Question

In a regular pentagon ABCDE, draw a diagonal BE and then find the measure of :

(i) $\angle BAE$

(ii) $\angle ABE$

(iii) $\angle BED$

Answer 1

(i) Since number of sides in the pentagon = 5

$\therefore$ Each exterior angle $=\frac{{360}^o}{5}={72}^o$

$\therefore \angle BAE={180}^o-{72}^o={108}^o$

(ii) In $\Delta ABE,AB=AE\therefore \angle ABE=\angle AEBBut\angle BAE+\angle ABE+\angle AEB={180}^o\therefore {108}^o+2\angle ABE={180}^o-{108}^o={72}^o\Rightarrow \angle ABE={36}^o$

(iii) Since $\angle AED={108}^o$ [$\because$ each interior angle $={108}^o$]

$\Rightarrow \angle AEB={36}^o\Rightarrow \angle BED={108}^o-{36}^o={72}^o$