I have been working on a problem for a few days now. This was a challenge problem on a lecture for Trigonometry. I managed to find an equation for the radius, but wasn't able to solve it.

Problem: Line $AB$ is drawn such that $\overline{AB} = 20$. Minor arc $AB$ is drawn with endpoints $AB$ such that the length of arc $AB$ is $21$. Find the area of the region bounded by the arc and the line.

Progress: First, draw the full diagram as shown below.

Full Diagram

We can compute the area of $\angle{ABC}$. Using the Law of Cosines, we have that \begin{align*} \cos C &= \frac{a^2+b^2-c^2}{2ab} \\ &= \frac{2r^2-400}{2r^2}\\ &= 1-\frac{200}{r^2} \end{align*}Therefore, $\angle ACB = \cos^{-1}\left(1-\frac{200}{r^2}\right)$. The length of the arc in terms of $\angle ACB$ is \begin{align*} &\frac{\angle ACB}{360}2\pi r \\ &= \frac{\angle ACB}{180}\pi r \\ &= \frac{\cos^{-1}\left(1-\frac{200}{r^2}\right)}{180}\pi r \end{align*}We know that \begin{align*} &21 = \frac{\cos^{-1}\left(1-\frac{200}{r^2}\right)}{180}\pi r \implies \\&\frac{3780}{\pi r} = {\cos^{-1}\left(1-\frac{200}{r^2}\right)} \implies\\ &\cos\left(\frac{3780}{\pi r}\right) = 1-\frac{200}{r^2} \end{align*} However, I couldn't solve this equation. Any help would be appreciated. I can also visualize a calculus approach involving finding the area between two curves, but I want to solve it in an elementary way if possible.

  • 2
    $\begingroup$ If $\theta$ is the angle at $C$ (in radians), then $\theta r= 21$ and $r \sin(\theta/2) = 10$. I used a computer to get $\theta \approx 1.0768$ and $r \approx 19.5018$, and the final area is $r^2\theta/2 - 10\sqrt{r^2-100}\approx 37.34$. As far as I can tell this is not something you can solve without some numerical approximation, but perhaps I am not being clever enough. $\endgroup$ – angryavian Aug 8 '17 at 20:30
  • $\begingroup$ Wolfram Alpha gave up on this with standard computation time, so I it is possible that you are right. $\endgroup$ – Nairit Sarkar Aug 8 '17 at 20:36
  • $\begingroup$ I actually did use Wolfram Alpha to solve $\theta / \sin(\theta/2) = 2.1$. $\endgroup$ – angryavian Aug 8 '17 at 20:57

As angryavian commented, the problem reduces to : find the zero of $$f(\theta)=\frac \theta {\sin\left(\frac\theta 2\right)}- 2.1\qquad \text{or}\qquad g(\theta)=\theta-2.1{\sin\left(\frac\theta 2\right)}$$ which can only be solved using numerical methods.

Let $x=\frac\theta 2$ to make the equation $$x=1.05 \sin(x)$$ and use, for an approximation, the magnificent $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ was proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (see here).

Skipping the trivial $x=0$, this leads to the quadratic $$4 x^2+\left(\frac{84}{5}-4 \pi \right) x+\pi\left(5 \pi -\frac{84 }{5}\right)=0$$ the positive solution of which being $$x=-\frac{21}{10}+\frac{\pi }{2}+\frac{1}{40} \sqrt{7056+3360 \pi -1600 \pi ^2}\approx 0.537445$$ making $$\theta \approx 1.07489$$ while the exact solution would be $\approx 1.07682$.

With this first result, we could use a Taylor expansion around $\theta=\frac \pi 3$ and get $$g(\theta)=\left(\frac{\pi }{3}-\frac{21}{20}\right)+\left(1-\frac{21 \sqrt{3}}{40}\right) \left(\theta-\frac{\pi }{3}\right)+\frac{21}{160} \left(\theta-\frac{\pi }{3}\right)^2+O\left(\left(\theta-\frac{\pi }{3}\right)^3\right)$$ Ignoring the higher order terms, another quadratic leading to $$\theta \approx 1.07683$$



Let $t $ be the angle $\angle ACB .$


$$21=rt $$


$$r\sin(t/2)=10 .$$ the area we want is the difference

$$S=\frac {r^2}{2}t-\frac{20.h}{2} $$ with $$h=r\cos(t/2) $$


$$S=\frac {21}{2}r-10\sqrt {r^2-100} $$

with $r $ satisfying $$r\sin (\frac {21}{2}r)=10$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.