In the figure (i) given below, straight lines AB and CD pass through the center O of a circle. If ∠OCE = 40° and ∠AOD = 75°, find the number of degrees in

(i) ∠CDE

(ii) ∠OBE.

(i) 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, the number of degrees in ∠CDE = 50.

(ii) From figure,

∠AOD + ∠DOB = 180° (∵ they form linear pair)
⇒ 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, the number of degrees in ∠OBE = 25.