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Question
In Fig, Chord ED is parallel to the diameter AC of the circle. Given ∠CBE = 65°, Calculate ∠ DEC.
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Solution
Consider the arc CDE. We find that ∠ CBE and ∠ CAE are the angles in the same segment of arc CDE.
∴ ∠ CAE = ∠ CBE
⇒ ∠ CAE = 65° ...( ∵ ∠ CBE = 65° )
Since AC is the diameter of the circle and the angle in a semicircle is a right angle.
Therefore, ∠ AEC = 90°.
Now, in Δ ACE, we have
∠ ACE + ∠ AEC + ∠ CAE = 180°
⇒ ∠ ACE + 90° + 65° = 180°
⇒ ∠ ACE = 25°
But ∠ DEC and ∠ ACE are alternate angles, because AC || DE.
∴ ∠ DEC = ∠ ACE = 25°.
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