OA = OB∴ ΔOAB is an isosceles triangle.⇒ ∠OAB = ∠OBA = 35°⇒ ∠AOB = 180° – 2 × 35° = 110°We know that radius from the CENTRE of the CIRCLE to the point of tangency is PERPENDICULAR to the tangent line.∴ In quadrilateral APBO,∠OAP = ∠OBP = 90°⇒ ∠APB = 360° – 2 × 90° – 110° = 70°

"> OA = OB∴ ΔOAB is an isosceles triangle.⇒ ∠OAB = ∠OBA = 35°⇒ ∠AOB = 180° – 2 × 35° = 110°We know that radius from the CENTRE of the CIRCLE to the point of tangency is PERPENDICULAR to the tangent line.∴ In quadrilateral APBO,∠OAP = ∠OBP = 90°⇒ ∠APB = 360° – 2 × 90° – 110° = 70°

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PA and PB are two tangents to a circle with centre O, from a point P outside the circle. A and B are points on the circle. If ∠OAB = 35°, then ∠APB is equal to∶

SRMJEEE Analytical Geometry in SRMJEEE 1 year ago

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Given, ∠ OAB = 35°In ΔOAB,Since OA = OB∴ ΔOAB is an isosceles triangle.⇒ ∠OAB = ∠OBA = 35°⇒ ∠AOB = 180° – 2 × 35° = 110°We know that radius from the CENTRE of the CIRCLE to the point of tangency is PERPENDICULAR to the tangent line.∴ In quadrilateral APBO,∠OAP = ∠OBP = 90°⇒ ∠APB = 360° – 2 × 90° – 110° = 70°

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Posted on 27 Oct 2024, this text provides information on SRMJEEE related to Analytical Geometry in SRMJEEE. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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