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 one year 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?
 one year 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?

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

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

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

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

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

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

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

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

AravindG
 one year ago
Best 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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