Question

In the circle with centre O, BO ⊥ diameter AD. Chord BC = chord DE. Find the angles p, q, r and ∠AOE.

Answer 1

Given:

• Circle with centre O.
• BO ⊥ Diameter AD.
• Chord BC = Chord DE.
• Find angles p, q, r and ∠AOE.

Step 1: Find angle q and angle p

Given that BO ⊥ AD, we know that ∠BOD = 90°.

The angle q is ∠BOC.

We are given that ∠COD = 24°. 

Since ∠BOD = ∠BOC + ∠COD, we can write: 90° = q + 24° 

q = 90° − 24° 

q = 66° 

The angle p is ∠OCD. Since OC and OD are radii of the same circle, ΔODC is an isosceles triangle with OC = OD. The angles opposite to the equal sides are also equal, so ∠OCD = ∠ODC = p. The sum of angles in ΔODC is 180°. ∠COD + ∠OCD + ∠ODC = 180°

24° + p + p = 180°

2p = 180° − 24° 

2p = 156° 

p = `156^°/2` 

p = 78° 

Step 2: Find angle r and angle AOE

Given that chord BC is equal to chord DE (BC = DE). In a circle, equal chords subtend equal angles at the center. Therefore, the angle subtended by chord BC at the center, ∠BOC, is equal to the angle subtended by chord DE at the center, ∠DOE.

We found that q = ∠BOC = 66°.

So, ∠DOE = 66°.

The angle r is ΔOED. Since OE and OD are radii of the same circle, ΔODE is an isosceles triangle with OE = OD. The angles opposite to the equal sides are also equal, so ∠OED = ∠ODE = r.

The sum of angles in ΔODE is 180°.

∠DOE + ∠OED + ∠ODE = 180° 

66° + r + r = 180° 

2r = 180° − 66°

2r = 114° 

r = `114^°/2`

r = 57°

To find ∠AOE, we use the fact that AD is a diameter, so ∠AOD = 180°.

∠AOD = ∠AOE + ∠DOE

180° = ∠AOE + 66°

∠AOE = 180° − 66°

∠AOE = 114°