[tex]\displaystyle\bf\\Se~da:\\\Delta ABC~~\text{Vezi desenul atasat.}\\m(\sphericalangle B)=90^o\\m(\sphericalangle C)=2\times m(\sphericalangle A)\\cateta~BC=2~cm\\\\Rezolvare:\\\text{Unghiurile ascutite ale triunghiului dreptunghic sunt complementare.}\\m(\sphericalangle C)+m(\sphericalangle A)=90^o\\2\times m(\sphericalangle A)+m(\sphericalangle A)=90^o\\3\times m(\sphericalangle A)=90^o\\m(\sphericalangle A)=\frac{90}{3}=30^o\\ m(\sphericalangle C)=2\times m(\sphericalangle A)=2\times30^o=60^o[/tex]
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[tex]\displaystyle\bf\\Cateta~BC=2~cm\\Cateta~BC~se~opune~\sphericalangle A~care~are~30^o\\\text{Conform teoremei unghiului de }30^o,~Cateta~BC=\frac{Ipotenuza~AC}{2}\\\\\implies~AC=2\times BC=2\times2=4~cm\\Cateta~AB~\text{o calculam cu T. Pitagora.}\\\\AB=\sqrt{AC^2-BC^2}=\sqrt{4^2-2^2}=\sqrt{16-4}=\sqrt{12}=2\sqrt{3}~cm\\\\Aria~\Delta ABC=\frac{Cateta1 \times Cateta2}{2}=\\\\=\frac{BC\times AB}{2}=\frac{2\times 2\sqrt{3}}{2}=\boxed{\bf2\sqrt{3}~cm^2}[/tex]
