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Question Number 174776 by Tawa11 last updated on 10/Aug/22

Answered by aleks041103 last updated on 11/Aug/22

Commented by aleks041103 last updated on 11/Aug/22

$$R=5$$$$r=?$$$$OA=R-r=5-r$$$$AB=r$$$$\Rightarrow OB=\sqrt{{(5-r)}^2-r^2}=\sqrt{5(5-2r)}$$$$tg\left(ACB\right)=\frac{r}{5+\sqrt{5(5-2r)}}$$$$CAistheanglebisectorofC,since$$$$Aisthecenterofacirclewhosetangents$$$$intersectatC.$$$$\Rightarrow ACB=15°$$$$nowifweknowtg\left(15°\right)=t^{-1}wecanfindr.$$$$5+\sqrt{5(5-2r)}=rt$$$$\Rightarrow {(rt-5)}^2=25-10r$$$$t^2{r}^2-10tr+10r=0$$$$\Rightarrow t^{2}r-10(t-1)=0$$$$\Rightarrow r=\frac{10(t-1)}{t^2}$$$$tg\left(30\right)=\frac{2tg\left(15\right)}{1-{tg}^2\left(15\right)}=\frac{2}{t(1-\frac{1}{t^2})}=\frac{2t}{t^2-1}=\frac{1}{\sqrt{3}}$$$$t^2-2\sqrt{3}t-1=0$$$$t_{1,2}=\frac{2\sqrt{3}\pm \sqrt{12+4}}{2}=\sqrt{3}\pm 2$$$$t>0$$$$\Rightarrow t=2+\sqrt{3}$$$$\Rightarrow r=\frac{10(1+\sqrt{3})}{7+4\sqrt{3}}$$

Commented by Tawa11 last updated on 11/Aug/22

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by peter frank last updated on 11/Aug/22

$$\mathrm{thank}\mathrm{you}$$

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