Question

O is the centre of the circle and chord AC and BD intersect at P such that angle APB is equal to 120 and angle pbc is equal to 15 find the value of angle ADB with diagram​

Answer 1

∠ADB = 105°  chord AC and BD intersect at P such that∠APB is equal to 120 and ∠PBC is equal to 15

Step-by-step explanation:

∠DBC = ∠PBC  (as P lies on BD)

=> ∠DBC = 15°

120° = ∠DAP + ∠ADP  ( Exterior angle of traingle = Sum of two opposite interior angles)

∠DAP = ∠DAC   as P lies on AC

∠ADP = ∠ADB   (as P lies on BD)

120° = ∠DAC + ∠ADB

∠DAC = ∠ DBC  ( angle subtended by same chord CD)

∠DAC = 15°

=> 120° =  15° + ∠ADB

=> ∠ADB = 105°

Learn more:

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in the given figure CAB = 25 degree find angle BDC,DBA,COB ...

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Answer 2

Step-by-step explanation:

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