# Law of sines

A triangle labelled with the components of the 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 an arbitrary triangle to the sines of its angles. According to the law,

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

where a, b, and c are the lengths of the sides of a triangle, and A, B, and C are the opposite angles (see the figure to the right), and D is the diameter of the triangle's circumcircle. When the last part of the equation is not used, sometimes the law is stated using the reciprocal:

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

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 formula gives two possible values for the enclosed angle, leading to an ambiguous case.

The law of sines is one of two trigonometric equations commonly applied to find lengths and angles in a general triangle, with the other being the law of cosines.

## Examples

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

Given: side a = 20, side c = 24, and angle C = 40°

Using the law of sines, we conclude that

$\frac{\sin A}{20} = \frac{\sin 40^\circ}{24}.$
$A = \arcsin\left( \frac{20\sin 40^\circ}{24} \right) \approx 32.39^\circ.$

Or another example of how to solve a problem using the law of sines:

If two sides of the triangle are equal to x and the length of the third side, the chord, is given as 100 feet and the angle C opposite the chord is given in degrees, then

$\angle A = \angle B = \frac{180^\circ-\angle C}{2}= 90-\frac{\angle C}{2}\!$

and

${x \over \sin A}={\mbox{chord} \over \sin C}\text{ or }{x \over \sin B}={\mbox{chord} \over \sin C}\,\!$

${\mbox{chord} \,\sin A \over \sin C} = x\text{ or }{\mbox{chord} \,\sin B \over \sin C} = x.\!$

## Numeric problems

Like the law of cosines, although the law of sines is mathematically true, it has problems for numeric use. Much precision may be lost if an arcsine is computed when the sine of an angle is close to one.

## Some applications

• The sine law can be used to prove the angle sum identity for sine when α and β are each between 0 and 90 degrees.
To prove this, make an arbitrary triangle with sides a, b, and c with corresponding arbitrary angles A, B and C. Draw a perpendicular to c from angle C. This will split the angle C into two different angles, α and β, that are less than 90 degrees, where we choose to have α to be on the same side as A and β be on the same side as B. Use the sine law identity that relates side c and side a. Solve this equation for the sine of C. Notice that the perpendicular makes two right angles triangles, also note that sin(A) = cos(α), sin(B) = cos(β) and that c = a sin(β) + b sin(α). After making these substitutions you should have sin(C) =sin(α + β) = sin(β)cos(α) + (b/a)sin(α)cos(α). Now apply the sine law identity that relates sides b and a and make the substitutions noted before. Now substitute this expression for (b/a) into the original equation for sin(α + β) and you will have the angle sum identity for α and β in terms of sine.
The only thing that was used in the proof that was not a definition was the sine law. Thus the sine law is equivalent to the angle sum identity when the angles sum is between 0 and 180 degrees and when each individual angle is between 0 and 90 degrees.

## The ambiguous case

When using the law of sines to solve triangles, there exists an ambiguous case where two separate triangles can be constructed (i.e., there are two different possible solutions to the triangle).

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

• The only information known about the triangle is the angle A and the sides a and b
• The angle A is acute (i.e., A < 90°).
• The side a is shorter than the side b (i.e., a < b).
• The side a is longer than the altitude of a right angled triangle with angle A and hypotenuse b (i.e., a > b sin A).

Given all of the above premises are true, the angle B may be acute or obtuse; meaning, one of the following is true:

$B = \arcsin {b \sin A \over a}\!$

or

$B= 180^\circ - \arcsin {b \sin A \over a}.$

## Relation to the circumcircle

In the identity

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

the common value of the three fractions is actually the diameter of the triangle's circumcircle.[1] It can be shown that this quantity is equal to

\begin{align} \frac{abc} {2S} & {} = \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}

where S is the area of the triangle and s is the semiperimeter

$s = \frac{a+b+c} {2}.$

The second equality above is essentially Heron's formula.

## Spherical case

In the spherical case, the formula is:

$\frac{\sin A}{\sin \alpha} = \frac{\sin B}{\sin \beta} = \frac{\sin C}{\sin \gamma}.$

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

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

$\lim_{\alpha \rightarrow 0} \frac{\sin \alpha}{\alpha} = 1$

and the same for ${\sin \beta}$ and ${\sin \gamma}$.

## 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.

## Unified formulation

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

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

The law of sines in constant curvature $K$ reads as[2]

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

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

Let $p_K(r)$ indicate the circumference of a circle of radius $r$ in a space of constant curvature $K$. Then $p_K(r)=2\pi \sin_K r$. 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.[3]

## 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 al-Khujandi, Abul Wafa Bozjani, Nasir al-Din al-Tusi and Abu Nasr Mansur.[4]

Al-Jayyani's The book of unknown arcs of a sphere in the 11th century introduced the general law of sines.[5] The plane law of sines was later described 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.[6]

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."[7] Regiomontanus was a 15th-century German mathematician.

## Derivation

Triangle ABC with line between Vertex C and side c used in the derivation or proof of the Sine Rule

Make a triangle with the sides a, b, and c, and angles A, B, and C. Draw the altitude from vertex C to the side across c; by definition it divides the original triangle into two right angle triangles. Mark the length of this line h.

It can be observed that:

$\sin A = \frac{h}{b}\text{ and } \sin B = \frac{h}{a}$

$b \sin A = h\text{ and } a \sin B = h \,$

$b \sin A = h = a\sin B \,$

Therefore

$b \sin A = a\sin B \,$

and

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

Doing the same thing with the line drawn between vertex A and side a will yield:

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

### Alternative derivation

Observe that the area $T$ of the triangle can be written as any of

$T = \frac{1}{2}bc \sin A = \frac{1}{2}ac \sin B = \frac{1}{2}ab \sin C\,.$

Multiplying these by $2/abc$ gives

$\frac{2T}{abc} = \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}\,.$

## An equation with sines for tetrahedra

A tetrahedron structure with vertices O, A, B, C. The product of the sines of ∠OAB, ∠OBC, ∠OCA appears on one side of the identity, and the product of the sines of ∠OAC, ∠OCB, ∠OBA on the other.

An equation involving sine functions and tetrahedra is as follows. For a tetrahedron with vertices O, A, B, C, it is true that

\begin{align} & {} \quad \sin\angle OAB\cdot\sin\angle OBC\cdot\sin\angle OCA \\ & = \sin\angle OAC\cdot\sin\angle OCB\cdot\sin\angle OBA. \end{align}

One may view the two sides of this identity as corresponding to clockwise and counterclockwise orientations of the surface.

Putting any of the four vertices in the role of O yields four such identities, but in a sense at most three of them are independent: If the "clockwise" sides of three of them are multiplied and the product is inferred to be equal to the product of the "counterclockwise" sides of the same three identities, and then common factors are cancelled from both sides, the result is the fourth identity. One reason to be interested in this "independence" relation is this: It is widely known that three angles are the angles of some triangle if and only if their sum is a half-circle. What condition on 12 angles is necessary and sufficient for them to be the 12 angles of some tetrahedron? Clearly the sum of the angles of any side of the tetrahedron must be a half-circle. Since there are four such triangles, there are four such constraints on sums of angles, and the number of degrees of freedom is thereby reduced from 12 to 8. The four relations given by this sines law further reduce the number of degrees of freedom, not from 8 down to 4, but only from 8 down to 5, since the fourth constraint is not independent of the first three. Thus the space of all shapes of tetrahedra is 5-dimensional.