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 $270^{\circ}$