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Question Number 185993 by ajfour last updated on 30/Jan/23

Answered by mr W last updated on 30/Jan/23

Commented by mr W last updated on 30/Jan/23

$$\sin \theta =\frac{s}{1}$$$$\cos \theta =\frac{2r}{s}\Rightarrow r=\frac{\cos \theta s}{2}=\frac{\sin \theta \cos \theta }{2}$$$$s=r+\frac{r}{\tan \frac{\theta }{2}}$$$$\frac{s}{r}=1+\frac{1}{\tan \frac{\theta }{2}}=\frac{2}{\cos \theta }$$$$lett=\tan \frac{\theta }{2}$$$$1+\frac{1}{t}=\frac{2(1+t^2)}{1-t^2}$$$$\frac{1+t}{t}=\frac{2(1+t^2)}{1-t^2}$$$$3t^3+t^2+t-1=0$$$$\Rightarrow t\approx 0.469396$$$$d=2r=\frac{2t(1-t^2)}{{(1+t^2)}^2}\approx 0.491$$

Commented by ajfour last updated on 31/Jan/23

$$yessir,thatsright!$$$$vnicelypresented.$$

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