Question

AB, BC and CD are three consecutive sides of a regular polygon. If angle BAC = ${20}^o$ ; find :

(i) its each interior angle,

(ii) its each exterior angle

(iii) the number of sides in the polygon.

Answer 1

$\because$ Polygon is regular

$\therefore$ AB = BC

$\Rightarrow \angle BAC=\angle BCA$ [$\angle s$ oop. to equal sides]

But $\angle BAC={20}^o\therefore \angle BCA={20}^{o}i.e.In\Delta ABC$,

$\angle B+\angle BAC+\angle BCA={180}^o\angle B+{20}^o+{20}^o={180}^o\angle B={180}^o-{40}^o\angle B={140}^o$

(i) each interior angle $={140}^o$

(ii) each exterior angle $={180}^o-{140}^o={40}^o$

(iii) Let no. of sides = n

$\therefore \frac{{360}^o}{n}={40}^{o}n=\frac{{360}^o}{{40}^o}=9,n=9\therefore \left(i\right){140}^o\left(ii\right)9$.