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Question Number 14145 by RasheedSindhi last updated on 28/May/17

Commented by RasheedSindhi last updated on 28/May/17

$$\mathrm{Prove}\mathrm{that}$$$$\alpha +\beta +\gamma +\delta +\xi =180°$$

Answered by ajfour last updated on 28/May/17

$$let\mathit{A}bevertexofpentagon$$$$oppositeto\mathit{\alpha }andsoon..$$$$\mathrm{\Sigma }A=3\pi and$$$$\pi -\delta =(\pi -A)+(\pi -B)$$$$\pi -\xi =(\pi -B)+(\pi -C)$$$$\pi -\alpha =(\pi -C)+(\pi -D)$$$$\pi -\beta =(\pi -D)+(\pi -E)$$$$\pi -\gamma =(\pi -A)+(\pi -A)$$$$\Rightarrow 5\pi -\mathrm{\Sigma }\alpha =10\pi -2\mathrm{\Sigma }A$$$$or5\pi -\mathrm{\Sigma }\alpha =10\pi -6\pi$$$$\mathrm{\Sigma }\mathit{\alpha }=\mathit{\pi }.$$

Commented by RasheedSindhi last updated on 28/May/17

$$\mathrm{Thank}\mathrm{you}\mathrm{sir}.$$

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