Question

In the given figure, A, B, C are 3 points on the circumference of a circle with centre O. Arc AB = 2 arc BC and ∠AOB = 108°.

Calculate:

1. ∠BOC
2. ∠OAB
3. ∠OAC
4. ∠BAC

Answer 1

Given:

• Points A, B, C lie on the circumference of a circle with center O.
• Arc AB = 2 × arc BC.
• ∠AOB = 108°.

Stepwise calculation:

i. Calculate ∠BOC:

Since arc AB = 2 × arc BC, the angle at the center subtending arc AB is twice the angle subtending arc BC.

Let ∠BOC = x.

Then ∠AOB = 2x.

Given ∠AOB = 108°

So, 2x = 108°

x = `108^circ/2`

x = 54°

Therefore, ∠BOC = 54°.

ii. Calculate ∠OAB:

OA = OB are radii, so triangle OAB is isosceles.

Sum of angles in triangle OAB = 180°.

∠AOB = 108°, so remaining two angles are equal.

Each of ∠OAB and ∠OBA equals:

`(180^circ - 108^circ)/2`

= `72^circ/2`

= 36°

Therefore, ∠OAB = 36°.

iii. Calculate ∠OAC:

First find ∠AOC.

Since ∠AOB = 108° and ∠BOC = 54°

Then ∠AOC = ∠AOB + ∠BOC

= 108° + 54°

= 162°

OA = OC are radii, so triangle OAC is isosceles.

Sum of angles in triangle OAC = 180°.

∠AOC = 162°, so each of ∠OAC and ∠OCA equals:

`(180^circ - 162^circ)/2`

= `18^circ/2`

= 9°

Therefore, ∠OAC = 9°.

iv. Calculate ∠BAC:

∠BAC = ∠OAB – ∠OAC

= 36° – 9°

= 27°