Question

In the given figure, AB = BC = CD and $\angle$ ABC = ${132}^o$. Calculate :

(i) $\angle$ AEB,

(ii) $\angle$ AED,

(iii) $\angle$ COD.

Answer 1

Given

(equal arcs subtend equal angles at the center of a circle)

(i) In cyclic quadrilateral

(opposite angles in a cyclic quadrilateral are supplementary)

Since (equal chords subtend equal angles on the circumference)

$\therefore \angle AEB=\frac{1}{2}\angle AEC={24}^o$

(ii) Since

\(\angle AEB= \angle BEC = \angle CED

$\therefore \angle AED={24}^0+{24}^0+{24}^0={72}^0$

(iii) subtends at the center and on the circumference