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Question
In the figure, m∠DBC = 58°. BD is the diameter of the circle. Calculate:
1) m∠BDC
2) m∠BEC
3) m∠BAC
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Solution
1) Given that BD is a diameter of the circle
The angle in a semicircle is a right angle.
∴ ∠BCD = 90°
In ΔBDC, by angle sum property, we have
∠DBC + ∠BCD + ∠BDC = 180°
⇒ 58° + 90° + ∠BDC = 180°
⇒ 148° + ∠BDC = 180°
⇒ ∠BDC = 180° - 148°
⇒ ∠BDC = 32°
2) quadrilateral BECD is a cyclic quadrilateral
⇒ ∠BEC + ∠BCD = 180° (Opposite angles are supplementary)
⇒ ∠BEC + 32° = 180°
⇒ ∠BEC = 148°
3) Angles in the same segment are equal.
⇒ ∠BAC = ∠BDC = 32°
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