No worries! We‘ve got your back. Try BYJU‘S free classes today! Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Try BYJU‘S free classes today! Open in App 2 Open in App Suggest Corrections 3 From an external point P, tangents PA and PB are drawn to a circle with centre O. If ∠PAB = 50°, then find ∠AOB. It is given that PA and PB are tangents to the given circle. ∴∠PAO=90° (Radius is perpendicular to the tangent at the point of contact.) Now ∠PAB=50° (Given) ∴∠OAB=∠PAO−∠PAB=90°−50°=40° In ∆OAB, OB = OA (Radii of the circle) ∴∠OAB=∠OBA=40° (Angles opposite to equal sides are equal.) Now ∠AOB+∠OAB+∠OBA=180° (Angle sum property) ⇒∠AOB=180°−40°−40°=100° Concept: Number of Tangents from a Point on a Circle Is there an error in this question or solution? |