# Determine radius of a circle given length of chord and subtended minor arclength.

In circle $$O$$, chord $$AB = 30$$ subtends minor arc $$AB = 12\pi$$. Determine radius $$OA$$.

Discussion: It seems that attaching a string to the ends of a straight stick (briefly disregarding a catenary and using reasonable lengths) fixes the circle formed by the string. Hence, we can determine its radius. At least it seems so.

The equation relating angle, radius, and arclength is given by $$\theta=\frac{\text{arclength}}{\text{radius}}=\frac{a}{r}=\frac{12\pi}{r}$$

The relationship between angles and side lengths of an isosceles triangle with given third side length and angle can be found by the equation representing the law of cosines: \begin{align} c^2&=a^2+b^2-2ab\cos C\\ (30)^2&=r^2+r^2-2(r)(r)\cos\theta\\ 900&=2r^2-2r^2\cos\theta \end{align}

You now have two equations relating $$\theta$$ and $$r$$. Can you take it from here?

• That’s where I hit the wall. Cos(12pi/r) ? Oct 18, 2019 at 20:07

Let $$\theta$$ be the sector angle and $$R$$ the radius. Establish the equations below,

$$\theta=\frac{12\pi}{R}, \>\>\>\>\> \sin\frac{\theta}{2} = \frac{15}{R}$$

Eliminate $$\theta$$ to get the equation for $$R$$,

$$R\sin\frac{6\pi}{R} = 15$$

Solve for $$R$$ numerically to get $$R =16.4$$.

• No surprise here. Solve numerically for R? What to do with sin(6pi/R) ? Oct 18, 2019 at 20:04
• @TrickHorse - There is no analytical solution for this equation. You may just use the Newton's approximation with an initial guess, say, 15. Oct 18, 2019 at 21:12
• Thx Quanto. I’ve not seen this problem before. Guess I see why. Appreciate the input. Trickhorse Oct 18, 2019 at 21:31