Correct option is (c) 55°
Given, ∠AOB = 110∘
In △AOB
OA = OB (Radius of the circle)
Thus, ∠OAB = ∠OBA (Isosceles triangle property)
Sum of angles of the triangle = 180∘
∠AOB + ∠OAB + ∠OBA = 180∘
110 + 2∠OBA = 180∘
∠OBA = 35∘
Since, PQ is a tangent touching the circle at B.
Thus, ∠OBQ = 90∘
Now, ∠ABQ + ∠OBA = 90∘
∠ABQ + 35° = 90∘
∠ABQ = 55∘