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chance

 
 
Reply Mon 8 Feb, 2010 01:44 am
In a equilateral triangle ABC a random point Q is taken.
Calculate the chance that the triangle ABQ is a obtuse angled triangle?


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Type: Discussion • Score: 0 • Views: 957 • Replies: 7
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fresco
 
  1  
Reply Mon 8 Feb, 2010 10:15 am
@whimsical,
Q must lie within a semi-circle radius c/2 where c is the length of a side of the equilateral triangle.
Area of semi-circle =Half Pi (c/2)^2
Area of whole equilateral triangle = c/2 . c sin 60 = c^2. (root3)/4

Hence p( ABQ obtuse)= 1st area/2nd area= Pi/2(root 3)=appx 0.91
fresco
 
  2  
Reply Tue 9 Feb, 2010 12:49 am
@fresco,
CORRECTION:

The semicircle is not fully within the triangle. It intersects the sides at c/2.
Therefore the smaller area is 1/3 area of semicircle plus 1/2 area of large triangle.
p(AQB obtuse) = [1/6 Pi .(c/2)^2 + 1/4. (c)^2. sin60]/[1/2(c)^2. sin60]
p(AQB obtuse) =[1/24 Pi+ 1/4 sin60] /[1/2 sin60]
fresco
 
  2  
Reply Tue 9 Feb, 2010 01:02 am
@fresco,
=0.81 appx
0 Replies
 
oolongteasup
 
  1  
Reply Tue 9 Feb, 2010 01:59 am
@whimsical,
1 upon root 3
oolongteasup
 
  1  
Reply Tue 9 Feb, 2010 02:28 am
@oolongteasup,
elegance? k

area of the equilateral triangle with sides k = k squared times root3 upon 4

area of the right triangle inside with hypotenuse k and sides i

as i squared plus i squared = k squared (pythagoras)

area of right triangle = half i squared therefore

area = k squared upon 4 (substitution)

(k squared upon 4) divided by (k times root 3 upon 4) is

one upon root 3

fresco
 
  2  
Reply Tue 9 Feb, 2010 08:45 am
@oolongteasup,
Its not a right angled triangle which defines the boundaries of the locus of Q. It's the area bounded by the semi circle radius k/2 and its intersection with the equilateral triangle side k. (Theorem: Angle in a semi-circle is a right angle)
oolongteasup
 
  0  
Reply Tue 9 Feb, 2010 06:34 pm
@fresco,
oops
0 Replies
 
 

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