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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.

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2 Answers 2

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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?

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  • $\begingroup$ That’s where I hit the wall. Cos(12pi/r) ? $\endgroup$
    – TrickHorse
    Oct 18, 2019 at 20:07
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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 $R =16.4$.

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

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