Sine law
Trigonometry 

Reference 
Laws and theorems 

Calculus 
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,
 $\backslash frac\{a\}\{\backslash sin\; A\}\; \backslash ,=\backslash ,\; \backslash frac\{b\}\{\backslash sin\; B\}\; \backslash ,=\backslash ,\; \backslash frac\{c\}\{\backslash sin\; C\}\; \backslash ,=\backslash ,\; D\; \backslash !$
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:
 $\backslash frac\{\backslash sin\; A\}\{a\}\; \backslash ,=\backslash ,\; \backslash frac\{\backslash sin\; B\}\{b\}\; \backslash ,=\backslash ,\; \backslash frac\{\backslash sin\; C\}\{c\}\; \backslash !$
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 nonenclosed 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.
Contents
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 h_{1}.
Using the definition of $\backslash textstyle\; \backslash sin\; \backslash alpha\; =\; \backslash frac\{\backslash text\{opposite\}\}\; \{\backslash text\{hypotenuse\}\}$ we see that for angle A on the right angle triangle ABR and C on R'BC we have:
 $\backslash sin\; A\; =\; \backslash frac\{h\_1\}\{c\}\backslash text\{;\; \}\; \backslash sin\; C\; =\; \backslash frac\{h\_1\}\{a\}$
Solving for h_{1}
 $h\_1\; =\; c\; \backslash sin\; A\backslash text\{;\; \}\; h\_1\; =\; a\; \backslash sin\; C\; \backslash ,$
Equating h_{1} in both expressions:
 $h\_1\; =\; c\; \backslash sin\; A\; =\; a\; \backslash sin\; C\; \backslash ,$
Therefore:
 $\backslash frac\{a\}\{\backslash sin\; A\}\; =\; \backslash frac\{c\}\{\backslash sin\; C\}.$
Doing the same thing from angle A to side a we call the altitude h_{2} and the two right angle triangles ABR and AR'C:
 $\backslash sin\; B\; =\; \backslash frac\{h\_2\}\{c\}\backslash text\{;\; \}\; \backslash sin\; C\; =\; \backslash frac\{h\_2\}\{b\}$
Solving for h_{2}
 $h\_2\; =\; c\; \backslash sin\; B\backslash text\{;\; \}\; h\_2\; =\; b\; \backslash sin\; C\; \backslash ,$
Therefore:
 $\backslash frac\{b\}\{\backslash sin\; B\}\; =\; \backslash frac\{c\}\{\backslash sin\; C\}$
Equating the $\backslash textstyle\; \backslash frac\{c\}\{\backslash sin\; C\}$ terms in both expressions above we have:
 $\backslash frac\{a\}\{\backslash sin\; A\}\; =\; \backslash frac\{b\}\{\backslash sin\; B\}\; =\; \backslash frac\{c\}\{\backslash sin\; C\}$
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.
Since we already have a right angle triangle we can use the definition of sine:
 $\backslash sin\; A\; =\; \backslash frac\{a\}\{c\}\; \backslash text\{;\; \}\; \backslash sin\; B\; =\; \backslash frac\{b\}\{c\}$
Solving for c:
 $c\; =\; \backslash frac\{a\}\{\backslash sin\; A\}\; \backslash text\{;\; \}\; c\; =\; \backslash frac\{b\}\{\backslash sin\; B\}$
Therefore:
 $\backslash frac\{a\}\{\backslash sin\; A\}\; =\; \backslash frac\{b\}\{\backslash sin\; B\}$
For the remaining angle C we need to remember that it is a right angle and $\backslash textstyle\; \backslash sin\; C\; =\; 1$ in this case. Therefore we can rewrite c = c · 1 as:
 $c\; =\; c\; \backslash sin\; C$
The above is equivalent to:
 $c\; =\; \backslash frac\{c\}\{\; \backslash sin\; C\}$
Equating c in both the equations above we again have:
 $\backslash frac\{a\}\{\backslash sin\; A\}\; =\; \backslash frac\{b\}\{\backslash sin\; B\}\; =\; \backslash frac\{c\}\{\backslash sin\; C\}$
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 h_{1} and create the two extended right triangles RBA' and RBC.
From the definition of sine we again have:
 $\backslash sin\; A\text{'}\; =\; \backslash frac\{h\_1\}\{c\}\backslash text\{;\; \}\; \backslash sin\; C\; =\; \backslash frac\{h\_1\}\{a\}$
We use identity $\backslash textstyle\; \backslash sin\; \backslash pi\; \; \backslash theta\; =\; \backslash sin\; \backslash theta$ to express $\backslash textstyle\; \backslash sin\; A\text{'}$ in terms of $\backslash textstyle\; \backslash sin\; A$. By definition we have:
 $A\; +\; A\text{'}\; =\; \backslash pi$
 $A\; =\; \backslash pi\; \; A\text{'}$
 $\backslash sin\; A\; =\; \backslash sin\; (\backslash pi\; \; A\text{'})\; =\; \backslash sin\; A\text{'}$
Therefore:
 $\backslash sin\; A\; =\; \backslash frac\{h\_1\}\{c\}\backslash text\{;\; \}\; \backslash sin\; C\; =\; \backslash frac\{h\_1\}\{a\}$
and
 $\backslash frac\{a\}\{\backslash sin\; A\}\; =\; \backslash frac\{c\}\{\backslash sin\; C\}$
We now draw an altitude from A calling it h_{2} and forming two right triangles ABR and AR'C.
From this we straightforwardly get:
 $\backslash sin\; B\; =\; \backslash frac\{h\_2\}\{c\}\backslash text\{;\; \}\; \backslash sin\; C\; =\; \backslash frac\{h\_2\}\{b\}$
and
 $\backslash frac\{b\}\{\backslash sin\; B\}\; =\; \backslash frac\{c\}\{\backslash sin\; C\}$
Equating the $\backslash textstyle\backslash frac\{\backslash sin\; C\}\{c\}$ in both equations above we again get:
 $\backslash frac\{a\}\{\backslash sin\; A\}\; =\; \backslash frac\{b\}\{\backslash sin\; B\}\; =\; \backslash frac\{c\}\{\backslash sin\; C\}$
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:
 $C\; =\; \backslash arcsin\; \{c\; \backslash sin\; A\; \backslash over\; a\}\; \backslash text\{\; or\; \}\; C\text{'}\; =\; \backslash pi\; \; \backslash arcsin\; \{c\; \backslash sin\; A\; \backslash over\; a\}$
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
 $\backslash frac\{\backslash sin\; A\}\{20\}\; =\; \backslash frac\{\backslash sin\; 40^\backslash circ\}\{24\}.$
 $A\; =\; \backslash arcsin\backslash left(\; \backslash frac\{20\backslash sin\; 40^\backslash circ\}\{24\}\; \backslash right)\; \backslash approx\; 32.39^\backslash circ.$
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
 $\backslash angle\; A\; =\; \backslash angle\; B\; =\; \backslash frac\{180^\backslash circ\backslash angle\; C\}\{2\}=\; 90\backslash frac\{\backslash angle\; C\}\{2\}\backslash !$
and
 $\{x\; \backslash over\; \backslash sin\; A\}=\{\backslash mbox\{chord\}\; \backslash over\; \backslash sin\; C\}\backslash text\{\; or\; \}\{x\; \backslash over\; \backslash sin\; B\}=\{\backslash mbox\{chord\}\; \backslash over\; \backslash sin\; C\}\backslash ,\backslash !$
 $\{\backslash mbox\{chord\}\; \backslash ,\backslash sin\; A\; \backslash over\; \backslash sin\; C\}\; =\; x\backslash text\{\; or\; \}\{\backslash mbox\{chord\}\; \backslash ,\backslash sin\; B\; \backslash over\; \backslash sin\; C\}\; =\; x.\backslash !$
Relation to the circumcircle
In the identity
 $\backslash frac\{a\}\{\backslash sin\; A\}\; \backslash ,=\backslash ,\; \backslash frac\{b\}\{\backslash sin\; B\}\; \backslash ,=\backslash ,\; \backslash frac\{c\}\{\backslash sin\; C\},\backslash !$
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
 $\backslash begin\{align\}$
\frac{abc} {2S} & {} = \frac{abc} {2\sqrt{s(sa)(sb)(sc)}} \\[6pt] & {} = \frac {2abc} {\sqrt{(a^2+b^2+c^2)^22(a^4+b^4+c^4) }}, \end{align}
where S is the area of the triangle and s is the semiperimeter
 $s\; =\; \backslash frac\{a+b+c\}\; \{2\}.$
The second equality above is essentially Heron's formula.
Spherical case
In the spherical case, the formula is:
 $\backslash frac\{\backslash sin\; A\}\{\backslash sin\; \backslash alpha\}\; =\; \backslash frac\{\backslash sin\; B\}\{\backslash sin\; \backslash beta\}\; =\; \backslash frac\{\backslash sin\; C\}\{\backslash sin\; \backslash gamma\}.$
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
 $\backslash lim\_\{\backslash alpha\; \backslash rightarrow\; 0\}\; \backslash frac\{\backslash sin\; \backslash alpha\}\{\backslash alpha\}\; =\; 1$
and the same for $\{\backslash sin\; \backslash beta\}$ and $\{\backslash sin\; \backslash gamma\}$.
 See also Spherical law of cosines and Halfside formula.
Hyperbolic case
In hyperbolic geometry when the curvature is −1, the law of sines becomes
 $\backslash frac\{\backslash sin\; A\}\{\backslash sinh\; a\}\; =\; \backslash frac\{\backslash sin\; B\}\{\backslash sinh\; b\}\; =\; \backslash frac\{\backslash sin\; C\}\{\backslash sinh\; c\}\; \backslash ,.$
In the special case when B is a right angle, one gets
 $\backslash sin\; C\; =\; \backslash frac\{\backslash sinh\; c\}\{\backslash sinh\; b\}\; \backslash ,$
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.
Unified formulation
Define a generalized sine function, depending also on a real parameter $K$:
 $\backslash sin\_K\; x\; =\; x\; \; \backslash frac\{K\; x^3\}\{3!\}\; +\; \backslash frac\{K^2\; x^5\}\{5!\}\; \; \backslash frac\{K^3\; x^7\}\{7!\}\; +\; \backslash cdots.$
The law of sines in constant curvature $K$ reads as^{[2]}
 $\backslash frac\{\backslash sin\; A\}\{\backslash sin\_K\; a\}\; =\; \backslash frac\{\backslash sin\; B\}\{\backslash sin\_K\; b\}\; =\; \backslash frac\{\backslash sin\; C\}\{\backslash sin\_K\; c\}\; \backslash ,.$
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 $p\_K(r)=2\backslash pi\; \backslash sin\_K\; r$. Therefore the law of sines can also be expressed as:
 $\backslash frac\{\backslash sin\; A\}\{p\_K(a)\}\; =\; \backslash frac\{\backslash sin\; B\}\{p\_K(b)\}\; =\; \backslash frac\{\backslash sin\; C\}\{p\_K(c)\}\; \backslash ,.$
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 alKhujandi, Abul Wafa Bozjani, Nasir alDin alTusi and Abu Nasr Mansur.^{[4]}
AlJayyani'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 alDīn alTūsī. 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 rightangled triangles in Book IV, and these solutions are in turn the bases for his solutions of general triangles."^{[7]} Regiomontanus was a 15thcentury German mathematician.
An equation with sines for tetrahedra
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 halfcircle. 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 halfcircle. 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 5dimensional.
See also
 Gersonides
 Halfside formula – for solving spherical triangles
 Law of tangents
 Mollweide's formula – for checking solutions of triangles
 Solution of triangles
 Surveying
References
External links
 Template:Springer
 cuttheknot
 Degree of Curvature
 Finding the Sine of 1 Degree