# Law of sines

(Redirected from Ambiguous case)
A triangle labelled with the components of the law of sines. Big A, B and C are the angles, and little a, b, c are the sides opposite them. (a opposite A, etc.)

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. However, calculating this may result in numerical error if an angle is close to 90 degrees. 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.

## Proof

There are three cases to consider in proving the law of sines. The first is when all angles of the triangle are acute. The second is when one angle is a right angle. The third is when one angle is obtuse.

### For acute triangles

We make a triangle with the sides a, b, and c, and angles A, B, and C. Then we draw the altitude from vertex B to side b; by definition it divides the original triangle into two right angle triangles: ABR and R'BC. Mark this line h1.

Triangle ABC with altitude from B drawn

Using the definition of $\textstyle \sin \alpha = \frac{\text{opposite}} {\text{hypotenuse}}$ we see that for angle A on the right angle triangle ABR and C on R'BC we have:

$\sin A = \frac{h_1}{c}\text{; } \sin C = \frac{h_1}{a}$

Solving for h1

$h_1 = c \sin A\text{; } h_1 = a \sin C \,$

Equating h1 in both expressions:

$h_1 = c \sin A = a \sin C \,$

Therefore:

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

Doing the same thing from angle A to side a we call the altitude h2 and the two right angle triangles ABR and AR'C:

Triangle ABC with altitude from A drawn
$\sin B = \frac{h_2}{c}\text{; } \sin C = \frac{h_2}{b}$

Solving for h2

$h_2 = c \sin B\text{; } h_2 = b \sin C \,$

Therefore:

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

Equating the $\textstyle \frac{c}{\sin C}$ terms in both expressions above we have:

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

### For right angle triangles

We make a triangle with the sides a, b, and c, and angles A, B, and C where C is a right angle.

Triangle ABC with right angle C.

Since we already have a right angle triangle we can use the definition of sine:

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

Solving for c:

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

Therefore:

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

For the remaining angle C we need to remember that it is a right angle and $\textstyle \sin C = 1$ in this case. Therefore we can rewrite c = c / 1 as:

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

Equating c in both the equations above we again have:

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

### For obtuse triangles

We make a triangle with the sides a, b, and c, and angles A, B, and C where A is an obtuse angle.In this case if we draw an altitude from any angle other than A the point where this line will touch the base of the triangle ABC will lie outside any of the lines a, b, or c. We draw the altitude from angle B, calling it h1 and create the two extended right triangles RBA' and RBC.

Obtuse triangle ABC with altitude drawn from B.

From the definition of sine we again have:

$\sin A' = \frac{h_1}{c}\text{; } \sin C = \frac{h_1}{a}$

We use identity $\textstyle \sin \pi - \theta = \sin \theta$ to express $\textstyle \sin A'$ in terms of $\textstyle \sin A$. By definition we have:

$A + A' = \pi$
$A = \pi - A'$
$\sin A = \sin (\pi - A') = \sin A'$

Therefore:

$\sin A = \frac{h_1}{c}\text{; } \sin C = \frac{h_1}{a}$

and

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

We now draw an altitude from A calling it h2 and forming two right triangles ABR and AR'C.

Obtuse triangle ABC with altitude from A

From this we straightforwardly get:

$\sin B = \frac{h_2}{c}\text{; } \sin C = \frac{h_2}{b}$

and

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

Equating the $\textstyle\frac{\sin C}{c}$ in both equations above we again get:

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

Proving the theorem in all cases.

## 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). In the case shown below they are triangle ABC and AB'C'.

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 A and the sides a and c
• The angle A is acute (i.e., A < 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 B, where h = c sin A (i.e., a > h).

Given all of the above premises are true, then either of the angles C or C' may produce a valid triangle; meaning, both of the following are true:

$C = \arcsin {c \sin A \over a} \text{ or } C' = \pi - \arcsin {c \sin A \over a}$

From there we can find the corresponding B and b or B' and b' if required, where b is the side bounded by angles A and C and b' bounded by A and 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:

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

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

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

## 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.

## 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.