Question

30.In the given figure. BD is the diameter of a circle. If 2DBC = 48°, find
(i)2BDC () ZBAC fi) 2BEC

Answer 1

Answer:

Given that BD is a diameter of the circle

Given that BD is a diameter of the circleThe angle in a semicircle is a right angle.

Given that BD is a diameter of the circleThe angle in a semicircle is a right angle.∴ ∠BCD = 90°

Given that BD is a diameter of the circleThe angle in a semicircle is a right angle.∴ ∠BCD = 90°In ΔBDC, by angle sum property, we have

Given that BD is a diameter of the circleThe angle in a semicircle is a right angle.∴ ∠BCD = 90°In ΔBDC, by angle sum property, we have∠DBC + ∠BCD + ∠BDC = 180°

Given that BD is a diameter of the circleThe 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°

Given that BD is a diameter of the circleThe 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°

Given that BD is a diameter of the circleThe 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°

Given that BD is a diameter of the circleThe 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°

Given that BD is a diameter of the circleThe 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

Given that BD is a diameter of the circleThe 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)

Given that BD is a diameter of the circleThe 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°

Given that BD is a diameter of the circleThe 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°

Given that BD is a diameter of the circleThe 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.

Given that BD is a diameter of the circleThe 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°