In the given figure, O is the centre of circle. Find ∠AQB, given that PA and PB are tangents to the circle and ∠APB = 75°.
CBSE Class 10 Sample Paper for 2023 Boards - Maths Standard
Question 2 Important
Question 3 Important
Question 4 Important
Question 5
Question 6 Important
Question 7
Question 8 Important
Question 9
Question 10 Important
Question 11 Important
Question 12
Question 13 Important
Question 14 Important
Question 15
Question 16 Important
Question 17
Question 18 Important
Question 19 [Assertion Reasoning] Important
Question 20 [Assertion Reasoning] Important
Question 21 Important
Question 22
Question 23 Important You are here
Question 24 (Choice 1)
Question 24 (Choice 2)
Question 25 (Choice 1)
Question 25 (Choice 2) Important
Question 26
Question 27 Important
Question 28 (Choice 1) Important
Question 28 (Choice 2)
Question 29 Important
Question 30 (Choice 1)
Question 30 (Choice 2) Important
Question 31
Question 32 (Choice 1) Important
Question 32 (Choice 2)
Question 33 Important
Question 34 (Choice 1)
Question 34 (Choice 2) Important
Question 35
Question 36 (i) [Case Based] Important
Question 36 (ii) Important
Question 36 (iii) (Choice 1)
Question 36 (iii) (Choice 2)
Question 37 (i) [Case Based] Important
Question 37 (ii) (Choice 1) Important
Question 37 (ii) (Choice 2)
Question 37 (iii)
Question 38 (i) [Case Based] Important
Question 38 (ii) (Choice 1)
Question 38 (ii) (Choice 2) Important
Question 38 (iii)
CBSE Class 10 Sample Paper for 2023 Boards - Maths Standard
Last updated at Dec. 13, 2024 by Teachoo
Question 23 In the given figure, O is the centre of circle. Find ∠AQB, given that PA and PB are tangents to the circle and ∠APB = 75°. We know that Radius is perpendicular to the tangent ∴ ∠ OAP = ∠ OBP = 90° In quadrilateral PAOB Sum of angles = 360° ∠ APB + ∠ OAP + ∠ OBP + ∠ AOB = 360° 75° + 90° + 90° + ∠ AOB = 360° 255° + ∠ AOB = 360° ∠ AOB = 360° − 255° ∠ AOB = 105° Also, We know that Angle subtended by an arc at the centre is double the angle subtended by the same arc at any point on the circle. ∴ ∠ AOB = 2 × ∠ AQB 105° = 2 × ∠ AQB 2 × ∠ AQB = 105° ∠ AQB = (105° )/2 ∠ AQB = 52.5°