In the given figure, AB is a side of a regular pentagon and BC is a side of a regular hexagon.

(i) ∠AOB

(ii) ∠BOC

(iii) ∠AOC

(iv) ∠OBA

(v) ∠OBC

(vi) ∠ABC

We know that,

The angle subtended by each side of an n-sided regular polygon at the center of circle = $\dfrac{360°}{n}$.

(i) Given,

AB is the side of the pentagon.

Angle subtended by each arm of the pentagon at the center of the circle is $\dfrac{360°}{5}$ = 72°.

Hence, ∠AOB = 72°.

(ii) Given,

BC is the side of the hexagon.

Angle subtended by each arm of the hexagon at the center of the circle is $\dfrac{360°}{6}$ = 60°.

Hence, ∠BOC = 60°.

(iii) From figure,

⇒ ∠AOC = ∠AOB + ∠BOC = 72° + 60° = 132°.

Hence, ∠AOC = 132°.

(iv) In △ OAB,

⇒ OA = OB

⇒ ∠OBA = ∠OAB = x (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OAB + ∠OBA + ∠AOB = 180°

⇒ x + x + 72° = 180°

⇒ 2x + 72° = 180°

⇒ 2x = 180° - 72°

⇒ 2x = 108°

⇒ x = $\dfrac{108°}{2}$

⇒ x = 54°

⇒ ∠OBA = 54°.

Hence, ∠OBA = 54°.

(v) In △ OBC,

⇒ OC = OB

⇒ ∠OBC = ∠OCB = y (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OBC + ∠OCB + ∠BOC = 180°

⇒ y + y + 60° = 180°

⇒ 2y + 60° = 180°

⇒ 2y = 180° - 60°

⇒ 2y = 120°

⇒ y = $\dfrac{120°}{2}$

⇒ y = 60°

⇒ ∠OBC = 60°.

Hence, ∠OBC = 60°.

(vi) From figure,

⇒ ∠ABC = ∠OBA + ∠OBC = 54° + 60° = 114°.

Hence, ∠ABC = 114°.