\sin \alpha - 3 \sin (90^o - \alpha) = 0\\
\sin \alpha - 3 \cos \alpha = 0 \\
\sin \alpha = 3 \cos \alpha \\
\frac{\sin \alpha}{\cos \alpha} = 3 \\
\tan \alpha = 3 \\
\alpha \approx 71^o 33' 54'' \\
\beta \approx 18^o 26' 5''