I believe you have to solve it by using law of cosines, connecting triangles with the origin.
Or anyway you prefer, I just need a solution.
+128474 The coordinates of C = (2cos (pi/12), 2sin(pi/12) = (2cos 15°, 2sin 15°)
cos (15°) = [ 1 + √3 ] sin (15°) = [√3 - 1 ]
_______ _______
2√2 2√2
So C = ( [ 1 + √3] / √2 , [ √3 - 1 ] / √2 )
The coordinates of A = (4cos (3pi/12) , 4 sin (3pi/12) = (4cos (pi/4) , 4sin (pi/ 4) ) =
( 4/√2, 4 / √2 )
The coordinates of B = (3cos (5pi/12) , 3 sin (5pi/12) = (3cos (75°) , 3sin (75°) ) =
( 3 [ √3 - 1] / [2√2] , 3[ 1 + √3]/ [2√2] )
So
AB^2 = ( 3 [ √3 - 1] / [2√2] - 4/√2)^2 + ( 3[ 1 + √3]/ [2√2] - 4/√2)^2 = 25 - 12√3 ≈ 4.22
AC^2 = (4/√2 - [ 1 + √3] / √2)^2 + ( 4/√2 - [ √3 - 1 ] / √2)^2 = 20 - 8√3 ≈ 6.14
BC^2 = ( 3 [ √3 - 1] / [2√2] - [ 1 + √3] / √2)^2 + ( 3[ 1 + √3]/ [2√2] - [ √3 - 1 ] / √2)^2 = 7
[ Because of all the nasty radicals....I used Wolfram Alpha to compute these ]
