Menu Close

Question-48676




Question Number 48676 by Tawa1 last updated on 27/Nov/18
Answered by tanmay.chaudhury50@gmail.com last updated on 27/Nov/18
sum of internal angles of n sided polygon is  (n−2)×180^o   so   120+110+2x+30=(5−2)×180  2x=540−260  2x=280  x=140^o
$${sum}\:{of}\:{internal}\:{angles}\:{of}\:{n}\:{sided}\:{polygon}\:{is} \\ $$$$\left({n}−\mathrm{2}\right)×\mathrm{180}^{{o}} \\ $$$${so}\: \\ $$$$\mathrm{120}+\mathrm{110}+\mathrm{2}{x}+\mathrm{30}=\left(\mathrm{5}−\mathrm{2}\right)×\mathrm{180} \\ $$$$\mathrm{2}{x}=\mathrm{540}−\mathrm{260} \\ $$$$\mathrm{2}{x}=\mathrm{280}\:\:{x}=\mathrm{140}^{{o}} \\ $$
Commented by Tawa1 last updated on 27/Nov/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Leave a Reply

Your email address will not be published.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