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.

Result

o =  15.6

Solution:

Φ1=360⋅44+4+4=120∘ Φ2=360⋅44+4+4=120∘ Φ3=360⋅44+4+4=120∘  a=2⋅3⋅sin⁡Φ12=5.2 b=2⋅3⋅sin⁡Φ22=5.2 c=2⋅3⋅sin⁡Φ32=5.2  o=a+b+c=15.6 \Phi_1 = 360 \cdot \dfrac{ 4} { 4+4+4 } = 120 ^\circ \ \\ \Phi_2 = 360 \cdot \dfrac{ 4} { 4+4+4 } = 120 ^\circ \ \\ \Phi_3 = 360 \cdot \dfrac{ 4} { 4+4+4 } = 120 ^\circ \ \\ \ \\ a = 2 \cdot 3 \cdot \sin \dfrac{\Phi_1}{2} = 5.2 \ \\ b = 2 \cdot 3 \cdot \sin \dfrac{\Phi_2}{2} = 5.2 \ \\ c = 2 \cdot 3 \cdot \sin \dfrac{\Phi_3}{2} = 5.2 \ \\ \ \\ o= a+b+c = 15.6 \ \\

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