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Template:Short description Template:About Template:Multiple image Template:Trigonometry In trigonometry, the law of sines (sometimes called the sine formula or sine rule) is a mathematical equation relating the lengths of the sides of any triangle to the sines of its angles. According to the law, <math display="block"> \frac{a}{\sin{\alpha}} \,=\, \frac{b}{\sin{\beta}} \,=\, \frac{c}{\sin{\gamma}} \,=\, 2R, </math> where Template:Math, and Template:Math are the lengths of the sides of a triangle, and Template:Math, and Template:Math are the opposite angles (see figure 2), while Template:Math is the radius of the triangle's circumcircle. When the last part of the equation is not used, the law is sometimes stated using the reciprocals; <math display="block"> \frac{\sin{\alpha}}{a} \,=\, \frac{\sin{\beta}}{b} \,=\, \frac{\sin{\gamma}}{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>Template:Cite web</ref>

Proof

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With the side of length Template:Mvar as the base, the triangle's altitude can be computed as Template:Math or as Template:Math. Equating these two expressions gives <math display=block>\frac{b}{\sin \beta} = \frac{c}{\sin \gamma}\,,</math> and similar equations arise by choosing the side of length Template:Mvar or the side of length Template:Mvar as the base of the triangle.

The ambiguous case of triangle solution

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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 Template:Math and Template:Math.

Template:Bi Template:Clear

Given a general triangle, the following conditions would need to be fulfilled for the case to be ambiguous:

If all the above conditions are true, then each of angles Template:Math and Template:Math produces a valid triangle, meaning that both of the following are true: <math display="block"> {\gamma}' = \arcsin\frac{c \sin{\alpha}}{a} \quad \text{or} \quad {\gamma} = \pi - \arcsin\frac{c \sin{\alpha}}{a}.</math>

From there we can find the corresponding Template:Math and Template:Math or Template:Math and Template:Math if required, where Template:Math is the side bounded by vertices Template:Math and Template:Math and Template:Math is bounded by Template:Math and Template:Math.

Examples

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

Example 1

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File:Law of sines (example 01).svg
Example 1

Given: side Template:Math, side Template:Math, and angle Template:Math. Angle Template:Math is desired.

Using the law of sines, we conclude that <math display="block">\frac{\sin \alpha}{20} = \frac{\sin (40^\circ)}{24}.</math> <math display="block"> \alpha = \arcsin\left( \frac{20\sin (40^\circ)}{24} \right) \approx 32.39^\circ. </math>

Note that the potential solution Template:Math is excluded because that would necessarily give Template:Math.

Example 2

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File:Law of sines (example 02).svg
Example 2

If the lengths of two sides of the triangle Template:Math and Template:Math are equal to Template:Math, the third side has length Template:Math, and the angles opposite the sides of lengths Template:Math, Template:Math, and Template:Math are Template:Math, Template:Math, and Template:Math respectively then <math display="block">\begin{align} & \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{align}</math>

Relation to the circumcircle

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In the identity <math display="block"> \frac{a}{\sin{\alpha}} = \frac{b}{\sin{\beta}} = \frac{c}{\sin{\gamma}},</math> the common value of the three fractions is actually the diameter of the triangle's circumcircle. This result 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">Template:Cite web</ref>

Proof

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File:Sinelaw radius (Greek angles).svg
Deriving the ratio of the sine law equal to the circumscribing diameter. Note that triangle Template:Math passes through the center of the circumscribing circle with diameter Template:Math.

As shown in the figure, let there be a circle with inscribed <math> \triangle ABC</math> and another inscribed <math> \triangle ADB</math> that passes through the circle's center Template:Math. The <math> \angle AOD</math> has a central angle of <math> 180^\circ</math> and thus Template:Nowrap by Thales's theorem. Since <math> \triangle ABD</math> is a right triangle, <math display="block"> \sin{\delta}= \frac{\text{opposite}}{\text{hypotenuse}}= \frac{c}{2R},</math> where <math display="inline"> R= \frac{d}{2}</math> is the radius of the circumscribing circle of the triangle.<ref name=":0" /> Angles <math>{\gamma}</math> and <math>{\delta}</math> lie on the same circle and subtend the same chord Template:Math; thus, by the inscribed angle theorem, Template:Nowrap Therefore, <math display="block"> \sin{\delta} = \sin{\gamma} = \frac{c}{2R}.</math>

Rearranging yields <math display="block"> 2R = \frac{c}{\sin{\gamma}}.</math>

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

Template:Equation box 1 = \frac{b}{\sin{\beta}} = \frac{c}{\sin{\gamma}}=2R.</math>}}

Relationship to the area of the triangle

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The area of a triangle is given by Template:Nowrap where <math>\theta</math> is the angle enclosed by the sides of lengths Template:Math and Template:Math. Substituting the sine law into this equation gives <math display="block">T=\frac{1}{2}ab \cdot \frac {c}{2R}.</math>

Taking <math>R</math> as the circumscribing radius,<ref>Template:CitationTemplate:Cbignore</ref>

Template:Equation box 1

It can also be shown that this equality implies <math display="block">\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 Template:Math is the area of the triangle and Template:Math is the semiperimeter Template:Nowrap

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 Template:Nowrap we have<ref>Mitchell, Douglas W., "A Heron-type area formula in terms of sines," Mathematical Gazette 93, March 2009, 108–109.</ref>

Template:Equation box 1

where <math>R</math> is the radius of the circumcircle: Template:Nowrap

Spherical law of sines

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

Suppose the radius of the sphere is 1. Let Template:Math, Template:Math, and Template:Math be the lengths of the great-arcs that are the sides of the triangle. Because it is a unit sphere, Template:Math, Template:Math, and Template:Math are the angles at the center of the sphere subtended by those arcs, in radians. Let Template:Math, Template:Math, and Template:Math be the angles opposite those respective sides. These are dihedral angles between the planes of the three great circles.

Then the spherical law of sines says: <math display="block">\frac{\sin A}{\sin a} = \frac{\sin B}{\sin b} = \frac{\sin C}{\sin c}.</math>

File:Spherical trigonometry vectors.svg

Vector proof

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Consider a unit sphere with three unit vectors Template:Math, Template:Math and Template:Math drawn from the origin to the vertices of the triangle. Thus the angles Template:Math, Template:Math, and Template:Math are the angles Template:Math, Template:Math, and Template:Math, respectively. The arc Template:Math subtends an angle of magnitude Template:Math at the centre. Introduce a Cartesian basis with Template:Math along the Template:Math-axis and Template:Math in the Template:Math-plane making an angle Template:Math with the Template:Math-axis. The vector Template:Math projects to Template:Math in the Template:Math-plane and the angle between Template:Math and the Template:Math-axis is Template:Math. Therefore, the three vectors have components: <math display="block">\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, Template:Math is the volume of the parallelepiped formed by the position vectors of the vertices of the spherical triangle Template:Math, Template:Math and Template:Math. This volume is invariant to the specific coordinate system used to represent Template:Math, Template:Math and Template:Math. The value of the scalar triple product Template:Math is the Template:Math determinant with Template:Math, Template:Math and Template:Math as its rows. With the Template:Math-axis along Template:Math the square of this determinant is <math display="block"> \begin{align} \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{align}</math> Repeating this calculation with the Template:Math-axis along Template:Math gives Template:Math, while with the Template:Math-axis along Template:Math it is Template:Math. Equating these expressions and dividing throughout by Template:Math gives <math display="block"> \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 Template:Mvar is the volume of the parallelepiped 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 <math display="block">\lim_{a \to 0} \frac{\sin a}{a} = 1</math> and the same for Template:Math and Template:Math.

File:Sine law spherical small.svg

Geometric proof

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Consider a unit sphere with: <math display="block">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 display="block">\begin{align}

 AD &= \sin c \\
 AE &= \sin b

\end{align}</math>

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

Combining them we have: <math display="block">\begin{align} \sin c \sin B &= \sin b \sin C \\ \Rightarrow \frac{\sin B}{\sin b} &=\frac{\sin C}{\sin c} \end{align}</math>

By applying similar reasoning, we obtain the spherical law of sines: <math display="block">\frac{\sin A}{\sin a} =\frac{\sin B}{\sin b} =\frac{\sin C}{\sin c} </math>

Template:See also

Other proofs

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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 display="block">\begin{align}

  \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{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">Template:Citation</ref> (see Figure 3 in this paper) to derive the sine law using elementary linear algebra and projection matrices.

Hyperbolic case

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

In the special case when Template:Math is a right angle, one gets <math display="block">\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.

Template:See also

The case of surfaces of constant curvature

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Define a generalized sine function, depending also on a real parameter <math>\kappa</math>: <math display="block">\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}.</math>

The law of sines in constant curvature <math>\kappa</math> reads as<ref name="mathworld"/> <math display="block">\frac{\sin A}{\sin_\kappa a} = \frac{\sin B}{\sin_\kappa b} = \frac{\sin C}{\sin_\kappa c} \,.</math>

By substituting <math>\kappa=0</math>, <math>\kappa=1</math>, and <math>\kappa=-1</math>, one obtains respectively <math>\sin_{0}(x) = x</math>, <math>\sin_{1}(x) = \sin x</math>, and <math>\sin_{-1}(x) = \sinh x</math>, that is, the Euclidean, spherical, and hyperbolic cases of the law of sines described above.<ref name="mathworld"/>

Let <math>p_\kappa(r)</math> indicate the circumference of a circle of radius <math>r</math> in a space of constant curvature <math>\kappa</math>. Then <math>p_\kappa(r)=2\pi\sin_\kappa(r)</math>. Therefore, the law of sines can also be expressed as: <math display="block">\frac{\sin A}{p_\kappa(a)} = \frac{\sin B}{p_\kappa(b)} = \frac{\sin C}{p_\kappa(c)} \,.</math>

This formulation was discovered by János Bolyai.<ref>Template:Cite book</ref>

Higher dimensions

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A tetrahedron has four triangular facets. The absolute value of the polar sine (Template:Math) of the normal vectors to the three facets that share a vertex of the tetrahedron, divided by the area of the fourth facet will not depend upon the choice of the vertex:<ref>Template:Cite journal</ref>

<math display="block">\begin{align} & \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{align}</math>

More generally, for an Template:Math-dimensional simplex (i.e., triangle (Template:Math), tetrahedron (Template:Math), pentatope (Template:Math), etc.) in Template:Math-dimensional Euclidean 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. Writing Template:Math for the hypervolume of the Template:Math-dimensional simplex and Template:Math for the product of the hyperareas of its Template:Math-dimensional facets, the common ratio is <math display="block">\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}.</math>

Note that when the vectors Template:Math, from a selected vertex to each of the other vertices, are the columns of a matrix Template:Mvar then the columns of the matrix <math>N = -V (V^TV)^{-1} \sqrt{\det{V^TV}} / (n-1)!</math> are outward-facing normal vectors of those facets that meet at the selected vertex. This formula also works when the vectors are in a Template:Mvar-dimensional space having Template:Math. In the Template:Math case that Template:Mvar is square, the formula simplifies to <math>N = -(V^T)^{-1} |\det{V}| / (n-1)!\,.</math>

History

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An equivalent of the law of sines, that the sides of a triangle are proportional to the chords of double the opposite angles, was known to the 2nd century Hellenistic astronomer Ptolemy and used occasionally in his Almagest.<ref>Template:Cite book</ref>

Statements related to the law of sines appear in the astronomical and trigonometric work of 7th century Indian mathematician Brahmagupta. In his Brāhmasphuṭasiddhānta, Brahmagupta expresses the circumradius of a triangle as the product of two sides divided by twice the altitude; 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.<ref>Template:Cite book Template:Pb Template:Cite book</ref> An equation even closer to the modern law of sines appears in Brahmagupta's Khaṇḍakhādyaka, in a method for finding the distance between the Earth and a planet following an epicycle; however, Brahmagupta never treated the law of sines as an independent subject or used it systematically for solving triangles.<ref>Template:Cite book Template:Pb Template:Cite book</ref>

The spherical law of sines is sometimes credited to 10th century scholars Abu-Mahmud Khujandi or Abū al-Wafāʾ (it appears in his Almagest), but it is given prominence in Abū Naṣr Manṣūr's Treatise on the Determination of Spherical Arcs, and was credited to Abū Naṣr Manṣūr by his student al-Bīrūnī in his Keys to Astronomy.<ref>Template:Cite book Template:Pb Template:Cite book</ref> Ibn Muʿādh al-Jayyānī's 11th-century Book of Unknown Arcs of a Sphere also contains the spherical law of sines.<ref name="MacTutor Al-Jayyani">Template:MacTutor</ref>

The 13th-century Persian mathematician Naṣīr al-Dīn al-Ṭūsī stated and proved the planar law of sines:<ref>Template:Cite web</ref>

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's Elements (a geometric version of the law 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.<ref>Template:Cite book</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>Template:Cite book</ref> Regiomontanus was a 15th-century German mathematician.

See also

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References

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Template:Ancient Greek mathematics