# 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 $270^{\circ}$

• If $12\pi=\dfrac34(2\pi r)$, you should be able to obtain the radius of the circle easily. You now want the length of the chord corresponding to the $90^\circ$ arc of this circle, which you should find easy to do with a bit of Pythagoras... (hint: the chord is the hypotenuse of an isosceles (why?) right triangle.) – J. M. is a poor mathematician Jun 9 '12 at 14:05

## 1 Answer

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.

• Thanks that makes sense. So from the origin a line goes to 270 degress which is equal to radius and acts as a base and another line goes from origin to 360 degrees which acts as a perpendicualr then we calculate the hypot. This definitely makes sense. But what if the question was changes and angle was not 90 degrees. I have seen some solutions in which a distance formula is used. – Rajeshwar Jun 9 '12 at 14:14
• There are two cases to consider: (i) the angle (your $270^\circ$) is $\le 180^\circ$ and (ii) the angle is $\gt 180^\circ$. In case (i), work with that angle, in case (ii) work with $360^\circ$ minus the angle. Call the smallet angle $\theta$. Draw the triangle $OAB$ with vertices the centre and the two endpoints. Drop a perpendicular from $O$ to $P$ on $AB$. Then $\frac{AP}{OA}=\sin(\theta/2)$. But $OA$ is the radius $r$. So $AP=r\sin(\theta/2)$ and therefore $AB=2r\sin(\theta/2)$. [This is actually how the sine function began, for solving exactly this problem.] – André Nicolas Jun 9 '12 at 14:24