Question

PA and PB are tangent to circle O and PD bisects BPA. if mAOC= 68 degrees, how much greater is mBCO than mOAD (the diagram is not to scale).
Could you please explain how you got your answer?
Thank you!

Answer 1

mAOC=mBOC=68
OB=OC so mCBO=mBCO=x
x+x+68=180
2x+68=180 x=56
mAOC+mDOA=180
68+mDOA=180
mDOA=112
OD=OA so mODA=mOAD=y
y+y+112=180
2y+112=180 y=34
mBCO-mOAD=x-y=56-34=22
do you understand??

Answer 2

@marsss Thank you so much! This really helped me.

Answer 3

welcome dude :)