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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

$$\text{You can't use 'macro parameter character \#' in math mode}$$

Answered by floor(10²Eta[1]) last updated on 25/Sep/22

$$\frac{\theta \pi +2\pi \alpha (1+\sqrt{2})}{180}=\mathrm{P}$$$$$$$$\frac{\mathrm{sen}\theta }{2+2\sqrt{2}}=\frac{\mathrm{sen}\alpha }{2+\sqrt{2}}$$$$\theta +2\alpha =180\Rightarrow \mathrm{sen}\theta =\mathrm{sen}(180-2\alpha )=\mathrm{sen}\left(2\alpha \right)$$$$\frac{\mathrm{sen}2\alpha )(2+\sqrt{2})}{2+2\sqrt{2}}=\mathrm{sen}\alpha (2+2\sqrt{2})$$$$2\cos \alpha +\sqrt{2}\cos \alpha =1+\sqrt{2}$$$$\cos \alpha =\frac{1+\sqrt{2}}{2+\sqrt{2}}\times \frac{2-\sqrt{2}}{2-\sqrt{2}}=\frac{\sqrt{2}}{2}\Rightarrow \alpha =45$$$$\mathrm{P}=\frac{90\pi (2+\sqrt{2})}{180}=\pi (1+\sqrt{2})$$

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

$$P=\frac{\pi }{2}(2+\sqrt{2})=\frac{\pi }{\sqrt{2}}(1+\sqrt{2})$$

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