Caut rezolvarea exercitiului 6. Rog seriozitate!
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[tex]\it \widehat A=\widehat B=\dfrac{\pi}{6} \Rightarrow \widehat C=\dfrac{2\pi}{3}\\ \\ \\ Th. \ sinusurilor \Rightarrow \dfrac{AB}{sinC}=2R \Rightarrow R=\dfrac{AB}{2sinC}=\dfrac{3}{2\sin\dfrac{2\pi}{3}}=\dfrac{3}{2\cdot\dfrac{\sqrt3}{2}}=\\ \\ \\ =\dfrac{3}{\sqrt3}=\dfrac{\sqrt3\cdot\sqrt3}{\sqrt3}=\sqrt3[/tex]