Problem
As shown in the figure, an oblique line $AB$ has been drawn on the fan, and the greater and smaller circles are inscribed on either side of $AB$.
When the diameter of the greater circle, and the lengths of $OD$ and $AD$ are known, find the diameter of the small circle and the length of the oblique line $AB$.
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Solution
Let $OD=a, \ AD=b$ and the diameter of the greater circle be $2r$.
Let $∠O’ OC=α, \ ∠O’ AC=β$ and the diameter of the smaller circle be $d$.
For the right-angled triangle $△O’ OC$,
$$\frac{O’ C}{O’ O}=\frac{r}{a+r}=sinα.$$
Using the trigonometric table, we can find the value of $∠O’ OC=α$, and therefore the value of $cosα$,
$$\frac{OC}{O’ O}=\frac{OC}{a+r}=cosα,$$
$$∴ \ OC=(a+r)×cosα.$$
Since $CA=OD+AD-OC$,
$$CA=a+b-(a+r)×cosα.$$
For the right-angled triangle $△O’ AC$,
$$\frac{O’ C}{CA}=\frac{r}{a+b-(a+r)×cosα}=tanβ.$$
The trigonometric table gives us the value of $∠O’ AC=β$, and therefore the values of $sin2β$ and $cos2β$.
For the right-angled triangle $△AOF$,
$$\frac{OF}{AO}=\frac{OF}{a+b}=sin2β,$$
$$∴ \ OF=(a+b)×sin2β.$$
Since $d=AO-OF$,
$$d=(a+b)×(1-sin2β ).$$
Also,
$$\frac{FA}{AO}=\frac{FA}{a+b}=cos2β,$$
$$∴ \ FA=(a+b)×cos2β.$$
Since $FA=FB$,
$$AB=2(a+b)×cos2β.$$
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Reference
Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), p.45; pp.325-326.