In the given figure, AD is a diameter. O is the centre of the circle. AD is parallel to BC and ∠CBD = 32°.
Find: ∠BED
Find: ∠BED
Exercise 17 (C) | Q 24.3 | Page 267
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In Δ OAB ,
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)
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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