In the given figure below, two circles C1 and C2 touch each other internally at P. Two lines PCA and PCB meet the circles C1 in C, D and C2 in A, B respectively. If ∠BDC=130∘, then the value of ∠ABP is equal to

50∘

80∘

100∘

120∘

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In the given figure below, two circles C1 and C2 touch each other internally at P. Two lines PCA and PCB meet the circles C1 in C, D and C2 in A, B respectively. If ∠BDC=130∘, then the value of ∠ABP is equal to

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Text solutionVerified

(a): ∠BDC=130∘ ∴∠CDP=180∘−130∘=50∘ CD∥AB ∴ ∠ABP=50∘=∠CDP

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Question Text	In the given figure below, two circles C1 and C2 touch each other internally at P. Two lines PCA and PCB meet the circles C1 in C, D and C2 in A, B respectively. If ∠BDC=130∘, then the value of ∠ABP is equal to
Answer Type	Text solution:1
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In the given figure below, two circles C1 and C2 touch each other internally at P. Two lines PCA and PCB meet the circles C1 in C, D and C2 in A, B respectively. If ∠BDC=130∘, then the value of ∠ABP is equal to

Text solutionVerified

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