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Question-176684




Question Number 176684 by peter frank last updated on 25/Sep/22
Commented by peter frank last updated on 25/Sep/22
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by Rasheed.Sindhi last updated on 25/Sep/22
See Q#176281
$${See}\:{Q}#\mathrm{176281} \\ $$
Answered by floor(10²Eta[1]) last updated on 25/Sep/22
((θπ+2πα(1+(√2)))/(180))=P    ((senθ)/(2+2(√2)))=((senα)/(2+(√2)))  θ+2α=180⇒senθ=sen(180−2α)=sen(2α)  ((sen2α)(2+(√2)))/(2+2(√2)))=senα(2+2(√2))  2cosα+(√2)cosα=1+(√2)  cosα=((1+(√2))/(2+(√2)))×((2−(√2))/(2−(√2)))=((√2)/2)⇒α=45  P=((90π(2+(√2)))/(180))=π(1+(√2))
$$\frac{\theta\pi+\mathrm{2}\pi\alpha\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\mathrm{180}}=\mathrm{P} \\ $$$$ \\ $$$$\frac{\mathrm{sen}\theta}{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}=\frac{\mathrm{sen}\alpha}{\mathrm{2}+\sqrt{\mathrm{2}}} \\ $$$$\theta+\mathrm{2}\alpha=\mathrm{180}\Rightarrow\mathrm{sen}\theta=\mathrm{sen}\left(\mathrm{180}−\mathrm{2}\alpha\right)=\mathrm{sen}\left(\mathrm{2}\alpha\right) \\ $$$$\frac{\left.\mathrm{sen2}\alpha\right)\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}=\mathrm{sen}\alpha\left(\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$$$\mathrm{2cos}\alpha+\sqrt{\mathrm{2}}\mathrm{cos}\alpha=\mathrm{1}+\sqrt{\mathrm{2}} \\ $$$$\mathrm{cos}\alpha=\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{2}}}×\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{2}}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\Rightarrow\alpha=\mathrm{45} \\ $$$$\mathrm{P}=\frac{\mathrm{90}\pi\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}{\mathrm{180}}=\pi\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$
Commented by peter frank last updated on 25/Sep/22
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by Tawa11 last updated on 25/Sep/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by BaliramKumar last updated on 25/Sep/22
P = (π/2)(2+(√2)) = (π/( (√2)))(1+(√2))
$${P}\:=\:\frac{\pi}{\mathrm{2}}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)\:=\:\frac{\pi}{\:\sqrt{\mathrm{2}}}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$

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