Difference between revisions of "Law of Sines"

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Given a [[triangle]] with sides of length a, b and c, opposite [[angle]]s of measure A, B and C, respectively, and a [[circumcircle]] with radius R, <math>\frac{a}{\sin{A}}=\frac{b}{\sin{B}}=\frac{c}{\sin{C}}=2R</math>.
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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.
  
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==Theorem==
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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 <cmath> is the circumradius:
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</cmath>\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R<math></math>
 
==Proof ==
 
==Proof ==
 
=== Method 1 ===
 
=== Method 1 ===
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<center><math> \sin \theta = \frac{\frac a2}R \Leftrightarrow \frac a{\sin\theta} = 2R. </math> </center>
 
<center><math> \sin \theta = \frac{\frac a2}R \Leftrightarrow \frac a{\sin\theta} = 2R. </math> </center>
  
The same holds for b and c thus establishing the identity.
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The same holds for <math>b</math> and <math>c</math>, thus establishing the identity.
  
 
<center>[[Image:Lawofsines.PNG]]  
 
<center>[[Image:Lawofsines.PNG]]  
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<math>\frac{1}{2}bc\sin A = \frac{1}{2}ac\sin B = \frac{1}{2}ab\sin C </math>
 
<math>\frac{1}{2}bc\sin A = \frac{1}{2}ac\sin B = \frac{1}{2}ab\sin C </math>
  
Multiplying the equation by <math>\frac{2}{abc} </math> yeilds:  
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Multiplying the equation by <math>\frac{2}{abc} </math> yields:  
  
 
<math>\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} </math>
 
<math>\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} </math>
  
==See also==
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==Problems==
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===Introductory===
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*If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?
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<center><math>
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\mathrm{(A) \ } 2
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\qquad \mathrm{(B) \ } 8/\sqrt{15}
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\qquad \mathrm{(C) \ } 5/2
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\qquad \mathrm{(D) \ } \sqrt{6}
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\qquad \mathrm{(E) \ }  (\sqrt{6} + 1)/2
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</math></center>
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([[University of South Carolina High School Math Contest/1993 Exam/Problem 29|Source]])
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===Intermediate===
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*[[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]])
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===Advanced===
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Let <math>ABCD </math> be a convex quadrilateral with <math>AB=BC=CD </math>, <math>AC \neqBD </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>.
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([[2007 BMO Problems/Problem 1|Source]])
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==See Also==
 
* [[Trigonometry]]
 
* [[Trigonometry]]
 
* [[Trigonometric identities]]
 
* [[Trigonometric identities]]

Revision as of 20:54, 1 February 2008

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.

Theorem

In triangle $\triangle ABC$, where $a$ is the side opposite $A$, $b$ opposite $B$, $c$ opposite $C$, and \[is the circumradius:\]\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$$ (Error compiling LaTeX. ! Missing $ inserted.)

Proof

Method 1

In the diagram below, circle $O$ circumscribes triangle $ABC$. $OD$ is perpendicular to $BC$. Since $\triangle ODB \cong \triangle ODC$, $BD = CD = \frac a2$ and $\angle BOD = \angle COD$. But $\angle BAC = 2\angle BOC$ making $\angle BOD = \angle COD = \theta$. Therefore, we can use simple trig in right triangle $BOD$ to find that

$\sin \theta = \frac{\frac a2}R \Leftrightarrow \frac a{\sin\theta} = 2R.$

The same holds for $b$ and $c$, thus establishing the identity.

Lawofsines.PNG

This picture could be replaced by an asymptote drawing. It would be appreciated if you do this.

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: $[ABC] = \frac{1}{2}ab\sin C$

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

$\frac{1}{2}bc\sin A = \frac{1}{2}ac\sin B = \frac{1}{2}ab\sin C$

Multiplying the equation by $\frac{2}{abc}$ yields:

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

Problems

Introductory

  • If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?
$\mathrm{(A) \ } 2 \qquad \mathrm{(B) \ } 8/\sqrt{15}  \qquad \mathrm{(C) \ } 5/2 \qquad \mathrm{(D) \ } \sqrt{6} \qquad \mathrm{(E) \ }  (\sqrt{6} + 1)/2$

(Source)

Intermediate

(Source)

Advanced

Let $ABCD$ be a convex quadrilateral with $AB=BC=CD$, $AC \neqBD$ (Error compiling LaTeX. ! Undefined control sequence.), and let $E$ be the intersection point of its diagonals. Prove that $AE=DE$ if and only if $\angle BAD+\angle ADC = 120^{\circ}$.

(Source)

See Also

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