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Aadarsh
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From a point "P", 2 tangents, PA and PB are drawn to a circle with centre "O". If OP is equal to diameter of circle, prove that triangle PAB is an equilateral triangle.
 2 years ago
 2 years ago
Aadarsh Group Title
From a point "P", 2 tangents, PA and PB are drawn to a circle with centre "O". If OP is equal to diameter of circle, prove that triangle PAB is an equilateral triangle.
 2 years ago
 2 years ago

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Aadarsh Group TitleBest ResponseYou've already chosen the best response.0
Please explain with diagram.
 2 years ago

nenadmatematika Group TitleBest ResponseYou've already chosen the best response.0
go angela go :D
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.0
Angela sister, ur figure is somewhat correct. But, we should first join OA and OB separately. Then, join AB as a separate linesegment for the triangle PAB.
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.0
I don't think it will help in getting the solution.
 2 years ago

siddhantsharan Group TitleBest ResponseYou've already chosen the best response.0
Okay I got it but my mouse is not working properly.. If somebody could redraw a good figure labelled??
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.0
Wait. I am drawing the figure.
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.0
dw:1330584831022:dw
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.0
The only thing I have to prove is : Prove that AQ = OQ = AO = r, where "r" is the radius. If u can prove AQ = radius, I can do the rest.
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.0
Anyone der? Please reply soon.
 2 years ago

Sandeez Group TitleBest ResponseYou've already chosen the best response.0
so you need a theorem proof?
 2 years ago

Sandeez Group TitleBest ResponseYou've already chosen the best response.0
do you know any of the angles?
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.0
No angle. But I know that angle OAP = 90 degrees. See, in triangle, OAQ, OA = OQ = radius So, it is an isosceles triangle. Now, if we prove that AQ = radius, then it becomes an equilateral triangle.
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.0
AQ will be equal to radius, that's what our teacher told. She said, line segment from A on hypotenuse OP at its midpoint will be equal to the radius. But, what's the justification? What reason should I write?
 2 years ago

Mani_Jha Group TitleBest ResponseYou've already chosen the best response.1
Take the point of intersection of AB and OP as I. Can you prove that OI=IQ=r/2, Aadarsh?
 2 years ago

Mani_Jha Group TitleBest ResponseYou've already chosen the best response.1
Take OI=x. IP is this, (2rx). Using pythagoras' theorem, find AI from triangle AIP(AP=r*sqrt3) and then from triangle AOI. Equate those two. You will get x=r/2. After that, prove the congruency of AOI and AQI. You will get AQ=r. Do I need to show you the steps?
 2 years ago

Mani_Jha Group TitleBest ResponseYou've already chosen the best response.1
Hey pls tell me if you got it
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.0
Can u show the steps plzzzzzzzzzzzzz?
 2 years ago

kunal Group TitleBest ResponseYou've already chosen the best response.2
its easy boss................... see..... we have PO=2OA => in triangle OAP, sin(angle APO) = OA/OP => sin(angel APO) = OA/2OA = 1/2 = sin 30 => angle APO = 30 also, angle APO= angle OPB, => angle APB = 60 also AP =PB => angle ABP = angle BAP by angle sum property, all angles are 60 then its a eqvilateral triangel
 2 years ago

Mani_Jha Group TitleBest ResponseYou've already chosen the best response.1
Will you accept that AP=r*sqrt3? \[AI ^{2}=PA ^{2}PI ^{2}\] =\[3r ^{2}(2rx)^{2}\] \[AI ^{2}=AO ^{2}OI ^{2}\] =\[r^{2}x ^{2}\] \[3r ^{2}(2rx)^{2}=r ^{2}x ^{2}\] Now solve this, u get x=r/2.
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.0
Thanks Kunal and Mani bhaiiiiiiii. Short and sweet proof. Cheers Kunal!!!!!!!!!
 2 years ago
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