Menu Close

Question-14145




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

Leave a Reply

Your email address will not be published. Required fields are marked *