Connecting outer diameter

OA = OD

∴∠ODA=∠OAD=∠DAC

∠∠DAC+∠ADC = 90。

∴∠ODA+∠ADC=90

That is, ∠ ODC = 90, OD⊥BC.

∫tanB = AC/BC = 3/4，AC=3

∴BC=4,AB=5

∴sinB=AC/AB=3/5

∴BO=AB-OA=5-r

SinB=OD/OB=r/(5-r)=3/5。

∴r= 15/8