You are watching: Circle divided into 3 equal parts

geometry one area angle

re-publishing

point out

follow

edited Mar 18 in ~ 20:50

Andrei

29.3k44 gold badges2121 silver- badges4646 bronze badges

request Mar 18 at 20:40

JonasHausJonasHaus

21

$endgroup$

1

include a comment |

## 4 answers 4

active earliest Votes

1

See more: Ground State Electron Configuration For Phosphorus, Error 403 (Forbidden)

$egingroup$

Given the angle $ heta$, split by the diameter comprise $B$, consider the complying with diagram:

$overlineBO$ is the line v the center and $overlineBA$ is the chord cutting turn off the lune who area us wish come compute.

The area that the one wedge subtended by $angle BOA$ is$$fracpi- heta2r^2 ag1$$The area that $ riangle BOA$ is$$frac12cdotoverbracersinleft(frac heta2 ight)^ extaltitudecdotoverbrace2rcosleft(frac heta2 ight)^ extbase=fracsin( heta)2r^2 ag2$$Therefore, the area that the lune is $(1)$ minus $(2)$:$$fracpi- heta-sin( heta)2r^2 ag3$$To acquire the area split into thirds, us want$$fracpi- heta-sin( heta)2r^2=fracpi3r^2 ag4$$which method we desire to solve$$ heta+sin( heta)=fracpi3 ag5$$whose solution deserve to be achieved numerically (e.g. Use $M=fracpi3$ and $varepsilon=-1$ in this answer)$$ heta=0.5362669789888906 ag6$$Giving us

**Numerical Details**