Question

In the given figure, AB and CD line segments pass through the centre O of the circle. If ∠OCE = 40°, ∠AOD = 75°, find ∠CDE and ∠OBE.

Answer 1

Given:

AB and CD pass through the centre O of the circle.

∠OCE = 40°

∠AOD = 75°

Step 1: Recognize that CD is a diameter

Since CD passes through the centre O of the circle, CD is a diameter.

This implies arc CED forms a semicircle.

Therefore, ∠CED = 90° (an angle in a semicircle is a right angle).

Step 2: Consider triangle CDE

In triangle CDE,

∠CDE + ∠DCE + ∠CED = 180°

Substitute the known values:

• ∠DCE = ∠OCE = 40° (given)
• ∠CED = 90°

So,

∠CDE + 40° + 90° = 180°

= ∠CDE = 180° − 130° = 50°

Step 3: Consider triangle DOB

For triangle DOB, note:

∠AOD = ∠OBD + ∠ODB (external angle = sum of two opposite internal angles)

∠AOD = 75°   ...[given]

∠ODB = ∠CDE = 50° 

75° = ∠OBD + 50°

= ∠OBD = 25°

Since ∠OBE = ∠OBD (E lies on the ray extending DB)

∠OBE = 25°