[tex]\text{Il calculam pe }\sin x\text{ din teorema fundamentala a trigonometriei.}\\
\boxed{\sin^2 x+\cos^2 x=1}\\
\sin^2 x=1-\cos^2 x\\
\sin^2 x=1-\dfrac{25}{169}\\
\sin^2 x=\dfrac{144}{169}\\
\\
\sin x=\pm \dfrac{12}{13}\\
\text{Dar }x\in\left(0,\dfrac{\pi}{2}\right),\text{deci e pozitiv.Prin urmare }\boxed{\sin x=\dfrac{12}{13}}[/tex]
