In the figure below, O is the center of the circle and A, B, C are points on the circle.

What are the angles of Δ ABC?

Here,

∠AOB = ∠AOC = 120°

Sum of a complete whole angle is 360°

∴ ∠AOB + ∠AOC + ∠BOC = 360°

120 + 120 + ∠BOC = 360°

∴ 240 + ∠BOC = 360°

∴ ∠BOC = 360-240

= 120°

In Δ OAB,

OA = OB = radius of circle

In a triangle with 2 equal sides, the angles opposite to equal sides are equal.

∴ ∠OAB = ∠OBA = y°

In a triangle,

Sum of all angles of a triangle is 180°

Hence, in Δ OAB,

∠OAB + ∠OBA + ∠AOB = 180°

∴ y + y + 120 = 180°

∴ 120 + 2y = 180

∴ 2y = 180-120

= 60

∴

∴ ∠OAB = ∠OBA = 30° …(eq)1

In Δ OAC,

OA = OC = radius of circle

In a triangle with 2 equal sides, the angles opposite to equal sides are equal.

∴ ∠OAC = ∠OCA = y°

In a triangle,

Sum of all angles of a triangle is 180°

Hence, in Δ OAB,

∠OAC + ∠OCA + ∠AOC = 180°

∴ y + y + 120 = 180°

∴ 120 + 2y = 180

∴ 2y = 180-120

= 60

∴

∴ ∠OAC = ∠OCA = 30° …(eq)2

In Δ OBC,

OB = OC = radius of circle

In a triangle with 2 equal sides, the angles opposite to equal sides are equal.

∴ ∠OCB = ∠OBC = y°

In a triangle,

Sum of all angles of a triangle is 180°

Hence, in Δ OCB,

∠OCB + ∠OBC + ∠COB = 180°

∴ y + y + 120 = 180°

∴ 120 + 2y = 180

∴ 2y = 180-120

= 60

∴

∴ ∠OAC = ∠OBC = 30° …(eq)3

From equations 1,2 and 3

In Δ ABC

∠A = ∠BAO + ∠OAC

= 30 + 30

= 60°

∠B = ∠OBA + ∠OBC

= 30 + 30

= 60°

∠C = ∠OCB + ∠OCA

= 30 + 30

= 60°