In a circle with centre O, chord AB subtends an angle θ with the tangent to the circle at point A. What is the measure of angle ∠ABO, where B is a point on the circle and OB is a radius?

Introduction / Context:
This question examines your understanding of the relationships between tangents, radii, chords and angles in a circle. A key idea in circle geometry is that the radius drawn to the point of tangency is perpendicular to the tangent line. Additionally, angles formed by a chord and a tangent relate closely to angles inside the circle. Here, you are asked to connect the angle between a chord and a tangent at one point of the circle to an angle involving the radius at another point, using properties of isosceles triangles and right angles.

Given Data / Assumptions:
• There is a circle with centre O.
• AB is a chord of the circle.
• A tangent at point A makes an angle θ with the chord AB.
• OA is a radius and is perpendicular to the tangent at A.
• OB is another radius, joining O to point B on the circle.
• We must find the angle ∠ABO, which is the angle at B between BA and BO.

Concept / Approach:
First, notice that the tangent at A is perpendicular to the radius OA, so the angle between OA and the tangent is 90°. The chord AB forms an angle θ with the tangent, so the angle between OA and AB can be obtained by subtracting θ from 90°. This angle at A between OA and AB is ∠OAB. Next, observe that OA and OB are both radii of the circle and hence are equal, so triangle AOB is isosceles. In an isosceles triangle, the base angles opposite the equal sides are equal, so ∠OAB equals ∠ABO. Therefore, once you find ∠OAB as 90° - θ, you can directly conclude the value of ∠ABO.

Step-by-Step Solution:
Step 1: Recognise that OA is perpendicular to the tangent at A, so the angle between OA and the tangent is 90°. Step 2: The chord AB makes an angle θ with the tangent at A, which means the angle between AB and the tangent is θ. Step 3: The angle between OA and AB, that is ∠OAB, is the difference between 90° and θ, so ∠OAB = 90° - θ. Step 4: Note that OA and OB are radii of the circle, so OA = OB, which makes triangle AOB an isosceles triangle. Step 5: In isosceles triangle AOB, the base angles at A and B are equal, so ∠OAB = ∠ABO. Step 6: Therefore, ∠ABO = 90° - θ.

Verification / Alternative check:
You can visualise triangle AOB: since the tangent at A is orthogonal to OA, and AB tilts away from the tangent by θ, it must be closer to OA by 90° - θ. Because triangle AOB is isosceles with OA = OB, the angle at B must mirror the angle at A. There is no need to know the actual numerical values of θ or any side lengths; the relationships between angles and equal sides determine the answer uniquely. This reasoning is consistent with general properties of isosceles triangles and tangents to circles.

Why Other Options Are Wrong:
Option θ would imply that ∠ABO equals the angle between the chord and the tangent directly, ignoring the 90° angle between the tangent and the radius, which is incorrect. Option 90° + θ would give an angle larger than 90° inside triangle AOB, which would conflict with the geometry of a triangle with a right angle already at A when considering OA and the tangent. Option 2 (180° - θ) is unrelated to the angle structure of triangle AOB and the given configuration, and does not follow from any standard theorem here. Only 90° - θ is consistent with all the circle and triangle properties involved.

Common Pitfalls:
A common mistake is to assume that the angle at B is simply θ or 90° + θ because students do not carefully track how many right angles are present and how the chord interacts with both the tangent and the radius. Another error is to forget that triangle AOB is isosceles, which is crucial for equating ∠OAB and ∠ABO. To avoid confusion, it is helpful to sketch the figure, mark the 90° angle between OA and the tangent, and then express all other angles in terms of θ, keeping in mind the equal side lengths OA and OB.

Final Answer:
The measure of angle ∠ABO is 90° - θ.