Question

In the given figure, AB is a side of regular pentagon and BC is a side of regular hexagon.
(i) ∠AOB
(ii) ∠BOC
(iii) ∠AOC
(iv) ∠OBA
(v) ∠OBC
(vi) ∠ABC

Answer 1

As given that AB is the side of a pentagon the angle subtended by each arm of the pentagon at the center of the circle is = `(360°)/5` = 72°

Thus angle ∠AOB = 72°

Similarly, as BC is the side of a hexagon hence the angle subtended by BC at the center is = `(360°)/6` i.e. 60°
∠BOC = 60°

Now ∠AOC = ∠AOB + ∠BOC =72° + 60° = 132°

The triangle thus formed, ΔAOB is an isosceles triangle with OA = OB as they are radii of the same circle.

Thus ∠OBA = ∠BAO as they are opposite angles of equal sides of an isosceles triangle.

The sum of all the angles of a triangle is 180°
so, ∠AOB + ∠OBA + ∠BAO = 180°
⇒ 2∠OBA + 72° = 180° as, ∠OBA = ∠BAO
⇒ 2∠OBA = 180° - 72°
⇒ 2∠OBA = 180°
⇒ 2∠OBA =54° 
as, ∠OBA = ∠BAO So,
∠OBA = ∠BAO = 54°

The triangle thus formed, ΔBOC is an isosceles triangle with OB = OC as they are radii of the same are.

Thus ∠OBC = ∠OCB as they are opposite angles of equal sides of an isosceles triangle.

The sum of all the angles of a triangle is 180°
so, ∠BOC + ∠OBC + ∠OCB = 180°
2∠OBC + 60° = 180° as , ∠OBC = ∠OCB
2∠OBC = 180° - 60°
2∠OBC = 120°
∠OBC = 60°
as ∠OBC = ∠OCB
So, ∠OBC = ∠OCB = 60°
∠ABC = ∠OBA + ∠OBC = 54° + 60°= 114°