Question

In this figure, m∠BDA = _____° and m∠BCA = _____°.

Answer 1

Find ∠BOA:

∠BOA + ∠BOA(reflect) = 360º (Angles at a point)

∠BOA + 250º = 360º

∠BOA = 360 - 250

∠BOA = 110º

Find ∠BDA:

2 ∠BDA = ∠BOA (Angle at the centre is twice the angle at the circumference)

2 ∠BDA = 110º

∠BDA = 110 ÷ 2

∠BDA = 55º

∠OBC = ∠OAC = 90º (tangent and the radius will always form a right angle)

Find ∠BCA:

∠BCA + CAO + AOB + OBC = 360º (Sum of angles in a quadrilateral is 360º)

∠BCA + 90º + 110º + 90º = 360º

∠BCA + 290º = 360º

∠BCA = 70º

Answer: In this figure, m∠BDA = 55° and m∠BCA = 70°.

Answer 2

Answer:Answer: In this figure, m∠BDA = 55° and m∠BCA = 70°.

Step-by-step explanation: