Can't understand this solution. I came across a problem which was already present on the internet.

If an arc with a length of $12\pi$ is $\frac{3}{4}$ of the circumference of the circle, what is the 
  shortest distance between the endpoints of the arc?

According to a certain site the solution is something like this

$$12\pi\left(\frac{4}{3}\right)=\text{circumference}=16\pi=2\cdot\text{radius}\cdot\pi$$
   $$\text{radius}=8$$
   $$x^2+y^2=64$$
   Let $x=0$ for the first endpoint and let $y=0$ for the other, then find the two points
      $(0,8)$ and $(8,0)$. Now find the distance between these two points:
      $$d=((0-8)^2+(8-0)^2)^{1/2}=(128)^{1/2}\qquad \text{Ans}$$

I on the other hand decided to take my own approach since I couldnt figure out what happened after the radius
Step 1:
$12\pi = (3/4)$ (Circumference)
Cirum $= 16\pi$ so radius of the circle in question is $8$
Step 2:
Since $16\pi = 360^{\circ}$ so
$12\pi$ is $270^{\circ}$  . 
Edited: From the suggestions i got from users here is how i would solve this:
Construct a line from the origin that goes to $270^{\circ}$ which is equal to radius and acts as a base and another line goes from origin to $360^{\circ}$ which acts as a perpendicualr then we calculate the hypotenuse (shortest distance). This definitely makes sense. But what if the question changes and angle is not $90^{\circ}$. I would appreciate it if someone could explain how to solve this using the distance formula as done above without the need of calculating $270^{\circ}$
 A: Draw a picture. By your calculation, the angle subtended at the origin by the arc is $270^\circ$. So  going from one endpoint of the arc to the other the short way around, the angle is $360^\circ-270^\circ=90^\circ$.
Then by the Pythagorean Theorem the distance between the endpoints is $\sqrt{8^2+8^2}$, or more simply $8\sqrt{2}$.
Added: Let $A$ and $B$ be on a circle with radius $r$ and centre $O$. There are two arcs joining $A$ and $B$, the "short" one and the "long" one. Let $\theta$ be the angle subtended at $O$ by the short arc.  So $\theta=\angle AOB$. We want to calculate $AB$.
Drop a perpendicular from $O$ to $AB$, meeting $AB$ at point $P$. Then $\angle AOP=\theta/2$.  We have $\frac{PA}{OA}=\frac{PA}{r}=\sin(\theta/2)$, so $PA=r\sin(\theta/2)$ and therefore
$$AB=2r\sin(\theta/2).$$
Remark: The need to find the chord when one knows the arc is precisely what led to the development that ultimately gave us the modern sine function. It was initially driven by the needs of astronomy (and astrology). The history of trigonometry is very interesting. Many of the trigonometric identities we know and love, and other more obscure ones, were used to speed up  "solving triangles," both plane and spherical.  
