Question

(a) In the figure (1) given below, if ZDBC = 58° and BD is a diameter of the circle,
calculate
(1) ZBDC
(ii) ZBEC
(iii) ZBAC
(b) In the figure (ii) given below, AB is parallel to DC, ZBCE = 80° and ZBAC = 250
Find :
(1) ZCAD
(ii) ZCBD
(iii) ZADC
(2014)
(2008)
D​

Answer 1

Answer:

(i) ∠DBC = 58˚ (given)

Now, BD is the diameter.

∠BCD = 90˚ (angle in a semicircle)

In △BDC

∠BDC + 90˚ + 58˚ = 180˚ (Sum of the angles of a triangle)

∴ ∠BDC = 180˚ – (90˚ + 58˚) = 32˚

(ii) BECD is a cyclic quadrilateral

∵ ∠BEC + ∠BDC = 180˚ (Oppo. Angles of a cyclic quadrilateral)

∴ ∠BEC = 180˚ - ∠BDC

= 180˚ – 32˚ = 148˚

(iii) ∠BAC = ∠BDC = 32˚ (angles in the same segment of a circle)