From a point P, two tangents PA and PB are drawn to a circle C (0, r). ∠If OP = 2r, then find ∠APB. What type of triangle is APB?

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Question 3 From a point P, two tangents PA and PB are drawn to a circle C (0, r). If OP = 2r, then find ∠𝐴𝑃𝐵. What type of triangle is APB? Let ∠ APB = 2θ By symmetry, ∠ APO = ∠ BPO = 1/2 ∠ APB ∠ APO = ∠ BPO = θ Also, We know that radius is perpendicular to tangent ∴ ∠ OAP = 90° In right triangle Δ APO sin θ = 𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 sin θ = 𝑂𝐴/𝑂𝑃 sin θ = 𝒓/𝟐𝒓 sin θ = 1/2 ∴ θ = 30° Thus, ∠ APB = 2θ = 2 × 30° = 60° Now, we need to tell what type of triangle is APB Now, we need to tell what type of triangle is APB In Δ APB Since tangents from external point is equal PA = PB Thus, ∠ PAB = ∠ PBA By Angle Sum property in Δ APB ∠ APB + ∠ PAB + ∠ PBA = 180° 60° + ∠ PAB + ∠ PAB = 180° 60° + 2∠ PAB = 180° 2∠ PAB = 180° − 60° 2∠ PAB = 120° ∠ PAB = (120° )/2 ∠ PAB = 60° Thus, in Δ APB ∠ PAB = ∠ PBA = 60° And, ∠ APB = 60° Since all angles are 60°, ∴ Δ APB is an equilateral triangle

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.