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

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

|dw:1359898971161:dw|
@Reaper534 would you kindly help me?
sd|dw:1359901648939:dw| can you now express

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question