Not another geometry, going to bed

Assume that the polygon with 27 sides has $n$ sides. So number of diagonals in this polygon are $\binom{n}{2} - n$ .

$\therefore \binom{n}{2} - n=27 \Rightarrow n=9$ .

And interior angle of a regular polygon of $n$ sides $=180^\circ -\frac{360^\circ}{n}$ .

Therefore for this polygon of $n$ sides, we have interior angle (putting $n=9$ ) equal to $140^\circ . \square$

Number of diagonals in an $n$ sided polygon is $\dfrac{n(n-3)}{2}$

$\begin{aligned} \dfrac{n(n-3)}{2} & = 27 \\ n^2 -3n &= 54 \\ (n-9)(n+6)&= 0 \\ \implies n&= 9 &[n>0] \end{aligned}$

The interior angle of an $n$ sided polygon is $\dfrac{(n-2)180}{n}$

$\begin{aligned} \implies &\dfrac{(9-2)180}{9} \\ & (7)(20) \\ &\boxed{140} \end{aligned}$

Number of diagonals in a n-sided polygon is: $\frac{n(n-3)}{2}$ Therefore, $\frac{n(n-3)}{2} = 27 \Longrightarrow n = 9$ The sum of interior angles in a n-sided polygon is: $180(n-2)$ , divide that by the number of sides, $n=9$ . We get: $\frac{180(9-2)}{9}=140$ .

No. of diagonals=((n-3) n)/2=27.this formula will give the value of n,i.e.,the no of sides of the polygon=9.In general the no. of diagonals follow the following pattern (n-3)+(n-3)+(n-4)+(n-5)+....+3+2+1.Now,each interior angle of regular polygon=((n-2) 180)/n=(7*180)/9=140