[tex]\text{Voi rezolva problema in doua metode: } \\ \\ Metoda\;1 \;\;\;(trigonometrica): \\ sin\,x= \sqrt{1-cos^2x}= \sqrt{1- (\frac{5}{13}) ^{2} }=\sqrt{1- \frac{5^2}{13^2}}= \\ \\ =\sqrt{ \frac{13^2}{13^2} -\frac{5^2}{13^2}}= \sqrt{ \frac{13^2-5^2}{13^2}}=\sqrt{ \frac{169-25}{13^2}}= \\ \\ =\sqrt{ \frac{144}{13^2}}=\sqrt{ \frac{12^2}{13^2}}=\boxed{ \frac{12}{13}}[/tex]
[tex]Metoda\;2 \;\;\;(geometrica): \\ cos\,x= \frac{5}{13} \\ Luam\; \Delta ABC \;dreptunghic \;in \;A \;\;\;\;\;\;\;(\;\ \textless \ A=90^o\;) \\ Ipotenuza \;BC = 13\;cm \\ Cateta\;AC=5\;cm \\ Cateta AB \;nu\; o\; stim \\ \ \textless \ C = x \\ cos\,x= \frac{cateta\;alaturata}{ipotenuza}= \frac{AC}{BC} = \frac{5}{13} \\ sin\,x = \frac{cateta\;opusa}{ipotenuza}= \frac{AB}{BC} \\ \text{Aflam pe AB cu Pitagora} \\ AB = \sqrt{BC^2 - AC^2} = \sqrt{13^2 - 5^2}=\sqrt{169 - 25}=\sqrt{144}=12\;cm \\ =\ \textgreater \ sin\,x = \boxed{ \frac{12}{13}}[/tex]
