Question Number 164712 by cortano1 last updated on 21/Jan/22
Answered by mr W last updated on 21/Jan/22
Commented by mr W last updated on 21/Jan/22
$${R}^{\mathrm{2}} =\left({b}+{c}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$${R}^{\mathrm{2}} =\left({b}+\mathrm{2}{c}\right)^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$$\left({b}+\mathrm{2}{c}\right)^{\mathrm{2}} +{c}^{\mathrm{2}} =\left({b}+{c}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$$\mathrm{4}{c}^{\mathrm{2}} +\mathrm{2}{bc}−{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{c}=\frac{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right){b}}{\mathrm{4}} \\ $$$${R}=\sqrt{\left({b}+\frac{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right){b}}{\mathrm{4}}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} }=\frac{{b}\sqrt{\mathrm{6}\sqrt{\mathrm{5}}+\mathrm{30}}}{\mathrm{4}} \\ $$$${a}={R}−{b} \\ $$$$\frac{{a}}{{b}}=\frac{{R}}{{b}}−\mathrm{1}=\frac{\sqrt{\mathrm{6}\sqrt{\mathrm{5}}+\mathrm{30}}}{\mathrm{4}}−\mathrm{1}\:\checkmark \\ $$
Commented by Tawa11 last updated on 21/Jan/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$