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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

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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