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Question Number 37871 by kunal1234523 last updated on 18/Jun/18

$$\mathrm{A}\mathrm{regular}\mathrm{pyramid}\mathrm{has}\mathrm{for}\mathrm{its}\mathrm{base}\mathrm{polygon}$$$$\mathrm{of}n\mathrm{sides},\mathrm{and}\mathrm{each}\mathrm{slant}\mathrm{face}\mathrm{consist}\mathrm{of}\mathrm{an}$$$$\mathrm{isosceles}\mathrm{triangle}\mathrm{of}\mathrm{vertical}\mathrm{angle}2\alpha .\mathrm{If}\mathrm{the}$$$$\mathrm{slant}\mathrm{faces}\mathrm{are}\mathrm{each}\mathrm{inclined}\mathrm{at}\mathrm{angle}\beta \mathrm{to}$$$$\mathrm{the}\mathrm{base},\mathrm{and}\mathrm{at}\mathrm{an}\mathrm{angle}2\gamma \mathrm{to}\mathrm{one}\mathrm{another}$$$$\mathrm{show}\mathrm{that}$$$$\cos \beta =\tan \alpha \cot \frac{\pi }{\mathrm{n}},\mathrm{and}\sin \gamma =\sec \alpha \cos \frac{\pi }{\mathrm{n}}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 18/Jun/18

Commented by ajfour last updated on 18/Jun/18

$$\cos \beta =\tan \alpha \cot \frac{\pi }{n}\left(icanprove\right)$$$$but\sin \gamma =\cos \frac{\pi }{n}\left(ithink\right).$$

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