Question

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

Answer 1

Consider the regular polygon as in figure,

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 ${140}^{\circ }$