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The law of sines (sines law, sine formula, sine rule), in trigonometry, is a statement about any triangle in a plane. Where the sides of the triangle are a, b and c and the angles opposite those sides are A, B and C, then the law of sines states:## Examples

## 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).## Derivation

Make a triangle with the sides a, b, and c, and angles A, B, and C. Draw the altitude from angle 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.### Determine an angle

For 2nd angle:
### Determine a side

## A law of sines for tetrahedra

## See also

## External links

- $frac\{a\}\{sin\; A\}\; =\; frac\{b\}\{sin\; B\}\; =\; frac\{c\}\{sin\; C\}\; =\; 2R$

where R is the radius of the triangle's circumcircle. This law is useful when computing the remaining sides of a triangle if two angles and a side are known, a common problem in the technique of triangulation. It can also be used when two sides and one of the non-enclosed angles are known; in this case, the formula may give two possible values for the enclosed angle. When this happens, often only one result will cause all angles to be less than 180°; in other cases, there are two valid solutions to the triangle (see the ambiguous case section of this article for further information).

It can be shown that

- $begin\{align\}$

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.

Here is an example of how to solve a problem using the law of sines:

Given: side a = 20, side c = 14, and angle C = 40 degrees

Using the law of sines, we know that :$tfrac\{a\}\{sin\; A\}\; =\; tfrac\{c\}\{sin\; C\}.$

Inserting the given values into the formula, we find that :$tfrac\{20\}\{sin\; A\}\; =\; tfrac\{14\}\{sin\; 40\}.$

Thus, angle A is equal to 66.67 degrees by taking the arcsine.

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' (30.48 m) and the angle C opposite to the chord is given in degrees, then $angle\; A\; =\; angle\; B\; =\; tfrac\{180-C\}\{2\}$ and

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

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

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 hypothenuse 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

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

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 angle A and side a will yield:

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

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

- $angle\; C\; =\; 180\; -\; A\; -\; B$

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

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

- $begin\{align\}\; \&$

sinangle OABcdotsinangle OBCcdotsinangle OCA &= sinangle OACcdotsinangle OCBcdotsinangle 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.

- PlainMath.Net-Area with sines Excellent tutorial on using the Law of Sines to find area.
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- The Law of Sines at cut-the-knot
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- PlainMath.Net- The Law of Sines Review of The Law of Sines
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Last updated on Thursday October 09, 2008 at 05:32:23 PDT (GMT -0700)

View this article at Wikipedia.org - Edit this article at Wikipedia.org - Donate to the Wikimedia Foundation

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