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Question Number 185324 by cherokeesay last updated on 20/Jan/23

Answered by som(math1967) last updated on 20/Jan/23

$$AC=\sqrt{2^2+1^2}=\sqrt{5}=BC$$$$r\times \frac{2+1+\sqrt{5}}{2}=\frac{1}{2}\times 2\times 1$$$$\Rightarrow r=\frac{2}{3+\sqrt{5}}$$$$R\times \frac{\sqrt{5}+\sqrt{5}+2}{2}=\frac{1}{2}\times 2\times 2$$$$R=\frac{4}{2(\sqrt{5}+1)}=\frac{2}{\sqrt{5}+1}$$$$1+\frac{r}{R}=1+\frac{\frac{2}{3+\sqrt{5}}}{\frac{2}{\sqrt{5}+1}}$$$$=1+\frac{\sqrt{5}+1}{3+\sqrt{5}}=\frac{4+2\sqrt{5}}{3+\sqrt{5}}=\frac{2(\sqrt{5}+2)(3-\sqrt{5)}}{9-5}$$$$=\frac{2(\sqrt{5}+1)}{4}=\frac{\sqrt{5}+1}{2}=\phi$$$$$$

Commented by som(math1967) last updated on 20/Jan/23

Commented by cherokeesay last updated on 20/Jan/23

$$Nice!thankyou.$$

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