${{\sin }^2\alpha +\left(\frac{24}{25}\right)}^2=1$
${\sin }^2\alpha +\frac{576}{625}=1$
${\sin }^2\alpha =\frac{49}{625}$   $|\sqrt{}$,   $\sin \alpha >0$, bo $\alpha$ jest kątem ostrym
$\sin \alpha =\frac{7}{25}$

$\mathrm{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{\frac{7}{25}}{\frac{24}{25}}=\frac{7}{25}\cdot \frac{25}{24}=\frac{7}{24}$