In this figure, m∠bda = *blank* ° and m∠bca =*blank* °.

m∠BDA =  55°  

m∠BCA = 70°

Step-by-step explanation:

In the given figure

Tangents AC and BC

Chords AD and BD

Radius OA and OB

arc ADB (major arc AB)= 250°

minor AB + major AB = 360°

     minor AB (∠AOB) = 110°

∠BDA = half of ∠AOB

(because angle makes at circle is half angle at center on same arc)

Thus, ∠BDA = 110/2 = 55°

∠OBC and ∠OAC are make 90° because radius perpendicular to tangent.

In a quadrilateral, AOBC

∠OBC + ∠OAC + ∠AOB + ∠BCA = 360°

Sum of angles of quadrilateral is 360°

90° + 90° + 110° + ∠BCA = 360°

                             ∠BCA = 360° - 90° - 90° - 110°

                              ∠BCA = 70°

Hence, m∠BDA =  55° and m∠BCA = 70°