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klimenkov
Group Title
Two points on the plane \(A\) and \(B\) are given. \(AB = 2\). \(C\) is a randomly picked point in the circle of the radius \(R\) with the center in the midpoint of \(AB\). What is the probability that the \(\triangle ABC\) has an obtuse angle?
 3 months ago
 3 months ago
klimenkov Group Title
Two points on the plane \(A\) and \(B\) are given. \(AB = 2\). \(C\) is a randomly picked point in the circle of the radius \(R\) with the center in the midpoint of \(AB\). What is the probability that the \(\triangle ABC\) has an obtuse angle?
 3 months ago
 3 months ago

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nipunmalhotra93 Group TitleBest ResponseYou've already chosen the best response.1
The angle will be obtuse if the point lies inside the smaller circle. So the ratio of the areas should be the answer. Sounds right?
 3 months ago

nipunmalhotra93 Group TitleBest ResponseYou've already chosen the best response.1
no wait
 3 months ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
dw:1401618750847:dw
 3 months ago

Miracrown Group TitleBest ResponseYou've already chosen the best response.3
points A and B are fixed and they are 2 units apart dw:1401618811902:dw the center of our circle is the midpoint of segment AB
 3 months ago

Miracrown Group TitleBest ResponseYou've already chosen the best response.3
dw:1401618885857:dw
 3 months ago

nipunmalhotra93 Group TitleBest ResponseYou've already chosen the best response.1
is the answerdw:1401618840569:dw
 3 months ago

nipunmalhotra93 Group TitleBest ResponseYou've already chosen the best response.1
the answer is ratio of the area of the shaded region to total area.
 3 months ago

Miracrown Group TitleBest ResponseYou've already chosen the best response.3
we don't know the actual numerical radius of this circle , so we use R it could be R < 1 , in which case A , B would be outside the circle it could be R > 1 , in which case A , B are inside the circle or with R = 1 , A and Bb are on the circle
 3 months ago

Miracrown Group TitleBest ResponseYou've already chosen the best response.3
not exactly sure yet Let's try the case with R = 1 , so that A and B are on the circle . dw:1401619069277:dw C must be a point "in" the circle, so I interpret that to be in the interior of the circle.
 3 months ago

Miracrown Group TitleBest ResponseYou've already chosen the best response.3
dw:1401619212117:dw
 3 months ago

nipunmalhotra93 Group TitleBest ResponseYou've already chosen the best response.1
@bswan this is obtuse dw:1401619197813:dw
 3 months ago

Miracrown Group TitleBest ResponseYou've already chosen the best response.3
There certainly "appears" to be an obtuse angle at C .
 3 months ago

Miracrown Group TitleBest ResponseYou've already chosen the best response.3
dw:1401619275225:dw
 3 months ago

nipunmalhotra93 Group TitleBest ResponseYou've already chosen the best response.1
@bswan dw:1401619255446:dw and this is obtuse. And hence the answer I previously proposed.
 3 months ago

nipunmalhotra93 Group TitleBest ResponseYou've already chosen the best response.1
It should be noted that the question is asking for an obtuse angle in the triangle. So the angles at A and B could be obtuse too. Which is possible only if C lies in the following shaded region: dw:1401619378172:dw
 3 months ago

myko Group TitleBest ResponseYou've already chosen the best response.1
dw:1401619217208:dw so I would say the probability is the area of double shaded area vs total area of bigger circle
 3 months ago

nipunmalhotra93 Group TitleBest ResponseYou've already chosen the best response.1
@myko we also need to include the area of the smaller circle.
 3 months ago

myko Group TitleBest ResponseYou've already chosen the best response.1
it is included
 3 months ago

myko Group TitleBest ResponseYou've already chosen the best response.1
bigger circle includes the small one
 3 months ago

nipunmalhotra93 Group TitleBest ResponseYou've already chosen the best response.1
I meant that the double shaded area must cover the smaller circle too.
 3 months ago

Miracrown Group TitleBest ResponseYou've already chosen the best response.3
so, it looks like it need to be qualified with the condition that R > 1 .
 3 months ago

nipunmalhotra93 Group TitleBest ResponseYou've already chosen the best response.1
@myko and @Miracrown dw:1401619632107:dw is angle C not obtuse here?
 3 months ago

myko Group TitleBest ResponseYou've already chosen the best response.1
nowyou need to calculate what is the angle in black and later integrate from negative to positive of this angle to find the areadw:1401619688850:dw
 3 months ago

myko Group TitleBest ResponseYou've already chosen the best response.1
△ABC means the angle at vertex B @nipunmalhotra93
 3 months ago

Miracrown Group TitleBest ResponseYou've already chosen the best response.3
C "is" obtuse for that diagram, yes
 3 months ago

nipunmalhotra93 Group TitleBest ResponseYou've already chosen the best response.1
@myko dude... that means Triangle ABC.
 3 months ago

myko Group TitleBest ResponseYou've already chosen the best response.1
oh, never mind. You right
 3 months ago

nipunmalhotra93 Group TitleBest ResponseYou've already chosen the best response.1
dw:1401619854190:dw
 3 months ago

nipunmalhotra93 Group TitleBest ResponseYou've already chosen the best response.1
np... that happens... :)
 3 months ago

Miracrown Group TitleBest ResponseYou've already chosen the best response.3
If you choose C to be in the blackshaded region , is where it appears that we will "not" get an obtuse angle for C . dw:1401619882929:dw
 3 months ago

myko Group TitleBest ResponseYou've already chosen the best response.1
so then R must include the condition\(AC^2+CB^2>4\) from one side and what I said befor from the other.
 3 months ago

myko Group TitleBest ResponseYou've already chosen the best response.1
<4 correction
 3 months ago

Miracrown Group TitleBest ResponseYou've already chosen the best response.3
I think you can just say that R > 1. If R <= 1 , the prob (obtuse angle) = 1 .
 3 months ago

Miracrown Group TitleBest ResponseYou've already chosen the best response.3
Very interesting problem, that was.
 3 months ago
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