Find the number of sides of a regular polygon, when each of its angles has a measure of

(i) Each interior angle $=\left(\frac{2 n-4}{n} \times 90\right)^{\circ}$

So, $\left(\frac{2 n-4}{n} \times 90\right)^{\circ}=160^{\circ}$

$\Rightarrow \frac{2 n-4}{n}=\frac{160^{\circ}}{90^{\circ}}$

$\Rightarrow \frac{2 n-4}{n}=\frac{16}{9}$

$\Rightarrow 18 n-36=16 n$

$\Rightarrow 2 n=36$

$\therefore n=18$

(ii) Each interior angle $=\left(\frac{2 n-4}{n} \times 90\right)^{\circ}$

So, $\left(\frac{2 n-4}{n} \times 90\right)^{\circ}=135^{\circ}$

$\Rightarrow \frac{2 n-4}{n}=\frac{135^{\circ}}{90^{\circ}}$

$\Rightarrow \frac{2 n-4}{n}=\frac{3}{2}$

$\Rightarrow 4 n-8=3 n$

$\therefore \mathrm{n}=8$

(iii) Each interior angle $=\left(\frac{2 n-4}{n} \times 90\right)^{\circ}$

So, $\left(\frac{2 n-4}{n} \times 90\right)^{\circ}=175^{\circ}$

$\Rightarrow \frac{2 n-4}{n}=\frac{175^{\circ}}{90^{\circ}}$

$\Rightarrow \frac{2 n-4}{n}=\frac{35}{18}$

$\Rightarrow 36 n-72=35 n$

$\therefore \mathrm{n}=72$

(iv) Each interior angle $=\left(\frac{2 n-4}{n} \times 90\right)^{\circ}$

So, $\left(\frac{2 n-4}{n} \times 90\right)^{\circ}=162^{\circ}$

$\Rightarrow \frac{2 n-4}{n}=\frac{162^{\circ}}{90^{\circ}}$

$\Rightarrow \frac{2 n-4}{n}=\frac{9}{5}$

$\Rightarrow 10 n-20=9 n$

$\therefore n=20$

(v) Each interior angle $=\left(\frac{2 n-4}{n} \times 90\right)^{\circ}$

So, $\left(\frac{2 n-4}{n} \times 90\right)^{\circ}=150^{\circ}$

$\Rightarrow \frac{2 n-4}{n}=\frac{150^{\circ}}{90^{\circ}}$

$\Rightarrow \frac{2 n-4}{n}=\frac{5}{3}$

$\Rightarrow 6 n-12=5 n$

$\therefore n=12$