In a circle with center O, two chords AB and CD intersect internally at point E. If ∠AEC = 95° and ∠AOD = 110°, then ∠BOC is equal to:
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Correct answer: D
Explanation
∠AEC = 95°
∠AOD = 110°
For the internal intersection of two chords,
∠AEC = 1/2(Arc AC + Arc BD)
Arc AC + Arc BD = 190°
∠AOD = 110° ⇒ Arc AD = 110°
Using the relationship regarding circle arcs,
Arc BC = 60°
And the central angle is equal to the arc.
Therefore, ∠BOC = 60°