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

Ex10.2, 3 last slide.jpg

  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 So, A is the correct answer

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.