Largest angle of the triangle

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

Correct answer:

C =  117.2796 °

Step-by-step explanation:

$$\begin{gathered} x=1 \\ a=2\cdot x=2\cdot 1=2 \\ b=3\mathrm{/}2\cdot x=3\mathrm{/}2\cdot 1={\displaystyle \frac{3}{2}}=1.5 \\ c=3\cdot x=3\cdot 1=3 \\ c>a>b \\ {c}^2=a^2+b^2-2\cdot a\cdot b\cdot \cos (C) \\ C={\displaystyle \frac{180\mathrm{°}}{\pi }}\cdot \arccos ({\displaystyle \frac{a^2+b^2-c^2}{2\cdot a\cdot b}})={\displaystyle \frac{180\mathrm{°}}{\pi }}\cdot \arccos ({\displaystyle \frac{2^2+1.5^2-3^2}{2\cdot 2\cdot 1.5}})=117.2796\text{°}=117\mathrm{°}16^{\mathrm{\prime }}47\mathrm{"} \end{gathered}$$

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