Ex 10.2, 3 - If tangents PA and PB from point P to circle - Theorem 10.2: Equal tangents from external point (numerical type)

  1. Chapter 10 Class 10 Circles
  2. Serial order wise

Transcript

Ex 10.2,3 Choose the correct option and give justification. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80 , then POA is equal to (A) 50 (B) 60 (C) 70 (D) 80 Given: PA and PB are tangents to circle & APB = 80 To find: POA Construction: Join OA,OB & OP Proof: Since PA is tangent, OA PA OAP = 90 In OAP & OBP OA = OB PA = PB OP = OP OAP OBP So, OPA = OPB So, OPA = / APB = 1/2 80 = 40 In OPA POA + OPA + OAP = 180 POA + 40 + 90 = 180 POA + 130 = 180 POA = 180 130 = 50

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