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{{About|the law of sines in trigonometry|the law of sines in physics|Snell's law}}
[[File:Acute Triangle.svg|thumb|right|A triangle labelled with the components of the law of sines. Capital {{math|''A''}}, {{math|''B''}} and {{math|''C''}} are the angles, and lower-case {{math|''a''}}, {{math|''b''}}, {{math|''c''}} are the sides opposite them. ({{math|''a''}} opposite {{math|''A''}}, etc.)]]
{{Trigonometry}}
In [[trigonometry]], the '''law of sines''', '''sine law''', '''sine formula''', or '''sine rule''' is an [[equation]] relating the [[lengths]] of the sides of a [[triangle]] (any shape) to the [[sine]]s of its angles. According to the law,

:<math> \frac{a}{\sin A} \,=\, \frac{b}{\sin B} \,=\, \frac{c}{\sin C} \,=\, d, </math>

where {{math|''a'', ''b''}}, and {{math|''c''}} are the lengths of the sides of a triangle, and {{math|''A'', ''B''}}, and {{math|''C''}} are the opposite angles (see the figure to the right), while {{math|''d''}} is the [[diameter]] of the triangle's [[circumcircle]]. When the last part of the equation is not used, the law is sometimes stated using the [[Multiplicative inverse|reciprocals]];

:<math> \frac{\sin A}{a} \,=\, \frac{\sin B}{b} \,=\, \frac{\sin C}{c}. </math>

The law of sines can be used to compute the remaining sides of a triangle when two angles and a side are known—a technique known as [[triangulation]]. It can also be used when two sides and one of the non-enclosed angles are known. In some such cases, the triangle is not uniquely determined by this data (called the ''ambiguous case'') and the technique gives two possible values for the enclosed angle.

The law of sines is one of two trigonometric equations commonly applied to find lengths and angles in scalene triangles, with the other being the [[law of cosines]].

The law of sines can be generalized to higher dimensions on surfaces with constant curvature.<ref name=mathworld>{{cite web|url=http://mathworld.wolfram.com/GeneralizedLawofSines.html|website=mathworld|title=Generalized law of sines}}</ref>

==Proof==
The area {{math|''T''}} of any triangle can be written as one half of its base times its height.  Selecting one side of the triangle as the base, the height of the triangle relative to that base is computed as the length of another side times the sine of the angle between the chosen side and the base.  Thus, depending on the selection of the base the area of the triangle can be written as any of:
:<math>T = \frac{1}{2}b(c \sin A) = \frac{1}{2}c(a \sin B) = \frac{1}{2}a(b \sin C)\,.</math>
Multiplying these by {{math|{{sfrac|2|''abc''}}}} gives
:<math>\frac{2T}{abc} = \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}\,.</math>

==The ambiguous case of triangle solution==
When using the law of sines to find a side of a triangle, an ambiguous case occurs when two separate triangles can be constructed from the data provided (i.e., there are two different possible solutions to the triangle). In the case shown below they are triangles {{math|''ABC''}} and {{math|''AB′C′''}}.

:[[File:PictureAmbitext.svg]]

Given a general triangle, the following conditions would need to be fulfilled for the case to be ambiguous:
* The only information known about the triangle is the angle {{math|''A''}} and the sides {{math|''a''}} and {{math|''c''}}.
* The angle {{math|''A''}} is [[Angle#Types of angles|acute]] (i.e., {{math|''A''}} < 90°).
* The side {{math|''a''}} is shorter than the side {{math|''c''}} (i.e., {{math|''a'' < ''c''}}).
* The side {{math|''a''}} is longer than the altitude {{math|''h''}} from angle {{math|''B''}}, where {{math|''h'' {{=}} ''c'' sin ''A''}} (i.e., {{math|''a'' > ''h''}}).

If all the above conditions are true, then each of angles {{math|''C''}} and {{math|''C′''}} produces a valid triangle, meaning that both of the following are true:

:<math> C' = \arcsin\frac{c \sin A}{a} \text{ or } C = \pi - \arcsin\frac{c \sin A}{a}.</math>

From there we can find the corresponding {{math|''B''}} and {{math|''b''}} or {{math|''B′''}} and {{math|''b′''}} if required, where {{math|''b''}} is the side bounded by angles {{math|''A''}} and {{math|''C''}} and {{math|''b′''}} bounded by {{math|''A''}} and {{math|''C′''}}.

Without further information it is impossible to decide which is the triangle being asked for.

==Examples==
The following are examples of how to solve a problem using the law of sines.

===Example 1===
Given: side {{math|''a'' {{=}} 20}}, side {{math|''c'' {{=}} 24}}, and angle {{math|''C'' {{=}} 40°}}. Angle {{math|''A''}} is desired.

Using the law of sines, we conclude that

:<math>\frac{\sin A}{20} = \frac{\sin 40^\circ}{24}.</math>

:<math> A = \arcsin\left( \frac{20\sin 40^\circ}{24} \right) \approx 32.39^\circ. </math>

Note that the potential solution {{math|1=''A'' = 147.61°}} is excluded because that would necessarily give {{math|1=''A'' + ''B'' + ''C'' > 180°}}.

===Example 2===
If the lengths of two sides of the triangle {{math|''a''}} and {{math|''b''}} are equal to {{math|''x''}}, the third side has length {{math|''c''}}, and the angles opposite the sides of lengths {{math|''a''}}, {{math|''b''}}, and {{math|''c''}} are {{math|''A''}}, {{math|''B''}}, and {{math|''C''}} respectively then

:<math>
\begin{align}
& A = B = \frac{180^\circ-C}{2}= 90^\circ-\frac{C}{2} \\[6pt]
& \sin A = \sin B = \sin \left(90^\circ-\frac{C}{2}\right) = \cos \left(\frac{C}{2}\right) \\[6pt]
& \frac{c}{\sin C}=\frac{a}{\sin A}=\frac{x}{\cos \left(\frac{C}{2}\right)} \\[6pt]
& \frac{c \cos \left(\frac{C}{2}\right)}{\sin C} = x
\end{align}
</math>

==Relation to the circumcircle==
In the identity

:<math> \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C},</math>

the common value of the three fractions is actually the [[diameter]] of the triangle's [[circumcircle]] which dates back to [[Ptolemy]].<ref>Coxeter, H. S. M. and Greitzer, S. L. ''Geometry Revisited''. Washington, DC: Math. Assoc. Amer., pp. 1–3, 1967</ref><ref name=":0">{{Cite web|url=http://www.pballew.net/lawofsin.html|title=Law of Sines|website=www.pballew.net|access-date=2018-09-18}}</ref>
[[File:Sinelaw radius.jpg|thumb|Deriving the ratio of the sine law equal to the circumscribing diameter. Note that triangle ADB passes through the center of the circumscribing circle.]]

=== Proof ===
As shown in the figure, let there be a circle with inscribed<math> \bigtriangleup ABC</math>  and another inscribed <math> \bigtriangleup ADB</math> that passes through the circle's center '''O'''. The <math> \angle AOD</math> has a [[central angle]] of ''<math> 180^\circ</math>'' and thus <math> \angle ABD = 90^\circ</math>. Since <math> \bigtriangleup ABD</math> is a right triangle,

:<math> \sin D= \frac{\text{opposite}}{\text{hypotenuse}}= \frac{c}{2R},</math>

where <math> R</math> is the radius of the circumscribing circle of the triangle.<ref name=":0" />
Angles <math> \angle C</math> and <math> \angle D</math> have the same [[central angle]] thus they are the same: <math> \angle C=\angle D</math>. Therefore,

:<math> \sin D= \sin C=\frac{c}{2R}.</math>

Rearranging yields

:<math> 2R=\frac{c}{\sin C}.</math>

Repeating the process of creating <math> \bigtriangleup ADB</math> with other points gives

{{equation box 1|equation=<math> \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}=2R.</math>}}

=== Relationship to the area of the triangle ===
The area of a triangle is given by <math>T=\frac{1}{2}ab \sin \theta</math>, where <math>\theta</math> is the angle enclosed by the sides of lengths {{math|''a''}} and {{math|''b''}}. Substituting the sine law into this equation gives

:<math>T=\frac{1}{2}ab \cdot \frac {c}{2R}.</math>

Taking <math>R</math> as the circumscribing radius,<ref>{{Citation|last=Mr. T's Math Videos|title=Area of a Triangle and Radius of its Circumscribed Circle|date=2015-06-10|url=https://www.youtube.com/watch?v=t6QNGDPG4Og|access-date=2018-09-18}}</ref>  

{{equation box 1|equation=<math>T=\frac{abc}{4R}.</math>}}

It can also be shown that this equality implies

:<math>\begin{align}
\frac{abc} {2T} & = \frac{abc} {2\sqrt{s(s-a)(s-b)(s-c)}} \\[6pt]
& = \frac {2abc} {\sqrt{{(a^2+b^2+c^2)}^2-2(a^4+b^4+c^4) }},
\end{align}</math>

where {{math|''T''}} is the area of the triangle and {{math|''s''}} is the [[semiperimeter]] <math>s = \frac{a+b+c}{2}.</math>

The second equality above readily simplifies to [[Heron's formula]] for the area.

The sine rule can also be used in deriving the following formula for the triangle’s area: Denoting the semi-sum of the angles' sines as <math>S =\frac {\sin A + \sin
B + \sin C}{2}</math>, we have<ref>Mitchell, Douglas W., "A Heron-type area formula in terms of sines," ''Mathematical Gazette'' 93, March 2009, 108–109.</ref>

{{equation box 1|equation=<math>T = D^{2} \sqrt{S(S-\sin A)(S-\sin B)(S-\sin C)}</math>}}

where <math>D</math> is the diameter of the circumcircle: <math>D=2R=\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}</math>.

==Curvature==
The law of sines takes on a similar form in the presence of curvature.

===Spherical case===
In the spherical case, the formula is:
:<math>\frac{\sin A}{\sin \alpha} = \frac{\sin B}{\sin \beta} = \frac{\sin C}{\sin \gamma}.</math>

Here, {{math|''α''}}, {{math|''β''}}, and {{math|''γ''}} are the angles at the center of the sphere subtended by the three arcs of the spherical surface triangle {{math|''a''}}, {{math|''b''}}, and {{math|''c''}}, respectively. {{math|''A''}}, {{math|''B''}}, and {{math|''C''}} are the surface angles opposite their respective arcs.

[[File:Spherical trigonometry vectors.svg|thumb|right|200px]]

=== Vector proof ===
Consider a unit sphere with three unit vectors {{math|'''OA'''}}, {{math|'''OB'''}} and {{math|'''OC'''}} drawn from the origin to the vertices of the triangle. Thus the angles {{math|''α''}}, {{math|''β''}}, and {{math|''γ''}} are the angles {{math|''a''}}, {{math|''b''}}, and {{math|''c''}}, respectively. The arc {{math|BC}} subtends an angle of magnitude {{math|''a''}} at the centre. Introduce a Cartesian basis with {{math|'''OA'''}} along the {{math|''z''}}-axis and {{math|'''OB'''}} in the {{math|''xz''}}-plane making an angle {{math|''c''}} with the {{math|''z''}}-axis. The vector {{math|'''OC'''}} projects to {{math|ON}} in the {{math|''xy''}}-plane and the angle between {{math|ON}} and the {{math|''x''}}-axis is {{math|''A''}}. Therefore, the three vectors have components:

:<math>\mathbf{OA} = \begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}, \quad \mathbf{OB} = \begin{pmatrix}\sin c \\ 0 \\ \cos c\end{pmatrix}, \quad \mathbf{OC} = \begin{pmatrix}\sin b\cos A \\ \sin b\sin A \\ \cos b\end{pmatrix}.</math>

The [[scalar triple product]], {{math|'''OA''' · ('''OB''' × '''OC''')}} is the volume of the [[parallelepiped]] formed by the position vectors of the vertices of the spherical triangle {{math|'''OA'''}}, {{math|'''OB'''}} and {{math|'''OC'''}}. This volume is invariant to the specific coordinate system used to represent {{math|'''OA'''}}, {{math|'''OB'''}} and {{math|'''OC'''}}. The value of the [[scalar triple product]] {{math|'''OA''' · ('''OB''' × '''OC''')}} is the 3&nbsp;×&nbsp;3 determinant with {{math|'''OA'''}}, {{math|'''OB'''}} and {{math|'''OC'''}} as its rows.  With the {{math|''z''}}-axis along {{math|'''OA'''}} the square of this determinant is
:<math>	
\begin{align}
\bigl(\mathbf{OA} \cdot (\mathbf{OB} \times \mathbf{OC})\bigr)^2 & = \bigl(\det (\mathbf{OA}, \mathbf{OB}, \mathbf{OC})\bigr)^2 \\[4pt]
& =  \left( \begin{vmatrix} 
0 & 0 & 1 \\ 	 	
\sin c & 0 & \cos c \\ 	
\sin b \cos A & \sin b \sin A & \cos b 	
\end{vmatrix} \right)^2 = \left(\sin b \sin c \sin A\right)^2.
\end{align}</math>
Repeating this calculation with the {{math|''z''}}-axis along {{math|'''OB'''}} gives {{math|(sin ''c'' sin ''a'' sin ''B'')<sup>2</sup>}}, while with the {{math|''z''}}-axis along {{math|'''OC'''}} it is {{math|(sin ''a'' sin ''b'' sin ''C'')<sup>2</sup>}}. Equating these expressions and dividing throughout by {{math|(sin ''a'' sin ''b'' sin ''c'')<sup>2</sup>}} gives 
:<math>
\frac{\sin^2 A}{\sin^2 a}=\frac{\sin^2 B}{\sin^2 b}=\frac{\sin^2 C}{\sin^2 c} = \frac{V^2}{\sin^2 a \sin^2 b \sin^2 c},
</math>
where {{mvar|V}} is the volume of the [[parallelepiped]] formed by the position vector of the vertices of the spherical triangle.

It is easy to see how for small spherical triangles, when the radius of the sphere is much greater than the sides of the triangle, this formula becomes the planar formula at the limit, since

:<math>\lim_{\alpha \rightarrow 0} \frac{\sin \alpha}{\alpha} = 1</math>

and the same for {{math|sin ''β''}} and {{math|sin ''γ''}}.

[[File:Sine law spherical small.svg|thumb|378x378px]]

=== Geometric proof ===
Consider a unit circle with:

: <math>OA = OB = OC = 1</math>

Construct point <math>D</math> and point <math>E</math> such that <math>\angle ADO = \angle AEO = 90^\circ</math>

Construct point <math>A'</math> such that <math>\angle A'DO = \angle A'EO = 90^\circ</math>

It can therefore be seen that <math>\angle ADA' = B</math> and <math>\angle AEA' = C</math>

Notice that <math>A'</math> is the projection of <math>A</math> on plane <math>OBC</math>. Therefore <math>\angle AA'D = \angle AA'E = 90^\circ</math>

By basic trigonometry, we have:

: <math>AD = \sin c 
</math>

: <math>AE = \sin b</math>

But <math>AA' = AD \sin B = AE \sin C </math>

Combining them we have:

: <math>\sin c \sin B = \sin b \sin C</math>

: <math>\frac{\sin B}{\sin b} =\frac{\sin C}{\sin c} </math>

By applying similar reasoning, we obtain the spherical law of sine:

: <math>\frac{\sin A}{\sin a} =\frac{\sin B}{\sin b} =\frac{\sin C}{\sin c} </math>

{{see also|Spherical trigonometry|Spherical law of cosines|Half-side formula}}

=== Other proofs ===
A purely algebraic proof can be constructed from the [[spherical law of cosines]].. From the identity <math>\sin^2 A=1-\cos^2 A</math> and the explicit expression for <math>\cos A</math> from the spherical law of cosines
:<math>
\begin{align}
   \sin^2\!A &=1-\left(\frac{\cos a  - \cos b\, \cos c}{\sin b \,\sin c}\right)^2\\
   &
   =\frac{(1-\cos^2\!b)(1-\cos^2\!c)-(\cos a  - \cos b\, \cos c)^2}
          {\sin^2\!b \,\sin^2\!c}\\
 \frac{\sin A}{\sin a}&=\frac{[1-\cos^2\!a-\cos^2\!b-\cos^2\!c+2\cos a\cos b\cos c]^{1/2}}{\sin a\sin b\sin c}.
\end{align}</math>
Since the right hand side is invariant under a cyclic permutation of <math>a,\;b,\;c</math> the spherical sine rule follows immediately.

The figure used in the Geometric proof above is used by and also provided in Banerjee<ref name="banerjee">{{Citation | last = Banerjee | first = Sudipto | date = 2004 | title = Revisiting Spherical Trigonometry with Orthogonal Projectors | journal = The College Mathematics Journal | volume = 35 | pages = 375–381 | publisher = Mathematical Association of America|postscript=[https://www.researchgate.net/publication/228849546_Revisiting_Spherical_Trigonometry_with_Orthogonal_Projectors Text online]}}</ref> (see Figure 3 in this paper) to derive the sine law using elementary linear algebra and projection matrices.

===Hyperbolic case===
In [[hyperbolic geometry]] when the curvature is −1, the law of sines becomes
:<math>\frac{\sin A}{\sinh a} = \frac{\sin B}{\sinh b} = \frac{\sin C}{\sinh c} \,.</math>

In the special case when {{math|''B''}} is a right angle, one gets
:<math>\sin C = \frac{\sinh c}{\sinh b} </math>

which is the analog of the formula in Euclidean geometry expressing the sine of an angle as the opposite side divided by the hypotenuse.

:See also [[hyperbolic triangle]].

===Unified formulation===
Define a generalized sine function, depending also on a real parameter {{math|''K''}}:

:<math>\sin_K x = x - \frac{K x^3}{3!} + \frac{K^2 x^5}{5!} - \frac{K^3 x^7}{7!} + \cdots.</math>

The law of sines in constant curvature {{math|''K''}} reads as<ref name="mathworld"/>

:<math>\frac{\sin A}{\sin_K a} = \frac{\sin B}{\sin_K b} = \frac{\sin C}{\sin_K c} \,.</math>

By substituting {{math|''K'' {{=}} 0}}, {{math|''K'' {{=}} 1}}, and {{math|''K'' {{=}} −1}}, one obtains respectively the Euclidean, spherical, and hyperbolic cases of the law of sines described above.

Let {{math|''p''<sub>''K''</sub>(''r'')}} indicate the circumference of a circle of radius {{math|''r''}} in a space of constant curvature {{math|''K''}}. Then {{math|''p''<sub>''K''</sub>(''r'') {{=}} 2π sin<sub>''K''</sub> ''r''}}. Therefore the law of sines can also be expressed as:

:<math>\frac{\sin A}{p_K(a)} = \frac{\sin B}{p_K(b)} = \frac{\sin C}{p_K(c)} \,.</math>

This formulation was discovered by [[János Bolyai]].<ref>{{cite book|last=Katok|first=Svetlana|authorlink=Svetlana Katok|title=Fuchsian groups|year=1992|publisher=University of Chicago Press|location=Chicago|isbn=0-226-42583-5|page=22}}</ref>

==Higher dimensions==
For an {{math|''n''}}-dimensional [[simplex]] (i.e., [[triangle]] ({{math|''n'' {{=}} 2}}), [[tetrahedron]] ({{math|''n'' {{=}} 3}}), [[pentatope]] ({{math|''n'' {{=}} 4}}), etc.) in {{math|''n''}}-dimensional [[Euclidean space]], the [[absolute value]] of the [[polar sine]] of the [[normal vector]]s of the [[face (geometry)|face]]s that meet at a [[vertex (geometry)|vertex]], divided by the hyperarea of the face opposite the vertex is independent of the choice of the vertex.  Writing {{math|''V''}} for the hypervolume of the {{math|''n''}}-dimensional simplex and {{math|''P''}} for the product of the hyperareas of its {{math|(''n''−1)}}-dimensional faces, the common ratio is

: <math>\frac{(nV)^{n-1}}{(n-1)! P}.</math>

For example, a tetrahedron has four triangular faces.  The absolute value of the polar sine of the normal vectors to the three faces that share a vertex, divided by the area of the fourth face will not depend upon the choice of the vertex:
:<math>
\begin{align}
& \frac{\bigl|\operatorname{psin}(\mathbf{n_2}, \mathbf{n_3}, \mathbf{n_4})\bigr|}{\mathrm{Area}_1} =
\frac{\bigl|\operatorname{psin}(\mathbf{n_1}, \mathbf{n_3}, \mathbf{n_4})\bigr|}{\mathrm{Area}_2} =
\frac{\bigl|\operatorname{psin}(\mathbf{n_1}, \mathbf{n_2}, \mathbf{n_4})\bigr|}{\mathrm{Area}_3} =
\frac{\bigl|\operatorname{psin}(\mathbf{n_1}, \mathbf{n_2}, \mathbf{n_3})\bigr|}{\mathrm{Area}_4} \\[4pt]
= {} & \frac{(3\operatorname{Volume}_\mathrm{tetrahedron})^2}{2!~\mathrm{Area}_1 \mathrm{Area}_2 \mathrm{Area}_3 \mathrm{Area}_4}\,.
\end{align}
</math>

==History==
According to [[Ubiratàn D'Ambrosio]] and [[Helaine Selin]], the spherical law of sines was discovered in the 10th century. It is variously attributed to [[Abu-Mahmud Khojandi]], [[Abu al-Wafa' Buzjani]], [[Nasir al-Din al-Tusi]] and [[Abu Nasr Mansur]].<ref name=Sesiano>Sesiano just lists al-Wafa as a contributor. Sesiano, Jacques (2000) "Islamic mathematics" pp. 137–157, in {{citation|title=Mathematics Across Cultures: The History of Non-western Mathematics|first1=Helaine|last1=Selin|first2=Ubiratan|last2=D'Ambrosio|year=2000|publisher=[[Springer Science+Business Media|Springer]]|isbn=1-4020-0260-2}}</ref>

[[Ibn Muʿādh al-Jayyānī]]'s ''The book of unknown arcs of a sphere'' in the 11th century contains the general law of sines.<ref name="MacTutor Al-Jayyani">{{MacTutor|id=Al-Jayyani|title=Abu Abd Allah Muhammad ibn Muadh Al-Jayyani}}</ref> The plane law of sines was later stated in the 13th century by [[Nasīr al-Dīn al-Tūsī]]. In his ''On the Sector Figure'', he stated the law of sines for plane and spherical triangles, and provided proofs for this law.<ref>{{cite book | first=J. Lennart | last=Berggren | title=The Mathematics of Egypt, Mesopotamia, China, India, and Islam: A Sourcebook | chapter=Mathematics in Medieval Islam | publisher=Princeton University Press | year=2007 | isbn=978-0-691-11485-9 | page=518 }}</ref>

According to [[Glen Van Brummelen]], "The Law of Sines is really [[Regiomontanus]]'s foundation for his solutions of right-angled triangles in Book IV, and these solutions are in turn the bases for his solutions of general triangles."<ref>Glen Van Brummelen (2009). "''[https://books.google.com/books?id=bHD8IBaYN-oC&pg=&dq&hl=en#v=onepage&q=&f=false The mathematics of the heavens and the earth: the early history of trigonometry]''". Princeton University Press. p.259. {{isbn|0-691-12973-8}}</ref> Regiomontanus was a 15th-century German mathematician.

==See also==
* [[Gersonides]]
* [[Half-side formula]]{{snd}} for solving [[spherical triangles]]
* [[Law of cosines]]
* [[Law of tangents]]
* [[Law of cotangents]]
* [[Mollweide's formula]]{{snd}} for checking solutions of triangles
* [[Solution of triangles]]
* [[Surveying]]

==References==
{{Reflist}}

==External links==
* {{springer|title=Sine theorem|id=p/s085520}}
* [http://www.cut-the-knot.org/proofs/sine_cosine.shtml#law The Law of Sines] at [[cut-the-knot]]
* [http://mysite.du.edu/~jcalvert/railway/degcurv.htm Degree of Curvature]
* [http://www.efnet-math.org/Meta/sine1.htm Finding the Sine of 1 Degree]
* [http://mathworld.wolfram.com/GeneralizedLawofSines.html Generalized law of sines to higher dimensions]
* [http://proofwiki.org/wiki/Law_of_Sines Law of Sines - ProofWiki]

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