In the given figure, the straight lines AB and CD pass through the centre O of the circle. If ∠AOD = 75° and ∠OCE = 40°, find :

(i) ∠CDE

(ii) ∠OBE.

(i) Since, AB and CD pass through the center, thus they are the diameters of circle.

In △CED,

∠CED = 90° (∵ angle in semicircle is 90°.)

We know that sum of angles of a triangle is 180°.

⇒ ∠CED + ∠DCE + ∠CDE = 180°.

⇒ 90° + 40° + ∠CDE = 180°

⇒ ∠CDE + 130° = 180°

⇒ ∠CDE = 180° - 130°

⇒ ∠CDE = 50°.

Hence, ∠CDE = 50°.

(ii) From figure,

⇒ ∠AOD + ∠DOB = 180° (Linear pairs)

⇒ 75° + ∠DOB = 180°

⇒ ∠DOB = 180° - 75°

⇒ ∠DOB = 105°.

In △DOB,

∠ODB = ∠CDE = 50°

We know that,

Sum of angles of a triangle is 180°.

⇒ ∠DOB + ∠ODB + ∠DBO = 180°.

⇒ 105° + 50° + ∠DBO = 180°

⇒ ∠DBO + 155° = 180°

⇒ ∠DBO = 180° - 155°

⇒ ∠DBO = 25°.

From figure,

∠OBE = ∠DBO

∴ ∠OBE = 25°.

Hence, ∠OBE = 25°.