Question

The adjoining diagram shows a Pentagon inscribed in a circle, center $O.$ Given $AB=BC=CD$ and $\angle ABC={132}^{\circ }$. Calculate the value of

(i) $\angle AEB$

(ii) $\angle AED$

(iii) $\angle COD$

Answer 1

Given $AB=BC=CD$ and $\angle ABC={132}^{\circ }$

Join BE, CE, OC and OD.

i) ABCE is a cyclic quadrilateral.

$\angle ABC+\angle AEC={180}^{\circ }$ [Since Opposite angles of cyclic quadrilateral is $180^\circ$]
$\Rightarrow {132}^{\circ }+\angle AEC={180}^{\circ }$
$\Rightarrow \angle AEC={180}^{\circ }-{132}^{\circ }={48}^{\circ }$
$\Rightarrow \angle AEC=\angle AEB+\angle BEC$

$\Rightarrow \angle AEC=\angle AEB+\angle AEB$ [Since $\angle BEC=\angle AEB$ since equal chords subtend equal angles at a point on the circumference]

$2\angle AEB={48}^{\circ }$

$\Rightarrow \angle AEB={24}^{\circ }$

(ii) $CD=AB$

$\Rightarrow \angle CED=\angle AEB={24}^{\circ }$ [Equal chords subtend equal angles at a point on the
circumference]

$\angle CED={24}^{\circ }$
$\angle AED=\angle AEC+\angle CED$
$={48}^{\circ }+{24}^{\circ }={72}^{\circ }$

(iii) $\angle COD=2\angle CED$ [Since Angle at the center is double the angle at a point on the circumference]

$=2\times {24}^{\circ }={48}^{\circ }$

$\angle COD={48}^{\circ }$