Teorema sinusurilor:
[tex]\it \dfrac{AC}{sinB} = \dfrac{BC}{sinA} \Rightarrow sinB = \dfrac{AC\cdot sinA}{BC} = \dfrac{8\cdot sin30^o}{4\sqrt2} = \dfrac{2\cdot\dfrac{1}{2}}{\sqrt2} =
\\\;\\ \\\;\\
= \dfrac{1}{\sqrt2} = \dfrac{\sqrt2}{2} \Rightarrow m(\hat{B}) = 45^o[/tex]
[tex]\it m(\hat{C}) = 180^o-(30^o+45^o) = 180^o-75^o = 105^o[/tex]
