Mathematics
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
Circles
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Answer
(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°.
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