## kryton1212 2 years ago In the figure, PB is a tangent to the circle at B. R is a point on AB such that RP is the angle bisector of <APB. <BAP=h and <APR=k. (a) Express <BRQ and <BQR in terms of h and k. (b) Is BQR an isos. triangle?

1. kryton1212

|dw:1359898971161:dw|

2. kryton1212

@Reaper534 would you kindly help me?

3. AravindG

sd|dw:1359901648939:dw| can you now express <BRQ in terms of h and k?

4. AravindG

@kryton1212 ?

5. kryton1212

<BRQ=h+k

6. kryton1212

why <BPQ=k?

7. AravindG

:) well done!

8. AravindG

now what about b part? Any idea ?

9. kryton1212

why <BQR=h+k????

10. AravindG

i didnt say <BQR=h+k

11. kryton1212

12. kryton1212

13. AravindG

can you tell how you got <BRQ as h+k?

14. kryton1212

ext. angle of triangle

@AravindG: <RBP is not 90 degrees since AB is not a diameter

16. AravindG

@BAdhi thanks for the spot ..I cant believe I fell for that mistake ! sorry for that @kryton1212

17. kryton1212

so what is <BQR?

18. hartnn

|dw:1359966959882:dw|

19. hartnn

for p, i saw triangle ABP and for <BRQ i saw triangle BRP

20. kryton1212

<BRQ=h+k (ext. angle of triangle)... let me see...

21. kryton1212

where is p?

22. hartnn

<BRQ=h+k (ext. angle of triangle) is better way :P

23. hartnn

p=<ABP

24. kryton1212

thanks:)

25. kryton1212

but the question is asking <BQR....

26. kryton1212

and why <RPB=k? AP is not a tangent

27. hartnn

RP is the angle bisector of <APB.

28. kryton1212

ok, i miss this-.-

29. shubhamsrg

BQR seems a bit complicated hmm

30. kryton1212

is BQR an isos. triangle? we need to prove that BQR is an isos. triangle

31. kryton1212

since the answer is also h+k..

32. shubhamsrg

|dw:1359969862634:dw| I am not too sure what use we make of the tangent! :|

33. kryton1212

i know now, <BAC=QBP=h (<s in alt segment.) therefore, <BQR=h+k (ext. angle of triangle) is it?

34. shubhamsrg

BAC = QBP ? How come ?

35. kryton1212

PB is a tangent, and <BAC=<QBR (angle in alt. segment)

36. kryton1212

|dw:1359970287522:dw|

37. shubhamsrg

P lies out side the circle, your logic is flawed.

38. kryton1212

but the theorem is this,....

39. shubhamsrg

Had P been any point on the circumference, then that'd been equal,

40. kryton1212
41. shubhamsrg

The condition is that AB should be a diameter.

42. shubhamsrg

Is it given In the ques ? That AB is diameter ?

43. kryton1212

not at all

44. shubhamsrg

Only then the theorem is valid.

45. shubhamsrg

Which is not the case here

46. kryton1212

Alternate Segment Theorem

47. shubhamsrg

As I said, those angles would be equal had AB been diameter.

48. shubhamsrg

Which is not our case.

49. kryton1212

are you sure? the alternate segment theorem is not about the diameter.

50. shubhamsrg

Wait, leme revise

51. shubhamsrg

Ohh I see ! It makes sense, I missed a step. Yep thats correct.

52. kryton1212

thanks :)))