Question Number 14145 by RasheedSindhi last updated on 28/May/17
Commented by RasheedSindhi last updated on 28/May/17
$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\alpha+\beta+\gamma+\delta+\xi=\mathrm{180}° \\ $$
Answered by ajfour last updated on 28/May/17
$${let}\:\boldsymbol{{A}}\:{be}\:{vertex}\:{of}\:{pentagon} \\ $$$${opposite}\:{to}\:\boldsymbol{\alpha}\:{and}\:{so}\:{on}.. \\ $$$$\Sigma{A}=\mathrm{3}\pi\:\:{and}\: \\ $$$$\pi−\delta=\left(\pi−{A}\right)+\left(\pi−{B}\right) \\ $$$$\pi−\xi=\left(\pi−{B}\right)+\left(\pi−{C}\right) \\ $$$$\pi−\alpha=\left(\pi−{C}\right)+\left(\pi−{D}\right) \\ $$$$\pi−\beta=\left(\pi−{D}\right)+\left(\pi−{E}\right) \\ $$$$\pi−\gamma=\left(\pi−{A}\right)+\left(\pi−{A}\right) \\ $$$$\Rightarrow\:\:\:\mathrm{5}\pi−\Sigma\alpha=\mathrm{10}\pi−\mathrm{2}\Sigma{A}\: \\ $$$${or}\:\:\:\:\mathrm{5}\pi−\Sigma\alpha=\mathrm{10}\pi−\mathrm{6}\pi \\ $$$$\:\:\:\:\:\Sigma\boldsymbol{\alpha}\:=\boldsymbol{\pi}\:. \\ $$
Commented by RasheedSindhi last updated on 28/May/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$