Circle Geometry: Tangent Angle Calculation

This problem requires us to find the angle formed by the intersection of two tangents to a circle. We are given a circle with centre O, a chord AB, and a point C located on the minor arc AB. Tangents are drawn to the circle at points A and B, and these tangents meet at a point P. The angle subtended by the major arc AB at point C on the circumference is given as ∠ACB = $108^\circ$. We need to determine the measure of the angle ∠APB.

Key Concepts in Circle Geometry

To solve this problem, we will utilize the following properties of circles:

• Angle Subtended by an Arc: The angle subtended by an arc at the centre of the circle is twice the angle subtended by the same arc at any point on the circumference.
• Cyclic Quadrilateral Property: In a cyclic quadrilateral (a quadrilateral whose vertices all lie on the circle), the sum of opposite angles is $180^\circ$.
• Tangent-Radius Property: A tangent to a circle is always perpendicular to the radius drawn to the point of contact.
• Quadrilateral Angle Sum: The sum of the interior angles of any quadrilateral is $360^\circ$.

Determining Angles Related to Chord AB

First, let's find the angle subtended by the minor arc AB at the circumference.

• We are given that point C is on the minor arc AB, and ∠ACB = $108^\circ$. This angle subtends the major arc AB.
• Let D be a point on the major arc AB. The quadrilateral ADBC consists of points lying on the circle, making it a cyclic quadrilateral.
• According to the cyclic quadrilateral property, opposite angles sum to $180^\circ$. Therefore, ∠ADB + ∠ACB = $180^\circ$.
• Substituting the given value: ∠ADB + $108^\circ = 180^\circ$.
• Solving for ∠ADB: ∠ADB = $180^\circ - 108^\circ = 72^\circ$. This is the angle subtended by the minor arc AB at the circumference.
• Now, we find the angle subtended by the minor arc AB at the centre O, which is ∠AOB. Using the property that the angle at the centre is double the angle at the circumference: ∠AOB = $2 \times \ang{ADB}$.
• ∠AOB = $2 \times 72^\circ = 144^\circ$.

Calculating the Angle Between Tangents (∠APB)

Next, we analyze the quadrilateral OAPB.

• Consider the quadrilateral OAPB, where O is the centre, A and B are points of tangency, and P is the intersection of tangents PA and PB.
• OA and OB are radii of the circle.
• PA is the tangent at A, and PB is the tangent at B.
• According to the Tangent-Radius Property, the radius is perpendicular to the tangent at the point of contact. Therefore, ∠OAP = $90^\circ$ and ∠OBP = $90^\circ$.
• The sum of the interior angles of the quadrilateral OAPB is $360^\circ$. So, ∠AOB + ∠OAP + ∠APB + ∠OBP = $360^\circ$.
• Substitute the known angle values: $144^\circ + 90^\circ + \ang{APB} + 90^\circ = 360^\circ$.
• Combine the known angles: $144^\circ + 180^\circ + \ang{APB} = 360^\circ$.
• $324^\circ + \ang{APB} = 360^\circ$.
• Solve for ∠APB: ∠APB = $360^\circ - 324^\circ$.
• ∠APB = $36^\circ$.

Conclusion

Based on the geometric properties and calculations, the angle ∠APB formed by the intersection of the tangents at A and B is $36^\circ$.