In the given figure 'o' is the centre of the circle. determine angle AQB and angle AMB, if PA and PB are tangents and, angle APB = 750 Share with your friends Share 21 Manbar Singh answered this We have, ∠APB = 75°Now, PA and PB are the tangents to the circle from P at A and B respectively and OA and OB are the radii of the circle.We know that tangent to a circle is always ⊥ to the radius at the point of contact.Hence, OA⊥PA and OB⊥PB.Now, ∠OAP = ∠OBP = 90°In quadrilateral AOBP,∠ABP + ∠OAP + ∠OBP + ∠AOB = 360°⇒75° + 90° + 90° + ∠AOB = 360°⇒255° + ∠AOB = 360°⇒∠AOB = 360° - 255°⇒∠AOB = 105°We know that angle subtended by an arc at the centre is double the angle subtended by the same arc at any point on the circle.Consider the minor arc AB, then∠AOB = 2∠AQB⇒∠AQB = 12∠AOB = 12×105°⇒∠AQB = 52.5°Since AQBM is a cyclic quadrilateral, then∠AMB + ∠AQB = 180° Opposite angles of cyclic quad are supplementary⇒∠AMB + 52.5° = 180°⇒∠AMB = 180° - 52.5°⇒∠AMB = 127.5° 51 View Full Answer Radhika Rani answered this here anglePAO and angle PBO are equal to 90 degreetherefore by angle sum property of quadilateral angle AOB is equal to 75+90 + 90 + AOB=360AOB = 105 degreenow half of AOB = AQB AQB = 52.5 degreereflex AOB = 360 - 105= 255half of reflex AOB = AMBAMB = 127.5 degree 31
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