nyuuzyou/wikis huggingface dataset:
Angle Between Two Straight Lines described by Quadratic Equation

== Theorem ==
Let $\LL_1$ and $\LL_2$ represent $2$ straight lines in the Cartesian plane which are represented by a quadratic equation $E$ in two variables:
:$a x^2 + b y^2 + 2 h x y + 2 g x + 2 f y + c = 0$
Then the angle $\psi$ between $\LL_1$ and $\LL_2$ is given by:
:$\tan \psi = \dfrac {2 \sqrt {h^2 - a b} } {a + b}$
== Proof ==
From Homogeneous Quadratic Equation for Straight Lines Parallel to those Passing through Origin, $\LL_1$ and $\LL_2$ are parallel respectively to the $2$ straight lines through the origin $\LL'_1$ and $\LL'_2$ represented by the homogeneous quadratic equation:
:$a x^2 + 2 h x y + b y^2$
From Angle Between Two Straight Lines described by Homogeneous Quadratic Equation, the angle $\psi$ between $\LL'_1$ and $\LL'_2$ is given by:
:$\tan \psi = \dfrac {2 \sqrt {h^2 - a b} } {a + b}$
As:
:$\LL'_1$ is parallel to $\LL_1$
:$\LL'_2$ is parallel to $\LL_2$
it follows that the angle $\psi$ between $\LL_1$ and $\LL_2$ is given by:
:$\tan \psi = \dfrac {2 \sqrt {h^2 - a b} } {a + b}$
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $17$.
Category:Straight Lines
Angle Between Vectors in Terms of Dot Product

== Theorem ==
The angle between two non-zero vector quantities can be calculated by:
:$\theta = \arccos \dfrac {\mathbf v \cdot \mathbf w} {\norm {\mathbf v} \norm {\mathbf w} }$
where:
:$\mathbf v \cdot \mathbf w$ represents the dot product of $\mathbf v$ and $\mathbf w$ 
:$\norm {\, \cdot \,}$ represents vector length
:$\arccos$ represents arccosine.
== Proof ==
== Also see ==
* Angle Between Non-Zero Vectors Always Defined
Category:Vector Algebra
Category:Dot Product
Angle Bisector Theorem

== Theorem ==
Let $\triangle ABC$ be a triangle.
Let $D$ lie on the base $BC$ of $\triangle ABC$.
Then the following are equivalent:
:$(1): \quad AD$ is the angle bisector of $\angle BAC$
:$(2): \quad BD : DC = AB : AC$
where $BD : DC$ denotes the ratio between the lengths $BD$ and $DC$.
=== Exterior Angle ===
== Proof ==
=== $(1)$ implies $(2)$ ===
:
Let $CE$ be drawn through $C$ parallel to $DA$.
Let $BA$ be produced so as to meet $CE$ at $E$.
From we have that:
:$\angle ACE = \angle CAD$
But :
:$\angle CAD = \angle BAD$
and so:
:$\angle DAB = \angle ACE$
From :
:$\angle BAD = \angle AEC$
But from above $\angle ACE = \angle BAD$, so:
:$\angle ACE = \angle AEC$
So from :
:$AC = AE$
Since $AD \parallel EC$, from :
:$BD : DC = BA : AE$
But $AE = AC$, so:
:$BD : DC = AB : AC$
=== $(2)$ implies $(1)$ ===
Now suppose $BD : DC = AB : AC$.
Join $AD$.
Using the same construction, from :
:$BD : DC = AB : AE$
From :
:$BA : AC = BA : AE$
So from :
:$AC = AE$
So from :
:$\angle AEC = \angle ACE$
But from :
:$\angle AEC = \angle BAD$
Also, from :
:$\angle ACE = \angle CAD$
Therefore:
:$\angle BAD = \angle CAD$
and so $AD$ has bisected $\angle BAC$.
== Sources ==
* : Book $\text{VI}$. Propositions
* 
Category:Angle Bisector Theorem
Category:Triangles
Category:Named Theorems
Category:Angle Bisectors
Angle Bisector Theorem/Exterior Angle

== Theorem ==
Let $\triangle ABC$ be a triangle.
Let $AB$ be produced past $A$ to $D$.
Let the external angle $CAD$ be bisected by $AE$ where $BE$ is $BC$ produced.
:
Then:
:$BE : EC = AB : AC$
== Proof ==
Construct $CF$ parallel to the angle bisector $AE$.
By Parallelism implies Equal Corresponding Angles:
:$\angle AFC = \angle DAE$
By Parallelism implies Equal Alternate Angles:
:$\angle ACF = \angle CAE$
Given that $\angle DAC$ is bisected:
:$\angle DAE = \angle CAE$
By :
:$\angle AFC = \angle ACF$
By Triangle with Two Equal Angles is Isosceles:
:$\triangle AFC$ is isosceles
By the definition of isosceles triangle:
:$AF = AC$
By Parallelism implies Equal Corresponding Angles:
:$\angle BAE = \angle BFC$
$\angle ABC$ is shared.
By Triangles with Two Equal Angles are Similar:
:$\triangle BFC \sim \triangle BAE$
By the definition of similar triangles:
:$AB : BE = AF : CE$
Substituting for $AF$:
:$AB : BE = AC : CE$
Rearrange to give:
:$AB : AC = BE : EC$
== Sources ==
* 
Category:Angle Bisector Theorem
Angle Bisector Vector

== Theorem ==
Let $\mathbf u$ and $\mathbf v$ be vectors of non-zero length.
Let $\norm {\mathbf u}$ and $\norm {\mathbf v}$ be their respective lengths.
Then $\norm {\mathbf u} \mathbf v + \norm {\mathbf v} \mathbf u$ is the angle bisector of $\mathbf u$ and $\mathbf v$.
== Geometric Proof 1 ==
== Geometric Proof 2 ==
== Algebraic Proof ==
Category:Angle Bisector Vector
Category:Vector Algebra
Category:Angle Bisectors
Category:Euclidean Geometry
Angle Bisector Vector/Algebraic Proof

== Theorem ==
== Proof ==
Let $\mathbf a = \norm {\mathbf u} \mathbf v + \norm {\mathbf v} \mathbf u$.
Then:
Comparing the two expressions gives us:
:$\cos \angle \mathbf u, \mathbf a = \cos \angle \mathbf a, \mathbf v$
Since the angle used in the dot product is always taken to be between $0$ and $\pi$ and cosine is injective on this interval (from Shape of Cosine Function):
:$\angle \mathbf u, \mathbf a = \angle \mathbf a, \mathbf v$
The result follows.
Category:Angle Bisector Vector
Angle Bisector Vector/Geometric Proof 1

== Theorem ==
== Proof ==
:
As shown above:
:Let $\gamma$ be the angle between $\mathbf u$ and $\mathbf v$.
:Let $\alpha$ be the angle between $\norm {\mathbf u} \mathbf v$ and $\norm {\mathbf v} \mathbf u$.
:Let $\beta$ be the angle between $\mathbf u$ and $\norm {\mathbf u} \mathbf v + \norm {\mathbf v} \mathbf u$.
Note that $\norm {\mathbf u} \mathbf v$ is $\mathbf v$ multiplied by the length of $\mathbf u$.
By Vector Times Magnitude Same Length As Magnitude Times Vector the vectors $\norm {\mathbf u} \mathbf v$ and $\norm {\mathbf v} \mathbf u$ have equal length.
So $\norm {\mathbf u} \mathbf v$, $\norm {\mathbf v} \mathbf u$ and $\norm {\mathbf u} \mathbf v + \norm {\mathbf v} \mathbf u$ make an isosceles triangle.
Therefore:
But since $\mathbf v$ and $\norm {\mathbf u} \mathbf v$ are parallel, we also have:
Thus $\gamma = 2 \beta$, and the result follows.
Category:Angle Bisector Vector
Angle Bisector Vector/Geometric Proof 2

== Theorem ==
== Proof ==
The vectors $\norm {\mathbf u} \mathbf v$ and $\norm {\mathbf v} \mathbf u$ have equal length from Vector Times Magnitude Same Length As Magnitude Times Vector.
Thus $\norm {\mathbf u} \mathbf v + \norm {\mathbf v} \mathbf u$ is the diagonal of a rhombus.
 
The result follows from Diagonals of Rhombus Bisect Angles.
Category:Angle Bisector Vector
Angle Bisectors are Harmonic Conjugates

== Theorem ==
Let $\LL_1$ and $\LL_2$ be straight lines which intersect at $O$.
Let $\LL_3$ and $\LL_4$ be the angle bisectors of the angles formed at the point of intersection of $\LL_1$ and $\LL_2$.
Then $\LL_3$ and $\LL_4$ are harmonic conjugates with respect to $\LL_1$ and $\LL_2$.
== Proof ==
Consider a straight line parallel to $\LL_4$ which intersects $\LL_1$, $\LL_2$ and $\LL_3$ at $L$, $M$ and $N$ respectively.
From Bisectors of Adjacent Angles between Straight Lines Meeting at Point are Perpendicular, $\LL_3$ is perpendicular to $\LL_4$.
Hence as $LM$ is parallel to $\LL_4$, $LM$ is perpendicular to $ON$, which is $\LL3$.
:
The triangle $\triangle OLM$ has:
:$\angle NOL = \angle NOM$
:$\angle ONL = \angle ONM$ as both are right angles
:$ON$ common
So $\triangle ONL$ and $\triangle ONM$ are congruent.
So $N$ is the midpoint of $LM$.
From Harmonic Range with Unity Ratio, the points $L$, $N$, $M$ and the point at infinity form a harmonic range.
Hence from Straight Line which cuts Harmonic Pencil forms Harmonic Range, the straight lines $\LL_1$, $\LL_2$, $\LL_3$ and $\LL_4$ form a harmonic pencil.
That is: $\LL_3$ and $\LL_4$ are harmonic conjugates with respect to $\LL_1$ and $\LL_2$.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $22$. Equation of the bisectors of the angles between the two straight lines $a x^2 + 2 h x y + b y^2 = 0$
Category:Angle Bisectors
Category:Harmonic Ranges
Angle Bisectors are Locus of Points Equidistant from Lines

== Theorem ==
Let $\LL_1$ and $\LL_2$ be straight lines in the plane.
The locus of points which are equidistant from $\LL_1$ and $\LL_2$ are the angle bisectors of $\LL_1$ and $\LL_2$.
== Proof ==
Let $A'SA$ and $B'SB$ be the straight lines $\LL_1$ and $\LL_2$ respectively, intersecting at the point $S$.
Let $E$ denote the set of points equidistant from both $\LL_1$ and $\LL_2$.
Let $F$ denote the set of points on the angle bisectors of $\LL_1$ and $\LL_2$.
We are to show that $E = F$.
:
First we show that $F \subseteq E$.
Let $P \in F$.
, let $P$ be on the angle bisector of $\angle ASB$.
Drop perpendiculars $PM$ from $P$ to $SA$ and $PN$ from $P$ to $SB$.
Because:
:$\angle PSM = \angle PSN$
:$\angle PMS = \angle PNS$
:$PS$ is common
we have that:
:$\triangle PSM = \triangle PSN$
and so:
:$PM = PN$
That is, $P$ is equidistant from both $\LL_1$ and $\LL_2$ and so:
:$P \in E$
The same argument is used to show that an arbitrary point $P'$ on the angle bisector of $\angle ASB'$ is also equidistant from both $\LL_1$ and $\LL_2$.
That is:
:$P' \in E$
Hence:
:$F \subseteq E$
We see immediately that $S \in E$.
As $S$ is on the intersection of $\LL_1$ and $\LL_2$, $S$ is also on their angle bisectors.
Hence $S \in F$.
Let $P \in E$ such that $P \ne S$.
We note that $P \notin \LL_1$ and $P \notin \LL_2$ as that would mean trivially that $P \notin E$.
Drop perpendiculars $PM$ from $P$ to $SA$ and $PN$ from $P$ to $SB$.
With reference to the diagram, note that:
:$PN = PM$
:$\angle PMS = \angle PNS = 90 \degrees$
:$PS$ is common
By Pythagoras's Theorem:
:$SN^2 + PN^2 = PS^2 = SM^2 + PM^2$
But as $PM = PN$ it follows that:
:$SM = SN$
Thus $\triangle PSM$ and $\triangle PSN$ are congruent.
This means:
:$\angle PSN = \angle PSM$
and so $P$ is on one of the angle bisectors of $\LL_1$ and $\LL_2$.
That is:
:$P \in F$
Hence by definition of subset:
:$E \subseteq F$
We have:
:$E \subseteq F$
and:
:$F \subseteq E$
and so by definition of set equality:
:$E = F$
Hence the result.
== Sources ==
* : $\text {IV}$. Pure Geometry: Plane Geometry: The incentre
Category:Angle Bisectors
Angle between Straight Lines in Plane

== Theorem ==
Let $L_1$ and $L_2$ be straight lines embedded in a cartesian plane, given by the equations:
Then the angle $\psi$ between $L_1$ and $L_2$ is given by:
:$\psi = \arctan \dfrac {m_1 - m_2} {1 + m_1 m_2}$
=== General Form ===
== Proof ==
:
Let $\psi_1$ and $\psi_2$ be the angles that $L_1$ and $L_2$ make with the $x$-axis respectively.
Then by the definition of slope:
and so:
== Also presented as ==
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $5$
* : $\text {III}$. Analytical Geometry: The Straight Line: The angle between two lines
* : $\S 10$: Formulas from Plane Analytic Geometry: Angle $\psi$ between Two Lines having Slopes $m_1$ and $m_2$: $10.9$
* : $\S 8$: Formulas from Plane Analytic Geometry: Angle $\psi$ between Two Lines having Slopes $m_1$ and $m_2$: $8.9.$
Category:Angle between Straight Lines in Plane
Category:Straight Lines
Category:Angles
Angle between Straight Lines in Plane/Also presented as

== Angle between Straight Lines in Plane: Also presented as == 
When presenting Angle between Straight Lines in Plane, some sources retain the form:
:$\tan \psi = \dfrac {\tan \psi_2 - \tan \psi_1} {1 + \tan \psi_1 \tan \psi_2}$
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $5$
* : $\text {III}$. Analytical Geometry: The Straight Line: The angle between two lines
Category:Angle between Straight Lines in Plane
Angle between Straight Lines in Plane/General Form

== Theorem ==
Let $L_1$ and $L_2$ be straight lines embedded in a cartesian plane, given in general form:
Then the angle $\psi$ between $L_1$ and $L_2$ is given by:
:$\tan \psi = \dfrac {l_1 m_2 - l_2 m_1} {l_1 l_2 + m_1 m_2}$
== Proof ==
From the general equation for the straight line:
Hence the slope of $L_1$ and $L_2$ are $-\dfrac {l_1} {m_1}$ and $-\dfrac {l_2} {m_2}$ respectively.
Let $\psi$ be the angle between $L_1$ and $L_2$, as suggested.
Then:
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $5$
Category:Angle between Straight Lines in Plane
Angle between Vector Quantities in terms of Direction Cosines

== Theorem ==
Let $\mathbf a$ and $\mathbf b$ be vector quantities embedded in Cartesian $3$-space
Let $\theta$ be the angle between $\mathbf a$ and $\mathbf b$.
Then:
:$\cos \theta = \lambda_a \lambda_b + \mu_a \mu_b + \nu_a \nu_b$
where $\lambda_a$, $\mu_a$ and $\nu_a$ are the direction cosines of $\mathbf a$ with respect to the $x$-axis, $y$-axis and $z$-axis respectively, and similarly for $\mathbf b$.
== Proof ==
Let $\mathbf r$ be an arbitrary vector quantity embedded in a Cartesian $3$-space.
From Components of Vector in terms of Direction Cosines:
where:
:$x$, $y$ and $z$ denote the components of $\mathbf r$ in the $\mathbf i$, $\mathbf j$ and $\mathbf k$ directions respectively.
:$r$ denotes the magnitude of $\mathbf r$.
Hence:
== Sources ==
* : Chapter $\text I$: Vector Algebra: $\S 2$. $(5)$
* 
* 
Category:Direction Cosines
Angle of Intersection of Circles equals Angle between Radii

== Theorem ==
Let $\CC$ and $\CC'$ be circles whose centers are at $C$ and $C'$ respectively.
Let $\CC$ and $\CC'$ intersect at $A$ and $B$.
The angle of intersection of $\CC$ and $\CC'$ is equal to the angle between the radii to the point of intersection.
== Proof ==
From Normal to Circle passes through Center, the straight line passing through the center of a circle is normal to that circle.
Hence the radii $CA$ and $C'A$ are perpendicular to the tangents to $\CC$ and $\CC'$ respectively.
:
Thus, with reference to the above diagram, we have that:
:$\angle FAC = \angle DAC'$
as both are right angles.
Hence:
Hence the result.
== Sources ==
* : Chapter $\text {III}$. The Circle: $15$. Angle of intersection of two circles
Category:Circles
Angle of Intersection of Circles is Equal at Both Points

== Theorem ==
Let $\CC$ and $\CC'$ be circles whose centers are at $C$ and $C'$ respectively.
Let $\CC$ and $\CC'$ intersect at $A$ and $B$.
The angle of intersection of $\CC$ and $\CC'$ at $A$ is equal to the angle of intersection of $\CC$ and $\CC'$ at $B$.
== Proof ==
Consider the two triangles $CAC'$ and $CBC'$.
:
By definition of radius:
:$CA = CB$ and $C'A = C'B$
The result follows from Angle of Intersection of Circles equals Angle between Radii.
== Sources ==
* : Chapter $\text {III}$. The Circle: $15$. Angle of intersection of two circles
Category:Circles
Angle of Spherical Triangle from Sides

== Theorem ==
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
:$\cos A = \cosec b \cosec c \paren {\cos a - \cos b \cos c}$
== Proof ==
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $5$. The cosine-formula.
Category:Spherical Trigonometry
Angle of Tangent to Radius in Polar Coordinates

== Theorem ==
Let $C$ be a curve embedded in a plane defined by polar coordinates.
Let $P$ be the point at $\polar {r, \theta}$.
Then the angle $\psi$ made by the tangent to $C$ at $P$ with the radial coordinate is given by:
:$\tan \psi = r \dfrac {\d \theta} {\d r}$
== Proof ==
:
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 3$: Families of Curves. Orthogonal Trajectories
Category:Analytic Geometry
Category:Polar Coordinates
Category:Tangents
Angles in Circles have Same Ratio as Arcs

== Theorem ==
== Proof ==
Let $ABC, DEF$ be equal circles.
Let $\angle BGC, \angle EHF$ be angles at their centres $G, H$.
Let $\angle BAC, \angle EDF$ be angles at their circumferences.
We need to show that:
: $BC : EF = \angle BGC : \angle EHF = \angle BAC : \angle EDF$
:
Let any number of consecutive arcs $CK, KL$ be made equal to the arc $BC$.
Also let any number of consecutive arcs $FM, MN$ be made equal to the arc $EF$.
Join $GK, GL, HM, HN$.
Then since $BC = CK = KL$ it follows from Angles on Equal Arcs are Equal that:
: $\angle BGC = \angle CGK = \angle KGL$
Therefore, whatever multiple the arc $BL$ is of $BC$, that multiple is also the angle $\angle BGL$ of the angle $\angle BGC$.
For the same reason, whatever multiple the arc $NE$ is of $EF$, that multiple is also the angle $\angle NHE$ of the angle $\angle EHF$.
Then from Angles on Equal Arcs are Equal:
: $BL = EN \implies \angle BGL = \angle EHN$
: $BL > EN \implies \angle BGL > \angle EHN$
: $BL < EN \implies \angle BGL < \angle EHN$
Then from :
: $BC : EF = \angle BGL : \angle EHF$
But from the Inscribed Angle Theorem:
: $\angle BGC : \angle EHF = \angle BAC : \angle DEF$
Hence the result.
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Circles
Angles in Same Segment of Circle are Equal

== Theorem ==
== Proof ==
:
Let $ABCD$ be a circle, and let $\angle BAD, \angle BED$ be angles in the same segment $BAED$.
Let $F$ be the center of $ABCD$, and join $BF$ and $FD$.
From the Inscribed Angle Theorem:
:$\angle BFD = 2 \angle BAD$
:$\angle BFD = 2 \angle BED$
So:
:$\angle BAD = \angle BED$
Hence the result.
== Sources ==
* : Book $\text{III}$. Propositions
* 
* $(1)$
* $(1)$
Category:Circles
Angles of Orthic Triangle of Acute Triangle

== Theorem ==
Let $\triangle ABC$ be an acute triangle with sides $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.
Let $\triangle DEF$ be the orthic triangle of $\triangle ABC$.
Then the angles of $\triangle DEF$ are $180 \degrees - 2 A$, $180 \degrees - 2 B$ and $180 \degrees - 2 C$.
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(48)$
Category:Angles of Orthic Triangle of Acute Triangle
Category:Orthic Triangles
Category:Acute Triangles
Angles of Orthic Triangle of Acute Triangle/Proof

== Theorem ==
== Proof ==
:
Let $H$ be the orthocenter of $\triangle ABC$.
The quadrilateral $\Box FHDB$ is cyclic.
That is, $\Box FHDB$ can be circumscribed.
Hence:
Similarly, $\Box DHEC$ is also a cyclic quadrilateral.
Then:
The same argument can be applied to $B$ and $C$.
Hence the result.
== Sources ==
* : $\text V$. Trigonometry: The pedal triangle
Category:Angles of Orthic Triangle of Acute Triangle
Angles of Orthic Triangle of Obtuse Triangle

== Theorem ==
Let $\triangle ABC$ be an obtuse triangle whose sides are $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.
Let $\angle A$ be the obtuse angle of $\triangle ABC$.
Let $\triangle DEF$ be the orthic triangle of $\triangle ABC$.
Then the angles of $\triangle DEF$ are $2 A - 180 \degrees$, $2 B$ and $2 C$.
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(50)$
Category:Angles of Orthic Triangle of Obtuse Triangle
Category:Orthic Triangles
Category:Obtuse Triangles
Angles of Orthic Triangle of Obtuse Triangle/Proof

== Theorem ==
== Proof ==
:
Let $H$ be the orthocenter of $\triangle ABC$.
The quadrilateral $\Box DBEA$ is cyclic.
That is, $\Box DBEA$ can be circumscribed.
Hence:
Similarly, $\Box BEFC$ is also a cyclic quadrilateral.
Then:
The same argument can be applied to $C$, where we find:
:$(4): \quad \angle EFD = 2 C$
Hence:
Hence the result.
== Sources ==
* : $\text V$. Trigonometry: The pedal triangle
Category:Angles of Orthic Triangle of Obtuse Triangle
Angles on Equal Arcs are Equal

== Theorem ==
In equal circles, angles standing on equal arcs are equal to one another, whether at the center or at the circumference of those circles.
== Proof ==
Let $ABC$ and $DEF$ be equal circles.
Let $\angle BGC$ and $\angle EHF$ stand on the equal arcs $BC$ and $EF$.
:
Suppose $\angle BGC \ne \angle EHF$.
Then one of them is bigger.
Suppose $\angle BGC > \angle EHF$.
On the straight line $BG$, construct $\angle BGK$ equal to $\angle EHF$.
From Equal Angles in Equal Circles, these angles stand on equal arcs.
So arcs $BK = EF$.
But $EF = BC$ and so $BK = BC$, which is impossible.
So $\angle BGC$ and $\angle EHF$ are not unequal, therefore $\angle BGC = \angle EHF$.
From the Inscribed Angle Theorem, $2 \angle BAC = \angle BGC$ and $2 \angle EDF = \angle EHF$.
Therefore $2 \angle BAC = \angle EHF$.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Circles
Angles with Parallel or Perpendicular Arms are Equal or Supplementary

== Theorem ==
Let $A$ and $B$ be angles such that their corresponding arms are either parallel or perpendicular.
Then $A$ and $B$ are either equal or supplementary.
== Proof ==
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.6$: Corollary
Category:Angles
Category:Supplementary Angles
Angular Momentum of Particle moving with Constant Circular Motion

== Theorem ==
Let $P$ be a particle of mass $m$ moving:
:with constant speed $v$
:in a circle of radius $r$ about a point $O$.
Then the angular momentum has magnitude $m v r$.
== Proof ==
The angular momentum $\mathbf L$ of a particle with respect to a point is given by the formula: 
:$\mathbf L = \mathbf r \times \mathbf p$
A particle moving in a circle of radius $r$, its position vector $\mathbf r$ at an angle $\theta$ from the positive $x$-axis in a counterclockwise direction can be represented as: 
:$\mathbf r = r \cos \theta \mathbf i + r \sin \theta \mathbf j$
The velocity $\mathbf v$ of the particle moving with constant speed is directed tangentially to the circular path.
$\mathbf v$ can be expressed in terms of the angle $\theta$:
:$\mathbf v = \dfrac {\d \mathbf r} {\d t} = \dfrac \d {\d t} (r \cos \theta \mathbf i + r \sin \theta \mathbf j) = r \frac {\d \theta} {\d t} \paren {-\sin \theta \mathbf i + \cos \theta \mathbf j}$
Define (angular speed):
:$\omega = \dfrac {\d \theta} {\d t}$ 
So:
:$\mathbf v = r \omega \paren {-\sin \theta \mathbf i + \cos \theta \mathbf j}$
Now substituting into the angular momentum equation: 
:$L$ = $r p$
Using the vectors we have:
:$\mathbf L = \paren {r \cos \theta \mathbf i + r \sin \theta \mathbf j} \times \paren {m r \omega \paren {-\sin \theta \mathbf i + \cos \theta \mathbf j} }$
This can be simplified using the properties of cross products:
:$\mathbf i \times \mathbf i = 0$
:$\mathbf j \times \mathbf j = 0$
:$\mathbf i \times \mathbf j = \mathbf k$
Using the distributive property of the cross product, we get:
:$\mathbf L = m r^2 \omega \paren {\cos \theta \mathbf i \times \cos \theta \mathbf j} + m r^2 \omega \paren {\sin \theta \mathbf j \times \paren {-\sin \theta \mathbf i} }$
This results in:
:$\mathbf L = m r^2 \omega \paren {\cos^2 \theta + \sin^2 \theta} \mathbf k = m r^2 \omega \mathbf k$
Recall the relationship: 
:$v = r \omega \implies \omega = \dfrac v r$
Substituting $\omega$ into the angular momentum gives: 
:$\mathbf L = m r^2 \paren {\dfrac v r} \mathbf k = m v r \mathbf k$
Thus, the magnitude of the angular momentum $\mathbf L$ is:
:$L = m v r$
== Sources ==
* 
* 
Category:Angular Momentum
Category:Kinematics
Angular Speed of Rotation of Plane of Oscillation of Foucault's Pendulum

== Theorem ==
Let $P$ be a Foucault pendulum.
The angular speed of rotation $\alpha$ of the plane of oscillation of $P$ is given by:
:$\alpha = \omega \sin \lambda$
where:
:$\lambda$ denotes the latitude on Earth at which $P$ is located
:$\omega$ denotes the angular speed of rotation of Earth.
== Proof ==
== Sources ==
* 
* 
Category:Foucault's Pendulum
Angular Velocity is Vector Quantity

== Theorem ==
The physical quantity that is angular velocity can be correctly handled as a vector.
== Proof ==
In order to show that angular velocity is a vector, it is sufficient to demonstrate that it fulfils the vector space axioms.
Specifically, all we need to do is demonstrate the following.
Let $\bsomega_1$ be the angular velocity of a body about an axis which passes through a fixed point $O$.
Let $P$ be a point in $B$ whose position vector with respect to $O$ is $\mathbf r$.
The tangential velocity of $P$ is $\bsomega_1 \times \mathbf r$.
Now let $\bsomega_2$ be the angular velocity about a different axis which also passes through $O$ at an angle to the first axis.
The tangential velocity of $P$ due to this angular velocity is $\bsomega_2 \times \mathbf r$.
But linear velocities combine as vectors.
Hence the total linear velocity of $P$ is given by:
Hence the motion of $P$ is the same as it would be due to an angular velocity which is the vector sum of the two components.
== Sources ==
* : Chapter $\text {IV}$: The Operator $\nabla$ and its Uses: $5$. Simple Examples of Curl (footnote $*$)
Category:Angular Velocity