Vezi desenul atasat.
[tex]\displaystyle\bf\\Se~da:\\\Delta ABC cu:\\AB=6~cm\\BC=8~cm\\AC=10~cm\\\\Se~cere:\\cos\,A=?\\cos\,B=?\\cos\,C=?\\\\Rezolvare:\\\\\textbf{Aplicam R.T. Pitagora pt. a afla natura triunghiului.}\\\\\sqrt{AB^2+BC^2}=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10=AC\\\\\implies~\Delta ABC~este~dreptunghic~cu~ipotenuza~AC~si~ \sphericalangle~B=90^o.\\\\cos A=\frac{c.a.}{ip}=\frac{AB}{AC}=\frac{6}{10}=\frac{3}{5}\\\\cos B=cos\,90^o=0\\\\cos C=\frac{c.a.}{ip}=\frac{BC}{AC}=\frac{8}{10}=\frac{4}{5}[/tex]
