Question

ABCO is a quadrilateral inscribed in the circle with centre O, Chord AB = Chord BC and ∠AOC = 136°. Calculate:

1. ∠AOB
2. ∠OAC
3. ∠OAB
4. ∠BAC

Answer 1

Step 1:

Since Chord AB = Chord BC, the angles subtended by these chords at the center are equal.

So, ∠AOB = ∠BOC

The sum of these angles equals ∠AOC: ∠AOB + ∠BOC = ∠AOC

Substitute and solve: 2 × ∠AOB = 136°

`∠AOB = 136^circ/2 = 68^circ`

Therefore, ∠BOC = 68°

Step 2:

In ΔOAC, OA = OC (radii).

Thus, ΔOAC is an isosceles triangle.

The base angles are equal: ∠OAC = ∠OCA

The sum of angles in ΔOAC is 180°: ∠OAC + ∠OCA + ∠AOC = 180°

Substitute and solve: 2 × ∠OAC + 136° = 180°

2 × ∠OAC = 180° – 136° = 44°

 ∠OAC = `44^circ/2` = 22°

Step 3:

In ΔOAB, OA = OB (radii)

Thus, ΔOAB is an isosceles triangle.

The base angles are equal: ∠OAB = ∠OBA

The sum of angles in ΔOAB is 180°: ∠OAB + ∠OBA + ∠AOB = 180°

Substitute and solve: 2 × ∠OAB + 68° = 180°

2 × ∠OAB = 180° – 68° = 112°

∠OAB = `112^circ/2` = 56°

Step 4:

∠BAC is the difference between ∠OAB and ∠OAC

∠BAC = ∠OAB – ∠OAC

Substitute the values: ∠BAC = 56° – 22° = 34°

The calculated angles are ∠AOB = 68°, ∠OAC = 22°, ∠OAB = 56° and ∠BAC = 34°.