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When $\theta$ is in radians, a central angle of $\theta$ creates a sector of the circle with area equal to $\frac12r^2\theta$. I started wondering what the area formula would be if the vertex of the angle was on the circumference rather than the center. But after searching on this site, I found that same question being asked. The answer was that there is no formula because a single value of $\theta$ can lead to different areas (rotate the angle without changing its measure to change sector area). So I have a new question. What is the maximum sector area in a circle, when the inscribed angle is $\theta$?

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Since the base for a given angle is fixed, we just need to maximize the height

enter image description here

which requires $AC=BC$.

According to the following sketch we need to find:

$$ A=\operatorname{Area}\left(\triangle{ABC}\right)+ \operatorname{Area}(\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{OAB} )-\operatorname{Area}\left(\triangle{OAB}\right)$$

enter image description here

or also

$$ A=2\operatorname{Area}\left(\triangle{OBC}\right)+ \operatorname{Area}(\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{OAB} )=$$

$$=2\cdot \frac12 \cdot 2r\cos \frac \theta 2 \cdot r \sin \frac \theta 2+\frac12 r^2\cdot 2\theta=\boxed{r^2\left(\sin \theta +\theta\right)}$$

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  • $\begingroup$ I think I understand everything except for one thing: how do we know that AC = BC? $\endgroup$
    – MumboJumbo
    Commented May 2, 2023 at 7:21
  • $\begingroup$ What is the maximum distance between the chord and the circumference, that is the height? $\endgroup$
    – user
    Commented May 2, 2023 at 7:39
  • $\begingroup$ @MumboJumbo The are for the circular segment $\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{AB}$ is fixed. Then we need to maximize the are of the triangle $\triangle{ABC}$ and since AB is fixed we need to maximize its height which is of course obtained when the point $C$ is such that $AC=AB$. It seems trivial but we can also prove this fact if you have some doubt on that. $\endgroup$
    – user
    Commented May 2, 2023 at 16:11
  • $\begingroup$ Is it because perpendicular bisector makes it an isosceles triangle? $\endgroup$
    – MumboJumbo
    Commented May 4, 2023 at 13:30
  • $\begingroup$ @MumboJumbo Yes of course the maximum is obtained for the perpendicular bisector! $\endgroup$
    – user
    Commented May 4, 2023 at 13:33
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Theoretically you could get $\theta$ closer and closer to a straight angle, and the area of the major circular sector (and circular segment) would get closer and closer to the entire circle.

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    $\begingroup$ It seems that the OP is looking for the maximum area for a fixed value for the angle $\theta$. $\endgroup$
    – user
    Commented May 1, 2023 at 10:17

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