# Another way to express the golden ratio mathematically

In this post I’m going to be proving that… $\varphi =\frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } }$

So, here I go… $x=\frac { 1+\sqrt { 5 } }{ 2 } \\ \\ \Rightarrow \quad { x }^{ 2 }=\frac { \left( 1+\sqrt { 5 } \right) }{ 2 } \cdot \frac { \left( 1+\sqrt { 5 } \right) }{ 2 } \\ \\ \Rightarrow \quad { x }^{ 2 }=\frac { 1+2\sqrt { 5 } +5 }{ 4 } \\ \\ \Rightarrow \quad { x }^{ 2 }=\frac { 6+2\sqrt { 5 } }{ 4 } \\ \\ \Rightarrow \quad { x }^{ 2 }-1=\frac { 6+2\sqrt { 5 } }{ 4 } -\frac { 4 }{ 4 } \\ \\ \Rightarrow \quad { x }^{ 2 }-1=\frac { 2+2\sqrt { 5 } }{ 4 } \\ \\ \Rightarrow \quad { x }^{ 2 }-1=\frac { 2 }{ 2 } \cdot \frac { \left( 1+\sqrt { 5 } \right) }{ 2 }$

Wait for it… $\Rightarrow \quad { x }^{ 2 }-1=1\cdot x\\ \\ \Rightarrow \quad { x }^{ 2 }-1=x\\ \\ \Rightarrow \quad { x }^{ 2 }=x+1\\ \\ \Rightarrow \quad \frac { { x }^{ 2 } }{ x } =\frac { x }{ x } +\frac { 1 }{ x } \\ \\ \Rightarrow \quad x=1+\frac { 1 }{ x } \\ \\ \Rightarrow \quad \frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ x } \\ \\ \Rightarrow \quad \frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ \left( 1+\frac { 1 }{ x } \right) } \\ \\ \Rightarrow \quad \frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ 1+\frac { 1 }{ x } } \\ \\ \Rightarrow \quad \frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ 1+\frac { 1 }{ \left( 1+\frac { 1 }{ x } \right) } } \\ \\ \therefore \quad \frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } }$

This expression for the golden ratio is quite common, however, before I produced this post – I think it would’ve been very hard to figure out how to derive it from scratch. There aren’t many quirky proofs like this one on the internet – I am quite certain. I hope you liked reading this post! 😀

# How to find a neat version of y when you have the equation of an ellipse

So, you have the equation of the ellipse but you need to completely isolate y. How would you go about doing this? Well, here is a fantastic example… ${ \left( \frac { x }{ a } \right) }^{ 2 }+{ \left( \frac { y }{ b } \right) }^{ 2 }=1\\ \\ \therefore \quad { \left( \frac { y }{ b } \right) }^{ 2 }=1-{ \left( \frac { x }{ a } \right) }^{ 2 }\\ \\ \therefore \quad \frac { { y }^{ 2 } }{ { b }^{ 2 } } =1-\frac { { x }^{ 2 } }{ { a }^{ 2 } } \\ \\ \therefore \quad { y }^{ 2 }={ b }^{ 2 }-\frac { { { b }^{ 2 }x }^{ 2 } }{ { a }^{ 2 } } \\ \\ \therefore \quad { y }^{ 2 }=\frac { { a }^{ 2 }{ b }^{ 2 } }{ { a }^{ 2 } } -\frac { { b }^{ 2 }{ x }^{ 2 } }{ { a }^{ 2 } } \\ \\ \therefore \quad { y }^{ 2 }=\frac { { a }^{ 2 }{ b }^{ 2 }-{ b }^{ 2 }{ x }^{ 2 } }{ { a }^{ 2 } } \\ \\ \therefore \quad { y }^{ 2 }=\frac { { b }^{ 2 }\left( { a }^{ 2 }-{ x }^{ 2 } \right) }{ { a }^{ 2 } } \\ \\ \therefore \quad { y }^{ 2 }=\frac { { b }^{ 2 } }{ { a }^{ 2 } } \cdot \left( { a }^{ 2 }-{ x }^{ 2 } \right) \\ \\ \therefore \quad y=\sqrt { \frac { { b }^{ 2 } }{ { a }^{ 2 } } } \cdot \sqrt { { a }^{ 2 }-{ x }^{ 2 } } \\ \\ \therefore \quad y=\frac { b }{ a } \cdot \sqrt { { a }^{ 2 }-{ x }^{ 2 } }$

This will come in handy if you’re trying to derive the area of an ellipse from absolute scratch.

# Deriving the formula for an ellipse

In this post, I’ll be demonstrating how one can derive the formula for an ellipse from absolute scratch.

To derive the formula for an ellipse, what we must first do is create a diagram like the one below. ** Click on the image above to see it in full size.

Now, the first thing we’ve got to acknowledge here is that: ${ D }_{ 1 }+{ D }_{ 2 }=2a$

What we’re basically saying is that D_1 + D_2 is equal to the length from -a to a in the diagram above.

This formula can be understood by watching the video below…

These photographs can also help the formula sink into your mind…

Ellipse Image 1: Ellipse Image 2: Now, look at the diagram at the top of this page once again…

What you will notice is that: ${ \left( c+x \right) }^{ 2 }+{ y }^{ 2 }={ D }_{ 1 }^{ 2 }\\ \\ \therefore \quad { D }_{ 1 }^{ 2 }={ c }^{ 2 }+2cx+{ x }^{ 2 }+{ y }^{ 2 }\\ \\ \therefore \quad { D }_{ 1 }=\sqrt { { c }^{ 2 }+2cx+{ x }^{ 2 }+{ y }^{ 2 } } \\ \\ { \left( c-x \right) }^{ 2 }+{ y }^{ 2 }={ D }_{ 2 }^{ 2 }\\ \\ \therefore \quad { D }_{ 2 }^{ 2 }={ c }^{ 2 }-2cx+{ x }^{ 2 }+{ y }^{ 2 }\\ \\ \therefore \quad { D }_{ 2 }=\sqrt { { c }^{ 2 }-2cx+{ x }^{ 2 }+{ y }^{ 2 } }$

If this is the case, we can say that: ** Click on the image of the workings to see it in full size.

Alright, so far so good… Now, it turns out – if you look at the diagram at the top of this page carefully, you will discover that: ${ b }^{ 2 }+{ c }^{ 2 }={ a }^{ 2 }\\ \\ \therefore \quad { c }^{ 2 }={ a }^{ 2 }-{ b }^{ 2 }$

And this ultimately means that: ${ a }^{ 4 }+\left( { a }^{ 2 }-{ b }^{ 2 } \right) { x }^{ 2 }={ a }^{ 2 }\left( { a }^{ 2 }-{ b }^{ 2 } \right) +{ a }^{ 2 }{ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }\\ \\ \therefore \quad { a }^{ 4 }+{ a }^{ 2 }{ x }^{ 2 }-{ b }^{ 2 }{ x }^{ 2 }={ a }^{ 4 }-{ a }^{ 2 }{ b }^{ 2 }+{ a }^{ 2 }{ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }\\ \\ -{ b }^{ 2 }{ x }^{ 2 }=-{ a }^{ 2 }{ b }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }\\ \\ { b }^{ 2 }{ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }={ a }^{ 2 }{ b }^{ 2 }\\ \\ \frac { { b }^{ 2 }{ x }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 } } +\frac { { a }^{ 2 }{ y }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 } } =\frac { { a }^{ 2 }{ b }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 } } \\ \\ \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\\ \\ { \left( \frac { x }{ a } \right) }^{ 2 }+{ \left( \frac { y }{ b } \right) }^{ 2 }=1$

The formula you see just above is the formula for an ellipse. You’ve derived it from scratch!!

# How to derive the formula for the area of an equilateral triangle

In this post I’ll be showing you how to derive the formula for the area of an equilateral triangle – in easy steps. In order to understand this derivation properly, you need to be familiar with Pythagoras’ theorem and also a few algebraic rules. What you’ll also need is a ruler, pair of compasses, a pencil and a sheet of paper.

Step 1: Put a point on a blank sheet of paper and name it A. Step 2: Put the needle of your compass on the point A and draw a circle around it. Step 3: Add a point B to this circle, on its edge. Step 4: Put the needle of your compass on the point B and your pencil on the point A. Step 5: Draw another circle with a radius the length AB. Step 6: Now add a few extra points to your drawing. Call these points C and D. Step 7: Connect the points A, B and C forming a triangle. Step 8: Draw a line going through the points C and D. Step 9: Where the line going through C and D intersects the triangle, place the point E. Step 10: Now look at your latest work very carefully… What you will notice is that the lengths AB, AC and BC are all equal to one another. This is because both the circles you drew – are exactly the same size. They each have radiuses equal in proportion. In simple terms, AB=AC=BC.

What you have to do now is name these lengths (r) for radius. Here’s the thing though, because the line going through C and D splits the triangle (equilateral, as each of its sides has the same length) down its middle, the length AE is equal to 1/2 x r, and similarly the length BE is equal to 1/2 x r. Together, the length AE + BE = AB = r. Step 11: Remember that I said that the line going through C and D splits the triangle down its middle. Also, notice that this exact line is perpendicular to the length AB. Now, because of this, at the point E, you’ve got two right angles. Name these two right angles big R.

[Knowing that these two angles are equal to 90 degrees is vital – because you’ll be able to use Pythagoras’ theorem to find the length CE.] Step 12: Find the length CE using Pythagoras’ theorem, Adjacent² + Opposite² = Hypotenuse². You will need this length to find the area of the equilateral triangle you’ve produced.

*Algebraic skills will be required from this point… ${ AE }^{ 2 }+{ CE }^{ 2 }={ AC }^{ 2 }\\ \\ \Rightarrow \quad { \left( \frac { 1 }{ 2 } r \right) }^{ 2 }+{ CE }^{ 2 }={ r }^{ 2 }\\ \\ \Rightarrow \quad { CE }^{ 2 }={ r }^{ 2 }-{ \left( \frac { 1 }{ 2 } r \right) }^{ 2 }\\ \\ \Rightarrow \quad { CE }^{ 2 }=\frac { 4r^{ 2 } }{ 4 } -\frac { { r }^{ 2 } }{ 4 } \\ \\ \Rightarrow \quad { CE }^{ 2 }=\frac { 3{ r }^{ 2 } }{ 4 } \\ \\ \Rightarrow \quad CE=\sqrt { \frac { 3{ r }^{ 2 } }{ 4 } } \\ \\ \therefore \quad CE=\frac { r\sqrt { 3 } }{ 2 }$ Step 13: Derive the formula for the area (A) of the equilateral triangle. Remember that the area of a right angled triangle is L x W x 1/2. $A=\frac { 1 }{ 2 } r\cdot \frac { r\sqrt { 3 } }{ 2 } \cdot \frac { 1 }{ 2 } +\frac { 1 }{ 2 } r\cdot \frac { r\sqrt { 3 } }{ 2 } \cdot \frac { 1 }{ 2 } \\ \\ =\frac { 1 }{ 8 } { r }^{ 2 }\sqrt { 3 } +\frac { 1 }{ 8 } { r }^{ 2 }\sqrt { 3 } \\ \\ =2\cdot \frac { 1 }{ 8 } { r }^{ 2 }\sqrt { 3 } \\ \\ =\frac { 1 }{ 4 } { r }^{ 2 }\sqrt { 3 }$

Presto!!! Keep in mind that you can transform the variable (r) into any variable you wish. This variable (r) is the length of each side of the equilateral triangle you were working with. The formula you’ve derived can be used to find the area of any equilateral triangle.

# Proof: Thales’ Theorem

In this post I’ll be demonstrating how you can prove that Thales’ Theorem is true. To follow the steps in this post (11 in total), what you will require is a ruler, pair of compasses and a pencil.

##### Step 1: Draw a random line on a sheet of paper. ##### Step 2: Place your compass needle on this line, and form a circle. ##### Step 3: Add 4 points to your drawing, as shown below… ##### Step 4: Name the points A, B, C and D as shown… ##### Step 5: Connect the points A, B and D together to form an isosceles triangle… ##### Step 6: Name the lines AB and BD the radius (r)… ##### Step 7: Since the lines AB and BD are equal to one another, it follows that the angles ∠BAD and ∠BDA are equivalent. This is because the angles below the apex of an isosceles triangle are equal. You must name these angles alpha (α). ##### Step 8: Now connect the points BC and CD together to form another isosceles triangle… ##### Step 9: The line BC is equal to r… Now label the line BC… ##### Step 10: Because the line BC and BD are both equal to r, the triangle BCD is an isosceles triangle. This means that the angles ∠BCD and ∠BDC must both be equivalent. Call these angles beta (β). ##### Step 11: Prove that the angle at point D is equal to 90 degrees.

Thales’ Theorem is as follows:

Because AC is the diameter of the circle you drew, the angle at the point D (α+β) must be equal to 90 degrees. In more specific and general terms, if you have the points A, C and D lying on a circle – and the line AC is in fact the diameter of this circle – then the angle at point D (α+β) must be a right angle.

Proof (which must be derived using the diagram you’ve created): All angles within a triangle (in 2 space) must add up to 180 degrees.

Mathematically, this means that: $\alpha +\alpha +\beta +\beta =180\\ \\ \Rightarrow \quad 2\alpha +2\beta =180\\ \\ \Rightarrow \quad 2\left( \alpha +\beta \right) =180\\ \\ \Rightarrow \quad \frac { 2\left( \alpha +\beta \right) }{ 2 } =\frac { 180 }{ 2 } \\ \\ \therefore \quad \alpha +\beta =90$

And as a result, Thales’ theorem must be true. The angle α+β is the angle at point D.

# Proof: Opposite angles formed when two lines intersect, are equal to one another

How can we prove that opposite angles (when two lines intersect) are in fact equal to one another?

Well, first of all – let’s draw a circle… We know that in a full circle, there are 360 degrees. This is an indisputable fact. Now, what happens if we split this circle in two with a straight line (going through its centre)? Well, each half of the circle (top and bottom) – will now contain 180 degrees. We know this because: $\frac { 360 }{ 2 } =180$

Ok, so far so good… Now, let’s draw another line through the circle (going through its centre) which intersects the first line we have drawn… As we can see, because we have done this, we now have 4 different angles. Let’s name the two angles which are situated in the top half of the circle α and β Earlier in this demonstration, we remarked that the top half of the circle (when it was split in two) contained 180 degrees. Mathematically and logically speaking, as this is the case, we must say that: $\alpha +\beta =180$

Great, now let’s name the angles in the bottom half of the circle x and y It follows, because the angles in the top half of the circle add up to 180 degrees, we must deduce that: $x+y=180$

So, it turns out we now have two useful equations: $\alpha +\beta =180$ $x+y=180$

Do we have enough to form our proof though? Unfortunately, not quite… We have to look at our most recent figure again, but this time from a different perspective…

You see, there are different top halves and bottom halves… • There exists top halves α+β and also… β+y
• There exists bottom halves x+y but also α+x

You may ask, why is this important? Well, here’s what’s crucial: $\beta +y=180$ $\alpha +x=180$

And now we have 4 different equations, 3 of which – will help us finally complete our proof. $\alpha +\beta =180$ $x+y=180$ $\beta +y=180$ $\alpha +x=180$

Here’s why we need these equations…

α+β and α+x are equivalent (180 degrees), so we can deduce that: $\alpha +\beta =\alpha +x\\ \\ \therefore \quad \beta =x$

*Subtract α from both sides of the equation.

α+β and β+y are equivalent (180 degrees), so we can deduce that: $\alpha +\beta =\beta +y\\ \\ \therefore \quad \alpha =y$

*Subtract β from both sides of the equation.

Hence, we’ve proven that: Opposite angles (when two lines intersect) are equal to one another.

β=x and α=y: # How to quickly double the area of a square (simple geometry lesson)

In this post, I’ll be demonstrating how you can quickly double the area of a square using a simple geometrical trick.

Let’s say you have an ordinary square, like the one below… Firstly, what you have to do is name the area of this square “A”… Then, what you do next is divide this square (diagonally) into 4 equal parts… After you have done this, you then name each part of this square “1/4 x A”… Remember that A is the entire area of the square, therefore 1/4 x A is a quarter of the area of the square.

Notice now, that to double the area of this square, all you have to do, is double the number of the 1/4 x A right angled triangles which currently exist – then configure them – like this… As you can see, you’ve now got eight of these 1/4 x A right angled triangles neatly configured… Not only are you left with a new square, double the size of your original square (follow the lines on the outside of the shape), but a handy equation, which proves that you doubled the area of the square you started off with… $8\times \frac { 1 }{ 4 } A=2A$ # Properties of C squared, Pythagorean Theorem

In this post, I’ll be writing about some peculiar properties of C squared in Pythagoras’ theorem.

Look at this diagram very carefully… *What are the weird properties of C^2..? It turns out that A1=A2 and A3=A4. A2 + A4 = C^2.

It turns out out that area A1 is equal to area A2, and that area A3 is equal to area A4:

A1 = A2

A3 = A4

This can be proven because:

1. ${ A }^{ 2 }+{ B }^{ 2 }={ C }^{ 2 }$
2. ${ x }^{ 2 }+{ D }^{ 2 }={ B }^{ 2 }$
3. ${ \left( C-x \right) }^{ 2 }+{ D }^{ 2 }={ A }^{ 2 }$

Now, due to the above: ${ D }^{ 2 }={ B }^{ 2 }-{ x }^{ 2 }\\ \\ { D }^{ 2 }={ A }^{ 2 }-{ \left( C-x \right) }^{ 2 }\\ \\ \therefore \quad { B }^{ 2 }-{ x }^{ 2 }={ A }^{ 2 }-{ \left( C-x \right) }^{ 2 }\\ \\ { B }^{ 2 }-{ x }^{ 2 }={ A }^{ 2 }-\left\{ { C }^{ 2 }-2Cx+{ x }^{ 2 } \right\} \\ \\ { B }^{ 2 }-{ x }^{ 2 }={ A }^{ 2 }-{ C }^{ 2 }+2Cx-{ x }^{ 2 }\\ \\ { B }^{ 2 }={ A }^{ 2 }-{ C }^{ 2 }+2Cx\\ \\ { B }^{ 2 }={ A }^{ 2 }-\left\{ { A }^{ 2 }+{ B }^{ 2 } \right\} +2Cx\\ \\ { B }^{ 2 }={ A }^{ 2 }-{ A }^{ 2 }-{ B }^{ 2 }+2Cx\\ \\ { B }^{ 2 }=-{ B }^{ 2 }+2Cx\\ \\ 2{ B }^{ 2 }=2Cx\\ \\ \therefore \quad { B }^{ 2 }=Cx\\ \\$

But… B^2 is actually the area A1 and Cx is the area A2, which means that A1=A2.

Now, if B^2=Cx, this means that: ${ A }^{ 2 }+Cx={ C }^{ 2 }\\ \\ \therefore \quad { A }^{ 2 }={ C }^{ 2 }-Cx\\ \\ { A }^{ 2 }=C\left( C-x \right) \\ \\$

However, A^2 is equal to the area A3, and C(C-x) is equal to the area A4 – which means that A3=A4. Hence, we’ve proven that:

A1=A2

A3=A4

Related:

2 ways to derive Pythagoras’ equation from scratch

# 2 ways to derive Pythagoras’ equation from scratch

The other day I discovered one more way to derive Pythagoras’ equation from scratch, completely by accident. I was deriving Pythagoras’ equation using the usual method, whilst navigating  a diagram similar to the one below, but without (B-A) measurements… *Note (regarding diagram above): x+y = 90 degrees

The usual method goes like this…

The area of the largest square is: ${ \left( A+B \right) }^{ 2 }$

It is also: $4\cdot \frac { 1 }{ 2 } AB+{ C }^{ 2 }$

Which means that: ${ \left( A+B \right) }^{ 2 }=4\cdot \frac { 1 }{ 2 } AB+{ C }^{ 2 }\\ \\ { A }^{ 2 }+2AB+{ B }^{ 2 }=2AB+{ C }^{ 2 }\\ \\ \therefore \quad { A }^{ 2 }+{ B }^{ 2 }={ C }^{ 2 }$

Now, when I added the lengths (B-A) to my diagram, which are included in the diagram above, I discovered a new way to derive Pythagoras’ equation…

I did this by focusing on the area C^2. It turns out that: $4\cdot \frac { 1 }{ 2 } AB+{ \left( B-A \right) }^{ 2 }={ C }^{ 2 }$

And since: ${ \left( B-A \right) }^{ 2 }\\ \\ ={ \left( A+B \right) }^{ 2 }-4AB\\ \\ ={ A }^{ 2 }+2AB+{ B }^{ 2 }-4AB\\ \\ ={ B }^{ 2 }-2AB+{ A }^{ 2 }$

I was able to say that: $4\cdot \frac { 1 }{ 2 } AB+\left\{ { B }^{ 2 }-2AB+{ A }^{ 2 } \right\} ={ C }^{ 2 }\\ \\ 2AB+{ B }^{ 2 }-2AB+{ A }^{ 2 }={ C }^{ 2 }\\ \\ \therefore \quad { A }^{ 2 }+{ B }^{ 2 }={ C }^{ 2 }$

Obviously, I was quite pleased. Have you discovered other ways in which to derive Pythagoras’ equation??

Related:

Video on how to come up with Pythagoras’s equation…

How To Come Up With Pythagoras’s Equation

# 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! 🙂