Question Number 176684 by peter frank last updated on 25/Sep/22
Commented by peter frank last updated on 25/Sep/22
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by Rasheed.Sindhi last updated on 25/Sep/22
$${See}\:{Q}#\mathrm{176281} \\ $$
Answered by floor(10²Eta[1]) last updated on 25/Sep/22
$$\frac{\theta\pi+\mathrm{2}\pi\alpha\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\mathrm{180}}=\mathrm{P} \\ $$$$ \\ $$$$\frac{\mathrm{sen}\theta}{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}=\frac{\mathrm{sen}\alpha}{\mathrm{2}+\sqrt{\mathrm{2}}} \\ $$$$\theta+\mathrm{2}\alpha=\mathrm{180}\Rightarrow\mathrm{sen}\theta=\mathrm{sen}\left(\mathrm{180}−\mathrm{2}\alpha\right)=\mathrm{sen}\left(\mathrm{2}\alpha\right) \\ $$$$\frac{\left.\mathrm{sen2}\alpha\right)\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}=\mathrm{sen}\alpha\left(\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$$$\mathrm{2cos}\alpha+\sqrt{\mathrm{2}}\mathrm{cos}\alpha=\mathrm{1}+\sqrt{\mathrm{2}} \\ $$$$\mathrm{cos}\alpha=\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{2}}}×\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{2}}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\Rightarrow\alpha=\mathrm{45} \\ $$$$\mathrm{P}=\frac{\mathrm{90}\pi\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}{\mathrm{180}}=\pi\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$
Commented by peter frank last updated on 25/Sep/22
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by Tawa11 last updated on 25/Sep/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by BaliramKumar last updated on 25/Sep/22
$${P}\:=\:\frac{\pi}{\mathrm{2}}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)\:=\:\frac{\pi}{\:\sqrt{\mathrm{2}}}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$