Angle subtended at the centre by a segment of a circle

Question

To find the area of segment of a circle, I used the following formula:
$\frac{r^2}{2} (\theta$ - $\sin \theta$)
But if the area is given and I want to find the angle $\theta$ how can I do that.
$\theta$ - $\sin \theta$ = $\frac{2A}{r^2}$
where A and r are the given values of area and radius of the circle respectively. Please answer this question as soon as possible any help would be appreciated.

Answer 1

It is enough to solve for $\theta\in[0,\pi]$, other ranges can be handled by symmetry.

For small angles, the Taylor expansion to the third order is

$$r:=\theta-\sin\theta\approx\frac{\theta^3}{6}$$ and can be used to find a simple approximation

$$\theta\approx \sqrt[3]{6r}.$$

And you get an even better approximation with

$$r\approx\frac{\theta^3}{\pi^2},\ \theta\approx \color{green}{\sqrt[3]{\pi^2r}}$$ which is exact at both ends of the range.

In blue the true curve, in green Taylor ($r$ as a function of $\theta$).

From this initial value, you can improve by Newton's iterations,

$$\theta'=\theta-\frac{\theta-\sin\theta-r}{1-\cos\theta}=\color{green}{\frac{\sin\theta-\theta\cos\theta+r}{1-\cos\theta}}$$

applied two or three times.

Below in magenta the first approximation and in green the first Newton's iterate overlaid on the exact curve ($\theta$ as a function of $r$).

Answer 2

As Lord Shark pointed out there is no closed form expression to calculate $\theta$. I am not aware of any series expansion either. What I can give you is a formula that can be used to determine an approximate value for $\theta$. The values obtained are good only for small $\theta$ (i.e. $\lt \frac{\pi}{2})$. $$\theta\approx\frac{x^3 + 6\sin x - 6x\cos x}{6\left(1-\cos x\right)},\space\space \rm{where} \space\space\space \it{x}\rm{=\sqrt[3]{\frac{12\it{A}}{\it{r}^2}}}\space\space and\space\space \theta\space\space is\space\space given\space\space in\space\space radians$$

It is up to you to take this formula or leave it depending on your requirements.

Now I will give the iteration formula based on Newton-Raphson method that can be used to improve $\theta$ in radians to a desired accuracy. $$\theta_{i+1} = \theta_{i} - \frac{\theta_{i} - \sin \theta_{i} - \frac{2A}{r^2}}{1 - \cos \theta_{i}},\space\space \rm{where} \space\space\space \it{\theta}\rm{_1=}\space\space\it{\theta}$$

I am adding the text given below at the request of Tanmay Gajapati. First of all, you don't seem foolish to me. I omitted this part from my answer assuming that you are already familiar with this type of calculations. Now I know that you are not, so here it is.

To use the $1^{st}$ formula, calculate the value of $x$ using the given area $A$ and radius $r$, e.g. if $r=10\space \rm{cm}\space$ and $\space A=1.18\space \rm{cm^2}$ $$x=\sqrt[3]{\frac{12\times 1.18}{10^2}}=0.52122$$ Then substitute this value of $x$ in the expression given for $\theta$ to obtain its value, e.g. $$\theta\approx \frac{0.52122^3+6\times \sin\left(0.52122\right)-6\times 0.52122\times\cos\left(0.52122\right)}{6\times\left(1-\cos\left(0.52122\right)\right)}=0.5236186\space\rm{rad}.$$

If you think the value obtained from the $1^{st}$ formula for $\theta$ is not accurate enough, you can use the $2^{nd}$ formula to improve it, e.g. $$\theta_2 = 0.5236186 - \frac{0.5236186 - \sin\left(0.5236186\right) - \frac{2\times 1.18}{10^2}}{1 - \cos\left(0.5236186\right)}=0.5236079\space\rm{rad}, $$ $$\theta_3 = 0.5236079 - \frac{0.5236079 - \sin\left(0.5236079\right) - \frac{2\times 1.18}{10^2}}{1 - \cos\left(0.5236079\right)}=0.5236079\space\rm{rad}. $$

As you can see from the last two values of $\theta$, it is no longer improving. Therefore, this is the value of the subtended angle $\theta$ at the center of the circle for the given values of $A$ and $r$.

Answer 3

Considering that you want to work with $\theta -\sin (\theta )$, use Taylor series to get $$\theta -\sin (\theta )=\frac{\theta ^3}{6}-\frac{\theta ^5}{120}+\frac{\theta ^7}{5040}-\frac{\theta ^9}{362880}+O\left(\theta ^{11}\right)$$ If you look at the plots of the lhs and rhs, you will not see any difference for $0 \leq \theta \leq \pi$. Taking that into account, letting $k=\frac{2A}{r^2}$, use series reversion to get $$\theta=t+\frac{t^3}{60}+\frac{t^5}{1400}+\frac{t^7}{25200}+\cdots \qquad \text{where} \qquad t=\sqrt[3]{6k}$$ Using, as @YNK did, $r=10$ and $A=1.18$ (that is to say $k=0.0236$) the above expansion will give $\theta=0.5236079073$ while the "exact" solution, obtained using Newton method, would be $0.5236079142$.