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Question
In the given figure, O is the centre of the circle. The tangents at B and D intersect each other at point P. If AB is parallel to CD and ∠ABC = 55°, find:
- ∠BOD
- ∠BPD
Solution
In the given figure, O is the centre of the circle AB || CD, ∠ABC = 55° tangents at B and D are drawn which meet at P.
∵ AB || CD
∴ ∠ABC = ∠BCD ...(Alternate angles)
∴ ∠ABC = 55° ...(Given)
i. Now arc BD subtands ∠BOD at the centre and ∠BCD at the remaining part of the circle.
∴ ∠BOD = 2∠BCD
= 2 × 55°
= 110°
ii. In quadrilateral OBPD,
∠OBP = ∠ODP = 90° ...(∵ BP and DP are tangents)
∴ ∠BOD + ∠BPD = 180°
`\implies` 110° + ∠BPD = 180°
`\implies` ∠BPD = 180° – 110° = 70°
Hence, ∠BOD = 110° and ∠BPD = 70°
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