Answer:
Blank 1 ⇒ supplementary
Blank 2 ⇒ complementary
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