$\phantom{X}d_n\,=\,\dfrac{\;n(n\,-\,3)\;}{2}\;=\;20\phantom{X} \;\Longleftrightarrow\;$ $\;n^2 - 3n - 40 = 0\;\Longleftrightarrow\;$ $\,\left\{\begin{array}{rcr} n\,=\,8\phantom{XX}\longleftarrow & {\text (resposta)\phantom{X}} \\ n\,=\,\cancel{-5}\;\longleftarrow& {\text (inadequado)} \\ \end{array} \right.\,$
se n = 8 o polígono é o OCTÓGONO
$\;A_i\,=\,\dfrac{\;S_i\;}{\;n\;}\;=\dfrac{\;180^o(n\,-\,2)\;}{n}\;=\;150^o\;\Longleftrightarrow$ $\;180^o\,\centerdot\,n\,-\,360^o\;=\;150^o\,\centerdot\,n\;\Longrightarrow$ $\;n\,=\,12\;$
$\,d_n\,=\,\dfrac{\,n\,\centerdot\,(n\,-\,3)\,}{2}\phantom{XX}n\,\in\,\mathbb{N}\,;\,n\,\geqslant\,3$
$\,\dfrac{\,n\,\centerdot\,(n\,-\,3)\,}{2}\,=\,15\;\Longleftrightarrow\;n^2\,-\,3n\,-\,30\,=\,0$
$\,n_1\,=\,\dfrac{\;9\,+ \sqrt{\,129\;}}{2}\phantom{XX}n_2 = \dfrac{\;9\,- \sqrt{\,129\;}}{2}\;\Longrightarrow$ $\;\,n\,\,\notin\,\,\mathbb{N}$
$\,130^o\,\lt\,\dfrac{\,180(n\,-\,2)\,}{n}\,\lt\,140^o\,$ que podemos então resolver como um sistema de inequações:
$\,\left\{\begin{array}{rcr} 130^o \lt \,\dfrac{\,180(n\,-\,2)\,}{n}\;&(I) \\ \dfrac{\,180(n\,-\,2)\,}{n}\,\lt\,140^o\;&(II) \end{array} \right.\,$Resolvento (I)
$\,130^o\,\lt\,\,\dfrac{\,180(n\,-\,2)\,}{n}\;\Longleftrightarrow$ $\;130n\,\lt\,180(n\,-\,2)\;\Longleftrightarrow$ $\;\boxed{\;n\,\gt\,7,2\;}\;(*)$
Resolvento (II)
$\,\dfrac{\,180(n\,-\,2)\,}{n}\lt\,140^o\;\Longleftrightarrow$ $\;180(n\,-\,2)\,\lt\,140n\;\Longleftrightarrow$ $\;\boxed{\;n\,\lt\,9\;}\;(**)$