Question

AB is the diameter of the circle with centre O. OM ⊥ AD and ON ⊥ BC, OM = ON. ∠AOD = 70°. Find:

1. ∠BOC
2. ∠DOC
3. ∠ODB

Answer 1

Step 1:

Since OM = ON and OM ⊥ AD, ON ⊥ BC, chords AD and BC are equidistant from the center.

Therefore, AD = BC

Step 2:

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

So, ∠BOC = ∠AOD

Given ∠AOD = 70°

Thus, ∠BOC = 70°

Step 3:

AOB is a straight line, so ∠AOD + ∠DOC + ∠BOC = 180°

Substitute the known values: 70° + ∠DOC + 70° = 180°

Calculate ∠DOC: ∠DOC = 180° – 140° = 40°

Step 4:

In ΔOAD, OA = OD (radii)

So, ΔOAD is an isosceles triangle

`∠ODA = ∠OAD = (180^circ - ∠AOD)/2`

`∠ODA = (180^circ - 70^circ)/2`

= `110^circ/2`

= 55°

In ΔODB, OD = OB (radii)

So, ΔODB is an isosceles triangle

∠ODB = ∠OBD

In ΔODB, ∠DOB = ∠DOC + ∠COB

= 40° + 70°

= 110°

`∠ODB = (180^circ - ∠DOB)/2`

= `(180^circ - 110^circ)/2`

= `70^circ/2`

= 35°

The measures of the angles are ∠BOC = 70°, ∠DOC = 40° and ∠ODB = 35°.