Question Number 71955 by Maclaurin Stickker last updated on 22/Oct/19
Commented by Maclaurin Stickker last updated on 22/Oct/19
$${What}\:{is}\:{the}\:{value}\:{of}\:\:\frac{{AB}}{{AC}}? \\ $$
Answered by mr W last updated on 22/Oct/19
$${AC}={R} \\ $$$${BC}=\mathrm{2}{r} \\ $$$$\frac{{r}}{{R}−{r}}=\mathrm{sin}\:\mathrm{22}.\mathrm{5}°=\frac{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}}=\lambda \\ $$$${r}=\frac{\lambda}{\mathrm{1}+\lambda}{R} \\ $$$${AB}={R}−\mathrm{2}{r}=\left(\frac{\mathrm{1}−\lambda}{\mathrm{1}+\lambda}\right){R} \\ $$$$\Rightarrow\frac{{AB}}{{AC}}=\frac{\mathrm{1}−\lambda}{\mathrm{1}+\lambda}=\frac{\mathrm{2}−\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}+\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}\approx\mathrm{0}.\mathrm{446} \\ $$
Commented by Maclaurin Stickker last updated on 22/Oct/19
$${what}\:{if}\:{it}\:{was}\:\frac{{AB}}{{BC}}\:? \\ $$
Commented by mr W last updated on 22/Oct/19
$${BC}=\mathrm{2}{r}=\frac{\mathrm{2}\lambda}{\mathrm{1}+\lambda} \\ $$$$\frac{{AB}}{{BC}}=\frac{\mathrm{1}−\lambda}{\mathrm{2}\lambda}=\frac{\mathrm{2}−\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}\approx\mathrm{0}.\mathrm{806} \\ $$