Question

${\huge{\mathcal{\fcolorbox{navy}{purple}{\color{white}{Question\:no-500}}}}}$
The diagram shows a Pentagon ABCDE inscribed in a circle, with centre O. Given AB = BC = CD and ∠ABC=132°. Calculate the value of
(i)∠AEB
(ii)∠AED
(iii)∠COD​

Answer 1

$\large\bf{\underline{\underline{According\: to \:the\: question,}}}$

$\boxed{\sf{∵AB=BC=CD}}$

$\bold{∠BCD=132°}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\pink{\bigstar}★ \large\underline{\boxed{\bf\pink{△AOB≅△BOC}}}$

$\mapsto\bf{BO=AO}$$\sf\color{red}(radius)$

$\mapsto\bf{BA=BC}$$\sf\color{red}(given)$

$\mapsto\bf{BO=CO}$$\sf\color{red}(radius)$

$\sf\bold\color{red}{\boxed{∴△AOB≅△BOC\: (by \:SSS\: rule)}}$

$\bf{so, \:∠ABC=∠OBC}$

$\bf{∠ABO= 132/2 =48°}$

$\bf{48°+∠AOE=180°}$

$\sf{∠AOE=132°}$

$\sf{In △AOE}$

$\sf{132° +∠OAE+∠OEA+=180°}$

∠OEA=∠OAE (∵OA & OE are radius)

$\sf{∠AEB=∠OAE=∠OEA= 48/2 =24°}$

∠AEB=∠OEA=24°

$\underline{\boxed{\sf\purple{Also,\: △BOC\:≅\:△\:COD}}}$

$\bold{∠COD=48°}$

Thankyou :)

Answer 2

Given:-

• ABCDE is a pentagon inscribed in a circle.
• AB = BC = CD
• ∠ABC = 132°  

step-by-step solution:-

• Let’s join points E & B and E & C.

Case (i): Measure of angle AEB

Since in cyclic quadrilateral ABCE, the sum of opposite angles is 180°

∴ ∠AEC + ∠ABC = 180°

⇒ ∠AEC = 180° - 132° = 48°

We have, AB = BC

So, ∠AEB = ∠BEC ….. [equal chords subtends equal angles]

∴ ∠AEB = ∠BEC = ½ * ∠AEC = ½ * 48° = 24°  

Thus , the measure of angle AEB = 24°

Case (ii): Measure of angle AED

Since we have, AB = BC = CD

∴ ∠AEB = ∠BEC = ∠CED = 24° …… [from case (i)]

Also, from the figure given, we can say,

∠AED = ∠ AEB + ∠BEC + ∠CED = 24°+ 24°+24° = 72°

Thus , the measure of angle AED = 72°

Case (iii): Measure of angle COD

→ From the figure, we can see that  

CD subtends ∠COD at the centre and ∠CED at. the circumference of the circle.

→ And, we know, that the central angle is twice. any inscribed angle subtended by the same chord, so,

∠COD = 2 * ∠CED = 2 * 24° = 48°

Thus, the measure of the angle COD = 48°.