[tex]\text{Il putem afla pe cos x din teorema fundamentala a trigonometriei.}\\
\boxed{\sin^2 x+\cos ^2x=1}\\
\sin^2x+\dfrac{25}{169}=1\\
\sin^2 x=1-\dfrac{25}{169}\\
\sin^2 x=\dfrac{144}{169}\\
\sin x=\pm \dfrac{12}{13}\\
x\in \left(0\textdegree,90\textdegree\right)\Rightarrow \sin x=\dfrac{12}{13}[/tex]
