Lines cb and ca are tangents to circle o at b and a. we can conclude that, for circle o, angle boa and angle bca are and angle boc and angle bco are blank 1 options: equal, complementary, supplementary blank 2 options: equal, complementary, supplementary

Blank 1 ⇒ supplementary

Blank 2 ⇒ complementary

Step-by-step explanation:

* Lets revise some facts to solve the problem

- The tangent to a circle is a line touch the circle at one point

- The tangent and the radius of a circle are perpendicular to each

 other at the point of contact

- If the two angles are supplementary, then the sum of their measure

 is 180°

- If the two angles are complementary, then the sum of their measure

 is 90°

* Now lets solve the question

∵ CB and CA are two tangents to the circle O at B and A

∵ OB and OA are radii

∴ OB ⊥ BC at point B and OA ⊥ AC at point A

∴ m∠OBC = 90° and m∠OAC = 90°

- In figure CBOA is a quadrilateral

∴ The sum of the measures of its interior angles is 360°

∵ m∠CBO + m∠BOA + m∠OAC + m∠BCA = 360°

∵ m∠CBO = 90° , m∠OAC = 90°

∴ 90° + m∠BOA + 90° + m∠BCA = 360°

∴ 180° + m∠BOA + m∠BCA = 360° ⇒ subtract 180 from both sides

∴ m∠BOA + m∠BCA = 180°

∴ ∠BOA and ∠BCA are supplementary

- In Δ CBO

∵ The sum of the measures of the interior angles of any triangle is 180°

∴ m∠BOC + m∠BCO + m∠CBO = 180°

∵ m∠CBO = 90°

∴ m∠BOC + m∠BCO + 90° = 180° ⇒ subtract 90° from both sides

∴ m∠BOC + m∠BCO = 90°

∴ ∠BOC and ∠BCO are complementary

step-by-step explanation:

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