# The quickest Sine Rule proof

In this post I’ll be demonstrating how to prove that the Sine Rule is true in the quickest manner possible.

First of all, let’s begin with writing down the 3 formulas which can be used to find the area of a triangle:

$A=\frac { b\cdot c\cdot \sin { \left( A \right) } }{ 2 }$

$A=\frac { a\cdot c\cdot \sin { \left( B \right) } }{ 2 }$

$A=\frac { a\cdot b\cdot \sin { \left( C \right) } }{ 2 }$

Now, let’s make the first two formulas above equivalent to one another…

$\frac { b\cdot c\cdot \sin { \left( A \right) } }{ 2 } =\frac { a\cdot c\cdot \sin { \left( B \right) } }{ 2 }$

Alright, now watch what happens when we multiply both sides of the equation by a handy expression…

$\frac { b\cdot c\cdot \sin { \left( A \right) } }{ 2 } \cdot \frac { 2 }{ c\cdot \sin { \left( A \right) } \cdot \sin { \left( B \right) } } =\frac { a\cdot c\cdot \sin { \left( B \right) } }{ 2 } \cdot \frac { 2 }{ c\cdot \sin { \left( A \right) \cdot \sin { \left( B \right) } } }$

If we do this, what we’re going to be left with is…

$\frac { b }{ \sin { \left( B \right) } } =\frac { a }{ \sin { \left( A \right) } }$

So far so good! Let’s now make these two area formulas equivalent to one another…

$\frac { a\cdot c\cdot \sin { \left( B \right) } }{ 2 } =\frac { a\cdot b\cdot \sin { \left( C \right) } }{ 2 }$

And now, let’s multiply both sides of the equation we’ve just created by a handy expression…

$\frac { a\cdot c\cdot \sin { \left( B \right) } }{ 2 } \cdot \frac { 2 }{ a\cdot \sin { \left( B \right) } \cdot \sin { \left( C \right) } } =\frac { a\cdot b\cdot \sin { \left( C \right) } }{ 2 } \cdot \frac { 2 }{ a\cdot \sin { \left( B \right) } \cdot \sin { \left( C \right) } }$

If we do this, what we’re going to be left with is…

$\frac { c }{ \sin { \left( C \right) } } =\frac { b }{ \sin { \left( B \right) } }$

And it turns out, because:

$\frac { b }{ \sin { \left( B \right) } } =\frac { a }{ \sin { \left( A \right) } }$

$\frac { c }{ \sin { \left( C \right) } } =\frac { b }{ \sin { \left( B \right) } }$

We can say that:

$\frac { a }{ \sin { \left( A \right) } } =\frac { b }{ \sin { \left( B \right) } } =\frac { c }{ \sin { \left( C \right) } }$

I’ve made a video related to this Sine Rule proof. You can watch it below if you wish.

Hope you enjoyed reading this post! 🙂

# Areas Of Triangles & The Sine Rule

Learn how to come up with the formula for areas of triangles, then use this formula to derive the sine rule formula.

Related:

The quickest Sine Rule proof