anonymous
  • 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
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1359898971161:dw|
anonymous
  • anonymous
@Reaper534 would you kindly help me?
AravindG
  • AravindG
sd|dw:1359901648939:dw| can you now express

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

AravindG
  • AravindG
@kryton1212 ?
anonymous
  • anonymous
anonymous
  • anonymous
why
AravindG
  • AravindG
:) well done!
AravindG
  • AravindG
now what about b part? Any idea ?
anonymous
  • anonymous
why
AravindG
  • AravindG
i didnt say
anonymous
  • anonymous
but the question ask
anonymous
  • anonymous
asks*
AravindG
  • AravindG
can you tell how you got
anonymous
  • anonymous
ext. angle of triangle
BAdhi
  • BAdhi
@AravindG:
AravindG
  • AravindG
@BAdhi thanks for the spot ..I cant believe I fell for that mistake ! sorry for that @kryton1212
anonymous
  • anonymous
so what is
hartnn
  • hartnn
|dw:1359966959882:dw|
hartnn
  • hartnn
for p, i saw triangle ABP and for
anonymous
  • anonymous
anonymous
  • anonymous
where is p?
hartnn
  • hartnn
hartnn
  • hartnn
p=
anonymous
  • anonymous
thanks:)
anonymous
  • anonymous
but the question is asking
anonymous
  • anonymous
and why
hartnn
  • hartnn
RP is the angle bisector of
anonymous
  • anonymous
ok, i miss this-.-
shubhamsrg
  • shubhamsrg
BQR seems a bit complicated hmm
anonymous
  • anonymous
is BQR an isos. triangle? we need to prove that BQR is an isos. triangle
anonymous
  • anonymous
since the answer is also h+k..
shubhamsrg
  • shubhamsrg
|dw:1359969862634:dw| I am not too sure what use we make of the tangent! :|
anonymous
  • anonymous
i know now,
shubhamsrg
  • shubhamsrg
BAC = QBP ? How come ?
anonymous
  • anonymous
PB is a tangent, and
anonymous
  • anonymous
|dw:1359970287522:dw|
shubhamsrg
  • shubhamsrg
P lies out side the circle, your logic is flawed.
anonymous
  • anonymous
but the theorem is this,....
shubhamsrg
  • shubhamsrg
Had P been any point on the circumference, then that'd been equal,
anonymous
  • anonymous
http://www.mathsrevision.net/gcse/pages.php?page=13
shubhamsrg
  • shubhamsrg
The condition is that AB should be a diameter.
shubhamsrg
  • shubhamsrg
Is it given In the ques ? That AB is diameter ?
anonymous
  • anonymous
not at all
shubhamsrg
  • shubhamsrg
Only then the theorem is valid.
shubhamsrg
  • shubhamsrg
Which is not the case here
anonymous
  • anonymous
Alternate Segment Theorem
shubhamsrg
  • shubhamsrg
As I said, those angles would be equal had AB been diameter.
shubhamsrg
  • shubhamsrg
Which is not our case.
anonymous
  • anonymous
are you sure? the alternate segment theorem is not about the diameter.
shubhamsrg
  • shubhamsrg
Wait, leme revise
shubhamsrg
  • shubhamsrg
Ohh I see ! It makes sense, I missed a step. Yep thats correct.
anonymous
  • anonymous
thanks :)))

Looking for something else?

Not the answer you are looking for? Search for more explanations.