Question

The angles of a polygon are in A.P with common difference $5^{\circ }$. If the smallest angle is ${120}^{\circ }$, find the number of sides of the polygon.

Answer 1

Sum of interior angles $=(n-2)\times {180}^o$

for $n$ sides

$\Rightarrow$$\frac{n}{2}(2a+(n-1\left)d\right)$

$\Rightarrow \frac{n}{2}(2\times 120+(n-1)\times 5^o)=(n-2)\times {180}^o$

$\Rightarrow$$\frac{n}{2}(240+5n-5)=180n-360$

$\Rightarrow$$240n+5n^2-5n=360n-720$

$\Rightarrow$$5n^2-125n+720=0$

$\Rightarrow$$n^2-25n+144=0$

$n=16,9$.