Question

O is the centre of the circle in which arc AB = 2 × arc BC. If ∠AOB = 104°, find

1. ∠BOC
2. ∠OAB
3. ∠OCA
4. ∠OBC

Answer 1

Step 1:

Since arc AB = 2 × arc BC, the central angles are in the same ratio.

∠AOB = 2 × ∠BOC

`∠BOC = (∠AOB)/2`

 ∠BOC = `104^circ/2` = 52°

Step 2:

In ΔOAB, OA = OB (radii), so it’s an isosceles triangle.

∠OAB = ∠OBA

`∠OAB = (180^circ - ∠AOB)/2`

`∠OAB = (180^circ - 104^circ)/2 = 76^circ/2 = 38^circ`

Step 3:

In ΔOBC, OB = OC (radii), so it’s an isosceles triangle.

∠OBC = ∠OCB

`∠OBC = (180^circ - ∠BOC)/2`

`∠OBC = (180^circ - 52^circ)/2 = 128^circ/2 = 64^circ`

Step 4:

In ΔOAC, OA = OC (radii), so it’s an isosceles triangle.

∠OAC = ∠OCA

∠AOC = ∠AOB + ∠BOC

∠AOC = 104° + 52° = 156°

`∠OCA = (180^circ - ∠AOC)/2`

`∠OCA = (180^circ - 156^circ)/2 = 24^circ/2 = 12^circ`

The angles are ∠BOC = 52°, ∠OAB = 38°, ∠OBC = 64° and ∠OCA = 12°.