Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Law of Sines"

(category)
(→‎Proof)
Line 3: Line 3:
 
==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> \angle BAC = 2\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  
+
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> \angle BAC = 2\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  
  
 
<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>

Revision as of 12:43, 20 October 2007

Given a triangle with sides of length a, b and c, opposite angles of measure A, B and C, respectively, and a circumcircle with radius R, $\frac{a}{\sin{A}}=\frac{b}{\sin{B}}=\frac{c}{\sin{C}}=2R$.

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 $\frac{2}{abc}$ yeilds:

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

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Law_of_Sines&oldid=18467"