In the given figure, BD is a side of regular hexagon, DC is a side of a regular pentagon and AD is a diameter. Calculate :

(i) ∠ADC,

(ii) ∠BDA,

(iii) ∠ABC,

(iv) ∠AEC.

Let O be the center of the circle.

Join BC, BO, CO and EO.

Since, BD is the side of a regular hexagon,

∴ ∠BOD = $\dfrac{360°}{6}$ = 60°.

Since, DC is the side of a regular pentagon,

∴ ∠COD = $\dfrac{360°}{5}$ = 72°.

In △BOD,

OB = OD [Radii of same circle]

∴ ∠OBD = ∠ODB = x (let) [Angles opposite to equal sides are equal]

⇒ ∠OBD + ∠ODB + ∠BOD = 180°

⇒ x + x + 60° = 180°

⇒ 2x = 180° - 60°

⇒ 2x = 120°

⇒ x = $\dfrac{120°}{2}$ = 60°.

(i) In △OCD,

⇒ OD = OC [Radii of same circle]

⇒ ∠ODC = ∠OCD = y (let) [Angles opposite to equal sides are equal]

⇒ ∠OCD + ∠ODC + ∠COD = 180°

⇒ y + y + 72° = 180°

⇒ 2y = 180° - 72°

⇒ 2y = 108°

⇒ y = $\dfrac{108°}{2}$ = 54°.

From figure,

∠ADC = ∠ODC = y = 54°.

Hence, ∠ADC = 54°.

(ii) From figure,

∠BDA = ∠BDO = 60°.

Hence, ∠BDA = 60°.

(iii) We know that,

Angle at the centre is twice the angle at remaining circumference.

Arc AC subtends ∠AOC at the centre and ∠ABC at the remaining part of the circle.

∴ ∠AOC = 2∠ABC

⇒ ∠ABC = $\dfrac{1}{2}$∠AOC

⇒ ∠ABC = $\dfrac{1}{2}$[∠AOD - ∠COD]

⇒ ∠ABC = $\dfrac{1}{2}$[180° - 72°]

⇒ ∠ABC = $\dfrac{1}{2} \times 108°$

⇒ ∠ABC = 54°.

Hence, ∠ABC = 54°.

(iv) In cyclic quadrilateral AECD,

⇒ ∠AEC + ∠ADC = 180°

⇒ ∠AEC + 54° = 180°

⇒ ∠AEC = 180° - 54°

⇒ ∠AEC = 126°.

Hence, ∠AEC = 126°.