Question Number 212801 by Spillover last updated on 24/Oct/24
Answered by som(math1967) last updated on 24/Oct/24
Commented by som(math1967) last updated on 24/Oct/24
$$\frac{{OL}}{{PL}}=\mathrm{tan}\:\mathrm{22}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${PL}=\mathrm{6}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right) \\ $$$${PL}={PM}=\mathrm{6}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right) \\ $$$$\:\frac{{MN}}{{PM}}=\mathrm{tan}\:\mathrm{30} \\ $$$${MN}=\frac{\mathrm{6}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)}{\:\sqrt{\mathrm{3}}} \\ $$$$\therefore\:{R}=\frac{\mathrm{6}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)}{\:\sqrt{\mathrm{3}}} \\ $$
Answered by mr W last updated on 24/Oct/24
$$\frac{{R}}{\mathrm{tan}\:\frac{\mathrm{60}°}{\mathrm{2}}}=\frac{\mathrm{6}}{\mathrm{tan}\:\frac{\mathrm{45}°}{\mathrm{2}}} \\ $$$$\Rightarrow{R}=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}=\mathrm{2}\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right) \\ $$