# Math Help - Proving the Law of Sines

1. ## 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.

2. 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..

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

Hence

$bSinC=cSinB$

$aSinC=cSinA$

$aSinB=bSinA$

Hence

$\frac{b}{SinB}=\frac{c}{SinC}\ or\ \frac{SinB}{b}=\frac{SinC}{c}$ etc

3. 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.

4. thanks a lot for the help guys.

5. Hello, GreenDay14!

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

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

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

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

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

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