# Difference between revisions of "Law of Sines"

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==Theorem== | ==Theorem== | ||

− | In triangle <math>\triangle ABC</math>, where <math>a</math> is the side opposite <math>A</math>, <math>b</math> opposite <math>B</math>, <math>c</math> opposite <math>C</math>, and <math></math> is the circumradius: | + | In triangle <math>\triangle ABC</math>, where <math>a</math> is the side opposite to <math>A</math>, <math>b</math> opposite to <math>B</math>, <math>c</math> opposite to <math>C</math>, and where <math>R</math> is the circumradius: |

− | < | + | <cmath>\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R</cmath> |

==Proof == | ==Proof == | ||

=== Method 1 === | === Method 1 === | ||

− | + | <asy> | |

+ | real r = 1; | ||

+ | pair A=(-1,5), B=(-4,-1), C=(4,-1), D, O; | ||

+ | O = circumcenter(A,B,C); | ||

+ | D = (B+C)/2; | ||

− | <center><math> \sin \theta = \frac{\frac a2}R \ | + | draw(C--A--B--C--O--B); |

+ | draw(O--D); | ||

+ | draw(circumcircle(A,B,C)); | ||

+ | |||

+ | label("$A$",A,(-1,1));label("$B$",B,(-1,-1));label("$C$",C,(1,-1));label("$D$",D,(0,-1));label("$O$",O,(0,1)); | ||

+ | label("$a/2$",(B+D)/2,(0,-1));label("$R$",(3*O+2*B)/5,(-1,1));label("$\theta$",O,(-0.8,-1.2));label("$\theta$",A,(0,-1.5)); | ||

+ | |||

+ | dot(A^^B^^C^^D^^O); | ||

+ | </asy> | ||

+ | |||

+ | In the diagram above, point <math> O </math> is the [[circumcenter]] of <math> \triangle ABC </math>. Point <math>D</math> is on <math>BC</math> such that <math> OD </math> is [[perpendicular]] to <math> BC </math>. Since <math> \triangle ODB \cong \triangle ODC </math>, <math> BD = CD = \frac a2 </math> and <math> \angle BOD = \angle COD </math>. But <math> 2\angle BAC = \angle BOC </math> making <math> \angle BOD = \angle COD = \theta </math>. We can use simple trigonometry in [[right triangle]] <math> \triangle BOD </math> to find that | ||

+ | |||

+ | <center><math> \sin \theta = \frac{\frac a2}R \iff \frac a{\sin\theta} = 2R. </math> </center> | ||

The same holds for <math>b</math> and <math>c</math>, thus establishing the identity. | The same holds for <math>b</math> and <math>c</math>, thus establishing the identity. | ||

− | |||

− | |||

− | |||

=== Method 2 === | === Method 2 === | ||

This method only works to prove the regular (and not extended) Law of Sines. | This method only works to prove the regular (and not extended) Law of Sines. | ||

− | The formula for the area of a triangle is | + | The formula for the area of a triangle is <math>[ABC] = \frac{1}{2}ab\sin C </math>. |

− | <math>[ABC] = \frac{1}{2}ab\sin C </math> | ||

Since it doesn't matter which sides are chosen as <math>a</math>, <math>b</math>, and <math>c</math>, the following equality holds: | Since it doesn't matter which sides are chosen as <math>a</math>, <math>b</math>, and <math>c</math>, the following equality holds: | ||

− | < | + | <cmath>\frac{1}{2}bc\sin A = \frac{1}{2}ac\sin B = \frac{1}{2}ab\sin C </cmath> |

− | Multiplying the equation by <math>\frac{2}{abc} </math> yields: | + | Assuming the triangle in question is nondegenerate, <math>abc \ne 0</math>. Multiplying the equation by <math>\frac{2}{abc} </math> yields: |

− | < | + | <cmath>\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} </cmath> |

==Problems== | ==Problems== | ||

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<center><math> | <center><math> | ||

− | \mathrm{(A) \ } | + | \mathrm{(A) \ 2} |

\qquad \mathrm{(B) \ } 8/\sqrt{15} | \qquad \mathrm{(B) \ } 8/\sqrt{15} | ||

\qquad \mathrm{(C) \ } 5/2 | \qquad \mathrm{(C) \ } 5/2 | ||

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([[University of South Carolina High School Math Contest/1993 Exam/Problem 29|Source]]) | ([[University of South Carolina High School Math Contest/1993 Exam/Problem 29|Source]]) | ||

+ | |||

===Intermediate=== | ===Intermediate=== | ||

*[[Triangle]] <math>ABC</math> has sides <math>\overline{AB}</math>, <math>\overline{BC}</math>, and <math>\overline{CA}</math> of [[length]] 43, 13, and 48, respectively. Let <math>\omega</math> be the [[circle]] [[circumscribe]]d around <math>\triangle ABC</math> and let <math>D</math> be the [[intersection]] of <math>\omega</math> and the [[perpendicular bisector]] of <math>\overline{AC}</math> that is not on the same side of <math>\overline{AC}</math> as <math>B</math>. The length of <math>\overline{AD}</math> can be expressed as <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are [[positive integer]]s and <math>n</math> is not [[divisibility | divisible]] by the [[square]] of any [[prime]]. Find the greatest [[integer]] less than or equal to <math>m + \sqrt{n}</math>. | *[[Triangle]] <math>ABC</math> has sides <math>\overline{AB}</math>, <math>\overline{BC}</math>, and <math>\overline{CA}</math> of [[length]] 43, 13, and 48, respectively. Let <math>\omega</math> be the [[circle]] [[circumscribe]]d around <math>\triangle ABC</math> and let <math>D</math> be the [[intersection]] of <math>\omega</math> and the [[perpendicular bisector]] of <math>\overline{AC}</math> that is not on the same side of <math>\overline{AC}</math> as <math>B</math>. The length of <math>\overline{AD}</math> can be expressed as <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are [[positive integer]]s and <math>n</math> is not [[divisibility | divisible]] by the [[square]] of any [[prime]]. Find the greatest [[integer]] less than or equal to <math>m + \sqrt{n}</math>. | ||

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([[Mock AIME 4 2006-2007 Problems/Problem 15|Source]]) | ([[Mock AIME 4 2006-2007 Problems/Problem 15|Source]]) | ||

===Advanced=== | ===Advanced=== | ||

− | Let <math>ABCD </math> be a convex quadrilateral with <math>AB=BC=CD </math>, <math>AC \ | + | Let <math>ABCD </math> be a convex quadrilateral with <math>AB=BC=CD </math>, <math>AC \neq BD </math>, and let <math>E </math> be the intersection point of its diagonals. Prove that <math>AE=DE </math> if and only if <math> \angle BAD+\angle ADC = 120^{\circ} </math>. |

([[2007 BMO Problems/Problem 1|Source]]) | ([[2007 BMO Problems/Problem 1|Source]]) | ||

+ | |||

==See Also== | ==See Also== | ||

* [[Trigonometry]] | * [[Trigonometry]] | ||

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* [[Geometry]] | * [[Geometry]] | ||

* [[Law of Cosines]] | * [[Law of Cosines]] | ||

+ | |||

+ | [[Category:Theorems]] | ||

[[Category:Trigonometry]] | [[Category:Trigonometry]] |

## Latest revision as of 00:37, 8 September 2018

The **Law of Sines** is a useful identity in a triangle, which, along with the law of cosines and the law of tangents can be used to determine sides and angles. The law of sines can also be used to determine the circumradius, another useful function.

## Contents

## Theorem

In triangle , where is the side opposite to , opposite to , opposite to , and where is the circumradius:

## Proof

### Method 1

In the diagram above, point is the circumcenter of . Point is on such that is perpendicular to . Since , and . But making . We can use simple trigonometry in right triangle to find that

The same holds for and , thus establishing the identity.

### Method 2

This method only works to prove the regular (and not extended) Law of Sines.

The formula for the area of a triangle is .

Since it doesn't matter which sides are chosen as , , and , the following equality holds:

Assuming the triangle in question is nondegenerate, . Multiplying the equation by yields:

## Problems

### Introductory

- If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?

(Source)

### Intermediate

- Triangle has sides , , and of length 43, 13, and 48, respectively. Let be the circle circumscribed around and let be the intersection of and the perpendicular bisector of that is not on the same side of as . The length of can be expressed as , where and are positive integers and is not divisible by the square of any prime. Find the greatest integer less than or equal to .

(Source)

### Advanced

Let be a convex quadrilateral with , , and let be the intersection point of its diagonals. Prove that if and only if .

(Source)