Largest angle of the triangle

Calculate the largest angle of the triangle whose sides have the sizes:
2a, 3/2a, 3a

Result

C =  117.28 °

Solution:

x=1 a=2⋅ x=2⋅ 1=2 b=3/2⋅ x=3/2⋅ 1=32=1.5 c=3⋅ x=3⋅ 1=3  c>a>b c2=a2+b2−2⋅ a⋅ b⋅ cos⁡(C) C=180∘π⋅arccos⁡(a2+b2−c22⋅ a⋅ b)=180∘π⋅arccos⁡(22+1.52−322⋅ 2⋅ 1.5)≐117.2796=117.28∘=117∘16′47"x = 1 \ \\ a = 2 \cdot \ x = 2 \cdot \ 1 = 2 \ \\ b = 3/2 \cdot \ x = 3/2 \cdot \ 1 = \dfrac{ 3 }{ 2 } = 1.5 \ \\ c = 3 \cdot \ x = 3 \cdot \ 1 = 3 \ \\ \ \\ c>a>b \ \\ c^2 = a^2+b^2 - 2 \cdot \ a \cdot \ b \cdot \ \cos(C) \ \\ C = \dfrac{ 180^\circ }{ \pi } \cdot \arccos (\dfrac{ a^2 + b^2 - c^2 }{ 2 \cdot \ a \cdot \ b } ) = \dfrac{ 180^\circ }{ \pi } \cdot \arccos (\dfrac{ 2^2 + 1.5^2 - 3^2 }{ 2 \cdot \ 2 \cdot \ 1.5 } ) \doteq 117.2796 = 117.28 ^\circ = 117^\circ 16'47"

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