the diagram O is the centre of the circle and OPA = 30° then ADB is (A) 60° (B) 120° (C) 45° (D) 90°
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2
Answer:
60° and 120°
Step-by-step explanation:
(i) Tangents drawn form an external point to the same circle are equal, i.e, PA=PB.
(ii) ∠APO=∠OPBand∠AOB+∠APB=180∘
(iii) ∠ACB=12∠AOB.
(iv) ACBD is a cyclic quadrilateral.
Answered by
3
Step-by-step explanation:
/_ADB = /_ACB
/_ACB = 180°-/_APB
= 180°-2/_OPA
= 180°-60
/_ACB = /_ADB = 120°
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