# Law of sines

In trigonometry, the law of sines, sine law, sine formula, or sine rule is an equation relating the lengths of the sides of any triangle to the sines of its angles. According to the law,

Law of Sines
Two triangles labelled with the components of the law of sines. α, β and γ are the angles associated with the vertices at capital A, B, and C, respectively. Lower-case a, b, and c are the lengths of the sides opposite them. (a is opposite α, etc.)
${\frac {a}{\sin {\alpha }}}\,=\,{\frac {b}{\sin {\beta }}}\,=\,{\frac {c}{\sin {\gamma }}}\,=\,2R,$ where a, b, and c are the lengths of the sides of a triangle, and α, β, and γ are the opposite angles (see figure 2), while R 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;
${\frac {\sin {\alpha }}{a}}\,=\,{\frac {\sin {\beta }}{b}}\,=\,{\frac {\sin {\gamma }}{c}}.$ 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 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.

## 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.

Ibn Muʿādh al-Jayyānī's The book of unknown arcs of a sphere in the 11th century contains the general law of sines. 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.

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." Regiomontanus was a 15th-century German mathematician.

## Proof

The area 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:

$T={\frac {1}{2}}b\left(c\sin {\alpha }\right)={\frac {1}{2}}c\left(a\sin {\beta }\right)={\frac {1}{2}}a\left(b\sin {\gamma }\right).$

Multiplying these by 2/abc gives
${\frac {2T}{abc}}={\frac {\sin {\alpha }}{a}}={\frac {\sin {\beta }}{b}}={\frac {\sin {\gamma }}{c}}\,.$

## 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 ABC and ABC′.

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 α and the sides a and c.
• The angle α is acute (i.e., α < 90°).
• The side a is shorter than the side c (i.e., a < c).
• The side a is longer than the altitude h from angle β, where h = c sin α (i.e., a > h).

If all the above conditions are true, then each of angles β and β′ 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 corresponding β and b or β′ and b′ if required, where b is the side bounded by vertices A and C and b′ is bounded by A and C′.

## Examples

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

### Example 1

Given: side a = 20, side c = 24, and angle γ = 40°. Angle α 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 solution α = 147.61° is excluded because that would necessarily give α + β + γ > 180°.

### Example 2

If the lengths of two sides of the triangle a and b are equal to x, the third side has length c, and the angles opposite the sides of lengths a, b, and c are α, β, and γ 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 the diameter of the triangle's circumcircle. This result dates back to Ptolemy.

Deriving the ratio of the sine law equal to the circumscribing diameter. Note that triangle ADB passes through the center of the circumscribing circle with diameter d.

### Proof

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 center O. The $\angle AOD$  has a central angle of $180^{\circ }$  and thus $\angle ABD=90^{\circ }$ . Since $\triangle ABD$  is a right triangle,

$\sin {\delta }={\frac {\text{opposite}}{\text{hypotenuse}}}={\frac {c}{2R}},$

where $R={\frac {d}{2}}$  is the radius of the circumscribing circle of the triangle. Angles ${\gamma }$  and ${\delta }$  have the same central angle thus they are the same: ${\gamma }={\delta }$ . 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 lengths a and b. Substituting the sine law into this equation gives

$T={\frac {1}{2}}ab\cdot {\frac {c}{2R}}.$

Taking $R$  as the circumscribing radius,

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

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}}

where T is the area of the triangle and s is the semiperimeter ${\textstyle s={\frac {a+b+c}{2}}.}$

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 ${\textstyle S={\frac {\sin A+\sin B+\sin C}{2}}}$ , we have

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

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

## The spherical law of sines

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 a, b, and c be the lengths of the great-arcs that are the sides of the triangle. Because it is a unit sphere, a, b, and c are the angles at the center of the sphere subtended by those arcs, in radians. Let A, B, and C 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:

${\frac {\sin A}{\sin a}}={\frac {\sin B}{\sin b}}={\frac {\sin C}{\sin c}}.$

### Vector proof

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

$\mathbf {OA} ={\begin{pmatrix}0\\0\\1\end{pmatrix}},\quad \mathbf {OB} ={\begin{pmatrix}\sin c\\0\\\cos c\end{pmatrix}},\quad \mathbf {OC} ={\begin{pmatrix}\sin b\cos A\\\sin b\sin A\\\cos b\end{pmatrix}}.$

The scalar triple product, OA ⋅ (OB × OC) is the volume of the parallelepiped formed by the position vectors of the vertices of the spherical triangle OA, OB and OC. This volume is invariant to the specific coordinate system used to represent OA, OB and OC. The value of the scalar triple product OA ⋅ (OB × OC) is the 3 × 3 determinant with OA, OB and OC as its rows. With the z-axis along OA 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 the z-axis along OB gives (sin c sin a sin B)2, while with the z-axis along OC it is (sin a sin b sin C)2. Equating these expressions and dividing throughout by (sin a sin b sin c)2 gives
${\frac {\sin ^{2}A}{\sin ^{2}a}}={\frac {\sin ^{2}B}{\sin ^{2}b}}={\frac {\sin ^{2}C}{\sin ^{2}c}}={\frac {V^{2}}{\sin ^{2}(a)\sin ^{2}(b)\sin ^{2}(c)}},$

where V 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

$\lim _{a\to 0}{\frac {\sin a}{a}}=1$

and the same for sin b and sin c.

### Geometric proof

Consider a unit sphere with:

$OA=OB=OC=1$

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

Construct point $A'$  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 $A'$  is the projection of $A$  on plane $OBC$ . Therefore $\angle AA'D=\angle AA'E=90^{\circ }$

By basic trigonometry, we have:

$AD=\sin c$

$AE=\sin b$

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

Combining them we have:

$\sin c\sin B=\sin b\sin C$

${\frac {\sin B}{\sin b}}={\frac {\sin C}{\sin c}}$

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

${\frac {\sin A}{\sin a}}={\frac {\sin B}{\sin b}}={\frac {\sin C}{\sin c}}$

### Other proofs

A purely algebraic proof can be constructed from the spherical 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 (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

${\frac {\sin A}{\sinh a}}={\frac {\sin B}{\sinh b}}={\frac {\sin C}{\sinh c}}\,.$

In the special case when B 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.

## The case of surfaces of constant curvature

Define a generalized sine function, depending also on a real parameter K:

$\sin _{K}x=x-{\frac {Kx^{3}}{3!}}+{\frac {K^{2}x^{5}}{5!}}-{\frac {K^{3}x^{7}}{7!}}+\cdots .$

The law of sines in constant curvature K reads as

${\frac {\sin A}{\sin _{K}a}}={\frac {\sin B}{\sin _{K}b}}={\frac {\sin C}{\sin _{K}c}}\,.$

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

Let pK(r) indicate the circumference of a circle of radius r in a space of constant curvature K. Then pK(r) = 2π sinK r. Therefore, the law of sines can also be expressed as:

${\frac {\sin A}{p_{K}(a)}}={\frac {\sin B}{p_{K}(b)}}={\frac {\sin C}{p_{K}(c)}}\,.$

This formulation was discovered by János Bolyai.

## Higher dimensions

For an n-dimensional simplex (i.e., triangle (n = 2), tetrahedron (n = 3), pentatope (n = 4), etc.) in n-dimensional Euclidean space, the absolute value of the polar sine (psin) 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 V for the hypervolume of the n-dimensional simplex and P for the product of the hyperareas of its (n − 1)-dimensional facets, the common ratio is

${\frac {(nV)^{n-1}}{(n-1)!P}}.$

For example, a tetrahedron has four triangular facets. The absolute value of the polar sine of the normal vectors to the three facets that share a vertex, divided by the area of the fourth facet will not depend upon the choice of the vertex:

{\begin{aligned}&{\frac {\left|\operatorname {psin} (\mathbf {n_{2}} ,\mathbf {n_{3}} ,\mathbf {n_{4}} )\right|}{\mathrm {Area} _{1}}}={\frac {\left|\operatorname {psin} (\mathbf {n_{1}} ,\mathbf {n_{3}} ,\mathbf {n_{4}} )\right|}{\mathrm {Area} _{2}}}={\frac {\left|\operatorname {psin} (\mathbf {n_{1}} ,\mathbf {n_{2}} ,\mathbf {n_{4}} )\right|}{\mathrm {Area} _{3}}}={\frac {\left|\operatorname {psin} (\mathbf {n_{1}} ,\mathbf {n_{2}} ,\mathbf {n_{3}} )\right|}{\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{aligned}}