Question

AB, BC and CD are three consecutive sides of a regular polygon. If angle BAC = ${20}^{\circ }$, find the number of sides in the polygon.

Answer 1

Give, $\angle BAC={20}^{\circ }$

In a regular polygon, all the sides are equal.

$\therefore AB=BC=CD$

In triangle $ABC$,

$\angle BAC=\angle BCA={20}^{\circ }$ -------angles opposite to equal sides in a triangle

$\angle ABC+\angle BAC+\angle BCA={180}^{\circ }$ ------sum of interior angles of the triangle

$\angle ABC+{20}^{\circ }+{20}^{\circ }={180}^{\circ }$

$\angle ABC={180}^{\circ }-{40}^{\circ }={140}^{\circ }$

The interior angle of the regular polygon is ${140}^{\circ }$

Each interior angle of a regular polygon $=\frac{\left[\right(2n-4)\times {90}^{\circ }]}{n}$

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

${140}^{\circ }=\frac{\left[\right(2n-4)\times {90}^{\circ }]}{n}$

${140}^{\circ }n={180}^{\circ }n-{360}^{\circ }$

${40}^{\circ }n={360}^{\circ }$

$n=9$

The regular polygon is a Nonagon.