We have a circle A and a line segment BQ is passing through the circle A such that it forms a triangle ABQ.
From the diagram, we have some information about the angles of triangle ΔABQ as follows :-
angle QAB = 36 degrees and angle BAQ = 54 degrees.
We know about the Triangle Sum theorem which states "The sum of all interior angles of any triangle is 180 degrees".
We can use this theorem to find the third angle of the triangle ΔABQ.
∠ABQ + ∠BQA + ∠QAB = 180 degrees
∠ABQ + 54° + 36° = 180°
∠ABQ + 90° = 180°
∠ABQ = 180° - 90°
∠ABQ = 90 degrees
If any line makes a right angle with the radius of any circle, then this line must be a tangent to that circle.
So, BQ would be a tangent line because angle ∠ABQ = 90 degrees. Hence, option A is correct.