CPhill

Notice that (2,3) will be the mid-point of both diameter endpoints

So..let (x, y) be the other diameter endpoint....and we have

(x   +  - 1) / 2 = 2             and        (y + -1) / 2 = 3

          multiply both sides by 2

x - 1     = 4                                 y - 1   =  6

                      add 1 to both sides

x =  5                                        y  =  7

So the other endpoint  is   (5, 7)

Cross-multiply and we get that

2xr - 2x^2 = 7     rearrange as

2x^2 - 2rx + 7 =  0

Single real solutions will occur when the dicriminant = 0

So

(2r)^2 - ( 4) (7) (2)  = 0

4r^2  -  56 = 0     

4r^2 =  56    divide both sides by 4

r^2 = 14      take both roots

r =  ±√14

And the product of these r's  =   (√14) ( - √14)  =    -14

OK....no prob...!!!

26 possible letters  x  10 possible digits  =

26 x 10   =

260 possible plates ....  

There isn't anything to  solve.....we just want to evaluate this :   tan (acos(3/x))

Let acos(3/x) = theta

So.....theta is an angle whose cosine = 3 / x

So.... using the  Pythagorean Theorem......y =  sqrt (x^2 - 3^2) = sqrt (x^2 - 9)

So..... tan (acos(3/x)) =  tan (theta)  =  y / x    =   sqrt (x^2 - 9) / 3

Thanks, Melody...!!!!

.....it took me long enough to make it in GeoGebra......this utility is still kinda' funky to use...IMHO....

Nice, asdf    >>>>>!!!!!

Just a rough pic for reference......definitely not exact!!!!

                     A

     E                            B

  D

                        C                                     

Since ABC = 128°.......then major arc CA is twice 128° = 256°

Therefore.......arc ABC =  360° - 256° = 104°

But.....since AB = BC....then each of these chords must span an arc of 52°

And since  AB = BC = CD = DE.....these equal chords each span an arc of 52°

So....arc EA  must be   360 - 4(52)  = 152°

And angle ACE is an inscrbed angle subtending EA....so it's measure is (1/2) measure arc EA = 76°

That would be true if E were the center of the circle.....but clearly....that's not the case....angle DEA will subtend an arc > 108°

Let us  let the center of the circle be at F

Drawing radii from F to both D and A  will result in right angles CDF  and  BAF

And ABCDF will form an irregular pentagon.....the sum of the interior angles of this pentagon = 540°

Angles DCB and ABC =  108°

So....we can find angle AFD as

540 - 108 - 108 - 90 - 90  = 144°

But AFD is a central angle in the circle....so....its measure = minor arc AD   =  144°