The-sum-of-all-but-one-of-the-interior-angles-of-a-convex-polygon-equals-2525-find-the-measure-of-the-exterior-angle-adjacent-to-the-remaining-interior-angle-can-you-please-help-if-possible-with

Question Number 55333 by Otchere Abdullai last updated on 21/Feb/19

$${The}\:{sum}\:{of}\:{all}\:{but}\:{one}\:{of}\:{the}\:{interior} \\ $$$${angles}\:{of}\:{a}\:{convex}\:{polygon}\:{equals}\: \\ $$$$\mathrm{2525}°\:.\:{find}\:{the}\:{measure}\:{of}\:{the} \\ $$$${exterior}\:{angle}\:{adjacent}\:{to}\:{the} \\ $$$${remaining}\:{interior}\:{angle}.\: \\ $$$${can}\:{you}\:{please}\:{help}\:{if}\:{possible}\:{with} \\ $$$${diagram}\: \\ $$

Answered by mr W last updated on 21/Feb/19

let′s say the polygon has n sides. α_n is the remaining interior angle. Σ_(k=1) ^(n−1) α_k +α_n =(n−2)×180° 2525°+α_n =(n−2)×180° ⇒α_n =(n−2)×180°−2525° convex polygon⇒ 0°<α_n <180° if n=16: α_n =−5°<0° if n=18: α_n =355°>180° ⇒n=17 ⇒α_n =(17−2)×180°−2525°=175° ok! exterior angle β_n =180°−α_n =5°

$${let}'{s}\:{say}\:{the}\:{polygon}\:{has}\:{n}\:{sides}. \\ $$$$\alpha_{{n}} \:{is}\:{the}\:{remaining}\:{interior}\:{angle}. \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\alpha_{{k}} +\alpha_{{n}} =\left({n}−\mathrm{2}\right)×\mathrm{180}° \\ $$$$\mathrm{2525}°+\alpha_{{n}} =\left({n}−\mathrm{2}\right)×\mathrm{180}° \\ $$$$\Rightarrow\alpha_{{n}} =\left({n}−\mathrm{2}\right)×\mathrm{180}°−\mathrm{2525}° \\ $$$${convex}\:{polygon}\Rightarrow\:\mathrm{0}°<\alpha_{{n}} <\mathrm{180}° \\ $$$${if}\:{n}=\mathrm{16}:\:\alpha_{{n}} =−\mathrm{5}°<\mathrm{0}° \\ $$$${if}\:{n}=\mathrm{18}:\:\alpha_{{n}} =\mathrm{355}°>\mathrm{180}° \\ $$$$\Rightarrow{n}=\mathrm{17} \\ $$$$\Rightarrow\alpha_{{n}} =\left(\mathrm{17}−\mathrm{2}\right)×\mathrm{180}°−\mathrm{2525}°=\mathrm{175}°\:{ok}! \\ $$$${exterior}\:{angle}\:\beta_{{n}} =\mathrm{180}°−\alpha_{{n}} =\mathrm{5}° \\ $$

Commented by mr W last updated on 21/Feb/19