Question

In the given circle with centre O, AB = BC = CD and ∠AOB = 40°. Calculate:

1. ∠AOD
2. ∠OAD
3. ∠BOD
4. ∠OBD

Answer 1

Step 1:

Since AB = BC = CD, the central angles subtended by these chords are equal.

∠BOC = ∠AOB = 40°

∠COD = ∠AOB = 40°

Step 2:

∠AOD = ∠AOB + ∠BOC + ∠COD

∠AOD = 40° + 40° + 40° = 120°

Step 3:

In ΔOAD, OA = OD (radii)

ΔOAD is an isosceles triangle

∠OAD = ∠ODA

`∠OAD = (180^circ - ∠AOD)/2`

`∠OAD = (180^circ - 120^circ)/2 = 60^circ/2 = 30^circ`

Step 4:

∠BOD = ∠BOC + ∠COD

∠BOD = 40° + 40° = 80°

Step 5:

In ΔOBD, OB = OD (radii)

ΔOBD is an isosceles triangle

∠OBD = ∠ODB

`∠OBD = (180^circ - ∠BOD)/2`

`∠OBD = (180^circ - 80^circ)/2 = 100^circ/2 = 50^circ`

The calculated angles are ∠AOD = 120°, ∠OAD = 30°, ∠BOD = 80° and ∠OBD = 50°.