Question

The interior angles of a polygon are in $AP$. The smallest angle is ${52}^o$ and the common difference is $8^o$ . Find the number of side of the polygon.

Answer 1

Let number of sides of the polygon is $n$.

Then sum of the interior angles of the polygon is $(n-2)\times {180}^o$......(1).

Also given the interior angles of the interior forms an A.P. whose first term is ${52}^o$ and common difference is $8^o$.

Then sum of its interior angles is $=\frac{n}{2}\times \{2\times {52}^o+(n-1)\times 8^o\}$......(2).

Then we've form (1) and (2) we get,

$\frac{n}{2}\times \{2\times {52}^o+(n-1)\times 8^o\}$$=(n-2)\times {180}^o$

or, $52n+4n(n-1)=180(n-2)$

or, $13n+n(n-1)=45(n-2)$

or, $n^2-33n+90=0$

or, $(n-3)(n-30)=0$

or, $n=3,30$.

So sides of polygon may be either $3$ or $30$.