# Proving the Law of Sines

• Feb 11th 2010, 05:56 PM
GreenDay14
Proving the Law of Sines
Hey Everyone,

I have a question here that asks:

Prove the Law of Sines for triangle ABC:

sinA/a = sinB/b = sinC/c

We have looked at all the side and angle axioms and up to euclidean geometry, any help here would be greatly appreciated. Thanks guys.
• Feb 11th 2010, 06:04 PM
Quote:

Originally Posted by GreenDay14
Hey Everyone,

I have a question here that asks:

Prove the Law of Sines for triangle ABC:

sinA/a = sinB/b = sinC/c

We have looked at all the side and angle axioms and up to euclidean geometry, any help here would be greatly appreciated. Thanks guys.

Hi GreenDay14.

If we do not have the triangle altitude on a particular base,
we can simply split the triangle into back to back right-angled triangles,
using any of the 3 sides as base.

Hence, there are 3 ways to write the triangle area..

$\displaystyle A=\frac{1}{2}abSinC=\frac{1}{2}acSinB=\frac{1}{2}b cSinA$

Hence

$\displaystyle bSinC=cSinB$

$\displaystyle aSinC=cSinA$

$\displaystyle aSinB=bSinA$

Hence

$\displaystyle \frac{b}{SinB}=\frac{c}{SinC}\ or\ \frac{SinB}{b}=\frac{SinC}{c}$ etc
• Feb 11th 2010, 06:17 PM
Random Variable
You can prove the Law of Sines by starting with a triangle consisting of vectors A,B, and C. This means means that A+B+C=0. Then take the cross product of both sides with vector A to get the first part of the relation. Do the same thing with vector B to get the second part of the relation.
• Feb 11th 2010, 06:22 PM
GreenDay14
thanks a lot for the help guys.
• Feb 11th 2010, 08:00 PM
Soroban
Hello, GreenDay14!

Quote:

Prove the Law of Sines for triangle ABC: .$\displaystyle \frac{\sin A}{a} \:=\: \frac{\sin B}{b} \:=\: \frac{\sin C}{c}$
Here is the classic textbook proof.
Code:

              C               *             *| *             * |  *         b *  |    *  a           *  |h      *         *    |        *         *    |          *       *      |            *     A * - - - * - - - - - - - * B               D

Draw altitude $\displaystyle CD$ to side $\displaystyle AB.$
. . Call it $\displaystyle h.$

In right triangle $\displaystyle CDA\!:\;\;\sin A \:=\:\frac{h}{b} \quad\Rightarrow\quad h \:=\:b\sin A$ .[1]

In right triangle $\displaystyle CDB\!:\;\;\sin B \:=\:\frac{h}{a} \quad\Rightarrow\quad h \:=\:a\sin B$ .[2]

Equate [1] and [2]: .$\displaystyle b\sin A \:=\:a\sin B \quad\Rightarrow\quad \boxed{\frac{\sin A}{a} \:=\:\frac{\sin B}{b}}$

In a similar fashion, we can prove that: .$\displaystyle \frac{\sin B}{b} \:=\:\frac{\sin C}{c}$