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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:
- ∠OBD
- ∠AOB
- ∠BED
Solution
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)
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