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Law of sines

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Property of all triangles on a Euclidean plane

This article is about the law of sines in trigonometry. For the law of sines in physics, seeSnell's law.

Law of Sines

Figure 1, Withcircumcircle

Figure 2, Without circumcircle

Two triangles labelled with components of the law of sines. The anglesα,β andγ are associated with the respective verticesA,B, andC; the respective sides of lengthsa,b, andc are opposite these (e.g., sidea is opposite vertexA with angleα).

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Intrigonometry, thelaw of sines,sine law,sine formula, orsine rule is anequation relating thelengths of the sides of anytriangle to thesines of itsangles. According to the law, ${\frac {a}{\sin {\alpha }}}\,=\,{\frac {b}{\sin {\beta }}}\,=\,{\frac {c}{\sin {\gamma }}}\,=\,2R,$ wherea,b, andc are the lengths of the sides of a triangle, andα,β, andγ are the opposite angles (see figure 2), whileR is theradius of the triangle'scircumcircle. When the last part of the equation is not used, the law is sometimes stated using thereciprocals; ${\frac {\sin {\alpha }}{a}}\,=\,{\frac {\sin {\beta }}{b}}\,=\,{\frac {\sin {\gamma }}{c}}.$ 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 astriangulation. 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 theambiguous 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 inscalene triangles, with the other being thelaw of cosines.

The law of sines can be generalized to higher dimensions on surfaces with constant curvature.^[1]

Proof

With the side of lengtha as the base, the triangle'saltitude can be computed asb sinγ or asc sinβ. Equating these two expressions gives ${\frac {\sin \beta }{b}}={\frac {\sin \gamma }{c}}\,,$ and similar equations arise by choosing the side of lengthb or the side of lengthc as the base of the triangle.

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 trianglesABC andABC′.

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 angleα and the sidesa andc.
The angleα isacute (i.e.,α < 90°).
The sidea is shorter than the sidec (i.e.,a <c).
The sidea is longer than the altitudeh from angleβ, whereh =c sinα (i.e.,a >h).

If all the above conditions are true, then each of anglesβ andβ′ produces a valid triangle, meaning that both of the following are true: ${\gamma }'=\arcsin {\frac {c\sin {\alpha }}{a}}\quad {\text{or}}\quad {\gamma }=\pi -\arcsin {\frac {c\sin {\alpha }}{a}}.$

From there we can find the correspondingβ andb orβ′ andb′ if required, whereb is the side bounded by verticesA andC andb′ is bounded byA andC′.

Examples

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

Example 1

Example 1

Given: sidea = 20, sidec = 24, and angleγ = 40°. Angleα is desired.

Using the law of sines, we conclude that ${\frac {\sin \alpha }{20}}={\frac {\sin(40^{\circ })}{24}}.$ $\alpha =\arcsin \left({\frac {20\sin(40^{\circ })}{24}}\right)\approx 32.39^{\circ }.$

Note that the potential solutionα = 147.61° is excluded because that would necessarily giveα +β +γ > 180°.

Example 2

Example 2

If the lengths of two sides of the trianglea andb are equal tox, the third side has lengthc, and the angles opposite the sides of lengthsa,b, andc areα,β, andγ respectively then ${\begin{aligned}&\alpha =\beta ={\frac {180^{\circ }-\gamma }{2}}=90^{\circ }-{\frac {\gamma }{2}}\\[6pt]&\sin \alpha =\sin \beta =\sin \left(90^{\circ }-{\frac {\gamma }{2}}\right)=\cos \left({\frac {\gamma }{2}}\right)\\[6pt]&{\frac {c}{\sin \gamma }}={\frac {a}{\sin \alpha }}={\frac {x}{\cos \left({\frac {\gamma }{2}}\right)}}\\[6pt]&{\frac {c\cos \left({\frac {\gamma }{2}}\right)}{\sin \gamma }}=x\end{aligned}}$

Relation to the circumcircle

In the identity ${\frac {a}{\sin {\alpha }}}={\frac {b}{\sin {\beta }}}={\frac {c}{\sin {\gamma }}},$ the common value of the three fractions is actually thediameter of the triangle'scircumcircle. This result dates back toPtolemy.^[2]^[3]

Proof

Deriving the ratio of the sine law equal to the circumscribing diameter. Note that triangleADB passes through the center of the circumscribing circle with diameterd.

As shown in the figure, let there be a circle with inscribed $\triangle ABC$ and another inscribed $\triangle ADB$ that passes through the circle's centerO. The $\angle AOD$ has acentral angle of $180^{\circ }$ and thus $\angle ABD=90^{\circ }$ , byThales's theorem. Since $\triangle ABD$ is a right triangle, $\sin {\delta }={\frac {\text{opposite}}{\text{hypotenuse}}}={\frac {c}{2R}},$ where ${\textstyle R={\frac {d}{2}}}$ is the radius of the circumscribing circle of the triangle.^[3] Angles ${\gamma }$ and ${\delta }$ lie on the same circle andsubtend the samechordc; thus, by theinscribed angle theorem, ${\gamma }={\delta }$ . Therefore, $\sin {\delta }=\sin {\gamma }={\frac {c}{2R}}.$

Rearranging yields $2R={\frac {c}{\sin {\gamma }}}.$

Repeating the process of creating $\triangle ADB$ with other points gives

${\frac {a}{\sin {\alpha }}}={\frac {b}{\sin {\beta }}}={\frac {c}{\sin {\gamma }}}=2R.$

Relationship to the area of the triangle

The area of a triangle is given by ${\textstyle T={\frac {1}{2}}ab\sin \theta }$ , where $\theta$ is the angle enclosed by the sides of lengthsa andb. Substituting the sine law into this equation gives $T={\frac {1}{2}}ab\cdot {\frac {c}{2R}}.$

Taking $R {\displaystyle R}$ as the circumscribing radius,^[4]

$T={\frac {abc}{4R}}.$

It can also be shown that this equality implies ${\begin{aligned}{\frac {abc}{2T}}&={\frac {abc}{2{\sqrt {s(s-a)(s-b)(s-c)}}}}\\[6pt]&={\frac {2abc}{\sqrt {{(a^{2}+b^{2}+c^{2})}^{2}-2(a^{4}+b^{4}+c^{4})}}},\end{aligned}}$ whereT is the area of the triangle ands is thesemiperimeter ${\textstyle s={\frac {1}{2}}\left(a+b+c\right).}$

The second equality above readily simplifies toHeron'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 ${\textstyle S={\frac {1}{2}}\left(\sin A+\sin B+\sin C\right)}$ , we have^[5]

$T=4R^{2}{\sqrt {S\left(S-\sin A\right)\left(S-\sin B\right)\left(S-\sin C\right)}}$

where $R {\displaystyle R}$ is the radius of the circumcircle: $2R={\frac {a}{\sin A}}={\frac {b}{\sin B}}={\frac {c}{\sin C}}$ .

Spherical law of sines

The spherical law of sines deals with triangles on a sphere, whose sides are arcs ofgreat circles.

Suppose the radius of the sphere is 1. Leta,b, andc be the lengths of the great-arcs that are the sides of the triangle. Because it is a unit sphere,a,b, andc are the angles at the center of the sphere subtended by those arcs, in radians. LetA,B, andC be the angles opposite those respective sides. These aredihedral angles between the planes of the three great circles.

Then the spherical law of sines says: ${\frac {\sin A}{\sin a}}={\frac {\sin B}{\sin b}}={\frac {\sin C}{\sin c}}.$

Vector proof

Consider a unit sphere with three unit vectorsOA,OB andOC drawn from the origin to the vertices of the triangle. Thus the anglesα,β, andγ are the anglesa,b, andc, respectively. The arcBC subtends an angle of magnitudea at the centre. Introduce a Cartesian basis withOA along thez-axis andOB in thexz-plane making an anglec with thez-axis. The vectorOC projects toON in thexy-plane and the angle betweenON and thex-axis isA. Therefore, the three vectors have components: $\mathbf {OA} ={\begin{pmatrix}0\\0\\1\end{pmatrix}},\quad \mathbf {OB} ={\begin{pmatrix}\sin c\\0\\\cos c\end{pmatrix}},\quad \mathbf {OC} ={\begin{pmatrix}\sin b\cos A\\\sin b\sin A\\\cos b\end{pmatrix}}.$

Thescalar triple product,OA ⋅ (OB ×OC) is the volume of theparallelepiped formed by the position vectors of the vertices of the spherical triangleOA,OB andOC. This volume is invariant to the specific coordinate system used to representOA,OB andOC. The value of thescalar triple productOA ⋅ (OB ×OC) is the3 × 3 determinant withOA,OB andOC as its rows. With thez-axis alongOA the square of this determinant is ${\begin{aligned}{\bigl (}\mathbf {OA} \cdot (\mathbf {OB} \times \mathbf {OC} ){\bigr )}^{2}&=\left(\det {\begin{pmatrix}\mathbf {OA} &\mathbf {OB} &\mathbf {OC} \end{pmatrix}}\right)^{2}\\[4pt]&={\begin{vmatrix}0&0&1\\\sin c&0&\cos c\\\sin b\cos A&\sin b\sin A&\cos b\end{vmatrix}}^{2}=\left(\sin b\sin c\sin A\right)^{2}.\end{aligned}}$ Repeating this calculation with thez-axis alongOB gives(sinc sina sinB)², while with thez-axis alongOC it is(sina sinb sinC)². Equating these expressions and dividing throughout by(sina sinb sinc)² gives ${\frac {\sin ^{2}A}{\sin ^{2}a}}={\frac {\sin ^{2}B}{\sin ^{2}b}}={\frac {\sin ^{2}C}{\sin ^{2}c}}={\frac {V^{2}}{\sin ^{2}(a)\sin ^{2}(b)\sin ^{2}(c)}},$ whereV is the volume of theparallelepiped formed by the position vector of the vertices of the spherical triangle. Consequently, the result follows.

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 $\lim _{a\to 0}{\frac {\sin a}{a}}=1$ and the same forsinb andsinc.

Geometric proof

Consider a unit sphere with: $OA=OB=OC=1$

Construct point $D {\displaystyle D}$ and point $E {\displaystyle E}$ such that $\angle ADO=\angle AEO=90^{\circ }$

Construct point $A^{'} {\displaystyle A'}$ such that $\angle A'DO=\angle A'EO=90^{\circ }$

It can therefore be seen that $\angle ADA'=B$ and $\angle AEA'=C$

Notice that $A^{'} {\displaystyle A'}$ is the projection of $A {\displaystyle A}$ on plane $O B C {\displaystyle OBC}$ . Therefore $\angle AA'D=\angle AA'E=90^{\circ }$

By basic trigonometry, we have: ${\begin{aligned}AD&=\sin c\\AE&=\sin b\end{aligned}}$

But $AA'=AD\sin B=AE\sin C$

Combining them we have: ${\begin{aligned}\sin c\sin B&=\sin b\sin C\\\Rightarrow {\frac {\sin B}{\sin b}}&={\frac {\sin C}{\sin c}}\end{aligned}}$

By applying similar reasoning, we obtain the spherical law of sine: ${\frac {\sin A}{\sin a}}={\frac {\sin B}{\sin b}}={\frac {\sin C}{\sin c}}$

See also:Spherical trigonometry,Spherical law of cosines, andHalf-side formula

Other proofs

A purely algebraic proof can be constructed from thespherical law of cosines. From the identity $\sin ^{2}A=1-\cos ^{2}A$ and the explicit expression for $\cos A$ from the spherical law of cosines ${\begin{aligned}\sin ^{2}\!A&=1-\left({\frac {\cos a-\cos b\,\cos c}{\sin b\,\sin c}}\right)^{2}\\&={\frac {\left(1-\cos ^{2}\!b\right)\left(1-\cos ^{2}\!c\right)-\left(\cos a-\cos b\,\cos c\right)^{2}}{\sin ^{2}\!b\,\sin ^{2}\!c}}\\[8pt]{\frac {\sin A}{\sin a}}&={\frac {\left[1-\cos ^{2}\!a-\cos ^{2}\!b-\cos ^{2}\!c+2\cos a\cos b\cos c\right]^{1/2}}{\sin a\sin b\sin c}}.\end{aligned}}$ Since the right hand side is invariant under a cyclic permutation of $a,\;b,\;c$ the spherical sine rule follows immediately.

The figure used in the Geometric proof above is used by and also provided in Banerjee^[6] (see Figure 3 in this paper) to derive the sine law using elementary linear algebra and projection matrices.

Hyperbolic case

Inhyperbolic geometry when the curvature is −1, the law of sines becomes ${\frac {\sin A}{\sinh a}}={\frac {\sin B}{\sinh b}}={\frac {\sin C}{\sinh c}}\,.$

In the special case whenB is a right angle, one gets $\sin C={\frac {\sinh c}{\sinh b}}$

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

The case of surfaces of constant curvature

Define a generalized sine function, depending also on a real parameter $\kappa$ : $\sin _{\kappa }(x)=x-{\frac {\kappa }{3!}}x^{3}+{\frac {\kappa ^{2}}{5!}}x^{5}-{\frac {\kappa ^{3}}{7!}}x^{7}+\cdots =\sum _{n=0}^{\infty }{\frac {(-1)^{n}\kappa ^{n}}{(2n+1)!}}x^{2n+1}.$

The law of sines in constant curvature $\kappa$ reads as^[1] ${\frac {\sin A}{\sin _{\kappa }a}}={\frac {\sin B}{\sin _{\kappa }b}}={\frac {\sin C}{\sin _{\kappa }c}}\,.$

By substituting $\kappa =0$ , $\kappa =1$ , and $\kappa =-1$ , one obtains respectively $\sin _{0}(x)=x$ , $\sin _{1}(x)=\sin x$ , and $\sin _{-1}(x)=\sinh x$ , that is, the Euclidean, spherical, and hyperbolic cases of the law of sines described above.^[1]

Let $p_{\kappa }(r)$ indicate the circumference of a circle of radius $r {\displaystyle r}$ in a space of constant curvature $\kappa$ . Then $p_{\kappa }(r)=2\pi \sin _{\kappa }(r)$ . Therefore, the law of sines can also be expressed as: ${\frac {\sin A}{p_{\kappa }(a)}}={\frac {\sin B}{p_{\kappa }(b)}}={\frac {\sin C}{p_{\kappa }(c)}}\,.$

This formulation was discovered byJános Bolyai.^[7]

Higher dimensions

Atetrahedron has four triangularfacets. Theabsolute value of thepolar sine (psin) of thenormal vectors to the three facets that share avertex of the tetrahedron, divided by the area of the fourth facet will not depend upon the choice of the vertex:^[8]

${\begin{aligned}&{\frac {\left|\operatorname {psin} (\mathbf {b} ,\mathbf {c} ,\mathbf {d} )\right|}{\mathrm {Area} _{a}}}={\frac {\left|\operatorname {psin} (\mathbf {a} ,\mathbf {c} ,\mathbf {d} )\right|}{\mathrm {Area} _{b}}}={\frac {\left|\operatorname {psin} (\mathbf {a} ,\mathbf {b} ,\mathbf {d} )\right|}{\mathrm {Area} _{c}}}={\frac {\left|\operatorname {psin} (\mathbf {a} ,\mathbf {b} ,\mathbf {c} )\right|}{\mathrm {Area} _{d}}}\\[4pt]={}&{\frac {(3~\mathrm {Volume} _{\mathrm {tetrahedron} })^{2}}{2~\mathrm {Area} _{a}\mathrm {Area} _{b}\mathrm {Area} _{c}\mathrm {Area} _{d}}}\,.\end{aligned}}$

More generally, for ann-dimensionalsimplex (i.e.,triangle (n = 2),tetrahedron (n = 3),pentatope (n = 4), etc.) inn-dimensionalEuclidean space, the absolute value of the polar sine of the normal vectors of the facets that meet at a vertex, divided by the hyperarea of the facet opposite the vertex is independent of the choice of the vertex. WritingV for the hypervolume of then-dimensional simplex andP for the product of the hyperareas of its(n − 1)-dimensional facets, the common ratio is ${\frac {\left|\operatorname {psin} (\mathbf {b} ,\ldots ,\mathbf {z} )\right|}{\mathrm {Area} _{a}}}=\cdots ={\frac {\left|\operatorname {psin} (\mathbf {a} ,\ldots ,\mathbf {y} )\right|}{\mathrm {Area} _{z}}}={\frac {(nV)^{n-1}}{(n-1)!P}}.$

Note that when the vectorsv₁, ...,v_n, from a selected vertex to each of the other vertices, are the columns of a matrixV then the rows of the matrixN = −⁠|detV|/(n−1)!⁠V⁻¹ are the outward-facing normal vectors of the facets that meet at the selected vertex.

History

An equivalent of the law of sines, that the sides of a triangle are proportional to thechords of double the opposite angles, was known to the 2nd century Hellenistic astronomerPtolemy and used occasionally in hisAlmagest.^[9]

Statements related to the law of sines appear in the astronomical and trigonometric work of 7th century Indian mathematicianBrahmagupta. In hisBrāhmasphuṭasiddhānta, Brahmagupta expresses the circumradius of a triangle as the product of two sides divided by twice thealtitude; the law of sines can be derived by alternately expressing the altitude as the sine of one or the other base angle times its opposite side, then equating the two resulting variants.^[10] An equation even closer to the modern law of sines appears in Brahmagupta'sKhaṇḍakhādyaka, in a method for finding the distance between the Earth and a planet following anepicycle; however, Brahmagupta never treated the law of sines as an independent subject or used it systematically for solving triangles.^[11]

The spherical law of sines is sometimes credited to 10th century scholarsAbu-Mahmud Khujandi orAbū al-Wafāʾ (it appears in hisAlmagest), but it is given prominence inAbū Naṣr Manṣūr'sTreatise on the Determination of Spherical Arcs, and was credited to Abū Naṣr Manṣūr by his studental-Bīrūnī in hisKeys to Astronomy.^[12]Ibn Muʿādh al-Jayyānī's 11th-centuryBook of Unknown Arcs of a Sphere also contains the spherical law of sines.^[13]

The 13th-century Persian mathematicianNaṣīr al-Dīn al-Ṭūsī stated and proved the planar law of sines:^[14]

In any plane triangle, the ratio of the sides is equal to the ratio of the sines of the angles opposite to those sides. That is, in triangle ABC, we have AB : AC = Sin(∠ACB) : Sin(∠ABC)

By employing the law of sines, al-Tusi could solve triangles where either two angles and a side were known or two sides and an angle opposite one of them were given. For triangles with two sides and the included angle, he divided them into right triangles that he could then solve. When three sides were given, he dropped a perpendicular line and then used Proposition II-13 of Euclid'sElements (a geometric version of thelaw of cosines). Al-Tusi established the important result that if the sum or difference of two arcs is provided along with the ratio of their sines, then the arcs can be calculated.^[15]

According toGlen Van Brummelen, "The Law of Sines is reallyRegiomontanus'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."^[16] Regiomontanus was a 15th-century German mathematician.

See also

Gersonides – Medieval Jewish philosopher
Half-side formula – for solvingspherical triangles
Law of cosines
Law of tangents
Law of cotangents
Mollweide's formula – for checking solutions of triangles
Solution of triangles
Surveying

References

^^a ^b ^c"Generalized law of sines".mathworld.
^Coxeter, H. S. M. and Greitzer, S. L.Geometry Revisited. Washington, DC: Math. Assoc. Amer., pp. 1–3, 1967
^^a ^b"Law of Sines".www.pballew.net. Archived from the original on December 29, 2002. Retrieved2018-09-18.
^Mr. T's Math Videos (2015-06-10),Area of a Triangle and Radius of its Circumscribed Circle,archived from the original on 2021-12-11, retrieved2018-09-18
^Mitchell, Douglas W., "A Heron-type area formula in terms of sines,"Mathematical Gazette 93, March 2009, 108–109.
^Banerjee, Sudipto (2004),"Revisiting Spherical Trigonometry with Orthogonal Projectors"(PDF),The College Mathematics Journal,35 (5), Mathematical Association of America:375–381,doi:10.1080/07468342.2004.11922099, archived fromthe original(PDF) on 2004-10-29
^Katok, Svetlana (1992).Fuchsian groups. Chicago: University of Chicago Press. p. 22.ISBN 0-226-42583-5.
^Eriksson, Folke (1978). "The law of sines for tetrahedra and n-simplices".Geometriae Dedicata.7 (1):71–80.doi:10.1007/bf00181352.
^Toomer, Gerald J., ed. (1998).Ptolemy's Almagest. Princeton University Press. pp. 7, fn. 10,462, fn. 96.
^Winter, Henry James Jacques (1952).Eastern Science.John Murray. p. 46.
Colebrooke, Henry Thomas (1817).Algebra, with Arithmetic and Mensuration from the Sanscrit of Brahmegupta and Bhascara. London:John Murray. pp. 299–300.
^Van Brummelen, Glen (2009).The Mathematics of the Heavens and the Earth. Princeton University Press. pp. 109–111.ISBN 978-0-691-12973-0.
Brahmagupta (1934).The Khandakhadyaka: An Astronomical Treatise of Brahmagupta. Translated by Sengupta, Prabodh Chandra. University of Calcutta.
^Sesiano, Jacques (2000). "Islamic mathematics". In Selin, Helaine; D'Ambrosio, Ubiratan (eds.).Mathematics Across Cultures: The History of Non-western Mathematics. Springer. pp. 137–157.ISBN 1-4020-0260-2.
Van Brummelen, Glen (2009).The Mathematics of the Heavens and the Earth. Princeton University Press. pp. 183–185.ISBN 978-0-691-12973-0.
^O'Connor, John J.;Robertson, Edmund F.,"Abu Abd Allah Muhammad ibn Muadh Al-Jayyani",MacTutor History of Mathematics Archive,University of St Andrews
^"Nasir al-Din al-Tusi - Biography".Maths History. Retrieved2025-03-10.
^Katz, Victor J. (2017-03-21).A History of Mathematics: An Introduction. Pearson. p. 315.ISBN 978-0-13-468952-4.
^Van Brummelen, Glen (2009).The Mathematics of the Heavens and the Earth: The Early History of Trigonometry.Princeton University Press. p. 259.ISBN 978-0-691-12973-0.

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