In the figure (ii) given below, AB is a diameter of the circle whose center is O. Given that ∠ECD = ∠EDC = 32°, calculate

(i) ∠CEF

(ii) ∠COF

(i) In △EDC,

∠ECD = ∠EDC = 32° (Given)

Since sum of angles of triangle = 180°.

⇒ ∠DEC + ∠ECD + ∠EDC = 180°
⇒ ∠DEC + 32° + 32° = 180°
⇒ ∠DEC = 180° - 64°
⇒ ∠DEC = 116°.

Since, ∠CEF and ∠DEC are linear pair,

∴ ∠CEF + ∠DEC = 180°
⇒ ∠CEF + 116° = 180°
⇒ ∠CEF = 180° - 116°
⇒ ∠CEF = 64°

Hence, ∠CEF = 64°.

(ii) ∠FDC = ∠EDC = 32°. (From figure)

Arc FC subtends ∠COF at center and ∠FDC at point D of circle so,

⇒ ∠COF = 2 ∠FDC
⇒ ∠COF = 2 × 32°
⇒ ∠COF = 64°

Hence, the value of ∠COF = 64°.