Q1. [1] § 10.1 Introduction § 10.2 Tangent to a Circle § 10.3 Number of Tangents from a Point on a Circle
In the given figure, $PA$ is a tangent from an external point $P$ to a circle with centre $O$. If $\angle POB = 115°$, then $\angle APO$ is equal to:
- A $25°$
- B $65°$
- C $90°$
- D $35°$
Previously asked in CBSE board exam
2025 30/1/1 Q16
Generated by claude-sonnet-4-6 · 2026-06-15 10:21 · grounding rag
Model Answer
Answer: (A) 25°
Since PA is a tangent, ∠OAP = 90°. ∠AOB = 180° − 115° = 65° (linear pair). In △OAP, ∠APO = 180° − 90° − 65° = 25°.
Explanation
- OB and OA are both radii, and B lies on the line PO extended, so ∠POA = 180° − 115° = 65° (linear pair).
- By Theorem 10.1, OA ⊥ PA, so ∠OAP = 90°.
- Angle sum in △OAP: ∠APO = 180° − 90° − 65° = 25°.
- Key theorem to cite: The tangent at any point of a circle is perpendicular to the radius through the point of contact (Theorem 10.1).
If a question refers to an image, map, graph or diagram that is not shown here, find the actual CBSE question paper on the CBSE website:
cbse.gov.in.