Question

In the circle with centre O, ∠BOD = 160° and chord BC = chord CD. Find ∠BOC, ∠OBD, ∠OBC and ∠CBD.

Answer 1

Given:

• Circle with centre O.
• ∠BOD = 160°.
• Chord BC = Chord CD.

Stepwise calculation:

1. Since chord BC = chord CD, the arcs BC and CD are equal, so angles subtended by these chords at the centre O are equal ∠BOC = ∠COD.

2. Given ∠BOD = 160° and because B, C and D lie on the circle:

∠BOD = ∠BOC + ∠COD

Since ∠BOC = ∠COD

160° = 2 × ∠BOC 

⇒ ∠BOC = 80°

3. Since OB = OC radii of circle, triangle OBC is isosceles, so ∠OBC = ∠OCB.

4. Using triangle sum in triangle OBC:

∠BOC + ∠OBC + ∠OCB = 180°

80° + ∠OBC + ∠OBC = 180°

2 × ∠OBC = 100°

⇒ ∠OBC = 50°

5. In A BOD, the sides OB and OD are radii, so OB = OD. This makes ABOD an isosceles triangle. The angles opposite the equal sides are equal, so ∠OBD = ∠ODB.

The sum of angles in ΔBOD is 180°: 

∠OBD + ∠ODB + ∠BOD = 180°

Substituting ∠OBD for ∠ODB and the given value of ∠BOD: 

2∠OBD +160° = 180°

2∠OBD = 180° − 160° = 20°

∠OBD = x = `20^°/2`

∠OBD = 10°

6. The angle CBD is the difference between ∠OBC and ∠OBD:

∠CBD = ∠OBC − ∠OBD

Using the values found in the previous steps:

∠CBD = 50° − 10°

∠CBD = 40°