Question

The adjoining figure shows a pentagon inscribed in a circle with centre O. Given AB = BC = CD and $\angle$ABC = ${130}^{\circ }$

Find (i) $\angle$AEB (ii) $\angle$AED (iii) $\angle$COD

• A. i) 25° ii) 75° iii) 50°
• B. i) 25° ii) 50° iii) 50°
• C. i) 25° ii) 90° iii) 50°
• D. i) 75° ii) 90° iii) 50°

Answer 1

Answer: (A), (C), (D)

The correct options are
A

i) 25° ii) 75° iii) 50°

C

i) 25° ii) 90° iii) 50°

D

i) 75° ii) 90° iii) 50°

i) ABCE is a cyclic quadrilateral.
$\angle$ABC + $\angle$AEC = ${180}^{\circ }$ [Since Opposite $\angle$s of cyclic quadrilateral is ${180}^{\circ }$]
$\Rightarrow$ ${130}^{\circ }$+ $\angle$AEC = ${180}^{\circ }$
$\Rightarrow$ $\angle$AEC = ${180}^{\circ }$ - ${130}^{\circ }$ = ${50}^{\circ }$
$\Rightarrow$ $\angle$AEC = $\angle$AEB + $\angle$BEC

$\Rightarrow$ $\angle$AEC = $\angle$AEB+ $\angle$AEB [Since $\angle$BEC = $\angle$AEB, since equal chords subtend equalangles at a point on thecircumference]

2$\angle$AEB = ${50}^{\circ }$

$\Rightarrow$ $\angle$AEB = ${25}^{\circ }$

(ii) CD = AB

$\Rightarrow$ $\angle$CED = $\angle$AEB = ${25}^{\circ }$ [Equal chords subtend equal angles at a point on the
circumference]

$\angle$CED = ${25}^{\circ }$
$\angle$AED = $\angle$AEC + $\angle$CED
= ${50}^{\circ }$ + ${25}^{\circ }$ = ${75}^{\circ }$

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

=2$\times$${25}^{\circ }$ = ${50}^{\circ }$

$\angle$COD = ${50}^{\circ }$