Express radius in terms of the length of a sector and its chord

Question

I have a problem that asks me to find the radius of a circle. I am given C, denoting the length of the circumference of a sector of the circle, and L, denoting the distance between the two ends of the sector, the chord. I am to express R in terms of C and L. I have narrowed down the eqation to, $$ 2R \sin \left(\frac{90C}{ \pi R}\right) = L $$ But I can't seem to make R the subject in the equation, please help.

Answer 1

According to my knowledge, the formula (1), in its original form, cannot be coerced to yield an equation, in which $R$ is the subject. But, we can concoct a method to separate $R$ from $C$ and $L$ after modifying this formula in a certain way. However, by doing so, we make the equation we derived a little less accurate.

$$L=2R\sin\left(\frac{C}{2R}\right) \tag{1}$$

Please note that we use $radians$ and not $degrees$ to measure angles. The derivation is started by expressing $\sin\left(\frac{C}{2R}\right)$ as a series. However, as shown below, we retain only the first four terms to obtain the required formula.

$$L=2R\left(\frac{C}{2R}-\frac{1}{6}\left(\frac{C}{2R}\right)^3+\frac{1}{120}\left(\frac{C}{2R}\right)^5-\frac{1}{5040}\left(\frac{C}{2R}\right)^7 \right)$$

This can be written as a cubic equation of $\left(\frac{C}{R}\right)^2$.

$$\left(\frac{C}{R}\right)^6-168\left(\frac{C}{R}\right)^4+13440\left(\frac{C}{R}\right)^2-322560\left(1-\frac{L}{C}\right)=0 $$

After this equation is solved to express $\left(\frac{C}{R}\right)^2$ in terms of $C$ and $L$, we can make $R$ the subject of a formula as shown below.

$$R\approx\frac{C}{\sqrt{56-p\times\sinh\left\{\frac{1}{3}\sinh^{-1}\left[q\left(\frac{11}{45}+\frac{L}{C}\right)\right]\right\}}} \tag{2},$$

where $$ \begin{matrix} p & = & 73.3212111192934401054087551,\space\space\space\space\space\space\space\space\space\space \\ q & = & 3.2732683535398851709761218413802. \\ \end{matrix} $$

The values of $R$ given by (2) are always larger than its exact value. If you want to test the accuracy of the formula (2), you can use the very-fast-converging formula (3), which is based on Newton-Raphson method, to find values of $R$ to compare with those of the former.

$$R_{n+1}=R_n\left(1-\frac{2R_n\sin\phi-L}{2R_n\sin\phi-C\cos\phi}\right),\space\space \rm{where}\space\space \mathit{\phi} =\frac{\mathit{C}}{2\mathit{R}}\space\space and \space\space \mathit{R}_0 =\frac{\mathit{L}}{2}\tag{3}$$

Instead of retaining the first four terms of the aforementioned series expansion, we could have kept the first three terms to obtain a quadratic equation in $\left(\frac{C}{R}\right)^2$, which can be used to derive another formula for $R$. The values of $R$ obtained from this formula are always smaller than its exact value and less accurate than the results of (2).