Triangle in circle

Vertices of the triangle ABC lies on a circle with radius 3 so that it is divided into three parts in the ratio 4:4:4.

Calculate the circumference of the triangle ABC.

Correct result:

o =  15.6

Solution:

$$\begin{gathered} {\mathrm{\Phi }}_1=360\cdot {\displaystyle \frac{4}{4+4+4}}=120^{\circ } \\ {\mathrm{\Phi }}_2=360\cdot {\displaystyle \frac{4}{4+4+4}}=120^{\circ } \\ {\mathrm{\Phi }}_3=360\cdot {\displaystyle \frac{4}{4+4+4}}=120^{\circ } \\ a=2\cdot 3\cdot \sin {\displaystyle \frac{{\mathrm{\Phi }}_1}{2}}=5.2 \\ b=2\cdot 3\cdot \sin {\displaystyle \frac{{\mathrm{\Phi }}_2}{2}}=5.2 \\ c=2\cdot 3\cdot \sin {\displaystyle \frac{{\mathrm{\Phi }}_3}{2}}=5.2 \\ o=a+b+c=15.6 \end{gathered}$$

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