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kryton1212
Group Title
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?
 one year ago
 one year ago
kryton1212 Group Title
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?
 one year ago
 one year ago

This Question is Closed

kryton1212 Group TitleBest ResponseYou've already chosen the best response.1
dw:1359898971161:dw
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.1
@Reaper534 would you kindly help me?
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.0
sddw:1359901648939:dw can you now express <BRQ in terms of h and k?
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.0
@kryton1212 ?
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.1
<BRQ=h+k
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.1
why <BPQ=k?
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.0
:) well done!
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.0
now what about b part? Any idea ?
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.1
why <BQR=h+k????
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.0
i didnt say <BQR=h+k
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.1
but the question ask <BQR
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.0
can you tell how you got <BRQ as h+k?
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.1
ext. angle of triangle
 one year ago

BAdhi Group TitleBest ResponseYou've already chosen the best response.1
@AravindG: <RBP is not 90 degrees since AB is not a diameter
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.0
@BAdhi thanks for the spot ..I cant believe I fell for that mistake ! sorry for that @kryton1212
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.1
so what is <BQR?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
dw:1359966959882:dw
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
for p, i saw triangle ABP and for <BRQ i saw triangle BRP
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.1
<BRQ=h+k (ext. angle of triangle)... let me see...
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.1
where is p?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
<BRQ=h+k (ext. angle of triangle) is better way :P
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.1
thanks:)
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.1
but the question is asking <BQR....
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.1
and why <RPB=k? AP is not a tangent
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
RP is the angle bisector of <APB.
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.1
ok, i miss this.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
BQR seems a bit complicated hmm
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.1
is BQR an isos. triangle? we need to prove that BQR is an isos. triangle
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.1
since the answer is also h+k..
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
dw:1359969862634:dw I am not too sure what use we make of the tangent! :
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.1
i know now, <BAC=QBP=h (<s in alt segment.) therefore, <BQR=h+k (ext. angle of triangle) is it?
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
BAC = QBP ? How come ?
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.1
PB is a tangent, and <BAC=<QBR (angle in alt. segment)
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.1
dw:1359970287522:dw
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
P lies out side the circle, your logic is flawed.
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.1
but the theorem is this,....
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
Had P been any point on the circumference, then that'd been equal,
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.1
http://www.mathsrevision.net/gcse/pages.php?page=13
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
The condition is that AB should be a diameter.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
Is it given In the ques ? That AB is diameter ?
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.1
not at all
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
Only then the theorem is valid.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
Which is not the case here
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.1
Alternate Segment Theorem
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
As I said, those angles would be equal had AB been diameter.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
Which is not our case.
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.1
are you sure? the alternate segment theorem is not about the diameter.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
Wait, leme revise
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
Ohh I see ! It makes sense, I missed a step. Yep thats correct.
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.1
thanks :)))
 one year ago
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