Law of sines: Definition from Answers.com

A triangle.

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This article is about the law of sines in trigonometry. For the law of sines in physics, see Snell's law.

In trigonometry, the law of sines (also known as the sines law, sine formula, or sine rule) is an equation relating the lengths of the sides of an arbitrary triangle to the sines of its angle. According to the law,

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

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). Sometimes the law is stated using the reciprocal of this equation:

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

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, the other being the law of cosines.

1 Examples
2 The ambiguous case
3 Relation to the circumcircle
4 Spherical case
5 History
6 Derivation
- 6.1 Determine an angle
- 6.2 Determine a side
7 A law of sines for tetrahedra
8 See also
9 External links

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

$\frac{\sin A}{20} = \frac{\sin 40^\circ}{24}.$

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

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

If two sides of the triangle are equal to R 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

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

and

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

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

The ambiguous case

When using the law of sines to solve triangles, under special conditions 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, where the angle A is not the included angle of the two sides (in the above image, the angle C is the included angle).
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:

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

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

Relation to the circumcircle

In the equation

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

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

$\begin{align} \frac{abc} {2S} & {} = \frac{abc} {2\sqrt{s(s-a)(s-b)(s-c)}} \\ & {} = \frac {2abc} {\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4) }}, \end{align}$

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 centre 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.

History

The spherical law of sines was discovered by the 10th century Persian mathematician Abu Nasr Mansur, whereas the plane law of sines was discovered by the 13th century Persian mathematician Nasīr al-Dīn al-Tūsī.

Derivation

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:

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

Therefore

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

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:

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

Determine an angle

For 2nd angle:

$\angle A = \sin^{-1} \Bigg( \left ( \frac{\sin B}{b} \right ) a \Bigg)$

for 3rd angle:

$\angle C = 180 - A - B$

Determine a side

$a = (\sin A) \left ( \frac{b}{\sin B} \right )$

A law of 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.

A corollary of the law of sines as stated above is that in a tetrahedron with vertices O, A, B, C, we have

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

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.

External links

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	Year 1461 (in Science & Technology)
	trigonometry (branch of mathematics)
	1617 (chronology)

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