Math Help - [Arcs, sectors and segments] Finding the arc

1. [Arcs, sectors and segments] Finding the arc

"A circle is modified so that it will measure a total of 30m around the major arc AB and the chord AB"

I don't know where to start.

2. Hello, Cthul

The perimeter of the shape formed by major arc $AB$ and chord $AB$ is 30 m.

Code:
* * *
*           *
*               *
*                 *

*         O         *
*         ♥         *
*    r  *120*  r    *
*       *
*  *           *  *
A ♥ - - - - - - - ♥ B
*           *
* * *

The circumference of the circle is: $2\pi r.$

The major arc has length: $\frac{2}{3}\times 2\pi r \:=\:\frac{4\pi}{3}r$

Using the Law of Cosines on $\Delta AOB\!:$

. . $AB^2 \:=\:r^2 + r^2 - 2(r)(r)\cos120^o \:=\:3r^2$

Hence: . $AB \:=\:r\sqrt{3}$

The perimeter is: . $\frac{4\pi}{3}r + r\sqrt{3} \:=\:30$

. . . . . . . . . . . . $\left(\frac{4\pi}{3} + \sqrt{3}\right)r \:=\:30$

Therefore: . $r \;=\;\dfrac{30}{\frac{4\pi}{3} + \sqrt{3}} \;=\;\dfrac{90}{4\pi + 3\sqrt{3}}\text{ meters}$

3. Originally Posted by Cthul
"A circle is modified so that it will measure a total of 30m around the major arc AB and the chord AB"

I don't know where to start.
Alternatively, split the triangle OAB into two identical right-angled triangles,
where O is the circle centre.

Then $|\angle OAB|=30^o$

$|AB|=2rcos30^o=r\sqrt{3}$

The length of major arc AB is $\frac{240}{360}{2\pi}r$

$30m=r\sqrt{3}+\frac{4{\pi}r}{3}$

4. Thanks to both, very helpful steps.