i dont know hiw to solve this ouestion
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∠BAD = 75 ....angle on the center is twice the angle on the circumference
∠BPD = 180-75 = 105 .......because BADP is a cyclic quadrilateral
∠BPD =∠ BCD ....because these angles are on the same segment
i hope it helps........
∠BPD = 180-75 = 105 .......because BADP is a cyclic quadrilateral
∠BPD =∠ BCD ....because these angles are on the same segment
i hope it helps........
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as DOB is twice the angle of DAB
∴DAB=150/2=75
taking ABCD as cyclic quadrilateral
DAB+DCB= 180° (∵ opp.angles of a cyclic quadrilateral is supplimentary)
∴DCB=180-75=105
DCB=DPB (∵ angles formed on the same segment of the circle are equal)
∴ DPB=105
∴DAB=150/2=75
taking ABCD as cyclic quadrilateral
DAB+DCB= 180° (∵ opp.angles of a cyclic quadrilateral is supplimentary)
∴DCB=180-75=105
DCB=DPB (∵ angles formed on the same segment of the circle are equal)
∴ DPB=105
Anonymous:
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