हिंदी

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 - Mathematics

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प्रश्न

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

योग

उत्तर

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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अध्याय 17: Circles - Exercise 17 (C) [पृष्ठ २६७]

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सेलिना Mathematics [English] Class 10 ICSE
अध्याय 17 Circles
Exercise 17 (C) | Q 24.1 | पृष्ठ २६७

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