Question

Find the number of sides of a regular polygon, when each of its angles has a measure of
(i) 160°
(ii) 135°
(iii) 175°
(iv) 162°
(v) 150°

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Each}\mathrm{interior}\mathrm{angle}={\left(\frac{2n-4}{n}\times 90\right)}^{°} \\ So,{\left(\frac{2n-4}{n}\times 90\right)}^{°}=160° \\ \Rightarrow \frac{2n-4}{n}=\frac{160°}{90°} \\ \Rightarrow \frac{2n-4}{n}=\frac{16}{9} \\ \Rightarrow 18n-36=16n \\ \Rightarrow 2n=36 \\ \therefore n=18 \\ \left(\mathrm{ii}\right)\mathrm{Each}\mathrm{interior}\mathrm{angle}={\left(\frac{2n-4}{n}\times 90\right)}^{°} \\ \mathrm{So},{\left(\frac{2n-4}{n}\times 90\right)}^{°}=135° \\ \Rightarrow \frac{2n-4}{n}=\frac{135°}{90°} \\ \Rightarrow \frac{2n-4}{n}=\frac{3}{2} \\ \Rightarrow 4n-8=3n \\ \therefore \mathrm{n}=8 \\ \left(\mathrm{iii}\right)\mathrm{Each}\mathrm{interior}\mathrm{angle}={\left(\frac{2n-4}{n}\times 90\right)}^{°} \\ \mathrm{So},{\left(\frac{2n-4}{n}\times 90\right)}^{°}=175° \\ \Rightarrow \frac{2n-4}{n}=\frac{175°}{90°} \\ \Rightarrow \frac{2n-4}{n}=\frac{35}{18} \\ \Rightarrow 36n-72=35n \\ \therefore \mathrm{n}=72 \\ \left(\mathrm{iv}\right)\mathrm{Each}\mathrm{interior}\mathrm{angle}={\left(\frac{2n-4}{n}\times 90\right)}^{°} \\ \mathrm{So},{\left(\frac{2n-4}{n}\times 90\right)}^{°}=162° \\ \Rightarrow \frac{2n-4}{n}=\frac{162°}{90°} \\ \Rightarrow \frac{2n-4}{n}=\frac{9}{5} \\ \Rightarrow 10n-20=9n \\ \therefore n=20 \\ \left(\mathrm{v}\right)\mathrm{Each}\mathrm{interior}\mathrm{angle}={\left(\frac{2n-4}{n}\times 90\right)}^{°} \\ \mathrm{So},{\left(\frac{2n-4}{n}\times 90\right)}^{°}=150° \\ \Rightarrow \frac{2n-4}{n}=\frac{150°}{90°} \\ \Rightarrow \frac{2n-4}{n}=\frac{5}{3} \\ \Rightarrow 6n-12=5n \\ \therefore n=12 \end{gathered}$$