Question

9. In the given figure , O is the centre of the circle and chord AC and BD intersect

at P such that APB= 1200 and PBC = 150

, find the value of ADB.​

Answer 1

Answer:

1. We know that ∠ACB=∠PCB
3. In △PCB
5. Using the angle sum property
7. ∠PCB+∠BPC+∠PBC=180
8. o
11. We know that ∠APB and ∠BPC are linear pair
13. By substituting the values
15. ∠PCB+(180
16. o
17. −110
18. o
19. )+25
20. o
21. =180
22. o
25. On further calculation
27. ∠PBC+70
28. o
29. +25
30. o
31. =180
32. o
35. ∠PCB+95
36. o
37. =180
38. o
41. By subtraction
43. ∠PCB=180
44. o
45. −95
46. o
49. So we get
51. ∠PCB=85
52. o
55. We know that the angles in the same segment of a circle equal
57. ∠ADB=∠ACB=85
58. o
61. Therefore, the value of ∠ADB is 85
62. o
64. solution