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Difference between revisions of "Law of Sines"

(→‎Method 1)
(→‎Advanced)
 
(7 intermediate revisions by 6 users not shown)
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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 where <math>R</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:
  
−
<math>\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R</math>
+
<cmath>\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R</cmath>
  
 
==Proof ==
 
==Proof ==
 
=== Method 1 ===
 
=== Method 1 ===
−
In the diagram below, circle <math> O </math> [[circumscribe]]s triangle <math> ABC </math>.  <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>.  Therefore, we can use simple trig in [[right triangle]] <math> BOD </math> to find that
+
<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 \Leftrightarrow \frac a{\sin\theta} = 2R. </math> </center>
+
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.
−
 
−
<center>[[Image:Lawofsines.PNG]]
 
−
{{asy replace}}
 
−
</center>
 
  
 
=== 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:  
  
−
<math>\frac{1}{2}bc\sin A = \frac{1}{2}ac\sin B = \frac{1}{2}ab\sin C </math>
+
<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:  
  
−
<math>\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} </math>
+
<cmath>\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} </cmath>
  
 
==Problems==
 
==Problems==
Line 37: Line 48:
  
 
<center><math>
 
<center><math>
−
\mathrm{(A) \ } 2
+
\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
Line 45: Line 56:
  
 
([[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>.
  
 
([[Mock AIME 4 2006-2007 Problems/Problem 15|Source]])
 
([[Mock AIME 4 2006-2007 Problems/Problem 15|Source]])
−
===Advanced===
+
===Olympiad===
−
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>.
+
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]]

Latest revision as of 15:19, 28 April 2024

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 to $A$, $b$ opposite to $B$, $c$ opposite to $C$, and where $R$ is the circumradius:

\[\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\]

Proof

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; 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 $O$ is the circumcenter of $\triangle ABC$. Point $D$ is on $BC$ such that $OD$ is perpendicular to $BC$. Since $\triangle ODB \cong \triangle ODC$, $BD = CD = \frac a2$ and $\angle BOD = \angle COD$. But $2\angle BAC = \angle BOC$ making $\angle BOD = \angle COD = \theta$. We can use simple trigonometry in right triangle $\triangle BOD$ to find that

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

The same holds for $b$ and $c$, 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 $[ABC] = \frac{1}{2}ab\sin C$.

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

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

Assuming the triangle in question is nondegenerate, $abc \ne 0$. 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)

Olympiad

Let $ABCD$ be a convex quadrilateral with $AB=BC=CD$, $AC \neq BD$, 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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