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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
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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 Δ OAP & Δ OBP OA = OB PA = PB OP = OP ∴ Δ OAP ≅ Δ OBP So, ∠ OPA = ∠ OPB So, ∠ OPA = 𝟏/𝟐∠ APB = 1/2 × 80° = 40° In Δ OPA ∠ POA + ∠ OPA + ∠ OAP = 180° ∠ POA + 40° + 90° = 180° ∠ POA + 130° = 180° ∠ POA = 180° – 130° = 50°

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