Răspuns :
[tex] \it m(\hat{C}) = 180^o-(45^o+30^o) = 105^o
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Th. \ sinusurilor \ \Rightarrow \dfrac{c}{sinC}=2R \Rightarrow \dfrac{4}{sin105^o} = 2R|_{:2} \Rightarrow
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\Rightarrow R=\dfrac{2}{sin105^o} \qquad (1)
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sin105^o = sin(180^o-105^o) =sin75^o = sin(45^o+30^o) =
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= sin45^ocos30^o+sin30^ocos45^o = \dfrac{\sqrt2}{2}\cdot \dfrac{\sqrt3}{2} + \dfrac{1}{2}\cdot \dfrac{\sqrt2}{2} =
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=\dfrac{\sqrt6+\sqrt2}{4} \qquad (2) [/tex]
[tex] \it (1),\ (2) \Rightarrow R=\dfrac{2}{\dfrac{\sqrt6+\sqrt2}{4}} =2\cdot \dfrac{4} {\sqrt6+\sqrt2} =\dfrac{ 2\cdot4\cdot(\sqrt6-\sqrt2)}{4} =
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= 2(\sqrt6-\sqrt2) [/tex]
