Mathematics
In the figure (i) given below, O is the centre of the circle. The tangents at B and D meet at P. If AB is parallel to CD and ∠ABC = 55°, find
(i) ∠BOD
(ii) ∠BPD.
Circles
ICSE
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Answer
(i) From figure,
∠BCD = ∠ABC (∵ alternate angles are equal)
∠BCD = 55°.
Arc BD subtends ∠BOD at the centre and ∠BCD at the remaining part of the circle.
∴ ∠BOD = 2∠BCD (∵ angle subtended at centre by an arc is double the angle subtended at remaining point of circle.)
∠BOD = 2 × 55° = 110°.
Hence, the value of ∠BOD = 110°
(ii) OB and OD are radius and, BP and DP are tangents to the circle.
∴ OB ⊥ BP and OD ⊥ DP.
In quadrilateral OBPD, sum of angles = 360°
∠BOD + ∠ODP + ∠OBP + ∠BPD = 360°
110° + 90° + 90° + ∠BPD = 360°
290° + ∠BPD = 360°
∠BPD = 360° - 290°
∠BPD = 70°.
Hence, the value of ∠BPD = 70°.
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