Question

In the given figure, AD is a diameter. O is the centre of the circle. AD is parallel to BC and ∠CBD = 32°.

Find:

1. ∠OBD
2. ∠AOB
3. ∠BED

Answer 1

i. AD is parallel to BC, i.e., OD is parallel to BC and BD is transversal

⇒ ∠ODB = ∠CBD = 32°   ...(Alternate angles)

In ΔOBD,

OD = OB   ...(Radii of the same circle)

⇒ ∠ODB = ∠OBD = 32°

ii. AD is parallel to BC, i.e., AO is parallel to BC and OB is transversal.

⇒ ∠AOB = ∠OBC   ...(Alternate angles)

⇒ ∠OBC = ∠OBD + ∠DBC

⇒ ∠OBC = 32° + 32°

⇒ ∠OBC = 64°

⇒ ∠AOB = 64°

iii. In ΔOAB,

OA = OB  ...(Radii of the same circle)

⇒ ∠OAB = ∠OBA = x  ...(say)

⇒ ∠OAB + ∠OBA + ∠AOB = 180°

⇒ x + x + 64° = 180°

⇒ 2x = 180° – 64°

⇒ 2x = 116°

⇒ x = 58°

⇒ ∠OAB = 58°

i.e. ∠DAB = 58°

⇒ ∠DAB = ∠BED = 58° ...(Angles inscribed in the same arc are equal)