Question

The number of sides in two regular polygons are as $5:4$ and the difference between their angles is ${60}^o,$ find the number of sides in the polygons.

Answer 1

Let $N,n(N>n)$ the sides of regular polygons so that each interior angle is $\frac{2N-4}{N}\frac{\pi }{2}$ and $\frac{2n-4}{n}\frac{\pi }{2}$ where as given $\frac{2N-4}{N}\frac{\pi }{2}-\frac{2n-4}{n}\frac{\pi }{2}={60}^o=\frac{\pi }{30}$

or $(1-\frac{2}{N})-(1-\frac{2}{n})=\frac{1}{30}$ or $\frac{1}{n}-\frac{1}{N}=\frac{1}{60}$

Also it is given that $\frac{N}{n}=\frac{5}{4}$ or $\frac{N}{5}=\frac{n}{4}=k$ say.

$\therefore \frac{1}{4k}-\frac{1}{5k}=\frac{1}{60}$ or $\frac{1}{20k}=\frac{1}{60}$ $\therefore k=3$
Hence the sides are $15$ and $12$ by $\left(2\right)$.