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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.

 Jun 6, 2019
 #1
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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 ]

 


cool cool cool

 Jun 6, 2019

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