In the given figure, O is the centre of the circle. The tangent at B and D intersect each other at point P.

If AB is parallel to CD and $∠$ ABC = $55^{o}$ , find :

(i) $∠$ BOD (ii) $∠$ BPD

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Solution

(i) $∠ B O D = 2 ∠ B C D \to ∠ B O D = 2 \times 55^{0} = 110^{0}$

ii) Since, BPDO is cyclic quadrilateral, opposite angles are supplementary.

$∠ B O D + ∠ B P D = 180^{0} \to ∠ B P D = 180^{0} - 110^{0} = 70^{0}$

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In the given figure, O is the centre of the circle. The tangent at B and D intersect each other at point P. If AB is parallel to CD and ∠ ABC = 55o, find : (i) ∠ BOD (ii) ∠ BPD

(i) ∠BOD=2∠BCD→∠BOD=2×550=1100 ii) Since, BPDO is cyclic quadrilateral, opposite angles are supplementary. ∠BOD+∠BPD=1800→∠BPD=1800−1100=700

In the given figure, O is the centre of the circle. The tangent at B and D intersect each other at point P.

If AB is parallel to CD and $∠$ ABC = $55^{o}$ , find :

(i) $∠$ BOD (ii) $∠$ BPD

(i) $∠ B O D = 2 ∠ B C D \to ∠ B O D = 2 \times 55^{0} = 110^{0}$

ii) Since, BPDO is cyclic quadrilateral, opposite angles are supplementary.

$∠ B O D + ∠ B P D = 180^{0} \to ∠ B P D = 180^{0} - 110^{0} = 70^{0}$