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

(i) ∠AOB

(ii) ∠OBC

(i) Given: AB is a side of a regular pentagon and BC is the side of a regular hexagon.

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

Thus, ∠AOB = 72°

Hence, ∠AOB = 72°.

(ii) Angle subtended by each side of the hexagon at the center of the circle = $\dfrac{360°}{6}$ = 60°

Thus, ∠BOC = 60°

OC = OB (radii of the same circle). So, Δ BOC is an isosceles triangle.

⇒ ∠OBC = ∠OCB

The sum of the angles in triangle BOC is 180°.

⇒ ∠OBC + ∠OCB + ∠COB = 180°

⇒ ∠OBC + ∠OBC + 60° = 180°

⇒ 2∠OBC + 60° = 180°

⇒ 2∠OBC = 180° - 60°

⇒ 2∠OBC = 120°

⇒ ∠OBC = $\dfrac{120°}{2}$

⇒ ∠OBC = 60°

Hence, ∠OBC = 60°.