Question

In the figure, m∠DBC = 58°. BD is the diameter of the circle. Calculate:

1) m∠BDC

2) m∠BEC

3) m∠BAC

Answer 1

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°