CPhill

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

So

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