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 $\theta$?
What’s the maximum area of a sector defined by an inscribed angle
geometry
Related Solutions
Here are some equal angle inscribed angles (they all subtend half the angle of the red arc). It is clear that the area of the last sector is properly contained in the areas of the others.
$\hspace{2.5cm}$
The area of the sector is the sum of the area between the arc and its chord (a constant area) and the area of the triangle, which is $\frac12$ the product of the length of the chord (a constant length) and the altitude of the triangle, which varies as the vertex moves around the circle.
In the diagram above $A$ is the centre of the circle and $CB$ is thus the diameter. Point $D$ is an arbitrary point in the circumference.
In $\triangle ACD$ $\angle CDA = \alpha$ since $AC = AD$
Thus
$$2 \cdot \alpha + \theta = 180 ^\circ$$
In $\triangle ADB$ $\angle ADB = \beta$ since $AB = AD$
Thus
$$2 \cdot \beta + (180^\circ - \theta) = 180 ^\circ$$
Add both these together
$$2 \cdot \alpha + 2 \cdot \beta + 180 ^\circ = 360 ^\circ$$
Which we can easily rearrange to show:
$$ \alpha + \beta = 90 ^\circ $$
We have thus proved $\angle CDB$ is a right angle as required.
Best Answer
Since the base for a given angle is fixed, we just need to maximize the height
which requires $AC=BC$.
According to the following sketch we need to find:
$$ A=\operatorname{Area}\left(\triangle{ABC}\right)+ \operatorname{Area}(\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{OAB} )-\operatorname{Area}\left(\triangle{OAB}\right)$$
or also
$$ A=2\operatorname{Area}\left(\triangle{OBC}\right)+ \operatorname{Area}(\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{OAB} )=$$
$$=2\cdot \frac12 \cdot 2r\cos \frac \theta 2 \cdot r \sin \frac \theta 2+\frac12 r^2\cdot 2\theta=\boxed{r^2\left(\sin \theta +\theta\right)}$$