Uhly pravidelného dvanásťuholníka
Aké veľké budú jednotlivé uhly daného pravidelného n-uholníka?
\( \normalsize n=12 \)
\( \large\alpha=\frac{\left(n-2\right)\cdot180\degree}{n}=\frac{\left(12-1\right)\cdot180\degree}{12}=\frac{11\cdot180\degree}{12}=\frac{1980\degree}{12}=165\degree \)
\( \large\alpha=\frac{\left(n-2\right)\cdot180\degree}{n}=\frac{\left(12-4\right)\cdot180\degree}{12}=\frac{8\cdot180\degree}{12}=\frac{1440\degree}{12}=120\degree \)
\( \large\alpha=\frac{\left(n-2\right)\cdot180\degree}{n}=\frac{\left(12-2\right)\cdot180\degree}{12}=\frac{10\cdot180\degree}{12}=\frac{1800\degree}{12}=150\degree \)
\( \large\alpha=\frac{\left(n-2\right)\cdot180\degree}{n}=\frac{\left(12-3\right)\cdot180\degree}{12}=\frac{9\cdot180\degree}{12}=\frac{1620\degree}{12}=135\degree \)