Question Number 75518 by aliesam last updated on 12/Dec/19
Answered by mr W last updated on 12/Dec/19
Commented by mr W last updated on 12/Dec/19
$${OB}={R} \\ $$$${BC}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${OA}={R}−\mathrm{1} \\ $$$${AB}=\mathrm{2} \\ $$$$\angle{ABC}=\mathrm{60}°=\frac{\pi}{\mathrm{3}} \\ $$$$\mathrm{cos}\:\angle{OBC}=\frac{\mathrm{3}}{\mathrm{2}{R}} \\ $$$$\mathrm{cos}\:\angle{ABO}=\frac{{R}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} −\left({R}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{2}×{R}}=\frac{\mathrm{3}+\mathrm{2}{R}}{\mathrm{4}{R}} \\ $$$$\angle{ABO}=\frac{\pi}{\mathrm{3}}−\angle{OBC} \\ $$$$\mathrm{cos}\:\angle{ABO}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\angle{OBC}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:\angle{OBC} \\ $$$$\Rightarrow\frac{\mathrm{3}+\mathrm{2}{R}}{\mathrm{4}{R}}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{2}{R}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\sqrt{\mathrm{4}{R}^{\mathrm{2}} −\mathrm{9}}}{\mathrm{2}{R}} \\ $$$$\Rightarrow\mathrm{2}{R}=\sqrt{\mathrm{3}\left(\mathrm{4}{R}^{\mathrm{2}} −\mathrm{9}\right)} \\ $$$$\Rightarrow\mathrm{8}{R}^{\mathrm{2}} =\mathrm{27} \\ $$$$\Rightarrow{R}=\frac{\mathrm{3}\sqrt{\mathrm{6}}}{\mathrm{4}}\approx\mathrm{1}.\mathrm{837} \\ $$