In the given figure, O is the centre of the given circle. AB is a side of a square, BC is a side of regular pentagon and CD is a side of regular hexagon. Find :

(i) ∠AOB

(ii) ∠AOC

(iii) ∠AOD

(iv) ∠BCD

Given:

• AB is a side of a square (4 sides).
• BC is a side of a regular pentagon (5 sides).
• CD is a side of a regular hexagon (6 sides).
• O is the centre of the circle.

Construction: Join OA, OB, OC and OD.

(i) Since AB is a side of a square inscribed in a circle, the angle subtended at the centre is:

∠AOB = $\dfrac{360°}{4}$

= 90°

Hence, ∠AOB = 90°.

(ii) BC is a side of a regular pentagon, so:

∠BOC = $\dfrac{360°}{5}$

= 72°

∠AOC = ∠AOB + ∠BOC = 90° + 72° = 162°

Hence, ∠AOC = 162°.

(iii) CD is a side of a regular hexagon, so:

∠COD = $\dfrac{360°}{6}$

= 60°

∠AOD = ∠AOB + ∠BOC + ∠COD = 90° + 72° + 60° = 222°

Hence, ∠AOD = 222°.

(iv) In Δ BOC,

• OB = OC (radii of same circle)
• ∠BOC = 72°

Using the angle sum property:

∠BOC + ∠BCO + ∠CBO = 180°

⇒ 72° + ∠OBC + ∠OBC = 180°

⇒ 72° + 2∠OBC = 180°

⇒ 2∠OBC = 180° - 72°

⇒ 2∠OBC = 108°

⇒ ∠OBC = $\dfrac{108°}{2}$

⇒ ∠OBC = 54°

Similarly, in Δ COD,

• OC = OD (radii of same circle)
• ∠COD = 60°

⇒ ∠DOC + ∠OCD + ∠CDO = 180°

⇒ 60° + ∠OCD + ∠OCD = 180°

⇒ 60° + 2∠OCD = 180°

⇒ 2∠OCD = 180° - 60°

⇒ 2∠OCD = 120°

⇒ ∠OCD = $\dfrac{120°}{2}$

⇒ ∠OCD = 60°

Now, ∠BCD = ∠OBC + ∠OCD

= 54° + 60° = 114°

Hence, ∠BCD = 114°.