Question

O is the centre of the circle. AB is a diameter.

Answer 1

Answer:

The value of a is 63° and the value of b is 27°

Step-by-step explanation:

In the given figure

∵ OA, OB, and OC are radii of the circle O

→ The radii of a circle are equal

∴ OA = OB = OC

In Δ AOC

∵ OA = OC

∴ Δ AOC is an isosceles triangle

→ Base angles of the isosceles triangle are equal

∵ ∠OAC and ∠OCA are the base angles of the triangle

∴ m∠OAC = m∠OCA

∵ m∠OAC = a°

∴ m∠OCA = a°

→ The sum of angles in any triangle is 180°

∵ a° + a° + m∠AOC = 180°

∵ m∠AOC = 54°

∴ a° + a° + 54 = 180°

→ Add the like terms

∴ 2a° + 54 = 180°

→ Subtract 54 from both sides

∴ 2a° = 126°

→ Divide both sides by 2

∵ m∠OAC = m∠OCA = 126 ÷ 2

∴ a° = 63

∴ The value of a is 63°

In ΔBOC

∵ OB = OC

∴ Δ BOC is an isosceles triangle

→ Base angles of the isosceles triangle are equal

∵ ∠OBC and ∠OCB are the base angles of the triangle

∴ m∠OBC = m∠OCB

∵ m∠OBC = b°

∴ m∠OCB = b°

∵ ∠AOC is an exterior angle of ΔBOC at the vertex O

→ The measure of the exterior angle equals the sum of the measure

    of the opposite interior angles to this vertex

∵ ∠OBC and ∠OCB are the opposite interior angle of ∠AOC

∴ m∠OBC + m∠OCB = m∠AOC

→ Substitute their measures

∵ b° + b° = 54°

→ Add the like terms

∴ 2b° = 54

→ Divide both sides by 2

∴ b° = 27

∴ The value of b is 27°