Question

O is the centre of the circle. Arc AB = Arc BC = Arc CD. If ∠OAB = 48°, find

1. ∠AOB
2. ∠BOD
3. ∠OBD

Answer 1

Step 1:

In ΔAOB, OA = OB because they are radii.

So, ΔAOB is an isosceles triangle.

Therefore, ∠OBA = ∠OAB = 48°

The sum of angles in ΔAOB is 180°

∠AOB = 180° – (∠OAB + ∠OBA)

∠AOB = 180° – (48° + 48°) = 180° – 96° = 84°

Step 2:

Since Arc AB = Arc BC = Arc CD, the angles subtended at the center are equal.

So, ∠AOB = ∠BOC = ∠COD = 84°

∠BOD = ∠BOC + ∠COD

∠BOD = 84° + 84° = 168°

Step 3:

In ΔOBD, OB = OD because they are radii.

So, ΔOBD is an isosceles triangle.

Therefore, ∠OBD = ∠ODB

The sum of angles in ΔOBD is 180°

∠OBD + ∠ODB + ∠BOD = 180°

2∠OBD + 168° = 180°

2∠OBD = 180° – 168° = 12°

∠OBD = `12^circ/2` = 6°

The measures are ∠AOB = 84°, ∠BOD = 168° and ∠OBD = 6°.