1)

BC² = 12
AC² = 8
AB² = 6 + 2 - 2V(12) = 8 -4V3

AB² = BC² + AC² - 2.BC.AC.cos(ACB)
8 - 4V3 = 12 + 8 -2.(2V3)(2V2).cos(ACB)
8 - 4V3 = 20 -8.V6.cos(ACB)
2 - V3 = 5 -2.V6.cos(ACB)
-3 - V3 = -2.V6.cos(ACB)
3 + V3 = 2.V6.cos(ACB)

cos(ACB) = (3+V3)/2V6

cos(ACB) = V3(V3+1)/(2.V2.V3)
cos(ACB) = (1+V3)/(2.V2)    (1)
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BC² = AC² + AB² - 2.AC.AB.cos(BAC)
12 = 8 + 8 - 4V3 - 2.2V2.(V6-V2).cos(BAC)
-4 = - 4V3 - 2.2V2.(V6-V2).cos(BAC)
-1 = -V3 - V2.(V6-V2).cos(BAC)
cos(BAC) = -(1-V3)/(2-V12)
cos(BAC) = -(1-V3)(2+V12)/[(2-V12)(2+V12)]
cos(BAC) = -(2+V12-2V3-V36)/[(2-V12)(2+V12)]
cos(BAC) = -(2+2V3-2V3-6)/(4-12)
cos(BAC) = 4/(-8)
cos(BAC) = - 1/2
-> angle(BAC) = 120°
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AC² = AB² + BC² - 2.AB.BC.cos(ABC)
8 = 8 - 4V3 + 12 - 2.(V6-V2)2V3.cos(ABC)
4V3 - 12 = -2.(V6-V2)2V3.cos(ABC)
V3 - 3 = -(V6-V2)V3.cos(ABC)
cos(ABC) = -(1-V3)/(V6-V2)
cos(ABC) = -(1-V3)(V6+V2)/[(V6-V2)(V6+V2)]
cos(ABC) = -(V6+V2-V18-V6)/(6-2)
cos(ABC) = -(V2-3V2)/4
cos(ABC) = 2V2/4
cos(ABC) = 1/V2
-> angle(ABC) = 45°
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Comme la somme des angles d'un triangle = 180°, on a:
angle(BAC) + angle(ACB) + angle(ABC) = 180°
120° + angle(ACB) + 45° = 180°
angle(ACB) = 15°

Avec(1) ->
cos(15°) = (1+V3)/(2.V2)
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2)
AH = AB.sin(ABC)
AH = (V6-V2).sin(45°)
AH = (V6-V2)/V2
AH = (V3 - 1)

Angle(KAC) = 120°/2 = 60°

Dans le triangleKAC:
AK/sin(ACB) = AC/sin(AKC)
AK = 2V2.sin(ACB)/sin(AKC)
AK = 2/sin(AKC)

angle(AKC) + angle(KAC)  + angle(ACK) = 180°
angle(AKC) + 60° + 15° = 180°
angle(AKC) = 105°
sin(AKC) = sin(105°) =sin(90°+15°) = cos(15°) =  (1+V3)/(2.V2)

AK = 2/[(1+V3)/(2.V2)]
AK = 4V2/(1+V3)
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Sauf distraction, vérifie.