Question

The difference any two consecutive interior angles of a polygon is $5^{\circ }$.If the smallest angle is ${120}^{\circ }$, find the number of the sides of the polygon.

• A. $5$
• B. $45$
• C. $25$
• D. $19$

Answer 1

Answer: (A), (D)

$5$
$19$

We know that the sum of all angles of a polygon with $n$ sides $={180}^0(n-2)$

We know that,

$S_n={180}^0(n-2)........\left(1\right)$

The angles of polygon will from an A.P.

Then common difference $d=5^0$

And first term $={120}^0$

let the side of polygon$=n$

Then,$S_n=\frac{n}{2}[2a+(n-1)d]........\left(2\right)$

By equation (1) and (2) to,

$\frac{n}{2}[2a+(n-1)d]={180}^0(n-2)$

$\Rightarrow \frac{n}{2}[2\times {120}^0+(n-1)5^0]={180}^0(n-2)$

$\Rightarrow n({240}^0+5^{0}n-5^0)={360}^{0}n-{720}^0$

$\Rightarrow 5n^2+235n-360n+720=0$

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

$\Rightarrow 5(n^2-25n+144)=0$

$\Rightarrow n^2-(16+9)n+144=0$

$\Rightarrow n^2-16n-9n+144=0$

$\Rightarrow n(n-16)-9(n-16)=0$

$\Rightarrow (n-16)(n-9)=0$

$\Rightarrow n=9,16$

Hence, this is the answer.