THEOREM in this FIGURE,OA and OB are perpendicular on Tangents PA and PB.Also ∠APB = 40°The sum of the ANGLES of a QUADRILATERAL = 360°In quadrilateral PABO-∠APB + ∠OAP + ∠AOB + ∠OBP = 360°⇒ 40° + 90° + ∠AOB + 90° = 360°⇒ ∠AOB = 140°In ΔAOB⇒ ∠OAB + ∠ABO + ∠AOB = 180°(∵ Sum of the angles of a triangle is 180°)⇒ ∠x + ∠x + 140° = 180°⇒ x = 20°

"> THEOREM in this FIGURE,OA and OB are perpendicular on Tangents PA and PB.Also ∠APB = 40°The sum of the ANGLES of a QUADRILATERAL = 360°In quadrilateral PABO-∠APB + ∠OAP + ∠AOB + ∠OBP = 360°⇒ 40° + 90° + ∠AOB + 90° = 360°⇒ ∠AOB = 140°In ΔAOB⇒ ∠OAB + ∠ABO + ∠AOB = 180°(∵ Sum of the angles of a triangle is 180°)⇒ ∠x + ∠x + 140° = 180°⇒ x = 20°

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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 ∠APB = 40°, then ∠OAB equal to∶

SRMJEEE Analytical Geometry in SRMJEEE 9 months ago

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Using THEOREM in this FIGURE,OA and OB are perpendicular on Tangents PA and PB.Also ∠APB = 40°The sum of the ANGLES of a QUADRILATERAL = 360°In quadrilateral PABO-∠APB + ∠OAP + ∠AOB + ∠OBP = 360°⇒ 40° + 90° + ∠AOB + 90° = 360°⇒ ∠AOB = 140°In ΔAOB⇒ ∠OAB + ∠ABO + ∠AOB = 180°(∵ Sum of the angles of a triangle is 180°)⇒ ∠x + ∠x + 140° = 180°⇒ x = 20°

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