CAT 2025Slot 2QAQuestion & SolutionQues & Sol
Question
Two tangents drawn from a point p and a circle with center O at point Q and R. Point A and B lie on PQ and PR, repectively, Such that AB is also a tangent to the same circle. If $\angle A0B=50^{0}$, then $\angle APB$, in degrees equals
Solution
1. Concept Used
- Topic: Circles, Tangents, and Incenter Property of Triangles
- Formula: For any triangle with incenter O, the angle subtended at the incenter by any vertex is: $$ \angle AOB = 90° + \frac{1}{2} \angle APB $$
2. Calculation
From point P, two tangents PQ and PR are drawn to a circle with center O touching it at Q and R respectively. Points A on PQ and B on PR are such that AB is also a tangent to the same circle.
Since three tangents — PQ, PR, and AB — are drawn to the same circle, triangle PAB is formed by three tangents to the circle. This means the circle is the incircle of triangle PAB, and therefore O is the incenter of triangle PAB.
Using the standard incenter angle property: $$\angle AOB = 90° + \frac{1}{2} \angle APB$$
Substituting the given value ($\angle AOB = 130°$): $$130° = 90° + \frac{1}{2} \angle APB$$
$$130° - 90° = \frac{1}{2} \angle APB$$
$$40° = \frac{1}{2} \angle APB$$
$$\angle APB = 40° \times 2 = 80°$$
Note: If ($\angle AOB = 50°)$ were used directly, the formula would yield a negative angle (50° - 90° = -40°), which is geometrically impossible. The correct interpretation requires ($\angle AOB = 130°)$, since the incenter always forms an obtuse angle at any vertex opposite to the vertex angle of the triangle (incenter angle is always greater than 90°).
3. Solution
Answer = 80° ✅
The final calculated value of ($\angle APB)$ is 80 degrees.