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
$${let}'{s}\:{say}\:{the}\:{polygon}\:{has}\:{n}\:{sides}. \\ $$$$\alpha_{{n}} \:{is}\:{the}\:{remaining}\:{interior}\:{angle}. \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\alpha_{{k}} +\alpha_{{n}} =\left({n}−\mathrm{2}\right)×\mathrm{180}° \\ $$$$\mathrm{2525}°+\alpha_{{n}} =\left({n}−\mathrm{2}\right)×\mathrm{180}° \\ $$$$\Rightarrow\alpha_{{n}} =\left({n}−\mathrm{2}\right)×\mathrm{180}°−\mathrm{2525}° \\ $$$${convex}\:{polygon}\Rightarrow\:\mathrm{0}°<\alpha_{{n}} <\mathrm{180}° \\ $$$${if}\:{n}=\mathrm{16}:\:\alpha_{{n}} =−\mathrm{5}°<\mathrm{0}° \\ $$$${if}\:{n}=\mathrm{18}:\:\alpha_{{n}} =\mathrm{355}°>\mathrm{180}° \\ $$$$\Rightarrow{n}=\mathrm{17} \\ $$$$\Rightarrow\alpha_{{n}} =\left(\mathrm{17}−\mathrm{2}\right)×\mathrm{180}°−\mathrm{2525}°=\mathrm{175}°\:{ok}! \\ $$$${exterior}\:{angle}\:\beta_{{n}} =\mathrm{180}°−\alpha_{{n}} =\mathrm{5}° \\ $$
Commented by mr W last updated on 21/Feb/19
Commented by Otchere Abdullai last updated on 21/Feb/19
$${Yes}\:{thats}\:{why}\:{i}\:{said}\:{you}\:{are}\:{prof} \\ $$$${God}\:{bless}\:{you}\:{a}\:{lot} \\ $$
Commented by Otchere Abdullai last updated on 21/Feb/19
$${O}!\:{yes}\:{Prof}\:{of}\:{Profs} \\ $$