PA and PB are tangent to circle O and PD bisects =BPA. if m=AOC = 68, how much greater is m=BCO than m=OAD?

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anonymous · Answer

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campbell_st · Answer

well 

Angle OPA = 22   (angle sum triangle & angle OAP = 90

Triangles BOP and AOP are congruent (SSS test)
                   OP = common side
                  OA  OB equal radii
                   AP = BP Tangents from an external point are equal

then angle OPB = 22  corresponding angles in congrent triangles

therefore angle APB = 44

BOC = 68  = AOC   corresponding angles in congruent triangles

Triangle BOC is isosceles 

therefore angle BCO = 112/2 = 56
Angle DOA = 180 - 68  supplementary to AOC

           DOA = 112

 Triangle DOA is isosceles OD and OA are radii

Then Angle OAD = 68/2   angle OAD = 34

then angle BCO = 56 anf OAD = 34   then BCO is 22 degree larger