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Question Number 177784 by mnjuly1970 last updated on 09/Oct/22

Answered by aleks041103 last updated on 09/Oct/22

Commented by aleks041103 last updated on 09/Oct/22

$$construct$$$$H\in {OO}^{\prime },s.t.BH\mathrm{\perp }{OO}^{\prime }$$$$\mathrm{\Delta }{OO}^{\prime }A(\measuredangle O=90°)\Rightarrow \measuredangle O^{\prime }+\measuredangle A=90°$$$$\Rightarrow \alpha +\beta =90°$$$$ButAOBisaquartercircle$$$$\Rightarrow \measuredangle {AO}^{\prime }B=90°$$$$But$$$$\measuredangle {OO}^{\prime }A+\measuredangle {AO}^{\prime }B+\measuredangle {BO}^{\prime }H=180°$$$$\Rightarrow \alpha +\measuredangle {BO}^{\prime }H=90°$$$$\Rightarrow \measuredangle {BO}^{\prime }H=\beta$$$$Butbyconstructionfor\mathrm{\Delta }{HO}^{\prime }B(\measuredangle H=90°)$$$$\Rightarrow \measuredangle {HO}^{\prime }B+\measuredangle O^{\prime }BH=90°$$$$\Rightarrow \measuredangle O^{\prime }BH=\alpha$$$$\Rightarrow \mathrm{\Delta }{O}^{\prime }OA=\mathrm{\Delta }{BHO}^{\prime }$$$$since:$$$$1.bothareright\mathrm{\Delta }$$$$2.\measuredangle O^{\prime }BH=\measuredangle {OO}^{\prime }A$$$$3.{AO}^{\prime }=O^{\prime }B=r$$$$\Rightarrow BH={OO}^{\prime }=\frac{R}{2}$$$$\measuredangle OHB=90°andOB=R$$$$\Rightarrow OH=\sqrt{R^2-{\left(\frac{R}{2}\right)}^2}=\frac{\sqrt{3}}{2}R$$$$\Rightarrow O^{\prime }H=OH-{OO}^{\prime }=\frac{\sqrt{3}-1}{2}R$$$$\Rightarrow O^{\prime }B=\sqrt{{\left(\frac{\sqrt{3}-1}{2}R\right)}^2+{\left(\frac{R}{2}\right)}^2}=$$$$=\frac{R}{2}\sqrt{{(\sqrt{3}-1)}^2+1}=\frac{\sqrt{5-2\sqrt{3}}}{2}R=r$$$$\Rightarrow \frac{r}{R}=\frac{\sqrt{5-2\sqrt{3}}}{2}$$

Commented by Tawa11 last updated on 09/Oct/22

$$\mathrm{Great}\mathrm{sir}.$$

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