aarjaneiro/planetmath huggingface dataset:
Bregman divergence

Bregman divergence

A

Bregman divergence

, or

Bregman distance

,

B

F

on a space

𝒳

⊆

ℝ

d

is defined for a

strictly convex

and

differentiable function

F

:

𝒳

→

ℝ

as

B

F

⁢

(

p

,

q

)

=

F

⁢

(

p

)

-

F

⁢

(

q

)

-

⟨

p

-

q

,

∇

⁡

F

⁢

(

q

)

⟩

,

(1)

where

⟨

p

,

q

⟩

=

p

T

⁢

q

denotes the inner product, and

∇

⁡

F

⁢

(

x

)

=

[

∂

⁡

F

∂

⁡

x

1

,

⋯

,

∂

⁡

F

∂

⁡

x

d

]

T

the

partial derivatives

.

Choosing

F

⁢

(

x

)

=

∑

i

=

1

d

x

i

2

yields the squared Euclidean distance

B

x

2

⁢

(

p

,

q

)

=

||

p

-

q

||

2

, and choosing

F

⁢

(

x

)

=

∑

i

=

1

d

x

i

⁢

log

⁡

x

i

yields the

relative entropy

, called the Kullback-Leibler divergence.

Title

Bregman divergence

Canonical name

BregmanDivergence

Date of creation

2013-03-22 19:11:38

Last modified on

2013-03-22 19:11:38

Owner

FrankTokyo (25936)

Last modified by

FrankTokyo (25936)

Numerical id

6

Author

FrankTokyo (25936)

Entry type

Definition

Classification

msc 51K05

Synonym

Bregman distance

Generated on Fri Feb 9 22:02:21 2018 by

LaTeXML
compass and straightedge construction of inverse point

compass and straightedge construction of inverse point

Let

c

be a circle in the

Euclidean plane

with center

O

and let

P

≠

O

. One can

construct

the

inverse point

P

′

of

P

using

compass

and straightedge.

If

P

∈

c

, then

P

=

P

′

. Thus, it will be assumed that

P

∉

c

.

The construction of

P

′

depends on whether

P

is in the interior of

c

or not. The case that

P

is in the interior of

c

will be dealt with first.

1.

Draw the ray

O

⁢

P

→

.

2.

Determine

Q

∈

O

⁢

P

→

such that

Q

≠

O

and

O

⁢

P

¯

≅

P

⁢

Q

¯

.

3.

Construct the

perpendicular bisector

of

O

⁢

Q

¯

in order to find one point

T

where it intersects

c

.

4.

Draw the ray

O

⁢

T

→

.

5.

Determine

U

∈

O

⁢

P

→

such that

U

≠

O

and

O

⁢

T

¯

≅

T

⁢

U

¯

.

6.

Construct the perpendicular bisector of

O

⁢

U

¯

in order to find the point where it intersects

O

⁢

P

→

. This is

P

′

.

Now the case in which

P

is not in the interior of

c

will be dealt with.

1.

Connect

O

and

P

with a

line segment

.

2.

Construct the perpendicular bisector of

O

⁢

P

¯

in order to determine the

midpoint

M

of

O

⁢

P

¯

.

3.

Draw an arc of the circle with center

M

and radius

O

⁢

M

¯

in order to find one point

T

where it intersects

C

. By Thales’ theorem, the angle

∠

⁢

O

⁢

T

⁢

P

is a

right angle

; however, it does not need to be drawn.

4.

Drop the

perpendicular

from

T

to

O

⁢

P

¯

. The point of

intersection

is

P

′

.

A justification for these constructions is supplied in the entry inversion of plane.

If you are interested in seeing the rules for compass and straightedge constructions, click on the provided.

Title

compass and straightedge construction of inverse point

Canonical name

CompassAndStraightedgeConstructionOfInversePoint

Date of creation

2013-03-22 17:13:17

Last modified on

2013-03-22 17:13:17

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

12

Author

Wkbj79 (1863)

Entry type

Algorithm

Classification

msc 51K99

Classification

msc 53A30

Classification

msc 51M15

Generated on Fri Feb 9 22:02:24 2018 by

LaTeXML
angle of view of a line segment

angle of view of a line segment

Let

P

⁢

Q

be a

line segment

and

A

a point not belonging to

P

⁢

Q

.  Let the magnitude of the angle

P

⁢

A

⁢

Q

be

α

.  One says that the line segment

P

⁢

Q

is seen from the point

A

in an angle of

α

; one may also speak of the

angle of view

of

P

⁢

Q

.

The locus of the points from which a given line segment

P

⁢

Q

is seen in an angle of

α

(with

0

<

α

<

180

∘

) consists of two

congruent

circular arcs having the line segment as the common chord and containing the

circumferential angles

equal to

α

.

Especially, the locus of the points from which the line segment is seen in an angle of

90

∘

is the circle having the line segment as its

diameter

.

Note.

The

explementary arcs

of the above mentioned two arcs form the locus of the points from which the segment

P

⁢

Q

is seen in the angle

180

∘

-

α

.

Title

angle of view of a line segment

Canonical name

AngleOfViewOfALineSegment

Date of creation

2013-03-22 17:34:11

Last modified on

2013-03-22 17:34:11

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

14

Author

pahio (2872)

Entry type

Topic

Classification

msc 51M04

Classification

msc 51F20

Related topic

CircumferentialAngleIsHalfCorrespondingCentralAngle

Related topic

ThalesTheorem

Related topic

CalculatingTheSolidAngleOfDisc

Related topic

ExampleOfCalculusOfVariations

Related topic

ProjectionOfRightAngle

Defines

angle of view

Generated on Fri Feb 9 22:02:29 2018 by

LaTeXML
area of spherical calotte by means of chord

area of spherical calotte by means of chord

Let the arc

P

⁢

R

of a circle with radius

r

rotate about the

diameter

P

⁢

Q

.  The

surface of revolution

is a

spherical calotte

with the height

h

.  If the of the chord

P

⁢

R

is

k

, we obtain from the

right triangle

P

⁢

Q

⁢

R

the proportion equation

h

k

=

k

2

⁢

r

,

i.e. the chord

k

is the central proportional of the height and the diameter.  Accordingly, we can substitute

2

⁢

r

⁢

h

=

k

2

to the expression

A

=

2

⁢

π

⁢

r

⁢

h

of the area of the spherical calotte derived in the parent entry (

http://planetmath.org/AreaOfSphericalZone

). Thus we have an alternative

A

=

π

⁢

k

2

(1)

for finding the area of a spherical calotte.

References

1

K. Väisälä

:

Geometria

.  Kymmenennen painoksen muuttamaton lisäpainos.  Werner Söderström Osakeyhtiö, Porvoo & Helsinki (1971).

Title

area of spherical calotte by means of chord

Canonical name

AreaOfSphericalCalotteByMeansOfChord

Date of creation

2013-03-22 18:19:20

Last modified on

2013-03-22 18:19:20

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

5

Author

pahio (2872)

Entry type

Derivation

Classification

msc 51M04

Synonym

alternative way to find area of spherical calotte

Related topic

ThalesTheorem

Related topic

SimilarityOfTriangles

Generated on Fri Feb 9 22:02:31 2018 by

LaTeXML
circumferential angle is half the corresponding central angle

circumferential angle is half the corresponding central angle

Consider a circle with center

O

and two distinct points on the circle

A

and

B

. If

C

is a third point on the circle not equal to either

A

or

B

, then the

circumferential angle

at

C

subtending the arc

A

⁢

B

is the angle

A

⁢

C

⁢

B

. Here, by arc

A

⁢

B

, we mean the arc of the circle that does not contain the points

C

.

Similarly, the

central angle

subtending arc

A

⁢

B

is the angle

A

⁢

O

⁢

B

. The central angle corresponds to the arc

A

⁢

B

measured on the same side of the circle as the angle itself. Note that if

A

⁢

B

is a

diameter

of the circle, then the central angle is

180

∘

.

Theorem 1

.

[Euclid, Book III, Prop. 20]

In any circle, a circumferential angle is half the size of the central angle subtending the same arc.

Proof.

There are actually several distinct cases. Consider

∠

⁢

B

⁢

A

⁢

C

in a circle with center

O

, and draw

A

⁢

O

,

B

⁢

O

,

C

⁢

O

as well as the chord containing both

A

and

O

:

In this case, the center of the circle lies between the arms of the circumferential angle. Now, since

A

⁢

O

=

O

⁢

B

,

△

⁢

A

⁢

O

⁢

B

is

isosceles

, and

∠

⁢

F

⁢

O

⁢

B

is an

exterior angle

. Thus

∠

⁢

F

⁢

O

⁢

B

=

∠

⁢

O

⁢

A

⁢

B

+

∠

⁢

O

⁢

B

⁢

A

=

2

⁢

∠

⁢

O

⁢

A

⁢

B

Similarly,

△

⁢

A

⁢

O

⁢

C

is isosceles, and

∠

⁢

F

⁢

O

⁢

C

=

∠

⁢

O

⁢

A

⁢

C

+

∠

⁢

O

⁢

C

⁢

A

=

2

⁢

∠

⁢

O

⁢

A

⁢

C

and it follows that

∠

⁢

B

⁢

O

⁢

C

=

∠

⁢

F

⁢

O

⁢

B

+

∠

⁢

F

⁢

O

⁢

C

=

2

⁢

∠

⁢

O

⁢

A

⁢

B

+

2

⁢

∠

⁢

O

⁢

A

⁢

C

=

2

⁢

∠

⁢

B

⁢

A

⁢

C

proving the result.

A second case is the case in which both arms of the angle lie to one side of the circle’s center:

The proof is

similar

to the previous case, except that the angle in question is the difference rather than the sum of two known angles. Here we see that both

△

⁢

A

⁢

O

⁢

B

and

△

⁢

A

⁢

O

⁢

C

are isosceles, so that again

∠

⁢

C

⁢

O

⁢

F

=

2

⁢

∠

⁢

O

⁢

A

⁢

C

∠

⁢

B

⁢

O

⁢

F

=

2

⁢

∠

⁢

O

⁢

A

⁢

B

Subtracting, we get

∠

⁢

C

⁢

O

⁢

B

=

∠

⁢

C

⁢

O

⁢

F

-

∠

⁢

B

⁢

O

⁢

F

=

2

⁢

∠

⁢

O

⁢

A

⁢

C

-

2

⁢

∠

⁢

O

⁢

A

⁢

B

=

2

⁢

∠

⁢

B

⁢

A

⁢

C

as desired.

The final case is the case in which one arm of the angle goes through the center of the circle. This is a degenerate form of the first case, and the same proof follows through except that one of the angles is zero.
∎

Title

circumferential angle is half the corresponding central angle

Canonical name

CircumferentialAngleIsHalfTheCorrespondingCentralAngle

Date of creation

2013-03-22 17:13:28

Last modified on

2013-03-22 17:13:28

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

14

Author

rm50 (10146)

Entry type

Theorem

Classification

msc 51M04

Related topic

AngleOfViewOfALineSegment

Related topic

RiemannSphere

Defines

circumferential angle

Defines

central angle

Generated on Fri Feb 9 22:02:33 2018 by

LaTeXML
common point of triangle medians

common point of triangle medians

Theorem.

The three medians (

http://planetmath.org/Median

) of a

triangle

intersect one another in one point, which divides each median in the ratio

2

:

1

.

Proof.

Let the medians of a triangle

A

⁢

B

⁢

C

be

A

⁢

D

,

B

⁢

E

and

C

⁢

F

.  Any median vector is the

arithmetic mean

of the side vectors emanating from the same vertex.  Using vectors, let us form three ways all beginning from the vertex

A

, the first going simply

2

/

3

of the median vector

A

⁢

D

→

( in the picture):

2

3

⁢

A

⁢

D

→

=

2

3

⋅

1

2

⁢

(

A

⁢

B

→

+

A

⁢

C

→

)

=

1

3

⁢

(

A

⁢

B

→

+

A

⁢

C

→

)

(1)

The second way goes first the side vector

A

⁢

B

→

and then

2

/

3

of the median vector

B

⁢

E

→

(green in the picture):

A

⁢

B

→

+

2

3

⁢

B

⁢

E

→

=

A

⁢

B

→

+

2

3

⋅

1

2

⁢

[

-

A

⁢

B

→

+

(

A

⁢

C

→

-

A

⁢

B

→

)

]

=

1

3

⁢

(

A

⁢

B

→

+

A

⁢

C

→

)

(2)

Similarly, the third way goes first the side vector

A

⁢

C

→

and then

2

/

3

of the median vector

C

⁢

F

→

(red in the picture):

A

⁢

C

→

+

2

3

⁢

C

⁢

F

→

=

A

⁢

C

→

+

2

3

⋅

1

2

⁢

[

-

A

⁢

C

→

+

(

A

⁢

B

→

-

A

⁢

C

→

)

]

=

1

3

⁢

(

A

⁢

B

→

+

A

⁢

C

→

)

(3)

Thus the ways (2) and (3), where one goes from

A

to another vertex and continues along the corresponding median

2

/

3

of its length, lead to the point

M

which is attained directly along

A

⁢

D

.  This means that all medians intersect in

M

.  The

distance

of

M

from any vertex is

2

/

3

of the corresponding median, and so the rest of the median is

1

/

3

of its length, i.e. the ratio of the parts of any median is

2

:

1

.

Title

common point of triangle medians

Canonical name

CommonPointOfTriangleMedians

Date of creation

2013-03-22 17:46:54

Last modified on

2013-03-22 17:46:54

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Theorem

Classification

msc 51M04

Related topic

MutualPositionsOfVectors

Related topic

ParallelogramPrinciple

Related topic

DifferenceOfVectors

Related topic

TriangleMidSegmentTheorem

Related topic

LengthsOfTriangleMedians

Generated on Fri Feb 9 22:02:35 2018 by

LaTeXML
convex angle

convex angle

A positive angle is   if it is at most 180 , i.e.

π

radians.  Cf. the convex set.

A

convex angle

α

is

•

straight

if

α

=

180

o

,

•

obtuse

if

90

o

<

α

<

180

o

,

•

right

if

α

=

90

o

,

•

acute

if

0

o

<

α

<

90

o

,

•

skew

if it is acute or obtuse.

Note.

The

angle between two curves

(

http://planetmath.org/ConformalMapping

) which intersect each other in a point

P

means the angle between the

tangent lines

of the curves in

P

; such an angle may always be chosen acute or right.

Title

convex angle

Canonical name

ConvexAngle

Date of creation

2013-03-22 14:48:07

Last modified on

2013-03-22 14:48:07

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

17

Author

pahio (2872)

Entry type

Definition

Classification

msc 51M04

Related topic

Quadrant

Related topic

SolidAngle

Related topic

OrthogonalCurves

Related topic

ComplementaryAngles

Related topic

ContourLine

Defines

straight angle

Defines

obtuse angle

Defines

right angle

Defines

acute angle

Defines

straight

Defines

obtuse

Defines

right

Defines

acute

Generated on Fri Feb 9 22:02:37 2018 by

LaTeXML
corollary of Morley’s theorem

corollary of Morley’s theorem

We describe here, informally, a limiting case of Morley’s theorem.

One of the vertices of the

triangle

A

⁢

B

⁢

C

, namely

C

, has been pushed
off to infinity. Instead of two segments

B

⁢

C

and

C

⁢

A

, plus
two trisectors between them, we now have four

parallel

and equally
spaced lines. The triangle

P

⁢

Q

⁢

R

is still

equilateral

, and the three
triangles

adjacent

to it are still

isosceles

, but one of those has become
equilateral. We have

A

⁢

Q

⋅

B

⁢

R

=

A

⁢

R

⋅

B

⁢

P

.

Title

corollary of Morley’s theorem

Canonical name

CorollaryOfMorleysTheorem

Date of creation

2013-03-22 13:46:22

Last modified on

2013-03-22 13:46:22

Owner

drini (3)

Last modified by

drini (3)

Numerical id

8

Author

drini (3)

Entry type

Corollary

Classification

msc 51M04

Generated on Fri Feb 9 22:02:39 2018 by

LaTeXML
corresponding angles in transversal cutting

corresponding angles in transversal cutting

The following

theorem

is valid in

Euclidean geometry

:

Theorem 1

.

If two lines (

ℓ

and

m

) are cut by a third line, called a

transversal

(

t

), and one pair of corresponding angles (e.g. (

http://planetmath.org/Eg

)

α

and

β

) are

congruent

, then the cut lines are

parallel

.

Its

converse theorem

is also valid in Euclidean geometry:

Theorem 2

.

If two

parallel lines

(

ℓ

and

m

) are cut by a transversal (

t

), then each pair of corresponding angles (e.g.

α

and

β

) are congruent.

Remark

.

The angle

β

in both theorems may be replaced with its

vertical angle

β

1

.  The angles

α

and

β

1

are called

alternate interior angles

of each other.

Corollary 1

.

Two lines that are

perpendicular

to the same line are parallel to each other.

Corollary 2

.

If a line is perpendicular to one of two parallel lines, then it is also perpendicular to the other.

Corollary 3

.

If the left sides of two

convex angles

are parallel (or alternatively perpendicular) as well as their right sides, then the angles are congruent.

References

1

K. Väisälä:

Geometria

. Kolmas painos. Werner Söderström Osakeyhtiö, Porvoo ja Helsinki (1971).

Title

corresponding angles in transversal cutting

Canonical name

CorrespondingAnglesInTransversalCutting

Date of creation

2013-03-22 17:15:12

Last modified on

2013-03-22 17:15:12

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

12

Author

Wkbj79 (1863)

Entry type

Theorem

Classification

msc 51M04

Classification

msc 51-01

Related topic

EuclideanAxiomByHilbert

Related topic

HarmonicMeanInTrapezoid

Defines

transversal

Defines

vertical angle

Defines

alternate interior angle

Generated on Fri Feb 9 22:02:41 2018 by

LaTeXML
cosines law

cosines law

Cosines Law

.

Let

a

,

b

,

c

be the sides of a

triangle

and

A

the angle opposite to

a

. Then

a

2

=

b

2

+

c

2

-

2

⁢

b

⁢

c

⁢

cos

⁡

A

.

Remark.

Cosines law is the generalised form of

Pythagorean theorem

, which latter concerns only the

right triangles

.

Title

cosines law

Canonical name

CosinesLaw

Date of creation

2013-03-22 11:42:35

Last modified on

2013-03-22 11:42:35

Owner

drini (3)

Last modified by

drini (3)

Numerical id

26

Author

drini (3)

Entry type

Theorem

Classification

msc 51M04

Classification

msc 05A15

Classification

msc 05A10

Classification

msc 05A16

Synonym

law of cosines

Related topic

SinesLaw

Related topic

PythagorasTheorem

Related topic

DerivationOfCosinesLaw

Related topic

CosinesLaw

Related topic

StewartsTheorem

Related topic

AlternativeProofOfTheSinesLaw

Related topic

SinesLawProof

Related topic

ProofOfStewartsTheorem

Generated on Fri Feb 9 22:02:42 2018 by

LaTeXML
Dehn’s theorem

Dehn’s theorem

We all know the

elementary

formula

to compute the area of a

triangle

: basis times height divided by two. This formula can be justified with a scissor type

argument

:
one divides the triangle into smaller

polygons

and rearranges these polygons to obtain a

rectangle

which should have the same area.

Can we use the same argument to compute the volume of a

pyramid

? This is
the third Hilbert’s problem. Quite surprisingly the answer is

negative

, as states the

theorem

below. This means that the formulae to compute the volume of

polyhedra

cannot be proved without a limiting process (for example using integrals).

Definition 1

.

We say that two polyhedra

P

and

Q

are

scissor-equivalent

if there exists a finite number

P

1

,

…

,

P

N

of polyhedra and

θ

1

,

…

,

θ

N

isometries

such that

1.

P

=

⋃

k

=

1

N

P

k

and

Q

=

⋃

k

=

1

N

θ

k

⁢

(

P

k

)

;

2.

P

j

∩

P

k

and

θ

j

⁢

(

P

j

)

∩

θ

k

⁢

(

P

k

)

have empty interior
for every

k

≠

j

The properties given above assure that two scissor-equivalent polyhedra
must have the same volume. It is also simple to prove that the scissor-equivalence is indeed an

equivalence relation

.

Theorem 1

.

The

regular tetrahedron

is not

scissor-equivalent

to any

parallelepiped

.

Title

Dehn’s theorem

Canonical name

DehnsTheorem

Date of creation

2013-03-22 16:18:04

Last modified on

2013-03-22 16:18:04

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

11

Author

paolini (1187)

Entry type

Theorem

Classification

msc 51M04

Classification

msc 52B45

Related topic

BanachTarskiParadox

Related topic

HilbertsProblems

Related topic

RegularTetrahedron3

Defines

scissor-equivalent

Generated on Fri Feb 9 22:02:44 2018 by

LaTeXML
determining from angles that a triangle is isosceles

determining from angles that a triangle is isosceles

The following

theorem

holds in any

geometry

in which ASA is valid. Specifically, it holds in both

Euclidean geometry

and

hyperbolic geometry

(and therefore in

neutral geometry

) as well as in

spherical geometry

.

Theorem 1

.

If a

triangle

has two

congruent

angles, then it is

isosceles

.

Proof.

Let triangle

△

⁢

A

⁢

B

⁢

C

have angles

∠

⁢

B

and

∠

⁢

C

congruent.

Since we have

•

∠

⁢

B

≅

∠

⁢

C

•

B

⁢

C

¯

≅

C

⁢

B

¯

by the reflexive property (

http://planetmath.org/

Reflexive

) of

≅

(note that

B

⁢

C

¯

and

C

⁢

B

¯

denote the same

line segment

)

•

∠

⁢

C

≅

∠

⁢

B

by the

symmetric

property (

http://planetmath.org/Symmetric

) of

≅

we can use ASA to conclude that

△

⁢

A

⁢

B

⁢

C

≅

△

⁢

A

⁢

C

⁢

B

. Since corresponding parts of congruent triangles are congruent, we have that

A

⁢

B

¯

≅

A

⁢

C

¯

. It follows that

△

⁢

A

⁢

B

⁢

C

is isosceles.
∎

In geometries in which ASA and SAS are both valid, the

converse theorem

of this theorem is also true. This theorem is stated and proven in the entry angles of an

isosceles triangle

.

Title

determining from angles that a triangle is isosceles

Canonical name

DeterminingFromAnglesThatATriangleIsIsosceles

Date of creation

2013-03-22 17:12:23

Last modified on

2013-03-22 17:12:23

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

7

Author

Wkbj79 (1863)

Entry type

Theorem

Classification

msc 51M04

Classification

msc 51-00

Related topic

AnglesOfAnIsoscelesTriangle

Generated on Fri Feb 9 22:02:47 2018 by

LaTeXML
dihedral angle

dihedral angle

Two distinct

half-planes

, emanating from a same line

l

, the space (

ℝ

3

) into two regions called

dihedral angles

.  The line

l

is the

edge

of the dihedral angle and the bounding half-planes are its

sides

.

The angle, which the sides of a dihedral planes separate from a

normal plane

of the edge, is the

normal section

of the dihedral angle.  Apparently, all normal sections are equal.  According to the of the normal section, the dihedral angle may be called acute, right, obtuse,

straight

, skew (

http://planetmath.org/ConvexAngle

), convex and concave.  Unlike the

angle between two planes

, a dihedral angle may be over 90 .

If two planes intersect each other and if one of the four dihedral angles formed is right, then also the others are right.  Then we say that the planes are

perpendicular

to each other.

Title

dihedral angle

Canonical name

DihedralAngle

Date of creation

2013-03-22 18:48:07

Last modified on

2013-03-22 18:48:07

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

9

Author

pahio (2872)

Entry type

Definition

Classification

msc 51M04

Related topic

NormalSection

Related topic

PerpendicularityInEuclideanPlane

Related topic

TrihedralAngle

Related topic

SolidAngle

Defines

edge

Defines

side

Defines

normal section

Defines

concave

Defines

perpendicular

Defines

perpendicular planes

Generated on Fri Feb 9 22:02:55 2018 by

LaTeXML
explementary

explementary

The

explementary arc

of an arc

a

of a circle is the arc forming together with

a

the full circle.

Two angles are called

explementary angles

of each other, if their sum is the

full angle

2

⁢

π

, i.e.

360

∘

. In the below picture, the

α

=

60

∘

of an

equilateral triangle

and its explementary angle

β

=

300

∘

(which is an of the

triangle

) are seen.

Title

explementary

Canonical name

Explementary

Date of creation

2013-03-22 17:34:35

Last modified on

2013-03-22 17:34:35

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Definition

Classification

msc 51M04

Classification

msc 51F20

Related topic

ComplementaryAngles

Defines

explementary angle

Defines

explementary arc

Defines

full angle

Generated on Fri Feb 9 22:03:09 2018 by

LaTeXML