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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!
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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??
@marsss Thank you so much! This really helped me.
welcome dude :)
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