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

Mathematics
- anonymous

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

Mathematics
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- anonymous

|dw:1359898971161:dw|

- anonymous

@Reaper534 would you kindly help me?

- AravindG

sd|dw:1359901648939:dw|
can you now express

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## More answers

- AravindG

@kryton1212 ?

- anonymous

- anonymous

why

- AravindG

:) well done!

- AravindG

now what about b part? Any idea ?

- anonymous

why

- AravindG

i didnt say

- anonymous

but the question ask

- anonymous

asks*

- AravindG

can you tell how you got

- anonymous

ext. angle of triangle

- BAdhi

@AravindG:

- AravindG

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

- anonymous

so what is

- hartnn

|dw:1359966959882:dw|

- hartnn

for p, i saw triangle ABP
and for

- anonymous

- anonymous

where is p?

- hartnn

- hartnn

p=

- anonymous

thanks:)

- anonymous

but the question is asking

- anonymous

and why

- hartnn

RP is the angle bisector of

- anonymous

ok, i miss this-.-

- shubhamsrg

BQR seems a bit complicated hmm

- anonymous

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

- anonymous

since the answer is also h+k..

- shubhamsrg

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

- anonymous

i know now,

- shubhamsrg

BAC = QBP ? How come ?

- anonymous

PB is a tangent, and

- anonymous

|dw:1359970287522:dw|

- shubhamsrg

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

- anonymous

but the theorem is this,....

- shubhamsrg

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

- anonymous

http://www.mathsrevision.net/gcse/pages.php?page=13

- shubhamsrg

The condition is that AB should be a diameter.

- shubhamsrg

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

- anonymous

not at all

- shubhamsrg

Only then the theorem is valid.

- shubhamsrg

Which is not the case here

- anonymous

Alternate Segment Theorem

- shubhamsrg

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

- shubhamsrg

Which is not our case.

- anonymous

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

- shubhamsrg

Wait, leme revise

- shubhamsrg

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

- anonymous

thanks :)))

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