Question

the diagram O is the centre of the circle and OPA = 30° then ADB is (A) 60° (B) 120° (C) 45° (D) 90°​

Answer 1

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.

Answer 2

Step-by-step explanation:

/_ADB = /_ACB

/_ACB = 180°-/_APB

= 180°-2/_OPA

= 180°-60

/_ACB = /_ADB = 120°