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Answer:

OA is perpendicular to AB [∵radius is perpendicular to tangent at the point of contact]

sin = opposite/hypotenuse

tan = opposite/adjacent

tanB = OA/AB

tan30° = 75/AB

1/√3 = 75/AB

AB = 75√3cm

sinB  = OA/OB

sin30° = 75/OB

1/2 = 75/OB

OB = 75 x 2

OB = 150cm

Comparing triangle BAO and BPQ

∠B is common

∠BAO = ∠BPO [corresponding angles]

ΔBAO~ΔBPQ by AA similarity

⇒BA/BP=AO/PQ=BO/BQ[sides of similar triangles are proportional]

OB = OQ+QB

150 = 75+QB

QB = 150-75

BQ = 75cm

In ΔQPB

sinB = PQ/BQ

sin30° = PQ/75cm

1/2 = PQ/75

PQ = 75/2

PQ = 37.5cm

BA/BP=AO/PQ

75√3/BP=75/37.5

BP = 37.5√3

AB = AP+BP

75√3=AP+37.5√3

AP = 37.5√3