NinetynineCAT 2025 Slot 2 QA Q10 Geometry
Question 10 of the Quantitative Ability (QA) section, CAT 2025 Slot 2, with CAT's official answer and a worked solution.
Two tangents drawn from a point \( P \) touch a circle with center \( O \) at points \( Q \) and \( R \). Points \( A \) and \( B \) lie on \( P Q \) and \( P R \), respectively, such that \( A B \) is also a tangent to the same circle. If \( \angle A O B = 50 ^ { \circ } \), then \( \angle A P B \), in degrees, equals
TITA: type in the answer. There are no options and no negative marking.
Answer: 80 Geometry TITA
Solution
- \(OA\) and \(OB\) bisect the exterior angles at \(A\) and \(B\) of triangle \(PAB\) (the circle is the excircle opposite \(P\)).
- So \(\angle AOB = 90^\circ - \frac{\angle APB}{2}\).
- \(50 = 90 - \frac{\angle APB}{2} \Rightarrow \angle APB = 80^\circ\).
Shortcut. Standard result: \(\angle AOB = 90^\circ - \frac{P}{2}\) for a tangent AB cutting the two tangents from P.