Mmm...I see what you mean....that question IS a little vague !!!!
See the new image
The line containing BF will intersect the quarter-circle at H
So BH = the radius of the quarter-circle = 4
And FH is just the radius of our small circle = r
So
BH - FH = 4 - r = BF
We have 15 people....choose any 2
C(15,2) = 105 ways
If we extended BF through to the edge of the quarter-circle.....it's length would = 4 ....but we are subtracting off the radius of the circle, r. So BF = 4 - r and it is the hypotenuse of right triangle FGB
Hope that makes sense !!!
x^2
x^2 -2x - 5 [ x^4 - 2x^3 + kx^2 ]
x^4 - 2x^3 - 5x^2
_________________
k + 5 must = 0
k = -5
In right triangle FGB
r^2 +(2 + r)^2 = (4-r)^2
r^2 + r^2 + 4r + 4 = r^2 - 8r + 16
r^2 + 12r - 12 = 0 complete the square on r
r^2 + 12r + 36 = 12 + 36
(r + 6)^2 = 48 take the positive root
r + 6 = sqrt (48)
r = sqrt (48) - 6 = 4sqrt (3) - 6 ≈ .928
It means that a question has been answered. However, you can still add your answer, particularly if you believe that an answer (or answers) might be incorrect. ( I don't always check every answer closely.)
Welcome aboard !!!
phi (99) = [ (99) / (3 * 11) ] * (3-1)(11-1) = (3 ) (2) (10) = 60
C(14, 2) = 91 ways
15 people, choose any 2
C(15, 2) = 105 ways
