Selection Post Phase XIV (Graduate)25 Sep 2026Shift 1Geometry II · Circles & Quadrilaterals

In a circle with centre O, PA and PB are tangents touching the circle at A and B respectively. Point C lies on the major arc AB. If ∠ACB = 50°, find ∠APB.

केंद्र O वाले एक वृत्त (Circle) में PA और PB स्पर्श रेखाएँ (Tangent) हैं जो वृत्त को क्रमशः A और B पर स्पर्श करती हैं। बिंदु (Point) C दीर्घ चाप (Arc) AB पर स्थित है। यदि ∠ACB = 50° है, तो ∠APB ज्ञात कीजिए।

OABPC50°
Answer और ExplanationAnswer and explanation

Answer: (A) 80°

केंद्र (Centre) पर कोण (Angle) = 2 × दीर्घ चाप (Major arc) पर कोण ⇒ ∠AOB = 2 × 50° = 100°
चतुर्भुज (Quadrilateral) OAPB में ∠OAP = ∠OBP = 90°, इसलिए ∠APB = 180° − ∠AOB
∠APB = 180° − 100° = 80°
✓
∠APB = 80°।
Angle at the centre = 2 × angle on the major arc ⇒ ∠AOB = 2 × 50° = 100°
In quadrilateral OAPB, ∠OAP = ∠OBP = 90°, so ∠APB = 180° − ∠AOB
∠APB = 180° − 100° = 80°
✓
∠APB = 80°.
SMART APPROACH

∠APB = 180° − 2∠ACB = 180° − 100° = 80°।

∠APB = 180° − 2∠ACB = 180° − 100° = 80°.

Source: Selection Post Phase XIV (Graduate) · 25 Sep 2026 · Shift 1। Solution: Prashant Sir, Vijayastra।Source: Selection Post Phase XIV (Graduate) · 25 Sep 2026 · Shift 1. Solution by Prashant Sir, from Vijayastra.