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Difference between revisions of "Law of Sines"

(→‎Method 1)
(→‎Method 1)
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==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>
 +
defaultpen(fontsize(8));
 +
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 ===

Revision as of 17:45, 1 June 2011

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 where $R$ is the circumradius:

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

Proof

Method 1

[asy] defaultpen(fontsize(8)); 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:

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

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. Unknown error_msg), 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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