Question

In this figure, mBDA=___ and mBCA=___.

Answer 1

∠BDA is inscribed angle
inscribed angle = 1/2 × central angle
the central angle is small ∠BOA, use ∠BOA that face the same direction as ∠BDA

Find the central angle ∠BOA
central angle = 360° - large ∠BOA
central angle = 360° - 250°
central angle = 110°

Find ∠BDA
inscribed angle = 1/2 × central angle
∠BDA = 1/2 × small ∠BOA
∠BDA = 1/2 × 110°
∠BDA = 55°


To find ∠BCA, use tetragon BOAC to solve the problem
The sum of interior angles in tetragon is 360°. Because CA and CB is tangent of the circle, the angles which is formed by the tangent is 90°. So ∠CAO and ∠CBO are both equal to 90°
∠BCA + ∠CAO + ∠CBO + small ∠BOA = 360°
∠BCA + 90° + 90° + 110° = 360°
∠BCA + 290° = 360°
∠BCA = 360° - 290°
∠BCA = 70°

THE ANSWER
m∠BDA = 55°
m∠BCA = 70°

Answer 2

MBDA= ADB and mBCA= ACB