Question

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

Calculate:

1. ∠BDC
2. ∠BEC
3. ∠BAC

Answer 1

∠DBC = 58°   ...(Given)

Now, BD is the diameter.

∴ ∠DCB = 90°   ...(Angle in a semicircle)

i. In ΔBDC,

∠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°  ...(Opposite angles of a cyclic quadrilateral)

∴ ∠BEC = 180° – ∠BDC

∴ ∠BEC = 180° – 32° = 148°

iii. ∠BAC = ∠BDC = 32°   ...(Angles in the same segment of a circle)