Geometry Practice Test 2

In ΔDBC,
∠BDC+∠C+∠CBD = 180°
∠BDC = 180° − (25 +100)
∠BDC = 55°
∠BOC = 2BDC
∠BOC = 110°
[angle made by the same chord lies on the center]

In ΔBOC,
OB = OC = r
⇒ ∠OBC = ∠OCB
∠BOC+∠OBC+∠OCB = 180°
110° + 2x = 180°
2x = 70°
x = 35°
∠OBC = 35°
∠OBC−∠DBC = ∠OBD
35° − 25° = 10°
∠OBD = 10°