Răspuns :
[tex] \it Teorema \ sinusurilor \Rightarrow \dfrac{AC}{sinB} = \dfrac{BC}{sinA} \Rightarrow \dfrac{12}{sin45^o} = \dfrac{BC}{sin60^o} \Rightarrow
\\ \\ \\
\Rightarrow BC=\dfrac{12\cdot sin60^o}{sin45^o} = \dfrac{12\cdot\dfrac{\sqrt3}{2}}{\dfrac{\sqrt2}{2}} =\dfrac{^{\sqrt2)}12\sqrt3}{\ \sqrt2} =\dfrac{12\sqrt6}{2} = 6\sqrt6 [/tex]
[tex] \it \dfrac{AC}{sinB} = 2R \Rightarrow \dfrac{12}{sin45^o}=2R|_{:2} \Rightarrow R = \dfrac{12}{\dfrac{\sqrt2}{2}} \Rightarrow
\\ \\ \\
\Rightarrow R = 12\cdot\dfrac{2}{\sqrt2} = 12\cdot\dfrac{\sqrt2\cdot\sqrt2}{\sqrt2} =12\sqrt2 [/tex]
