Question

AB and CD intersect at the centre O of the circle given in the above diagram. If ∠EBA = 33° and ∠EAC = 82°, find.

1. ∠BAE
2. ∠BOC
3. ∠ODB

Answer 1

Given:

AB and CD are straight lines intersecting at the center O. This means AB is a diameter.

∠EBA = 33°

∠EAC = 82°

a. Find ∠BAE

1. Since AB is a diameter, the angle it subtends at the circumference is a right angle. Therefore, ∠AEB = 90°.

2. In ΔAEB, the sum of interior angles is 180°.

3. ∠BAE = 180° – 90° – 33°

4. ∠BAE = 57°

b. Find ∠BOC

1. First, let’s find ∠BAC. We know ∠EAC = 82° and we just found ∠BAE = 57°.

∠BAC = ∠EAC – ∠BAE

= 82° – 57°

= 25°

2. The angle subtended by an arc at the center is twice the angle subtended by it at any point on the circumference.

3. For arc BC, the angle at the center is ∠BOC and the angle at the circumference is ∠BAC.

∠BOC = 2 × ∠BAC

∠BOC = 2 × 25°

∠BOC = 50°

c. Find ∠ODB

1. Since AB is a straight line, we can find ∠AOC:

2. ∠DOB and ∠AOC are vertically opposite angles, so ∠DOB = 130°.

3. Now look at ΔDOB. Points D, E, B lie on a straight line, so ∠DBO is the same as ∠EBA = 33°.

4. In ΔDOB, the sum of angles is 180°:

∠ODB = 180° – ∠DOB – ∠DBO

∠ODB = 180° – 130° – 33°

∠ODB = 17°