Question

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?

Answer 1

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