PA and PB are tangents drawn to a circle of centre O from an external point P. Chord AB makes an angle of 30° with the radius at the point of contact. If length of the chord is 6 cm, find the length of the tangent PA and the length of the radius OA

PA and PB are tangents drawn to a circle of centre O from an external - CBSE Class 10 Sample Paper for 2024 Boards - Maths Standard

part 2 - Question 29 (Choice 1) - CBSE Class 10 Sample Paper for 2024 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards - Class 10
part 3 - Question 29 (Choice 1) - CBSE Class 10 Sample Paper for 2024 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards - Class 10
part 4 - Question 29 (Choice 1) - CBSE Class 10 Sample Paper for 2024 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards - Class 10
part 5 - Question 29 (Choice 1) - CBSE Class 10 Sample Paper for 2024 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards - Class 10 part 6 - Question 29 (Choice 1) - CBSE Class 10 Sample Paper for 2024 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards - Class 10

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PA and PB are tangents drawn to a circle of centre O from an external point P. Chord AB makes an angle of 30° with the radius at the point of contact. If length of the chord is 6 cm, find the length of the tangent PA and the length of the radius OA ∠PAB = 90° − 30 ° ∴ ∠PAB = 60° Also, since Tangents from external point are equal PA = PB And we know angles opposite to equal sides are equal Thus, ∠ PAB = ∠ PBA = 60° In Δ APB By Angle Sum property ∠ APB + ∠ PAB + ∠ PBA = 180° ∠ APB + 60° + 60° = 180° ∠ APB + 120° = 180° ∠ APB = 180° − 120° ∠ APB = 60° Thus, in Δ APB ∠ PAB = ∠ PBA ∠ APB = 60° Since all angles are 60°, ∴ Δ APB is an equilateral triangle And, in equilateral triangle, all sides are equal ∴ PA = AB PA = 6 cm Finding radius OA Since ∠APB = 60° By symmetry ∠ OPA = (∠ 𝐴𝑃𝐵)/2 = (60° )/2 ∠ OPA = 30° Using Trigonometry in right Δ AOP tan P = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝑃)/█(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝑃@ ) tan 30° = 𝑶𝑨/𝑷𝑨 1/√3 = 𝑂𝐴/6 0A = 𝟔/√𝟑 OA = (2 × 3)/√3 OA = 2√𝟑 cm

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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 and Computer Science at Teachoo