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anonymous
 3 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?
anonymous
 3 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?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359898971161:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Reaper534 would you kindly help me?

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but the question ask <BQR

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ext. angle of triangle

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

AravindG
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0for p, i saw triangle ABP and for <BRQ i saw triangle BRP

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0<BRQ=h+k (ext. angle of triangle)... let me see...

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but the question is asking <BQR....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and why <RPB=k? AP is not a tangent

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is BQR an isos. triangle? we need to prove that BQR is an isos. triangle

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0since the answer is also h+k..

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i know now, <BAC=QBP=h (<s in alt segment.) therefore, <BQR=h+k (ext. angle of triangle) is it?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0PB is a tangent, and <BAC=<QBR (angle in alt. segment)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359970287522:dw

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but the theorem is this,....

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

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

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Alternate Segment Theorem

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0are you sure? the alternate segment theorem is not about the diameter.

shubhamsrg
 3 years 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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