Question Number 173025 by mnjuly1970 last updated on 05/Jul/22
Answered by mr W last updated on 06/Jul/22
Commented by mr W last updated on 06/Jul/22
$$\left({r}_{\mathrm{1}} +{r}\right)^{\mathrm{2}} −\left({r}_{\mathrm{1}} −{r}\right)^{\mathrm{2}} =\left(\mathrm{2}−{r}_{\mathrm{1}} \right)^{\mathrm{2}} −{r}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$${rr}_{\mathrm{1}} =\mathrm{1}−{r}_{\mathrm{1}} \\ $$$$\Rightarrow{r}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{1}+{r}} \\ $$$${CM}^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +{r}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$${OC}=\mathrm{2}−{r}_{\mathrm{2}} \\ $$$${ME}=\sqrt{\mathrm{1}^{\mathrm{2}} +{r}^{\mathrm{2}} } \\ $$$${CE}={r}+{r}_{\mathrm{2}} \\ $$$${CM}^{\mathrm{2}} ={ME}^{\mathrm{2}} +{CE}^{\mathrm{2}} −\mathrm{2}×{ME}×{CE}×\frac{{r}}{{ME}} \\ $$$$\mathrm{1}^{\mathrm{2}} +{r}_{\mathrm{2}} ^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +{r}^{\mathrm{2}} +\left({r}+{r}_{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}{r}\sqrt{{r}+{r}_{\mathrm{2}} } \\ $$$${r}+{r}_{\mathrm{2}} =\sqrt{{r}+{r}_{\mathrm{2}} } \\ $$$${r}+{r}_{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow{r}_{\mathrm{2}} =\mathrm{1}−{r} \\ $$$$\frac{{r}_{\mathrm{1}} }{{r}_{\mathrm{2}} }=\frac{\mathrm{1}}{\left(\mathrm{1}+{r}\right)\left(\mathrm{1}−{r}\right)}=\frac{\mathrm{7}}{\mathrm{6}} \\ $$$${r}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{7}} \\ $$$${r}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}} \\ $$$$\alpha=\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \frac{{r}}{\mathrm{1}}=\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}}=\mathrm{tan}^{−\mathrm{1}} \frac{\sqrt{\mathrm{7}}}{\mathrm{3}}=\mathrm{41}.\mathrm{4}° \\ $$
Commented by Tawa11 last updated on 11/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$