polar moment of inertia of L-section
The polar moment of inertia, describes the rigidity of a cross-section against torsional moment, likewise the planar moments of inertia described above, are related to flexural bending. The calculation of the polar moment of inertia around an axis z-z (perpendicular to the section), can be done with the Perpendicular Axes Theorem:
where the , , the moments of inertia around axes x-x and y-y, respectively, which are mutually perpendicular to z-z and meet at a common origin.
The dimensions of moment of inertia are .
Elastic section modulus
The elastic section modulus of any cross section around centroidal axis x-x, describes the response of the section under elastic flexural bending. It is defined as:
where , the moment of inertia of the section around x-x axis and , the distance from centroid of a given section fiber (that is parallel to the axis). For the angle section, due to its unsymmetry, the is different for a top fiber (at the tip of the vertical leg) or a bottom fiber (at the base of the horizontal leg). Normally, the more distant fiber (from centroid) is considered when finding the elastic modulus. This happens to be at the tip of the vertical leg (for bending around x-x). Using the possibly bigger , we get the smaller , which results in higher stress calculations, as will be shown shortly after. This is usually preferable for the design of the section. Therefore:
where the "min" or "max" designations are based on the assumption that , which is valid for any angle section.
Similarly, for the elastic section modulus , relative to the y-y axis, the minimum elastic section modulus is found with:
where the "min" designation is based on the assumptions that , which again is valid, for any angle section.
If a bending moment is applied on axis x-x, the section will respond with normal stresses, varying linearly with the distance from the neutral axis (which under elastic regime coincides to the centroidal x-x axis). Over the neutral axis the stresses are by definition zero. Absolute maximum stress will occur at the most distant fiber, with magnitude given by the formula:
From the last equation, the section modulus can be considered for flexural bending, a property analogous to cross-sectional A, for axial loading. For the latter, the normal stress is F/A.
The dimensions of section modulus are .
Plastic section modulus
The plastic section modulus is similar to the elastic one, but defined with the assumption of full plastic yielding of the cross section, due to flexural bending. In that case, the whole section is divided in two parts, one in tension and one in compression, each under uniform stress field. For materials with equal tensile and compressive yield stresses, this leads to the division of the section into two equal areas, , in tension and , in compression, separated by the neutral axis. This axis is called plastic neutral axis, and for non-symmetric sections, is not the same with the elastic neutral axis (which again is the centroidal one). The plastic section modulus is given by the general formula:
where the distance of the centroid of the compressive area from the plastic neutral axis and the respective distance of the centroid of the tensile area .
Around x axis
For the case of an angle cross-section, the plastic neutral axis for x-x bending, can be found by either one of the following two equations:
which becomes:
where , the distance of the plastic neutral axis from the bottom end of the section. The first equation is valid when the plastic neutral axis passes through the vertical leg, while the second one when it passes through the horizontal leg. Generally, it can't be known which equation is relevant beforehand.
Once the plastic neutral axis is determined, the calculation of the centroids of the compressive and tensile areas becomes straightforward. For the first case, that is when the axis crosses the vertical leg, the plastic modulus can be found like this:
which becomes:
where .
For the second case, that is when the axis passes through the horizontal leg, the plastic modulus is found with equation:
which can be simplified to:
where .
Around y axis
The plastic section modulus around y axis can be found in a similar way. If we orient the L-section, so that the vertical leg becomes horizontal, then the resulting shape is similar in form with the originally oriented one. Thus, the derived equations should have the same form, as found for the x-axis. We only have to swap for and vice-versa. This way, the exact position of the plastic neutral axis is given by the following formula:
where , the distance of the plastic neutral axis from the left end of the section. The first equation is valid when the plastic neutral axis passes through the horizontal leg, while the second one when it passes through the vertical leg (see figure below).
For the first case, that is when the y-axis crosses the horizontal leg, the plastic modulus is found by the formula:
where .
For the second case, that is when the y-axis crosses the vertical leg, the plastic modulus is found by the formula:
where .
Radius of gyration
Radius of gyration Rg of a cross-section, relative to an axis, is given by the formula:
where I the moment of inertia of the cross-section around the same axis and A its area. The dimensions of radius of gyration are . It describes how far from centroid the area is distributed. Small radius indicates a more compact cross-section. Circle is the shape with minimum radius of gyration, compared to any other section with the same area A.
Angle (L) section formulas
The following table, lists the main formulas for the mechanical properties of the angle (L) cross section.
Angle (L) section formulas
Quantity
Formula
Area:
Perimeter:
Centroid:
Moments of inertia around centroid
Principal axes and moments of inertia:
Elastic modulus:
Plastic modulus:
Plastic neutral axis:
(distances from bottom or left)
where:
Related pages
Properties of a Rectangular TubeProperties of I/H sectionProperties of unequal I/H sectionMoment of Inertia of an AngleAll Cross Section tools
Their L-shaped cross-section and sturdy composition make them important for a wide range of applications, including framing and reinforcement, bridge construction, and warehouse support. In this comprehensive guide, we will go over the fundamentals of steel angles, including their types, sizes, applications, and rates.
Use the section from your Hood design and calculate the section modulus using the formula S= I/y. Section Modulus: The section modulus (S) is geometry property of the cross section used for designing beams and flexural members. It does not represent anything physically.
In essence, the neutral axis is a line produced by the intersection of the undeformed cross-section plane and the deformed surface as shown in Fig. 3.23C. This produces a surface of zero strain oriented along the longitudinal axis of the beam-type member.
Depending upon the relationship between the plane and the slant surface, the cross-section or also called conic sections (for a cone) might be a circle, a parabola, an ellipse or a hyperbola. From the above figure, we can see the different cross sections of cone, when a plane cuts the cone at a different angle.
The major cross section elements considered in the design of streets and highways include the pavement surface type, cross slope, lane widths, shoulders, roadside or border, curbs, sidewalks, driveways, and medians.
They are formed by bending a single angle in a piece of steel. Angle Steel is 'L' shaped; the most common type of Steel Angles are at a 90 degree angle. The legs of the "L" can be equal or unequal in length. Steel angles are used for various purposes in a number of industries.
For the circular shapes, Sx = Ix/R (Figures 1.48c and 1.48d). In each case, the moment of inertia is divided by half the cross-sectional height, or thickness. From Equations 1.7 and 1.10, it can be seen that the section modulus for a rectangular cross section is Sx = (BH3/12)/(H/2) = BH2/6.
The following steps outline how to calculate the area of an L-shaped figure using the formula A = (l1 * w1) + (l2 * w2). First, determine the length of the first rectangle (l1) in units. Next, determine the width of the first rectangle (w1) in units. Next, determine the length of the second rectangle (l2) in units.
The section modulus (Z) of the cross-sectional shape is significant in designing beams. It is a direct measure of the strength of the beam. A beam that has a larger section modulus than another will be stronger and capable of supporting greater loads.
Resisting the applied Bending Moment acting on the section. The importance of knowing how to find the neutral axis is if you want to calculate the exact stress/strain at every point of the beam section.
Neutral axis is perpendicular to the plane of the loads. If stresses are linear and within the yield stress, the neutral axis passes through the centroid (that is, the neutral axis is one of the centroidal axes).
By definition, the section modulus (Sx) of a beam with a symmetric section equals its second moment of area divided by half its depth at the extreme fiber. The section modulus will help determine the cross-section shape of a beam as discussed in the Chapter 9.
Section properties involve the mathematical properties of structural shapes. They are of great use in structural analysis and design. Note that these properties have nothing to do with the strength of the material, but are based solely on the shape of the section.
A cross section is the shape that is created by cutting straight through a figure. Cross sections can be anything from points to lines to two-dimensional shapes. Parallel and perpendicular are used to describe the direction of cross sections. | 677.169 | 1 |
Hint: Trisection means dividing a line segment in three equal parts or dividing a line segment in the ratio 1:2 and 2:1 internally.In this question we have been given the ratio is 2:1. Start by considering 2 points which trisects the given line. | 677.169 | 1 |
NCERT Solutions For Class 6 Maths, Chapter 4, Exercise 4.3
NCERT Solutions For Class 6 Maths, Chapter 4, Basic Geometrical Ideas, Exercise 4.3 is all about study of points. A point determines a location and it is usually denoted by a capital letters in given figure. Exercise 4.3 class 6 maths, Basic Geometrical Ideas has total three questions to study. | 677.169 | 1 |
Area of a Triangle
This article was ported from my old Wordpress blog here, If you see any issues with the rendering or layout, please send me an email.
Problem: What is the area of the triangle within the rectangle?
Solution: In a moment of inspiration, we draw the following additional line:
Now the answer is obvious. Once we split the rectangle into two smaller rectangles, the sides of the triangle become diagonals of their respective rectangles. The diagonals obviously split each of the two smaller rectangles into halves, where one half lies inside our original triangle. Clearly then, the interior of the whole triangle is half the area of its enclosing rectangle.
Notice that this proof holds no matter where the apex of the triangle is within the rectangle.
This proof is probably the simplest example of an elegant proof. While at first the answer is not trivial, the mathematician introduces different perspective. Here, we realize that the area of a complex figure can be determined from the sum of the areas of its parts, and that we are free to choose those parts as best suits our proof. Additional such inspirations, which experienced mathematicians call "methods," include picking nice lines of symmetry, rotational axes, or translations of our figures to make a statement become trivial. We invite the reader, in traditional mathematical fashion, to come up with another proof of the area of a triangle using one of these different methods. Remember, the art is in the argument. | 677.169 | 1 |
What Is Sine, Cosine, and Tangent؟
Immerse yourself in the realm of trigonometry! Uncover the concepts of sine, cosine, and tangent - the fundamental elements for solving right triangles.
You have likely noticed the buttons on your calculator and may have played around with them before. In contrast, if you are reading this now, your curiosity has been piqued, and you couldn't resist finding out more.
What are the definitions of sine, cosine, and tangent?
Welcome to the wonderful world of trigonometry, where the fascinating properties of angles and triangles are explored.
Imagine being on top of a mountain, enjoying the beautiful view, or watching an airplane flying through the clear blue sky. These simple pleasures can be understood using the three fascinating mathematical functions of trigonometry - sine, cosine, and tangent.
Let's get into it.
The Fantastic Trio of Trigonometry: Sine, Cosine, and Tangent
Think of trigonometry like an old treasure chest brimming with valuable ideas that continue to be highly respected. The trio of sine, cosine, and tangent are considered the most valuable gems within this treasure chest.
Think of a triangle as a family unit, with one side representing the parent and the other two sides representing the children.
The sine of an angle can be defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse, while the cosine of an angle is the ratio of the adjacent side to the hypotenuse.
The tangent (tan) represents the relationship between sin and cos, similar to the bond between two siblings.
The trigonometric functions were first developed by ancient civilizations such as the Babylonians and Greeks, who recognized the special properties of triangles in solving problems related to astronomy, navigation, and construction.
Over time, these basic ideas evolved into the advanced field of trigonometry we know today, giving us a fresh perspective on angles and triangles.
In summary, let's ensure we are all in agreement.
Sine (sin): Picture a triangle as a mountain, with the summit as an acute angle. The sine of this angle is the ratio of the mountain's height (opposite side) to the distance from the summit to the base camp (hypotenuse).
In a right-angled triangle, the cosine of an angle represents the distance from the base to the hypotenuse, similar to the distance from the foot of a mountain to the base camp.
Tangent (tan) is a well-known mathematical ratio that compares the height of a mountain to its base (opposite side to adjacent side). It is also equal to the ratio of sine to cosine.
These trigonometric functions are not just impressive because of their connection to triangles, but they are also fundamental in various fields such as physics, engineering, computer graphics, and signal processing. They are Vital tools in our scientific arsenal.
Trigonometry is a valuable tool for mathematicians with many functions.
Trigonometry is incredibly versatile, playing a Vital role in various fields such as architecture, technology, and finance. The trio of sine, cosine, and tangent have been instrumental in driving numerous advancements and breakthroughs.
Think about architects who create stunning buildings with accuracy and imagination. They rely on trigonometry (sine and cosine) to build roof structures and determine the slope of a staircase.
In the advanced realm of 3D modeling, trigonometric functions determine the movement and transformation of objects in the digital world.
Trigonometry is not only used in architecture but also applies to various other scientific disciplines.
Sine, cosine, and tangent have played a significant role in shaping our knowledge of the universe and how we engage with it, from building pyramids to exploring Mars with rovers.
The Enduring Legacy of Trigonometry
The ancient civilizations of Babylon, Greece, and India have left a lasting impact on the field of trigonometry. They were the first to understand the significance of triangles and use trigonometry to address problems in various fields such as astronomy, navigation, and construction.
Early Greek mathematicians such as Hipparchus created initial forms of the sine function, and later scholars like Ptolemy built upon this work to establish spherical trigonometry as a field of study.
Trigonometry progressed as various individuals, including Islamic scholars, Indian mathematicians, and Renaissance scholars like Regiomontanus, contributed to its development.
Even in the present day, the influence of trigonometry's key players - sine, cosine, and tangent - can still be seen in designing impressive buildings, making advancements in space technology, and developing innovative computer graphics.
Conclusion
To sum up, sine, cosine, and tangent function as important tools in trigonometry, similar to the keys on a grand piano that come together to create a beautiful harmony. They have significantly influenced our comprehension of the world, simplifying difficult problems and uncovering the concealed patterns in our surroundings.
The an Important roles of mathematics showcase its flexibility, being utilized in a variety of fields such as architecture, physics, and computer graphics.
Their tale demonstrates the strength of human curiosity and inventiveness, proving that the mathematical concepts of sine, cosine, and tangent will always have an impact on future innovations and discoveries.
Let's honor these unrecognized heroes as we keep exploring the exciting world of trigonometry | 677.169 | 1 |
Geometry and trigonometry are two pivotal branches of mathematics that often interlock in their study of shapes, sizes, and the properties of space. In my exploration of these subjects, I've come to appreciate how geometry provides a broad canvas, addressing various figures and spatial relationships, while trigonometry focuses intently on the specific properties of triangles, particularly the right-angled ones.
Both play crucial roles in applications ranging from engineering to physics and understanding their differences not only enriches my mathematical knowledge but also enhances my practical problem-solving skills.
Join me as we uncover the distinct nature of each field and how they contribute to our grasp of the mathematical world around us – you might just see the angles and lines of your environment in a whole new light.
Main Differences Between Geometry and Trigonometry
The main differences between Geometry and Trigonometry are based on their scope, application, and the specific aspects of shapes they examine. While both fields are branches of mathematics that deal with shapes and space, they focus on different elements and principles.
Scope of Study:
Geometry: Encompasses the properties and relations of points, lines, surfaces, and solids.
Geometry: Used in a variety of contexts, from architectural design to the computation of spatial areas.
Trigonometry: Essential in fields that require angle measurements, such as surveying, astronomy, and physics.
Understanding these distinctions helps us appreciate how each discipline contributes uniquely to the field of mathematics and its real-world applications.
Fundamental Concepts of Geometry and Trigonometry
Geometry is a branch of mathematics that I understand to be concerned with the properties and relations of points, lines, surfaces, solids, and higher-dimensional analogs. It's broadly categorized into two:
Plane Geometry: Deals with shapes such as circles, triangles, rectangles, squares, and polygons that can be drawn on a flat surface.
Geometry uses Euclid's axioms as foundational principles, which serve as the bedrock of theorems about the attributes of shapes and figures. It involves various measurements like:
Property
Measurement
Length
Meters, Feet
Area
Square units
Volume
Cubic units
In geometry, I apply algebra and arithmetic to calculate dimensions, while the position and properties of space occupy a central role, defining the size and shape of objects.
Trigonometry, on the other hand, specifically focuses on the study of triangles, engaging with their angles and sides. The fundamental concepts here involve the trigonometric functions, which relate the ratios of sides of right triangles to their angles. These functions include:
Sine (sin)
Cosine (cos)
Tangent (tan)
Cotangent (cot)
Secant (sec)
Cosecant (csc)
These tools allow me to determine unknown lengths and angles of triangles. Here's a brief overview of how these functions look like in a right-angled triangle context:
Function
Description
Ratio
Sine
Opposite side over the hypotenuse
sin(θ) = o/h
Cosine
Adjacent side over the hypotenuse
cos(θ) = a/h
Tangent
Opposite side over adjacent side
tan(θ) = o/a
The connection between geometry and trigonometry is fundamental as trigonometry is a subset of geometric principles, focusing narrowly yet profoundly on the properties of triangles, a shape essential in the geometric universe.
Applications of Geometry and Trigonometry
Geometry and trigonometry are fundamental in various fields. In construction, geometry aids in determining the material requirements and spatial dimensions for buildings. Trigonometry is critical for calculating angles and distances, making it invaluable for creating stable structures.
When it comes to architecture, I apply geometry to design aesthetically pleasing and functional spaces. Trigonometry allows me to understand how to best use light and shadows within a design, enhancing the visual impact and functionality of the spaces I create.
In surveying, my tasks often involve measuring large tracts of land. Utilizing trigonometry, I calculate distances and angles between points on a plot, which is essential for accurate mapping and planning.
Navigation relies heavily on trigonometry as well. When I navigate, I often use it to find the shortest path between two points on a spherical surface, like the Earth, which is vital for air and sea travel.
Astronomy uses both branches to observe and interpret the universe's wonders. Geometry helps me understand the shapes and relative positions of celestial bodies, while trigonometry is vital for calculating their distances and movements.
In physics, these mathematical fields are indispensable. They allow me to model forces, motion, and energy patterns. Trigonometry, in particular, is integral in wave functions and harmonic motion.
Lastly, probability in both geometry and trigonometry enhances my understanding, such as predicting outcomes based on geometric distributions.
Field
Geometry Use
Trigonometry Use
Construction
Material and space calculation
Angle and distance measurements
Architecture
Designing spaces and structures
Optimizing light and shadow
Surveying
Accurate mapping and land measurements
Calculating distances and plot angles
Navigation
Plotting courses on spherical surfaces
Determining shortest paths
Astronomy
Interpreting shapes and positions of celestial bodies
Calculating distances and movements
Physics
Modeling forces and structures
Analyzing wave functions and motion
Probability
Predicting geometric distributions
Assessing outcomes based on angles
Geometry and trigonometry serve as foundational tools across these varied applications, each complementing the other to address complex problems in practical, real-world scenarios.
Conclusion
In summarizing, I find that geometry and trigonometry serve as two distinct, yet interrelated, branches of mathematics. Geometry is the broader discipline, focusing on the properties and relations of points, lines, shapes, and spaces. It's here where we explore the expanse of forms – from the simplest line to the most complex n-dimensional shapes.
On the other hand, trigonometry hones in on the measurement of triangles, particularly the right-angled ones. It's a powerful tool, which I've come to appreciate, for understanding the relationships between the sides and angles of these three-cornered figures. Trigonometry often proves indispensable in fields such as engineering and physics due to its precise nature in dealing with angles and distances.
I've learned that while one can study geometry without delving deep into trigonometry, understanding the latter requires a good grasp of the geometric fundamentals. They complement each other, with trigonometry often considered a subset of the larger geometric framework.
In my study and application of these mathematical fields, I've observed the unique role each plays in both theoretical and practical contexts. Their significance is undeniable in shaping the way we understand the world around us, construct buildings, navigate across the oceans, and even send satellites into space. | 677.169 | 1 |
A Royal Road to Geometry: Or, an Easy and Familiar Introduction to the ...
Right Angles, at the Center (Def. 11.) the Arks AD, DB, being each a fourth part of the whole Circumference, or half the Semi-circumference: hence, a Right Angle is faid to be of 90 Degrees.
3. If the Ark AD be bifećted in E, and EC be drawn, the Angles ACE, ECD, will be each of 45 deg. half ACD a Right Angle, or 90 Degrees.
And, if the Ark DB be trifected, at F & G, (i. e. divided into three equal parts), and FC be drawn, the Ark DF containing 30, and FGB 60 deg. the Angle DCF is faid to be an Angle of 30, and FCB an Angle of 60 deg. By which means, an Angle of any quantity may be obtained, or measured.
4. On the fame Center, C, with any other Radius, as Ca, let the Arch aed fb be drawn, which is alfo a Semicircle.
It is very obvious, that it is alfo divided into the fame number of parts, and in the fame proportion, as the Arch AEDFB; for it is bifected in d, and ad is again bisected in e, and db is alfo trifected at f and g; wherefore, AD, ad are each a fourth; ED, ed an eighth; BF, bf, a fixth; and FD, fd, a twelfth part of their respective Circles; and the Angles ACD, aCd; ECD, eCd, &c. are the fame in both.
From all which, it is clear, that, Angles may be formed or measured by an Ark or Circle of any Radius. And alfo, that equal Arks of the fame, or of equal Circles, or an equal number of degrees in a Circle of any Radius, will form equal Angles at the Center.
If you would have an Angle of 60 degrees at the point C, of the line BC.
With
With any Radius, at pleasure, defcribe the Ark BD, cutting the Line BC in B; with the fame Radius, on the Center B, cut the Ark BD at F, and draw CF.
It is very clear, that the Angle BCF would be the fame, if a lefs Radius had been taken, as C b.
For, draw the Chord Lines F B and fb, each will be equal to the Radius of its refpective Circle; and, the Triangles CFB, Cfb are equilateral; whofe Angles are of 60 degrees each (Cor. 1. 9. 1.); for the Arks BF and bf are, each a fixth part of the whole Circumference of their respective Circles, of which C is the common Center, (fee Prop. 11. 4.); and confequently, each contains 60 degrees on the circumference of that Circle, of which it is a Part; whofe Radius is CB or C b.
A defcription of the Inftrument called a PROTRACtor, with the application of it, in measuring and making Angles, of any known quantity or measure, may not be improper in this place. It is of fpecial ufe in Surveying, in drawing Plans of any piece of Ground for building on, &c. or of Buildings, already erected, being readier and more exact than a Line of Chords.
The Protractor is a Semicircle, divided on its Limb or Semi-circumference, AEDFB, into 180 equal parts; having a fmall Notch at C, the Center. See the laft Figure. Some Protractors have a Scale, added to the Semicircle, which are the beft and readieft in ufe.
In measuring an Angle, apply the Diameter, i. e. the Edge or Right Line AB, to either Side of the Angle ACE, with the Vertex, C, of the Angle, at the center of the Protractor; and, where-ever the Side CE, cuts the Limb or circular edge of the Inftrument, obferve how many De grees there are from A to E, the Ark intercepted between the Sides AC and CE, of the Angle ACE.
If
If it contains 45 or 50, or whatever number of Degrees it happens to be, (as 45 by the Figure) the Angle ACE is of fo many Degrees. If it had cut the Arch at D, as ACD, it is a Right Angle; and if beyond D, as ACF, it is obtufe ; the Complement FCB, i. e. the Ark FB, being fubtracted from the Arch of the Semicircle, ADB, or 180 degrees, gives the quantity of the Angle ACF.
2. If it is required to lie down or make an Angle of some known Quantity (as 45 deg.) at the Point C, of the Line AB.
Apply the edge of the Protractor, as above, with the Center, C, at the Point given; make a Mark or Point at E; take away the Inftrument, and draw E C.
Thus, may any Angle whatever be laid down on Paper.
A Scale of equal Parts is nothing more than a Right Line divided into any number of equal Parts, at pleasure.
Each Part may reprefent any measure you please, as an Inch, a Foot, a Yard, &c.; for, being equal, whatever measure any Object or Figure contains, in length and breadth, a fimilar Figure may be conftructed, on a Plane, having or containing the fame number of Divifions, each way, on the Scale, as the real Object contains of Feet, Yards, &c. One of thofe Parts is generally fubdivided into parts of the next inferior denomination, or into tenths and hundredths, denoting the Decimal Parts.
Obferve, that the Divifions on the Scale, (whatever measure is reprefented by them,) muft always be adapted to the Proportion you would delineate any Object, or form a Defign. See the Appendix (Page 15 and 16) for the conftruction of Scales.
N. B. A pair of Compaffes or Dividers, a Drawing Pen, and a ftreight Ruler, are all the Utenfils that are requifite, in Plane Geometry.
The Board or Paper, on which we draw any geometrical Figure, is fuppofed to be a Plane.
ABBREVIATIONS &c. EXPLAINED.
When, what is fup
And all the reft. posed to follow may be readily understood. Id eft. That is. When, what has been faid requires to be further explained.
&c.
Et cætera.
i. e.
viz.
e. g.
Videlicet.
To wit; or, that is to say. When any thing advanced is given in Grofs, which is more particularly specified, as follows after. Exempli gratia. For instance; or, for the fake of example. When an Example is to be given of what is advanced.
N. B. Nota Bene.
Mark well. That is, take particular
notice of that Paragraph.
QE. F. Quod erat faciendum. Which was required to
be done.
Diag.
Diagonal.
Par. AB.
or P.AB.
Rect. AB.
}ParallelogramAB.
Rectangle AB, or ABCD, &c.
R. Ang. ABC. Right Angle ABC, &c.
Trap.
Trapezium.
Pent.
Pentagon, &c.
Note. When it is required to join any two Points, it is meant, that a Right Line be drawn between them, i. e. from one Point to the other.
AB
ABBREVIATIONS, by way of Reference.
Poft. 1. or 2. Refers to the firft or second Poftulate; where it is requested that fuch or fuch things may
Def. 6. or 7. Or, Def.6. 3. or 6. 5. &c.
Ax. 3.
be granted.
Refers to the fixth or feventh Definition in the general Introduction. But, when there are two numbers, it refers to the 6th Definitions of the 3d or 5th Book.
Refers to the third Axiom for Demonftration, arifing from felf-evident properties of things.
2. Th. P. A. Refers to the fecond Article in the Theory of Plane Angles, for illuftration or Proof. Pr. 1. or 2. &c. Refers to the firft or fecond Problem, for the conftructing of fome Figure, &c.
P. 1. or 2. &c. Refers to the first or fecond Proposition of that Book, for proof of the Affertion. To the fecond Propofition of the third Book. To the 20 Corollary, of the 4th Propofition of the first Book.
P. 2. 3. &c.
C. 2. 4. I.
Hyp.
Sup.
Con.
That the thing is fo by the Hypothefis, or is
given in the Premifes.
That it is fo by Suppofition, only.
That it is fo by Construction; i. e. the thing was formed or made fo.
Conf. confequently. Th. therefore. Wh. wherefore, it is fo.
Note. When there is but one Number within Parenthesis or otherwife, as (15.) &c. it refers to the fifteenth Problem or Theorem of that fame Book. But if there be two Numbers, as (10. 1.) or (12. 3.) &c. it refers to the 10th Prop. of the first Book, or to the 12th of the third, &c. | 677.169 | 1 |
Right Triangle Trig Worksheet
Right Triangle Trig Worksheet. Corbettmaths – angles in degrees and radians, discover the coterminal angles for the indicated angles, and optimistic and unfavorable coterminal angles with this assemblage of reference and coterminal angles worksheets.
This is a two-page, 6-question worksheet that involves angles of elevation and depression, SOH-CAH-TOA and proper triangles in word problems. A mixture of sin, cos & tan drilling exercise. A few isosceles triangles at the end as extension.
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Cos
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Follow With Sin
This worksheet stack consists of ample workout routines to follow conversion between levels, minutes and seconds. We are a search engine for worksheets on the internet – like google/bing.
Conditional trigonometric equations stand true for some substitute values. Trigonometric solutions that fall in the interval 0 ≤ x ≤ 2π are known as the first options.
We do not host any worksheets on our webservers unless acknowledged so or we now have the permission of the original writer of the worksheet to host or it was created in-house. Search 50,000+ worksheets, curated by experts, created by academics and aligning to mainstream curriculums. Nagwa is an academic expertise startup aiming to help academics train and college students be taught.
In this worksheet, we'll follow discovering and specific the values of the three trigonometric ratios—sine, cosine, and tangent—for a given angle in a proper triangle. This handout offers college students with an outline for notes that the instructor can guide them by way of.
Proper Triangles And Trigonometry Graphic Organizer
Included here are fundamental identities like quotient, reciprocal, cofunction and Pythagorean identities, sum and difference identities, sum-to-product, product-to-sum, double angle and half angle identities and ample trig expression to be simplified, proved and verified utilizing the trigonometric formulas. Determine the reference angles in levels and radians, discover the coterminal angles for the indicated angles, and optimistic and adverse coterminal angles with this assemblage of reference and coterminal angles worksheets.
Interactive sources you'll be able to assign in your digital classroom from TPT. Practice with Sin/Cos/Tan Practice using the Sin/Cos/Tan buttons on your calculator.
Pupils practise figuring out the graphs for sine, cosine and tangent capabilities earlier than utilizing them to resolve equations and to determine transformations on this glorious trigonometric worksheet. From a given record of equations, pupils establish and choose the right reply to align to every graph. Engage your college students with this interactive drag and drop exercise for Google Slides to strengthen understanding of proper triangle trigonometry.
Find the Indicated Side / Perimeter of a Right Triangle Using Trigonometric Ratios. Utilize this array of two-part worksheets, whose first includes workout routines to search out the labeled facet and the 2nd deals with discovering the perimeter. The printables can be found in customary and metric items.
1) u + v 1) 2) u + w 2) 3) v + w 3) 4) u – v 4) 5) w – v 5) Find the magnitudes of the horizontal and vertical components for the vector v with the given … Displaying all worksheets related to – Trigonometry Triangles. To specifically and accurately measure the size of an angle in degrees, it is additional broken down into degrees, minutes and seconds.
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Corbettmaths – This is part 3 of 3 of a trigonometry evaluation. It covers the inverse trigonometric capabilities after which discovering missing angles of right angled tr.
Trigonometric identities help in simplifying trigonometric expressions. Trigonometric identities involving the Pythagorean theorem are probably the most commonly used ones.
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Draw the indicated angle on the coordinate plane, measure the angles in the quadrant and characterize as levels and radians and a lot more. This is a worksheet working angle of elevation and angle of melancholy word issues. The reply to the riddle is "Fo' Drizzle" which permits for simply checking scholar solutions.
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Students will use inverse trigonometric features to solve for 16 missing angles. This is a great exercise for breakout rooms, an task check, exit card or exit ticket, verify for understanding, quiz and more! Assign straight through your Google Classroom or create a Google LTI project for Canvas LMS.
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With this set of evaluating trigonometric capabilities worksheets at your disposal, you have not any dearth of follow workout routines. Begin with substituting the specified x-values in trigonometric capabilities and clear up for f. Allied angle worksheets here enclose exercises like discovering the precise worth of the trigonometric ratio offering angle measures in degrees or radians, evaluating trig ratios of allied angles and proving the trigonometric statements to mention just some. | 677.169 | 1 |
A circle's center is at #(2 ,4 )# and it passes through #(7 ,6 )#. What is the length of an arc covering #(15pi ) /8 # radians on the circle?
If the circle has a center at #(2,4)# and passes through #(7,6)# then it has a radius of
#color(white)("XXX")r=sqrt((7-2)^2+(6-4)^2)=sqrt(25+4) =sqrt(29)#
and a diameter of
#color(white)("XXX")d=2r=2sqrt(29)#
The circumference of the circle would be
#color(white)("XXX")"Circumference"_circ=pid = 2sqrt(29)pi#
The complete circle's circumference is covered by an arc of #2pi=(16pi)/8#
An arc of #(15pi)/8# will cover #((15pi)/8)/((16pi)/8)=15/16# of #"Circumference"_circ#
To find the length of an arc covering ( \frac{15\pi}{8} ) radians on the circle, we need to first find the radius of the circle using the given points.
Given:
Center of the circle: ( (2, 4) )
Point on the circle: ( (7, 6) )
Using the distance formula between two points, we can find the radius:
[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ]
Substituting the given points:
[ \text{Distance} = \sqrt{(7 - 2)^2 + (6 - 4)^2} ]
[ = \sqrt{5^2 + 2^2} ]
[ = \sqrt{25 + 4} ]
[ = \sqrt{29} ]
Now, the length of an arc on a circle is given by:
[ \text{Length of Arc} = r \times \text{angle in radians} ]
Given:
( \text{Angle} = \frac{15\pi}{8} )
( r = \sqrt{29} )
[ \text{Length of Arc} = \sqrt{29} \times \frac{15\pi}{8} ]
[ = \frac{15\pi \sqrt{29}}{8} ]
So, the length of an arc covering ( \frac{15\pi}{8} ) radians on the circle is ( \frac{15\pi \sqrt{29}}{8 | 677.169 | 1 |
Page 1Page 2Page 3Page 4Page 5212
MATHEMATICS
Therefore AQ
2
= AN
2
+ NQ
2
... (2)
From (1) and (2), we have
PQ
2
= PA
2
+ AN
2
+ NQ
2
Now PA = y
2
– y
1
, AN = x
2
– x
1
and NQ = z
2
– z
1
Hence PQ
2
= (x
2
– x
1
)
2
+ (y
2
– y
1
)
2
+ (z
2
– z
1
)
2
Therefore PQ =
2
1 2
2
1 2
2
1 2
) ( ) ( ) ( z z y y x x - + - + -
This gives us the distance between two points (x
1
, y
1
, z
1
) and (x
2
, y
2
, z
2
).
In particular, if x
1
= y
1
= z
1
= 0, i.e., point P is origin O, then OQ =
2
2
2
2
2
2
z y x + +
,
which gives the distance between the origin O and any point Q (x
2
, y
2
, z
2
).
Example 3 Find the distance between the points P(1, –3, 4) and Q (– 4, 1, 2).
Solution The distance PQ between the points P (1,–3, 4) and Q (– 4, 1, 2) is
PQ =
2 2 2
) 4 2 ( ) 3 1 ( ) 1 4 ( - + + + - -
= 4 16 25 + +
=
45
=
3 5
units
Example 4 Show that the points P (–2, 3, 5), Q (1, 2, 3) and R (7, 0, –1) are collinear.
Solution We know that points are said to be collinear if they lie on a line.
Now, PQ = 14 4 1 9 ) 5 3 ( ) 3 2 ( ) 2 1 (
2 2 2
= + + = - + - + +
QR = 14 2 56 16 4 36 ) 3 1 ( ) 2 0 ( ) 1 7 (
2 2 2
= = + + = - - + - + -
and PR = 14 3 126 36 9 81 ) 5 1 ( ) 3 0 ( ) 2 7 (
2 2 2
= = + + = - - + - + +
Thus, PQ + QR = PR. Hence, P, Q and R are collinear.
Example 5 Are the points A (3, 6, 9), B (10, 20, 30) and C (25, – 41, 5), the vertices
of a right angled triangle?
Solution By the distance formula, we have
AB
2
= (10 – 3)
2
+ (20 – 6)
2
+ (30 – 9)
2
= 49 + 196 + 441 = 686
BC
2
= (25 – 10)
2
+ (– 41 – 20)
2
+ (5 – 30)
2
= 225 + 3721 + 625 = 4571
2024-25
Ans. Three-dimensional geometry is a branch of mathematics that deals with the study of objects in three-dimensional space. It includes the analysis of points, lines, planes, and shapes in three dimensions, using concepts such as distance, angles, and coordinates.
2. How is three-dimensional geometry different from two-dimensional geometry?
Ans. Two-dimensional geometry deals with objects in a plane, whereas three-dimensional geometry involves objects in space. In two-dimensional geometry, we consider points, lines, and shapes on a flat surface, while in three-dimensional geometry, we consider objects that have length, width, and height.
3. What are the main components of three-dimensional geometry?
Ans. The main components of three-dimensional geometry are points, lines, and planes. A point represents a specific position in space, a line is a collection of points that extends infinitely in two opposite directions, and a plane is a flat surface that extends infinitely in all directions.
4. How do we calculate distance between two points in three-dimensional space?
Ans. To calculate the distance between two points (x₁, y₁, z₁) and (x₂, y₂, z₂) in three-dimensional space, we can use the distance formula:
Distance = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]
This formula is derived from the Pythagorean theorem by extending it to three dimensions.
5. How can we determine the equation of a plane in three-dimensional geometry?
Ans. To determine the equation of a plane in three-dimensional geometry, we need to know a point on the plane and the normal vector to the plane. If a point (x₁, y₁, z₁) lies on the plane and the normal vector to the plane is represented by (a, b, c), then the equation of the plane is given by:
a(x - x₁) + b(y - y₁) + c(z - z₁) = 0
This equation represents all the points (x, y, z) that lie on the planeImport | 677.169 | 1 |
If $$3 \hat{j}, 4 \hat{k}$$ and $$3 \hat{j}+4 \hat{k}$$ are the position vectors of the vertices $$A, B, C$$ respectively of $$\triangle A B C$$, then the position vector of the point in which the bisector of $$\angle \mathrm{A}$$ meets $$\mathrm{BC}$$ is
A
$$\frac{5}{3} \hat{\mathrm{j}}-4 \hat{\mathrm{k}}$$
B
$$5 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}$$
C
$$5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$$
D
$$\frac{5}{3} \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$$
4
MHT CET 2021 22th September Evening Shift
MCQ (Single Correct Answer)
+2
-0
If the vectors $$2 \hat{i}-\hat{j}-\hat{k} ; \hat{i}+2 \hat{j}-3 \hat{k}$$ and $$3 \hat{i}+\lambda \hat{j}+5 \hat{k}$$ are coplanar, then the value of $$\lambda$$ is | 677.169 | 1 |
R2 – Geometry and Trigonometry
This post covers the use of geometric angles and trigonometric functions in R2 Wave Numbers.
R2 Geometric Angles
Wikipedia defines a geometric angle as 'The figure formed by two rays, called the sides of the angle, sharing a common endpoint, called the vertex of the angle.' It is the amount of turn between the two rays. In the same way as for classical math, Wave Numbers measures angles in degrees or radians .
Wikipedia defines a radian as 'one radian is the angle subtended at the centre of a circle by an arc that is equal in length to the radius'.
In the Wave Number unit circle the length of the radian arc is 1^. A circle represents 360o and the circumference of a circle is 2πr^, so radians convert to angles by multiplying by 180o and dividing by π^. A ^ radian is a counterclockwise angle and a v is a clockwise angle.
The radius is a counter, in other words it has magnitude but does not have an Opposite Sign. Because of this, it is not used as a standalone value in an equation.
π^ represents a counterclockwise rotation of 180^o in radians. It is approximately 3.14^ radians. πv represents a counterclockwise rotation of 180vo. It is approximately 3.14v radians.
R2 Trigonometric Functions
R2 Trigonometric functions such as sine, cosine, tangent etc. are essential to rotation. They are based on a right angle triangle with:
sine = opposite/hypotenuse
cosine = adjacent/hypotenuse and
the tangent = opposite/adjacent.
Counters represent these functions because the functions are ratios.
Trigonometric functions apply to angles. Here are some of the most common sine and cosines for angles in R2 Wave Numbers.
Angle
Degree
Angle
Degree
Sine
Cosine
0
0
2πv
360v
0
1
π^/6
30^
11πv/6
330v
½
√3/2
π^/4
45^
7πv/4
315v
√2/2
√2/2
π^/3
60^
5πv/3
300v
√3/2
½
π^/2
90^
3πv/2
270v
1
0
2π^/3
120^
4πv/3
240v
√3/2
–½
3π^/4
135^
5πv/4
225v
√2/2
–√2/2
5π^/6
150^
7πv/6
210v
½
–√3/2
π^
180^
πv
180v
0
–1
7π^/6
210^
5πv/6
150v
–½
–√3/2
5π^/4
225^
3πv/4
135v
–√2/2
–√2/2
4π^/3
240^
2πv/3
120v
–√3/2
–½
3π^/2
270^
πv/2
90v
–1
0
5π^/3
300^
πv/3
60v
–√3/2
½
7π^/4
315^
πv/4
45v
–√2/2
√2/2
11π^/6
330^
πv/6
30v
–½
√3/2
2π
360^
0
0
0
1
Pythagoras Theorem
Calculate the distance from the origin or radius using Pythagoras's theorem as √(|a|2 + |b|2). | 677.169 | 1 |
Elementary Geometry for College Students (7th Edition)
by
Alexander, Daniel C.; Koeberlein, Geralyn M.
Answer
False
Work Step by Step
A quick trick I learned is that the longest side is the angle that does include both letters. For example, $\angle$A does not include letter B or C, therefore $\overline{BC}$ is the longest side.
45$^{\circ}$, 65$^{\circ}$, 70$^{\circ}$
$\angle$B $\angle$C $\angle$A
Side length $\overline{BC}$ is the longest side because the largest angle does not include the letters B or C in the triangle. | 677.169 | 1 |
Delve into the fascinating world of geometry with our comprehensive 30 60 90 triangle worksheet answer key. This key unlocks the secrets of trigonometry, providing a roadmap to understanding the properties and applications of these special triangles.
Discover the intricate relationships between side lengths and angles, delve into real-world scenarios where 30-60-90 triangles play a pivotal role, and uncover the historical significance that has shaped our understanding of geometry.
Understanding the 30-60-90 Triangle
A 30-60-90 triangle is a right triangle with angles measuring 30°, 60°, and 90°. It has specific geometric properties that make it useful in various applications, such as architecture, engineering, and trigonometry.The relationship between the side lengths and angles in a 30-60-90 triangle is as follows:
The side opposite the 30° angle is half the length of the hypotenuse.
The side opposite the 60° angle is √3 times the length of the shorter side.
The hypotenuse is twice the length of the shorter side.
Worksheet Analysis
The worksheet includes various problems related to 30-60-90 triangles, providing an opportunity to apply the concepts discussed earlier. Solving these problems will reinforce your understanding and enhance your problem-solving skills.
Worksheet
Refer to the attached worksheet for the specific problems and instructions.
Answer Key
The answer key is provided to assist you in checking your solutions and identifying areas for improvement. Carefully review the answers to ensure your understanding of the concepts and calculations.
30-60-90 triangles, with their unique properties, find applications in various real-world scenarios. Their simplicity and predictable relationships make them useful for solving practical problems.
Surveying and Architecture
In surveying, 30-60-90 triangles are used to determine distances and heights of objects indirectly. By measuring the angle of elevation or depression and one side of the triangle, surveyors can calculate the unknown dimensions.
Architects utilize 30-60-90 triangles to create balanced and aesthetically pleasing designs. The proportions of these triangles ensure harmonious relationships between different elements of a structure.
Navigation
In navigation, 30-60-90 triangles are used for triangulation, a technique to determine the location of a point by measuring angles from two known points. This method is commonly employed in maritime navigation and surveying.
Sailors use 30-60-90 triangles to determine the direction and distance to a destination by measuring the angle between the shoreline and a landmark.
Carpentry and Construction
Carpenters and construction workers use 30-60-90 triangles to ensure accurate measurements and angles in framing, roofing, and other construction tasks.
The 30-60-90 triangle's properties help determine the length of rafters, the angle of roof slopes, and the stability of structures.
Special Ratios and Proportions
30-60-90 triangles exhibit unique ratios and proportions that simplify problem-solving. These ratios stem from the triangle's specific angle measures and side lengths.
The most fundamental ratio in a 30-60-90 triangle is the 1:√3:2 ratio, which relates the lengths of the three sides. The shortest side (opposite the 30° angle) has a length of 1 unit, the side opposite the 60° angle has a length of √3 units, and the hypotenuse (opposite the 90° angle) has a length of 2 units.
Using Ratios to Solve Problems
These ratios can be applied to various problem-solving scenarios. For instance, if you know the length of one side, you can determine the lengths of the other two sides using the appropriate ratio.
Example: If the shortest side of a 30-60-90 triangle measures 6 units, the length of the side opposite the 60° angle is 6√3 units, and the hypotenuse is 12 units.
Additionally, the 30-60-90 triangle possesses a special altitude ratio. The altitude drawn from the vertex of the right angle to the hypotenuse divides the hypotenuse into a 1:√3 ratio. This ratio is particularly useful in geometry and trigonometry.
Historical Context
30-60-90 triangles have a rich history dating back to ancient civilizations.
In ancient Egypt, around 2000 BC, these triangles were used in the construction of pyramids and other structures. The Egyptians understood the special properties of 30-60-90 triangles and used them to create accurate measurements and angles.
Pythagoras' Theorem
Around 500 BC, the Greek mathematician Pythagoras discovered the famous theorem that bears his name. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.
$a^2 + b^2 = c^2$
This theorem was a major breakthrough in geometry and allowed mathematicians to calculate the lengths of sides in right triangles, including 30-60-90 triangles.
Visual Aids
To enhance the understanding of 30-60-90 triangles, we present visual aids that summarize their key properties and illustrate their geometric relationships.
The following table concisely presents the essential characteristics of 30-60-90 triangles:
Property
Value
Opposite side to 30° angle
Half the length of the hypotenuse
Adjacent side to 30° angle
Half the length of the hypotenuse multiplied by √3
Hypotenuse
Twice the length of the shortest side
Area
One-fourth the square of the hypotenuse
Illustrations: Geometric Relationships
The geometric relationships within 30-60-90 triangles can be visually represented through illustrations:
Figure 1:A right triangle with angles measuring 30°, 60°, and 90°. The shortest side is labeled "a," the opposite side to the 30° angle is labeled "b," the adjacent side to the 30° angle is labeled "c," and the hypotenuse is labeled "h."
Figure 2:A diagram demonstrating the special ratios and proportions within a 30-60-90 triangle. The opposite side is shown as half the length of the hypotenuse, while the adjacent side is shown as half the length of the hypotenuse multiplied by √3.
Final Thoughts
Through this comprehensive exploration, you'll master the intricacies of 30 60 90 triangles, empowering you to tackle any trigonometry challenge with confidence. Let this answer key be your guide as you embark on a journey of mathematical discovery.
Detailed FAQs
What is the most important property of a 30 60 90 triangle?
The most important property is the relationship between the side lengths, where the hypotenuse is always twice the length of the shorter leg.
How can I use the answer key to solve problems?
The answer key provides step-by-step solutions to the worksheet problems, guiding you through the process of applying the properties of 30 60 90 triangles.
What are some real-world applications of 30 60 90 triangles?
30 60 90 triangles are used in architecture, engineering, and surveying to determine heights, distances, and angles. | 677.169 | 1 |
In the attached figure, parallel lines \(l\) and \(m\) are cut by a transversal line \(n\), forming angles labeled \(1\), \(2\), \(3\), \(4\), and \(5\) as shown in the attached diagram. If angle \(1\) measures \(125\) degrees, what is the sum of angles \(3\) and \(5\)? | 677.169 | 1 |
Chord
One might wonder why we care about chords, circles, and the like, but there''s a reason: We need to be as accurate and as specific as possible in the world. So what exactly is a "chord?" We''ll soon find out.
Chords, explained
A chord is a line that passes through a circle or curved line. While the chord is straight, the segment that it passes through is not. A chord can pass through one or multiple curved segments or arcs.
When a chord passes through a circle, it creates "endpoints" on the boundary or perimeter of the circle.
Visualizing chords
Take a look at the following diagram to visualize how chords work:
Here we can see a total of two chords passing through the circle:
AB and AC
You might already be familiar with one type of chord called a diameter. A diameter is a chord that passes through the center of the circle. As such, this is the longest possible chord of a circle.
Fun facts about chords
There are a number of theorems or "rules" that involve chords. For example:
Two minor arcs are congruent only if their corresponding chords are congruent
Flashcards covering the Chord
Practice tests covering the Chord
Pair your student with a private math instructor to help them cover chords
Understanding early concepts like chords is critical, and tutors can help hammer home key lessons that might have been misunderstood during class time. Tutors can explain these concepts in new ways depending on your student''s hobbies or learning style. They can also challenge more advanced students with fun exercises if they''re getting bored in class. Reach out to Varsity Tutors today, and we''ll pair your student with an appropriate | 677.169 | 1 |
triangle, parallelograms, incidence --- hints
Use the following steps to construct a figure.
1) Pick non-collinear points A, B, C [these can be moved]
2) Pick point M inside triangle A.B.C [point M can be moved]
3) Locate point M_c such that quadrilateral A.M.B.M_c is a parallelogram.
4) Locate point M_b such that quadrilateral A.M.C.M_b is a parallelogram.
5) Locate point M_a such that quadrilateral B.M.C.M_a is a parallelogram. | 677.169 | 1 |
On teaching, math and other bloggable matters
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Geometry Journal, #2
I put a rectangle into Geogebra. It actually took me a few tries to really nail down what the problem was saying, but once I did I started dragging the points around. I noticed that the angle made by the two perpendiculars was constant also. Cool! After noodling around with the diagram I understand why:
I started drawing a lot of lines. This is another one of those diagrams that has a bajillion similar triangles. I was playing with this while my kids were going to sleep, so it was about 30 minutes of lazy drawing and redrawing. I found a lot of congruent triangles, some I was pretty sure would help me, but I couldn't make them work. Here was my favorite congruent triangle pair:
I grew frustrated with the computer and (following my takeaway from the previous post, actually!) I took out some paper and started writing equations. I was thinking about how I would get an expression for EF + EG (the two perpendiculars) and I had the idea that you could do that using similar triangles.
I took the two triangles above and set them similar to each other. Awful quality I know, sorry:
Here is that picture, cleaned up a bit:
The point is that if those triangles are similar, then those perpendiculars are in ratio also. Some fiddling tells us that if that's so, the sum of the perpendiculars is equal to:
Is EF a function of x, so that the x's cancel out? If so, we would be done. Call that yellow angle in the top right corner theta.
And that works. The sum of perpendiculars is equal to .
***
Here is the book's solution:
Here's the point:
It's a nice solution. When I was looking for a constant I was trying to do something like this — slide a segment over using a rectangle. The problem was I couldn't get the right structure when I was trying this, I always ended up stuck with a trapezoid.
The big strategy here is "when you're looking to prove something is constant, you can try to construct the thing out of both pieces." I did try that first, but ended up stumped so I moved to a different approach.
Otherwise, I'm not sure what else to take away from this solution. The move of using an isosceles triangle to prove both sides are congruent is sweet. | 677.169 | 1 |
Clear Quartz
Platonic solids are the five geometric solids whose faces are all identical, regular polygons meeting at the same three-dimensional angles. Also known as the five regular polyhedra, they consist of the tetrahedron (or pyramid), cube, octahedron, dodecahedron, and icosahedron. Pythagoras probably knew the tetrahedron, cube, and dodecahedron. According to Euclid, the octahedron and icosahedron were first discussed by the Athenian mathematician Theaetetus. However, the entire group of regular polyhedra owes its popular name to the great Athenian philosopher Plato | 677.169 | 1 |
geometry triangle angle sum worksheet answers
Geometry Triangle Angles Sum | 677.169 | 1 |
Rotations date period graph the image of the figure using the transformation given. Rotations worksheet 1 date find the coordinates of the vertices of each figure after the given transformation. 1 rotation 180 about the origin x y b r y g.
Click on the link s below for resources by concept. Sheet 1 answer key graph the new position of each point after rotating it about the origin. Atl hlu sr aieg7h ot4s b yrbe zszedr9vwe fdi.
Counterclockwise rotation of 150 around point p step 1 draw pa. Clockwise rotation 5 5 4 3 2 1 1 2 3 4 5 5. A p a 110 p a c b p a c b a point and its image are both the same distance from the center of rotation.
Rotations reflections and translations. Then use a protractor to draw a ray that forms a 150 angle with pa. Answer key sheet 1 write a rule to describe each rotation.
Two axis rotation activity rotations practice worksheet answer key 1 rotations practice worksheet answer key worksheet instructions study how the object in the top line is rotated. Checklist for unit 1 study guide for unit 1 study guide key for unit 1 calculators may be used in this unit but students must know how to compute answers without calculators as well as with. | 677.169 | 1 |
Solving for Angles in the Oblique Triangle (Math Problem Sample)
Instructions:
Solve for α in the oblique triangle ABC; AB = 30; AC = 15 and angle B = 20° 1. Type out the two equations substituting the numbers from the diagram. 2. Type out the Law of Sines set of relationships and type out the most appropriate version to use the Law of Cosines for this solution. 3. Solve for a using both methods (show step by step work) Solution
Solve for α in the oblique triangle ABC; AB = 30; AC = 15 and angle B = 20° 1 Type out the two equations substituting the numbers from the diagram. 2 Type out the Law of Sines set of relationships and type out the most appropriate version to use the Law of Cosines for this solution. 3 Solve for a using both methods (show step by step work) Solution 1 The sum of all angles of a triangle adds up to 180o A+B+C = 180, and B = 20 A+20+C = 180 A + C = 160
Other Topics:
Description: The Pythagoras theorem is an important mathematical relation that connects the geometry of a triangle. It gives the links between the sides of a triangle; right-angled triangle. The three sides of a right-angle triangle include: the two legs meeting at the right angle, and the longest diagonal (hypotenuse) ... | 677.169 | 1 |
Question4: Classroom activity (Constructing the 'square root spiral'): Take a large sheet of paper and construct the 'square root spiral' in the following fashion. Start with a point O and draw a line segment OP1 of unit length. Draw a line segment P1P2 perpendicular to OP1 of unit length. Now draw a line segment P2P3 perpendicular to OP2. Then draw a line segment P3P4 perpendicular to OP3. | 677.169 | 1 |
User Forum
Which of the following is a true statement about a cube? (a) It has a greater number of edges than vertices. (b) It has the same number of vertices as faces. (c) It has a greater number of faces than vertices. (d) It has a greater number of faces than edges. | 677.169 | 1 |
"ŚŽŖšŠ 53 ... square of CD . Wherefore , if a straight line , & c . Q.E.D. PROP . VII . - THEOREM . If a straight line be divided into any two parts , the squares of the whole line and of one of the parts , are equal to twice the rectangle contained ...
"ŚŽŖšŠ 54 ... twice the rectangle AB , BC , together with the square of AC . Wherefore , if a straight line , & c . Q.E.D. PROP . VIII . - THEOREM . If a straight line be divided into any two parts , four times the rectangle contained by the whole ... | 677.169 | 1 |
DS Geometry Questions
The lesson focuses on strategies for tackling Geometry Questions in Data Sufficiency on the GRE, emphasizing the importance of not trusting diagrams, understanding the necessity of given lengths to find lengths, and enhancing visual imagination skills to envision various shapes.
Don't trust diagrams in Data Sufficiency questions; they may not accurately represent the information given and can lead to incorrect assumptions.
To find a specific length in a geometry question, you must be provided with another specific length; angle measurements or ratio information alone are insufficient.
Enhancing visual imagination skills is crucial for envisioning the different shapes that could fit a given set of constraints, which is essential for answering Data Sufficiency questions effectively.
Practicing with different geometric constraints and attempting to draw multiple shapes that satisfy those constraints can significantly improve problem-solving skills in geometry.
Combining different pieces of information correctly is key to determining the sufficiency of data provided in these types of questions.
Chapters
00:00
The Deceptive Nature of Diagrams
05:08
The Principle of Length in Geometry Questions
06:37
Developing Visual Imagination for Geometry
The GMAT in its current form will be offered through early 2024. Starting in Q4 of 2023, a new, shorter version of the exam called the GMAT Focus will become available.
This lesson applies to the current GMAT only since the GMAT Focus edition does not test Geometry. | 677.169 | 1 |
We know that a right triangle is a triangle having a right
angle, where the side opposite the right angle is the hypotenuse,
and the perpendicular sides are the legs of the right triangle.
The Pythagorean theorem gives the relationship between the
lengths of the sides of a right triangles.
In the case where you know only the measure lengths of the sides
of a triangle, you need to test these measures. If one of the sides
of the triangle has a square measure equal to the sum of the square
measures of two other sides, then this side is called the
hypotenuse and opposite to this side is a 90 degree angle, which is
a right angle. So, you can say that this triangle is a right
triangle.
Pythagorean triple are very helpful to determine a right
triangle, such as:
(3, 4, 5), (5,12,13), (8, 15, 17), (7, 24, 25), and (20, 21,
29).
Wiki User
∙ 15y ago
This answer is:
Add your answer:
Earn +20 pts
Q: How do you know if a triangle is a right triangle using the pythagorean therom? | 677.169 | 1 |
Class 8 Maths Chapter 3 Understanding Quadrilaterals
NCERT Solutions For Class 8 Maths Chapter 3 Understanding Quadrilaterals3. How many sides does a regular polygon have if the measure of an exterior angle is 24°? Solution:
Each exterior angle = sum of exterior angles/Number of angles
24°= 360/ Number of sides
⇒ Number of sides = 360/24 = 15
Thus, the regular polygon has 15 sides.
4. How many sides does a regular polygon have if each of its interior angles is 165°?
Solution:
Interior angle = 165°
Exterior angle = 180° – 165° = 15°
Number of sides = sum of exterior angles/ exterior angles
⇒ Number of sides = 360/15 = 24
Thus, the regular polygon has 24 sides.
5.
a) Is it possible to have a regular polygon with measure of each exterior angle as 22°?
b) Can it be an interior angle of a regular polygon? Why?
Solution:
a) Exterior angle = 22°
Number of sides = sum of exterior angles/ exterior angle
⇒ Number of sides = 360/22 = 16.36
No, we can't have a regular polygon with each exterior angle as 22° as it is not divisor of 360.
b) Interior angle = 22°
Exterior angle = 180° – 22°= 158°
No, we can't have a regular polygon with each exterior angle as 158° as it is not divisor of 360.
6.
a) What is the minimum interior angle possible for a regular polygon? Why?
b) What is the maximum exterior angle possible for a regular polygon?
Solution:
a) Equilateral triangle is regular polygon with 3 sides has the least possible minimum interior angle because the regular with minimum sides can be constructed with 3 sides at least.. Since, sum of interior angles of a triangle = 180°
9. In the above figure both RISK and CLUE are parallelograms. Find the value of x.
Solution:
∠K + ∠R = 180° (adjacent angles of a parallelogram are supplementary)
⇒ 120° + ∠R = 180°
⇒ ∠R = 180° – 120° = 60°
also, ∠R = ∠SIL (corresponding angles)
⇒ ∠SIL = 60°
also, ∠ECR = ∠L = 70° (corresponding angles)
x + 60° + 70° = 180° (angle sum of a triangle)
⇒ x + 130° = 180°
⇒ x = 180° – 130° = 50°
10. Explain how this figure is a trapezium. Which of its two sides are parallel? (Fig 3.32)
Solution:
When a transversal line intersects two lines in such a way that the sum of the adjacent angles on the same side of transversal is 180° then the lines are parallel to each other.
Here, ∠M + ∠L = 100° + 80° = 180°
Thus, MN || LK
As the quadrilateral KLMN has one pair of parallel line therefore it is a trapezium. MN and LK are parallel lines.
11. Find m∠C in Fig 3.33 if AB || DC ?
Solution:
m∠C + m∠B = 180° (angles on the same side of transversal)
⇒ m∠C + 120° = 180°
⇒ m∠C = 180°- 120° = 60°
12. Find the measure of ∠P and ∠S if SP || RQ ? in Fig 3.34. (If you find m∠R, is there more than onemethod to find m∠P?)
Solution:
∠P + ∠Q = 180° (angles on the same side of transversal)
⇒ ∠P + 130° = 180°
⇒ ∠P = 180° – 130° = 50°
also, ∠R + ∠S = 180° (angles on the same side of transversal)
⇒ 90° + ∠S = 180°
⇒ ∠S = 180° – 90° = 90°
Thus, ∠P = 50° and ∠S = 90°
Yes, there are more than one method to find m∠P.
PQRS is a quadrilateral. Sum of measures of all angles is 360°.
Since, we know the measurement of ∠Q, ∠R and ∠S.
∠Q = 130°, ∠R = 90° and ∠S = 90°
∠P + 130° + 90° + 90° = 360°
⇒ ∠P + 310° = 360°
⇒ ∠P = 360° – 310° = 50°
Exercise 3.4
Page: 55
1. State whether True or False.
(a) All rectangles are squares.
(b) All rhombuses are parallelograms.
(c) All squares are rhombuses and also rectangles.
(d) All squares are not parallelograms.
(e) All kites are rhombuses.
(f) All rhombuses are kites.
(g) All parallelograms are trapeziums.
(h) All squares are trapeziums.
Solution:
(a) False.
Because, all square are rectangles but all rectangles are not square.
(b) True
(c) True
(d) False.
Because, all squares are parallelograms as opposite sides are parallel and opposite angles are equal.
(e) False.
Because, for example, a length of the sides of a kite are not of same length.
(f) True
(g) True
(h) True
2. Identify all the quadrilaterals that have.
(a) four sides of equal length (b) four right angles
Solution:
(a) Rhombus and square have all four sides of equal length.
(b) Square and rectangle have four right angles.
3. Explain how a square is.
(i) a quadrilateral (ii) a parallelogram (iii) a rhombus (iv) a rectangle
Solution
(i) Square is a quadrilateral because it has four sides.
(ii) Square is a parallelogram because it's opposite sides are parallel and opposite angles are equal.
(iii) Square is a rhombus because all the four sides are of equal length and diagonals bisect at right angles.
(iv)Square is a rectangle because each interior angle, of the square, is 90°
4. Name the quadrilaterals whose diagonals.
(i) bisect each other (ii) are perpendicular bisectors of each other (iii) are equal
Solution
(i) Parallelogram, Rhombus, Square and Rectangle
(ii) Rhombus and Square
(iii)Rectangle and Square
5. Explain why a rectangle is a convex quadrilateral.
Solution
Rectangle is a convex quadrilateral because both of its diagonals lie inside the rectangle.
6. ABC is a right-angled triangle and O is the mid-point of the side opposite to the right angle. Explain why O is equidistant from A, B and C. (The dotted lines are drawn additionally to help you).
Solution
AD and DC are drawn so that AD || BC and AB || DC
AD = BC and AB = DC
ABCD is a rectangle as opposite sides are equal and parallel to each other and all the
interior angles are of 90°.
In a rectangle, diagonals are of equal length and also bisects each other.
Hence, AO = OC = BO = OD
Thus, O is equidistant from A, B and C | 677.169 | 1 |
ftraight line at right angles to a given ftraight line , from a given point in the fame . T & Let AB be a given ftraight ...
УелЯдб 23 ... lines can- A not have a common fegment . T PROP . XII . PROB . E D B O draw a ftraight line perpendicular to a given straight line of an unlimited length , from a given point without it . Let AB be the given ftraight line , which may be ...
УелЯдб 29 ... given straight lines , but any two whatever of these must be greater than the third " . Let A , B , C be the three ... line DE terminated at the point D , but un- limited towards E , and make a DF equal to A , FG to B , and GH ...
УелЯдб 30 ... straight lines KF , FG , GK are equal to the three A , B , C : And therefore the triangle KFG has its three fides KF , FG , GK equal to the three given ... line , to make a rectilineal angle equal to a given rectilineal angle . Let AB be | 677.169 | 1 |
Determine how many points on the unit circle have \(-\dfrac13\) as their \(x\)-coordinate. Indicate these on the graph.
Determin how many points on the unit circle have \(-\dfrac13\) as their \(y\)-coordinate. Indicate these on the graph.
In this section, we will examine this type of revolving motion around a circle. To do so, we need to define the type of circle first, and then place that circle on a coordinate system. Then we can discuss circular motion in terms of the coordinate pairs.
Defining Sine, Cosine, and Tangent ratios for any angle
To define our trigonometric ratios, we begin by drawing a unit circle (a circle of radius \(1\) centered at the origin \((0,0)\)).
Recall that the x- and y-axes divide the coordinate plane into four quarters called quadrants. We label these quadrants to mimic the direction a positive angle would sweep. The four quadrants are labeled I, II, III, and IV.
For an acute angle \(t,\) we can label the intersection of the terminal side and the unit circle as by its coordinates, \((x,y)\). Since the circle has radius \(1\), using our trigonometric ratios we see that the triangle in orange below has base \(\cos t\) and height \(\sin t\). The coordinates \(x\) and \(y\) therefore can be related to the angle \(t\):\(x= \cos t\) and \(y= \sin t\). Also, \(\tan t=\dfrac{\sin t}{\cos t}\).
UNIT CIRCLE
A unit circle has a center at \((0,0)\) and radius \(1\). Form the angle with measure \(t\) with initial side coincident with the \(x\)-axis.
Let \((x,y)\) bepoint where the terminal side of the angle and unit circle meet. Then \((x,y)=(\cos t,\sin t)\). Further, \(\tan t=\dfrac{\sin t}{\cos t}\).
Defining Sine, Cosine, and Tangent for any Angle
Now that we have our unit circle labeled, we can learn how the \((x,y)\) coordinates relate to the angle \(t\). For an acute angle, the sine ratio \(\sin t\) isthe \(y\)-coordinate of the point where the corresponding terminal side of the angle intersects the unit circle and the cosine ratio \(\cost\) isthe \(x\)-coordinate of the point where the corresponding terminal side of the angle intersects the unit circle.We extend this definition to all angles. So, we definethe sine ratio \(\sin t\)to bethe \(y\)-coordinate of the point where the corresponding terminal side of the angle intersects the unit circle and the cosine ratio \(\cost\) to bethe \(x\)-coordinate of the point where the corresponding terminal side of the angle intersects the unit circle. Thetangent ratio\(\tan t\) is\(\dfrac{\sin t}{\cos t}\).
We may enclose the angle inparentheses or not depending on how clear the expression is:\(\sin t\) is the same as \(\sin (t)\) and \(\cos t\) is the same as \(\cos (t)\). When in doubt, use the extra parentheses when entering calculations into a calculator or computer.
SINE,COSINE, and TANGENTRATIOS
If \(t\) is an angle measurementand a point \((x,y)\) is both on the unit circle and the terminal side of the angle in standard position, then
The cosine of \(90^{\text{o}}\) is 0; the sine of \(90^{\text{o}}\) is 1.
Try It \(\PageIndex{2}\)
Find cosine and sine of the angle \(\pi\).
Answer
\(\cos (\pi)=−1, \sin (\pi)=0\)
The Pythagorean Identity
Now that we can define sine and cosine, we will learn how they relate to each other and the unit circle. Recall that the equation for the unit circle is \(x^2+y^2=1\). Because \(x= \cos t\) and \(y=\sin t\), we can substitute for \( x\) and \(y\) to get \(\cos ^2 t+ \sin ^2 t=1.\) This equation, \( \cos ^2 t+ \sin ^2 t=1,\) is known as the Pythagorean Identity. See below:
We can use the Pythagorean Identity to find the cosine of an angle if we know the sine, or vice versa. However, because the equation yields two solutions, we need additional knowledge of the angle to choose the solution with the correct sign. If we know the quadrant where the angle is, we can easily choose the correct solution.
How To: Given the sine of some angle t and its quadrant location, find the cosine of t
Substitute the known value of \(\sin (t)\) into the Pythagorean Identity.
Solve for \( \cos (t)\).
Choose the solution with the appropriate sign for the \(x\)-values in the quadrant where \(t\) is located.
Example \(\PageIndex{3}\)
If \(\sin (t)=\dfrac{3}{7}\) and \(t\) is in the second quadrant, find \( \cos (t)\).
Solution
If we drop a vertical line from the point on the unit circle corresponding to \(t\), we create a right triangle, from which we can see that the Pythagorean Identity is simply one case of the Pythagorean Theorem.
Because the angle is in the second quadrant, we know the \(x\)-value is a negative real number, so the cosine is also negative. So
\[ \cos (t)=−\dfrac{2\sqrt{10}}{7} \nonumber \]
Try It \(\PageIndex{4}\)
If \(\cos (t)=\dfrac{24}{25}\) and \(t\) is in the fourth quadrant, find \( \sin (t)\).
Answer
\(\sin (t)=−\dfrac{7}{25}\)
Finding Sines,Cosinesand Tangent Ratios of Special Angles
When possible, we would like to be precise about evaluation of the sine and cosine ratios. We give an example here.
Example \(\PageIndex{5}\)
a) Find the value of \(\sin 150^\text{o}\)
b) Find the value of \(\cos\dfrac{7\pi}{4}\).
Solution
a) We first draw a picture and identify a triangle (while there are two natural possibilities, wewilluse the triangle shown below (often called "the reference triangle)).
Then determine the reference angle indicated in red. To do this we note that the straight angle (black angle plus the red angle is \(180^\text{o}\). So the red angle measures \(30^\text{o}\). Then we use our knowledge of the 30-60-90 triangle (choosing the one with hypotenuse 1) and note the lengths on the diagram.
Finally, making note of the position of the triangle, we see that the \(x\)-coordinate must be negative and the \(y\)-coordinate must be positive, we find
So we conclude, in particular, that \(\sin 150^\text{o}=\dfrac12\).
Note that we could also use the standard triangle (using the circle of radius 2)and use signed lengths and form the sine ratio as before:
b) We first draw the angle together with the unit circle and label the intersection:
We determine the value of the red angle (noting that the black angle plus the red angle measures \(2\pi\). Weconstruct a right triangle that will help us determine the coordinates of the labeledintersection (here we will use, again, what is often called 'the reference triangle').
Now, we recall the 45-45-90 triangle from memory (or rederive it) and scale it to have thehypotenusehave length1.
Now, we note from the location of the point that the \(x\)-coordinate must be positiveand the \(y\)-coordinate must be negative. We determine the coordinates of the labeled intersection.
It follows that \(\cos\dfrac{7\pi}{4}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\).
As in part a) we could also use the standard triangle (using the circle of radius 2)and use signed lengths and form the sine ratio as before:
To find a reference angle, draw a sketch noting in which quadrant your terminal side is and use your familiarity with right angles and straight angles to figure out the reference angle. Try these exercises using the above example as a model.
Using a Calculator to Find Sine and Cosine
To find the cosine and sine of angles other than the special angles, we turn to a computer or calculator. Be aware: Most calculators can be set into "degree" or "radian" mode, which tells the calculator the units for the input value. When we evaluate \( \cos (30^\text{o})\) on our calculator, it will evaluate it as the cosine of 30 degrees if the calculator is in degree mode, or the cosine of 30 radians if the calculator is in radian mode.
Example \(\PageIndex{7}\): Using a Graphing Calculator to Find Sine and Cosine
a) Evaluate \( \sin150^\text{o}\) using a calculator or computer.
b) Evaluate \(\cos \dfrac{7\pi}{4}\)
Solution
a) This depends on your calculator. But we know the answer is \(\dfrac12\). Make sure the mode on the calculator is set to 'degree'.
b)This depends on your calculator. But we know the answer is \(\dfrac{1}{\sqrt{2}}\) so if you square your answer you should get \(\dfrac12\) and your answer should be positive. Make sure the mode on the calculator is set to 'radian'.
Try It \(\PageIndex{8}\)
Evaluate \(\sin(-50)\).
Answer
Approximately -7.66
Writing Exercises \(\PageIndex{9}\)
Explain how \(\cos(300{\text{o}})\) can be evaluated using the unit circle.
Give four angles \(\theta, \alpha,\beta\) and \(\gamma\) for which \(\cos(300^{\text{o}})=\cos(\theta)=\cos(\alpha)=\cos(\beta)=\cos(\gamma)\)
Explain what a reference triangle is and how you can create it. Is there another triangle that would give you the same benefit?
If you know \(\sin(x)=\dfrac{4}{5}\), how can you find \(\cos(x)\) and \(\tan(x)\) without first finding \(x\). How might you use similar triangles/figures to make use of a circle of radius \(5\) instead of the unit circle?
Exit Problem
Evaluate \(\cos(-210^{\text{o}})\) and \(\sin(-210^{\text{o}})\). Support your answer with an appropriate picture of the unit circle. Then evaluate \(\tan(210^{\text{o}})\).
Key Concepts
Finding the function values for the sine and cosine begins with drawing a unit circle, which is centered at the origin and has a radius of 1 unit.
Using the unit circle, the sine of an angle \(t\) equals the \(y\)-value of the endpoint on the unit circle of an arc of length \(t\) whereas the cosine of an angle \(t\) equals the \(x\)-value of the endpoint.
The sine and cosine values are most directly determined when the corresponding point on the unit circle falls on an axis.
When the sine or cosine is known, we can use the Pythagorean Identity to find the other. The Pythagorean Identity is also useful for determining the sines and cosines of special angles.
Calculators and graphing software are helpful for finding sines and cosines if the proper procedure for entering information is known.
The domain of the sine and cosine functions is all real numbers.
The range of both the sine and cosine functions is \([−1,1]\).
The sine and cosine of an angle have the same absolute value as the sine and cosine of its reference angle.
The signs of the sine and cosine are determined from the \(x\)- and \(y\)-values in the quadrant of the original angle.
An angle's reference angle is the size angle, \(t\), formed by the terminal side of the angle \(t\) and the horizontal axis.
Reference angles can be used to find the sine and cosine of the original angle.
Reference angles can also be used to find the coordinates of a point on a circle. | 677.169 | 1 |
Theorem:In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.
How do you teach children about right angles?
And that's acute angle what acute angle a right angle is exactly 90 degrees it's just right an obtuse angle is greater than 90 degrees. But less than 180 degrees.
What is a right angle Year 2?
Right angles
A right angle is a quarter of a full turn. It measures 90°.
How do you teach angles in a fun way?
Teaching Angles: 5 Activities for your Classroom
Door Angles. Using strong white tape suitable for the floor, show the angles to which the door may be opened like in the picture.
Angles in a Name.
Angles BINGO.
Masking Tape on Tables.
Time for Angles.
Geometry with Emile.
How do you solve right triangle theorem?
Trigonometry: How to Solve Right Triangles – YouTube
What are the 3 angles of a right triangleWhat is a right angle definition for kids?
Kids Definition of right angle
: an angle formed by two lines that are perpendicular to each other : an angle of 90 degrees.
Why is it called a right angle?
A right angle was described in ancient geometry as the meeting of two right, ie straight, lines, with regard to dimensional axes.
Why is it called right angle?
Right, meaning "correct", and right, meaning "straight", do have the same root, but "right angle" derives from the second rather than the first. A right angle was described in ancient geometry as the meeting of two right, ie straight, lines, with regard to dimensional axes.
What is called right angle?
Definition of right angle
: the angle bounded by two lines perpendicular to each other : an angle of 90° or ¹/₂ π radians.
How do you explain angles to kids?
Angles – Types and definition – Mathematics for kids – YouTube
How do you teach students about angles?
Looking for Angles in Letters
Have students use a ruler to create the letter of their first name. Then, with the ruler, draw random lines within their letter and colour as they wish. Students then need to find as many angles as they can and measure each angle! Simple, fun and worthwhile!
What are the three 3 sides of the right triangle solve a 45 45 90 triangle?
45-45-90 Special Right Triangles – YouTube
How do you solve a right angle?
How to find the angle of a right triangle
sin(α) = a / c so α = arcsin(a / c) (inverse sine)
cos(α) = b / c so α = arccos(b / c) (inverse cosine)
tan(α) = a / b so α = arctan(a / b) (inverse tangent)
cot(α) = b / a so α = arccot(b / a) (inverse cotangent)
What are the properties of right angle triangle?
Right Angle Triangle Properties
The side opposite angle of 90° is the hypotenuse. The hypotenuse is always the longest side. The sum of the other two interior angles is equal to 90°. The other two sides adjacent to the right angle are called base and perpendicular.
How do you introduce a right angle?
Right Angles – YouTube
What is the symbol of right angle?
symbol ∟
When two straight lines intersect each other at 90˚ or are perpendicular to each other at the intersection, they form the right angle. A right angle is represented by the symbol ∟.
Why is right angle important?
A triangle that has a right angle is also known as a right triangle, to which you can apply special properties such as the Pythagorean Theorem. Right angles are important to recognize and to understand, since they're used everywhere from geometry to trigonometry to real-life applications.
What is a right angle for children?
A right angle is an angle that measures 90˚ (degrees). It is also known as a 'quarter turn' because it is a quarter of a full turn, which measures 360˚. A right angle is represented by a small square inside the angle. An acute angle is one that measures LESS than 90˚.
Why do we need to learn angles for kids?
Why is it important to learn about measuring angles? Understanding what angles are, how they work, and how to measure them are important. They help us build better streets and cities, tell time using the sun and shadows, and make it possible to measure how far away the planets and stars are.
Why do kids need to learn angles?
Together with length, angle is possibly the most important tool used to describe shapes in construction, design and navigation. Angle also plays an important role in developing students' understanding of geometry.
How do you teach angles in math?
Intro to Angles for Kids: Understanding Angles for Children – YouTube
What is a right triangle called?
A right triangle (American English) or right-angled triangle (British), or more formally an orthogonal triangle , formerly called a rectangled triangle (Ancient Greek: ὀρθόςγωνία, lit. 'upright angle'), is a triangle in which one angle is a right angle (that is, a 90-degree angle) or two sides are perpendicular.
What is the 3 4 5 Triangle rule?
The 3:4:5 triangle is the best way I know to determine with absolutely certainty that an angle is 90 degrees. This rule says that if one side of a triangle measures 3 and the adjacent side measures 4, then the diagonal between those two points must measure 5 in order for it to be a right triangle. | 677.169 | 1 |
A Ruler is used for measuring the _____________of line segments.
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B
Angle
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C
Intersection
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D
Edge
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Solution
The correct option is A
Length
A Ruler has a flat surface with four straight edges. It has small divisions of cm on one long edge. Similarly the other long edge is divided into inches. Generally in geometry we measure the lengths in cm and m. | 677.169 | 1 |
Home » Two circles with centers A and B, and radii 5 cm and 3 cm, touch each other internally. If the perpendicular bisector of the segment AB meets the bigger circle in P and Q; find the length of PQ.
Two circles with centers A and B, and radii 5 cm and 3 cm, touch each other internally. If the perpendicular bisector of the segment AB meets the bigger circle in P and Q; find the length of PQ. | 677.169 | 1 |
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