Question

A chord with a length of `2` runs from `pi/12` to `pi/8` radians on a circle. What is the area of the circle?

Answer 1

`{2pi}/{1 - cos (pi/24)}`

Explanation:

Lost my answer in a tab crash twice.

I was plotting this, which is still in my clipboard:

(x^2+y^2-1)(y - x tan(pi/6) ) ( y(cos(pi/6) -1)-sin(pi/6)(x-1) )(x-cos(pi/6))=0

graph{(x^2+y^2-1)(y - x tan(pi/6) ) ( y(cos(pi/6) -1)-sin(pi/6)(x-1) )(x-cos(pi/6))=0 [-0.636, 1.469, -0.303, 0.75]}

I had another version with `pi/24` that didn't render very well.

Anyway, we have a sector of angle

`theta = pi/8 - pi/12 = pi/24`.

The chord `c` forms an isosceles triangle with two radii of length `r`. The endpoints of the chord are `(r,0)` and `(r cos theta , r sin theta)` and the third vertex is the origin. The length of the chord `c` satisfies

`c^2 = (r cos theta - r )^2 + (r sin theta - 0)^2`

`c^2 = r ^2 (cos^2 theta - 2 cos theta +1 + sin^ theta )`

`c^2 = r^2 (2 - 2 cos theta )`

We're interested in the area `A` of the circle,

`A = \pi r^2 = {pi c^2}/{2-2cos theta}`

Plugging in the numbers

`A = {pi(2^2)}/{2-2cos (pi/24)} = {2pi}/{1 - cos (pi/24)}`

We can actually get a nice radical form for `cos(pi/24)` but I'll spare you.

You'll have to work the calculator for yourself as well.

I should check but gotta go.