Check the picture below.
[tex]sin(\theta )=\cfrac{\stackrel{opposite}{30\cdot sin(24^o)}}{\underset{hypotenuse}{50}}\implies sin^{-1}[sin(\theta )]=sin^{-1}\left[ \cfrac{30\cdot sin(24^o)}{50} \right] \\\\\\ \theta =sin^{-1}\left[ \cfrac{30\cdot sin(24^o)}{50} \right]\implies \theta \approx 14.1^o[/tex]
